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Robin M. Zavod, PhD Editor-in-Chief, Currents in Pharmacy Teaching and Learning Professor of Pharmaceutical Sciences Midwestern University Chicago College of Pharmacy Downers Grove, Illinois

Marc W. Harrold, PhD Professor of Medicinal Chemistry Duquesne University Mylan School of Pharmacy Pittsburgh, Pennsylvania

Any correspondence regarding this publication should be sent to the publisher, American Society of Health-System Pharmacists, 7272 Wisconsin Avenue, Bethesda, MD 20814, attention: Special Publishing. The information presented herein reflects the opinions of the contributors and advisors. It should not be interpreted as an official policy of ASHP or as an endorsement of any product. Because of ongoing research and improvements in technology, the information and its applications contained in this text are constantly evolving and are subject to the professional judgment and interpretation of the practitioner due to the uniqueness of a clinical situation. The editors and ASHP have made reasonable efforts to ensure the accuracy and appropriateness of the information presented in this document. However, any user of this information is advised that the editors and ASHP are not responsible for the continued currency of the information, for any errors or omissions, and/or for any consequences arising from the use of the information in the document in any and all practice settings. Any reader of this document is cautioned that ASHP makes no representation, guarantee, or warranty, express or implied, as to the accuracy and appropriateness of the information contained in this document and specifically disclaims any liability to any party for the accuracy and/or completeness of the material or for any damages arising out of the use or non-use of any of the information contained in this document.

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ISBN: 978-1-58528-464-1



10 9 8 7 6 5 4 3 2 1

DEDICATION To our students, whose creative problem solving has not only taunted and teased us, but has also inspired us to develop review questions that provide meaningful guidance during the structure evaluation journey.

TABLE OF CONTENTS Preface .......................................................................................................................................................................................... vii Acknowledgments.......................................................................................................................................................................... ix

PART I QUESTIONS Section 1 General Self Assessment 1.1 1.2 1.3 1.4 1.5 1.6 1.7

Functional Group Characteristics and Roles........................................................................................................................ 1 Identifying Acidic and Basic Functional Groups................................................................................................................... 7 Solving pH/pKa Problems................................................................................................................................................... 11 Salts and Solubility............................................................................................................................................................ 13 Drug Binding Interactions.................................................................................................................................................. 17 Stereochemistry and Drug Action...................................................................................................................................... 21 Drug Metabolism............................................................................................................................................................... 23

Section 2 Whole Molecule Drug Evaluation 1.8 1.9 1.10 1.11 1.12 1.13 1.14 1.15 1.16 1.17 1.18 1.19 1.20 1.21 1.22 1.23 1.24 1.25 1.26 1.27

Aliskiren............................................................................................................................................................................ 27 Aripiprazole....................................................................................................................................................................... 31 Cefprozil............................................................................................................................................................................ 35 Cetirizine........................................................................................................................................................................... 39 Chlorpropamide and Other Sulfonylureas ......................................................................................................................... 41 Dabigatran Etexilate.......................................................................................................................................................... 45 Fenofibrate and Gemfibrozil.............................................................................................................................................. 47 Fluvoxamine...................................................................................................................................................................... 51 Haloperidol........................................................................................................................................................................ 55 Hydrocortisone.................................................................................................................................................................. 57 Levothyroxine.................................................................................................................................................................... 61 Lidocaine........................................................................................................................................................................... 65 Montelukast and Zafirlukast.............................................................................................................................................. 69 Phenobarbital and Other Barbiturates............................................................................................................................... 71 Pravastatin and Fluvastatin............................................................................................................................................... 73 Quinapril........................................................................................................................................................................... 75 Rivastigmine...................................................................................................................................................................... 79 Sitagliptin.......................................................................................................................................................................... 83 Sorafenib........................................................................................................................................................................... 87 Zanamivir and Oseltamivir ............................................................................................................................................... 91

PART 2 ANSWERS Section 3 General Self Assessment 2.1 Functional Group Characteristics and Roles...................................................................................................................... 95 2.2 Identifying Acidic and Basic Functional Groups................................................................................................................. 99

TABLE OF CONTENTS 2.3 2.4 2.5 2.6 2.7

Solving pH/pKa Problems................................................................................................................................................ 105 Salts and Solubility......................................................................................................................................................... 109 Drug Binding Interactions............................................................................................................................................... 113 Stereochemistry and Drug Action................................................................................................................................... 117 Drug Metabolism ........................................................................................................................................................... 123

Section 4 Whole Molecule Drug Evaluation 2.8 2.9 2.10 2.11 2.12 2.13 2.14 2.15 2.16 2.17 2.18 2.19 2.20 2.21 2.22 2.23 2.24 2.25 2.26 2.27

Aliskiren......................................................................................................................................................................... 129 Aripiprazole.................................................................................................................................................................... 137 Cefprozil......................................................................................................................................................................... 143 Cetirizine........................................................................................................................................................................ 149 Chlorpropamide and Other Sulfonylureas ...................................................................................................................... 153 Dabigatran Etexilate....................................................................................................................................................... 159 Fenofibrate and Gemfibrozil........................................................................................................................................... 165 Fluvoxamine................................................................................................................................................................... 169 Haloperidol..................................................................................................................................................................... 173 Hydrocortisone............................................................................................................................................................... 177 Levothyroxine................................................................................................................................................................. 183 Lidocaine........................................................................................................................................................................ 189 Montelukast and Zafirlukast........................................................................................................................................... 193 Phenobarbital and Other Barbiturates............................................................................................................................ 199 Pravastatin and Fluvastatin............................................................................................................................................ 205 Quinapril........................................................................................................................................................................ 211 Rivastigmine................................................................................................................................................................... 217 Sitagliptin....................................................................................................................................................................... 223 Sorafenib........................................................................................................................................................................ 227 Zanamivir and Oseltamivir ............................................................................................................................................ 233

INDEX .................................................................................................................................................................................... 239

vii

PREFACE If our suspicions are correct, you’ve heard the phrases, “Keep up the good work—you’ll get better with practice!” and “Keep practicing and you’ll see that it gets easier.” Regardless of whether it is a coach, teacher, clergy member, scout leader, friend, or family member providing this encouragement, the person receiving this advice is likely learning a new skill, technique, or language. With practice, soccer goals are scored, training wheels are removed, piano and ballet recitals are executed, and tents are pitched. Through self assessment, one is able to both relish the “victory,” as well as consider how he or she could have improved. When learning a new language, another common expression is, “If you don’t use it, you’ll lose it.” The Basic Concepts in Medicinal Chemistry textbook served to lay the “language” foundation for drug structure evaluation. Whereas written/spoken language is based on phonemic awareness and phonetics, the language of medicinal chemistry is based on functional group identification and evaluation. These functional groups “speak” to biological targets via key binding interactions, with stimulation or inhibition of key physiological and biochemical processes as the response. Mastery of this language will help facilitate a learner’s journey into understanding structure activity relationships—the next level of structure evaluation. The Basic Concepts in Medicinal Chemistry chapters discussed physicochemical properties, provided clinically relevant examples and, through the review problems, offered an opportunity to practice new structure evaluation skills. With all of the variations of practice makes perfect ringing in our ears, Marc and I agreed that one way to help cement the concepts in the textbook chapters is to offer learners a chance to engage in self assessment. Better yet, we knew that the next step was to put all of the functional group pieces together to support whole molecule evaluation. This workbook is split into two sections with two goals in mind. Our first goal is to help the learner reinforce his or her foundational knowledge. Additional practice problems focus on the following eight fundamental types of evaluation: 1. Identification of functional groups that contribute to water solubility and those that contribute to lipid solubility and the need for a balance between the two 2. The role of electron withdrawing and donating functional groups on adjacent atoms and functional groups 3. Identification of acidic and basic functional groups and related ionization under physiologically relevant conditions 4. Use of the Henderson-Hasselbalch equation to solve both qualitative and quantitative pH and pKa problems 5. Formation of inorganic and organic salts of specific functional groups and the value of those salts 6. Interaction chemistry between a drug molecule and its biological target for drug action 7. Spatial orientation of functional groups 8. Metabolic routes of drug activation, inactivation, and/or elimination Our second goal, and the natural next step in the learning process, is to have the learner put all of his or her evaluation skills together and fully assess all of the functional groups within a given drug molecule. Twenty contemporary drugs that span an equivalent number of drug classes were selected as representative models for whole molecule evaluation. Section questions challenge the learner to anticipate what could happen to a drug molecule (e.g., what metabolites could form), as well as to provide an explanation for an observed pharmacodynamic (e.g., mechanism of action) or pharmacokinetic parameter (e.g., percent orally bioavailability) based on the information gleaned from the structure evaluation process.

viii

PREFACE

(cont’d)

Although the workbook and associated on-line content (available to instructors only) are designed to allow learners to conduct a self assessment of their knowledge, other audiences are also likely to derive benefit. Those individuals who teach organic chemistry to pre-health profession students may find the use of clinically relevant agents in these problems valuable in enhancing student engagement. Pharmacy faculty tasked with “dusting the rust” off of prerequisite course content may find the workbook a valuable alternative to graded homework assignments. For those pharmacy faculty who contribute to an integrated sequence and have precious little time to review these fundamental concepts, we offer this workbook (and previous textbook) as a solution to quickly, but thoroughly translate organic chemistry concepts into chemistry meaningful for the pharmacy practitioner. We hope that you find the format of the workbook amenable to modification and that you enjoy solving a variety of problems for which drug structure evaluation can contribute. We are thankful to have had the opportunity to contribute to learner self assessment and for those faculty who appreciate the absolute need for practice prior to mastery.

Robin M. Zavod Marc W. Harrold

ix

ACKNOWLEDGMENTS

The writing and publishing of this text could not have been accomplished without the hard work and the support of others. We would like to thank the following individuals from ASHP who provided us with this opportunity and who were extremely valuable in answering our many queries: Ruth Bloom, Robin Coleman, and Johnna Hershey. Your thoughtful edits and suggestions were exceptionally helpful.

Over my lifetime several individuals have asked me if I think that fast or accurate is more important. My response has always been, “Who cares about fast, if it isn’t accurate?” A slow, methodical approach to problem solving has always served me well. Working through practice problems is one method to cement problem solving skills. I would like to acknowledge both my eighth grade chemistry teacher GH and my parents for instilling the value of “accurate.” I would also like to thank Marc for putting up with all of my varied challenges through this self-assessment book and our former textbook projects.—RZ

I am thankful for the opportunity to enhance our initial text and to expand its potential use. I would like to thank Robin for her original idea for this self-assessment text; my colleagues at Duquesne University who have provided valuable suggestions; my wife Barbara and my family for their love and support; and God for His many blessings.—MH

Part 1 QUESTIONS

Section 1 - General Self Assessment 1.1 1.2 1.3 1.4 1.5 1.6 1.7

Functional Group Characteristics and Roles.................... 3 Identifying Acidic and Basic Functional Groups............... 7 Solving pH/pKa Problems............................................... 11 Salts and Solubility........................................................ 13 Drug Binding Interactions.............................................. 17 Stereochemistry and Drug Action.................................. 21 Drug Metabolism........................................................... 23

Section 1 General Self Assessment

Structures for 1.1 and 2.1 The bold has been removed.

1.1 Functional Group Characteristics and Roles The exact same structures are in both of these chapters, but I have provided two

copies in the event you wanted that. 1. Shown below are the structures of warfarin, phenytoin, bromfenac, and salmeterol. a. Identify each of the boxed functional groups. A

F C E D

Warfarin

B

Bromfenac

Phenytoin

H J

I

K

L

Salmeterol Box

Functional Group Name

A B C D E

O

F

OH

G H I

D

B

A

C

O

H 2N

S O

C H3 N H

Sulfamethoxazole

Ibuprofen

J K L

3

N

O

G

4

Medicinal Chemistry Self Assessment

Structures for 1.1 and 2.1

The bold has been removed. The exact same structures are in both of these chapters, but I have provided two b. For each of the functional groups you identified, indicate if it is hydrophilic or hydrophobic in character. Also provide a brief explanation for your response. copies in the event you wanted that. Box

Hydrophilic or Hydrophobic

A B C

A

F

D

C

E

E

F G Warfarin H

D B

Bromfenac

Phenytoin

I

G

J K L

H J

I

K

L

Salmeterol

2. Shown below are the structures of ibuprofen and sulfamethoxazole. Four functional groups have been highlighted. Based on their electronic properties AND their relative positions in the molecule, identify if they are electron withdrawing or electron donating. Additionally, identify if this effect is due to resonance or induction.

O OH

Ibuprofen



D

B

A

C

O

H 2N

S O

C H3 N H

N

O

Sulfamethoxazole

Page1of2

overall evaluation of how each change will affect the chemical properties of imipramine. 1.1 Functional Group Characteristics and Roles

5

3. Shown below is the structure of imipramine as well as three analogs. Evaluate each analog and N provide an overall evaluation of how each change will affect the chemical properties of imipramine.

Imipramine

N

C H3

C H3

Imipramine N

N

Analog A

N

N

Cl

CH3

C H3

N

Analog B

C H3

Analog A

O

C H3

N

Analog C

C H3

Analog B

C H3

C H3

Analog C

4. Shown below is the structure of a tetrapeptide that is part of a larger protein receptor. The side chains of the four amino acids have been boxed. 4. Shown below is the structure of a tetrapeptide that is part of a larger protein receptor. The side chains of the four amino acids have been boxed. O

O

R1

H N

O

OH N H

N H O

N H2

O N H

R2 O

HO

a. Identify the four amino acids that comprise this tetrapeptide sequence.

b. For each amino acid, identify the key chemical properties of its highlighted side chain.



Answers can be found in Section 2.1 [this will be linked to section 2.1 title]

Page2of2

Note: The questions in this chapter are related to Chapter 2 in Harrold MW and Zavod RM, Basic Concepts in Medicinal Chemistry, American Society of Health-System Pharmacists, ©2013.

Section 1 General Self Assessment

1.2 Identifying Acidic and Basic Functional Groups

1.2

1. For each of the drugs or experimental drugs shown below, identify all of the acidic and basic functional 1. For each of the drugs or experimental drugs shown below, identify all of the acidic and basic functional groups. groups.

Experimental oral anticoagulant

Bromfenac

Sorbinil Experimental antidiabetic agent

5. Shown below is the structure of clonidine, an D adrenergic agonist that can be used to treat hypertension. Drug Name

Acidic Functional Groups

Basic Functional Groups

Clonidine contains a guanidine functional group (highlighted in bold) that has a pKa of 8.3. Other guanidine Experimental Oral Anticoagulant

functional Bromfenac groups, such as that seen with arginine are much more basic with a pKa of 12.5. Provide a chemical Sorbinil explanation for this difference. Experimental Antidiabetic Agent

pKa = 12.5

pKa = 8.3

Clonidine

Arginine 7

1.2 8

Medicinal Chemistry Self Assessment

1. For each of the drugs or experimental drugs shown below, identify all of the acidic and basic functional groups. 2. Using any one of the acidic functional groups that you identified in question 1, provide an explanation as to why the functional group is acidic. Also provide a similar type of analysis for any one of the basic functional groups that you identified in question 1.

3. Using the structures from question 1, modify all of the acidic functional groups to show their ionized forms and in the table below identify the normal pKa range for the specific functional group. Drug Name Experimental oral anticoagulant Acidic Functional Group

NormalBromfenac pKa Range

Experimental Oral Anticoagulant Bromfenac Sorbinil Experimental Antidiabetic Agent

4. Using the structures from question 1, modify all basic functional groups to show their ionized forms and in the table below identify the normal pKa range for the specific functional group. Sorbinil Drug Name

Basic Functional Group

Experimental antidiabetic agent Normal pKa Range

Experimental Oral Anticoagulant Bromfenac Sorbinil below is the structure of clonidine, an D adrenergic agonist that can be used to treat hypertension. 5. Shown Experimental Antidiabetic Agent

Clonidine contains a guanidine functional group (highlighted in bold) that has a pKa of 8.3. Other guanidine functional groups, such as that seen with arginine are much more basic with a pKa of 12.5. Provide a chemical 5. Shown below is the structure of clonidine, an α2 adrenergic agonist that can be used to treat hypertension. Clonidine contains a guanidine functional group (highlighted in bold) that has a pKa=8.3. explanation for this difference. Other guanidine functional groups, such as that seen with arginine, are much more basic with a pKa=12.5. Provide a chemical explanation for this difference. pKa = 12.5

pKa = 8.3

Clonidine



Arginine

Page1of2

1.2 Identifying Acidic and Basic Functional Groups

9

6. For each of the drug molecules shown below, determine if it is an acidic drug molecule, a basic drug molecule,

6. For each of the drug molecules shown below, determine if it is an acidic drug molecule, a basic drug

an amphoteric molecule, or drug a nonelectrolyte. molecule, drug an amphoteric molecule, or a nonelectrolyte. O N H

H 2N

N H

N N H

Lomitapide

O

COOH

O

NH

CF3

N O

NH

S

O

C H3

H N

CF3

C H3

Argatroban HO

OH

O

C H3

H

Cl C H3

NH

O

H O

H3C

OCH3

COOH

H 2N

N N

N H

N H2

N H2

Amiloride

HO

Pravastatin

S O N

O

C H3 O

CH2 N(CH3 )2

Diltiazem Drug Molecule

Acid/Base Character of Drug Molecule

Lomitapide Argatroban Pravastatin Amiloride Diltiazem

Answers can be found in Section 2.2 [this will be linked to section 2.2 title]

Page2of2 Note: The questions in this chapter are related to Chapter 3 in Harrold MW and Zavod RM, Basic Concepts in

Medicinal Chemistry, American Society of Health-System Pharmacists, ©2013.

Section 1 General Self Assessment

1.3 Solving pH/pKa Problems 1. Shown below are the structures of cefotaxime, nitrofurantoin, atenolol, and ezetimibe. Each of these drug 1. Shown below are the structures of cefotaxime, nitrofurantoin, atenolol, and ezetimibe. Each of these drug moleculescontains contains one ionizable functional group. The pKa values provided. molecules one ionizable functional group. The pK havehave beenbeen provided. a values a. Match the pKa values provided with the appropriate functional groups. For each functional group, b. For the each functional indicate whetherititiswould identify name of thegroup group and whether acidicbeorprimarily basic. ionized or primarily unionized at a b. Forfor each functional indicate whether it would primarily primarily at a cefotaxime at agroup plasma pH of 7.4, nitrofurantoin atbe a urinary pH ionized of 6.1, aor stomach pH unionized of 1.8. stomach pH=1.8, a urinary pH=6.1, or a plasma pH=7.4. Provide an explanation for your responses for cefotaxime at a plasma pH=7.4, nitrofurantoin at a urinary pH=6.1, and atenolol at a stomach pH=1.8. 

Cefotaxime pKa = 3.4

Nitrofurantoin Nitrofurantion pKa = 7.1

Atenolol pKa = 9.6

Drug (pKa Value)

Ezetimibe pKa = 10.2

Stomach (pH=1.8)

Urine (pH=6.1)

Plasma (pH=7.4)

Cefotaxime (3.4) 3. Shown below is the structure of natamycin. It contains two functional groups that could be potentially ionized. Nitrofurantoin (7.1)

The pKa values for natamycin are 4.6 and 8.4.

Atenolol (9.6)

Ezetimibe (10.2)

H3C

OH

O

O O

OH

OH

O

COOH

11

O

O

C H3

12

Medicinal Chemistry Self Assessment

Atenolol pKa = 9.6

Ezetimibe pKa = 10.2 2. In the previous question, we examined four pKa values in three different environments for a total of 12 different scenarios. Which of these 12 scenarios allow you to use the Rule of Nines to calculate the percent of ionization of the functional group in the specific environment? Identify the specific scenarios and use the Rule of Nines to calculate the percent of the functional group that would be ionized. 3. Shown below is the structure of natamycin. It contains two functional groups that could be potentially ionized.

The pKa values for natamycin are 4.6 and 8.4. 3. Shown below is the structure of natamycin. It contains two functional groups that could be potentially ionized. The pKa values for natamycin are 4.6 and 8.4.

H3C

OH

O

O O

OH

OH

O

O

COOH

O

C H3

Natamycin HO

OH N H2

a. Match the pKa values provided to the appropriate functional groups and identify if the functional group is acidic or basic. b. Using the Henderson-Hasselbalch equation, calculate the percent ionization that occurs for each of these functional groups at an intestinal pH=6.2. 4. The most basic functional group present within the structure of ranitidine has a pKa value of 8.2. Identify this

functional group and calculate the pH that is necessary for this functional group to be 70% ionized.

Page1of2

4. The most basic functional group present within the structure of ranitidine has a pKa=8.2. Identify this functional group and calculate the pH that is necessary for this functional group to be 70% ionized.

Ranitidine

Answers can be found in Section 2.3 [this will be linked to section 2.3 title] Note: The questions in this chapter are related to Chapter 4 in Harrold MW and Zavod RM, Basic Concepts in Medicinal Chemistry, American Society of Health-System Pharmacists, ©2013.

Section 1 General Self Assessment

1.4 Salts and Solubility

1.4 molecules Author Query 1. Some drug can beResponse formulated as either a potassium salt or hydrochloride salt. Evaluate all of the acidic and basic functional groups in each of the drug molecules drawn below and fill in the grid with the appropriate information. OH

H N

H 2N CH 3

HO H

OCH3

OH

H

N O

O

Cl Arformoterol

Baclofen

O N H 3C

F N

O H N

N N

N

N

CO2 H

N

HN

Irbesartan

Ciprofloxacin

H 3C HO

CH 3

H

N

CH 3 OH NH 2

OH

OH OH O

O

Tetracycline

  13



O



14

Medicinal Chemistry Self Assessment

Name of Drug Molecule

Is the Potassium Name of Functional Group That Can Form Salt Acidic, Basic, or Neutral? a Potassium Salt

Is the Hydrochloride Salt Acidic, Basic, or Neutral?

Name of Functional Group That Can Form a Hydrochloride Salt

2. Each of these drug molecules will treat a particular ailment by either managing a symptom or modulating a biochemical pathway. In either case, the drug has to get to its biological target. Using your knowledge about functional group character, describe how both of the drugs get to their respective 2. Eachtargets. of theseConsider drug molecules will treat a particular ailment by either managing a symptom or modubiological the concepts of solubility, absorption, distribution, and route of administration lating a biochemical pathway. In either case, the drug has to get to its biological target. Using your knowledge about functional group character, describe how in your answer. [pH (stomach) = 1; pH (intestine) = 8; pH (plasma) = both 7.4] of the drugs get to their respective biological targets. Consider the concepts of solubility, absorption, distribution, and route of administration in your answer. [pH (stomach) = 1; pH (intestine) = 8; pH (plasma) = 7.4] NCH3

Cl OH

O

O

N

CH2OH

O

CONH2

Scopolamine

Loperamide

Biological Targets and Routes of Administration 3. Based on your µstructural evaluation, provide a rationale for whyoral each of these drugs cannot be Loperamide: opioid receptors located in the large intestine; administration

delivered via an oralmuscarinic route of administration, or why there limited in absorption of thenervous drug when it is Scopolamine: acetylcholine receptors (M1)islocated the peripheral systems; transdermal administration administered orally. 3. Based on your structural evaluation, provide a rationale for why each of these drugs cannot be delivered via an oral route of administration, or why there is limited absorption of the drug when it is administered orally. Insulin Structure

PO3Na2 S

S

PO3Na2

H2N

A Chain

O H Asn Tyr Cys Asn COOH H2N Gly Ile Val Glu Gln Cys Cys Thr Ser Ile Cys Ser Leu Tyr Gln Leu Glu S H2 N Phe Val Asn Gln His

Insulin

Alendronate

S

Leu Cys Gly Ser His

Leu Val Glu Ala Leu Tyr

S

Leu Val Cys

B chain Gly Glu Arg Gly

HOOC



S

Thr30

Lys29 Pro28

Thr27

Tyr

Phe Phe

nistered orally. 1.4 Salts and Solubility

PO3Na2 PO3Na2

H2N OH

Alendronate Insulin 4. Based on your structural evaluation of embeconazole, answer the following questions: 4. Based on your structural evaluation of embeconazole, answer the following questions:

Embeconazole

Can it be formulated as: a. a topical ointment or cream?

YES NO

b. an aqueous solution?

YES NO

c. Which structural features and corresponding characteristics did you consider when answering each of these questions?

Answers can be found in Section 2.4 [this will be linked to section 2.4 title] Note: The questions in this chapter are related to Chapter 5 in Harrold MW and Zavod RM, Basic Concepts in Medicinal Chemistry, American Society of Health-System Pharmacists, ©2013.

15

Section 1 General Self Assessment

1.5 Drug Binding Interactions

1. Both delapril and lisinopril are inhibitors of angiotensin converting enzyme (biological target) and as such are valuable drugs used in the management of hypertension. The active form of these drugs requires the presence of two carboxylic acids or two functional groups that can participate in similar types of interactions at physiological pH. One of the carboxylic acids interacts with an active site arginine residue, and the other interacts with a Zn+2 atom. Both of these interactions are critical for drug action. Do the following: a. Modify the molecules below to show the form of the active drug at physiological pH (pH=7.4). b. Identify the type1.5 ofAnswer interaction possible between the carboxylic acids and the residues present within to Author Queries the biological target.



Zn+2

Delapril Delapril



 Zn+2

Lisinopril Lisinopril



17

18

Medicinal Chemistry Self Assessment

2. Tolterodine (Detrol®), fesoterodine (Toviaz®), and oxybutynin (Ditropan®) are anticholinergic agents 2. Tolterodine (Detrol®), fesoterodine (Toviaz®), and oxybutynin (Ditropan®) are anticholinergic agents used inin thethe treatment of overactive bladder. Based on the evaluation process described in used treatment of overactive bladder. Based onstructure the structure evaluation process described in Chapter 6* and previous chapters,* “read” each of these drug molecules and determine how each of these drug molecules interact via hydrogen bonding with the muscarinic receptor. Fill in the grid provided with your answers.

Tolterodine



Fesoterodine





Oxybutynin

Function Function Name of Functional Group

Name of ALL Anticholinergics That Contain Functional Group

One amino acid that interacts (via side chain) with the functional group via a H-bond donor, H-bond acceptor, complementary H-bonding interaction; NONE is a possible answer both or neither

1.5 Drug Binding Interactions

19

3. Each of the three odorant molecules drawn below produces a unique scent interaction with the 3. Each of the three odorant molecules drawn below produces a unique scent on on interaction with the olfactory receptors. Unlike most biological targets for drug action, olfactory receptors typically have olfactory Unlike most biologicalmolecules targets forand drug receptors typically have an an affinityreceptors. for a wide variety of odorant canaction, adoptolfactory unique conformations to enhance the affinity of a given odorant for the receptor. Based on the structural features found in each molephysiological pH? Indicate which functional group(s) can participate in each of the interactions identified. cule, what type of interactions are possible with the olfactory receptors at physiological pH? Indicate which functional group(s) can participate in each of the interactions identified.

Sclareol (herbal scent)



Sclareol Interaction Type

Nerolidol (green (greenwoody woody scent) scent)

Vanillin

Vanillin Functional Group

Nerolidol

Interaction Type

Functional Group

Interaction Type

Functional Group

van der Waals, van der Waals,sweet, bitter, and umami 4. Waals, There are five basic flavors van thatderour taste receptors detect: salty, sour, Hydrophobic Hydrophobic Hydrophobic

(savory). locatedbond on taste buds that are on our tongue, soft palate, epiglottis, and Hydrogen bond The taste receptors areHydrogen Hydrogen bond (acceptor and donor) (acceptor) (acceptor and donor) Ion–dipole (as the dipole)

Hydrogen bond (donor)

CO2 HIon–dipole (as the dipole)

H 2N

CO2 H

Cucumber flavor

Glutamic Acid

Green Pepper Flavor O

N

N

O HO HO

N

P O

O

HO

N

OH

Shitaki Mushrooms Inosinic Acid

H

Sclareol (herbal scent) Vanillin  20 Medicinal Chemistry Self Assessment

Nerolidol (green woody scent)

4. There are five basic flavors that our taste receptors detect: salty, sour, sweet, bitter, and umami (savory). The taste receptors are located on taste buds that are on our tongue, soft palate, epiglottis, and are upper Sour andour salty flavors are mediated by ionsour, channels, sweet, bitter, 4. There fiveesophagus. basic flavors that taste receptors detect: salty, sweet,whereas bitter, and umami and umami flavors are derived from activation of the respective G-protein coupled receptor. The umami is activated by L-amino acids, interactions (savory). The receptor taste receptors are located on taste budsspecifically that are onglutamate. our tongue,Determine soft palate,which epiglottis, and are possible with the side chain of glutamate (at physiological pH) and then determine if any of the following flavor molecules can interact with and activate this receptor. CO2 H

H 2N

CO2 H

Cucumber flavor

Glutamic Acid

Green Pepper Flavor O

N

N

O HO HO

N

P O

O

HO

H

N

OH

Shitaki Mushrooms Inosinic Acid

Answers can be found in Section 2.5 [this will be linked to section 2.5 title] Note: References to chapters are to Chapter 6 in Harrold MW and Zavod RM, Basic Concepts in Medicinal Chemistry, American Society of Health-System Pharmacists, ©2013.

Section 1 General Self Assessment

1.6 Stereochemistry and Drug Action

1.6 Stereochemistry and Drug Action

1. Shown below are the structures of acebutolol, estradiol, cefamandole, and nifedipine. For each of these compounds, identify all chiral centers. 1. Shown below are the structures of acebutolol, estradiol, cefamandole, and nifedipine. For each of these compounds, identify all chiral centers.

O

N

O

OH

OH

HO

HN

Estradiol

O

Acebutolol H3C

O

OH

S

N H O

C H3 N

S

N CO2 H

N

H N

H3 CO2 C

CO2 CH3 NO2

N

N

Cefamandole

Nifedipine

2. For each of the four drug molecules shown in question 1: a. Identify if it can have an enantiomer. Provide an explanation for your response. b. Identify if it can have a diastereomer. Provide an explanation for your response. c. Identify if it can have a geometric isomer. Provide an explanation for your response.

21

C H3

22

Medicinal Chemistry Self Assessment

3. Shown below is the structure of fluvastatin, an HMG-CoA reductase inhibitor used to lower plasma LDL levels. Fluvastatin contains two chiral centers, designated as A and B. Using the structure of fluvastatin and the CahnMedicinal Chemistry Self-Assessment Book: Batch Two 3. Shown below is the structure of fluvastatin, an HMG-CoA reductase inhibitor used to lower plasma Chapters 1.6 and 2.6 LDL levels. Fluvastatin contains the twoR/S chiral centers, designated and 2.centers. Using the structure of Ingold-Prelog (CIP) system, determine configurations for each ofasits1 chiral fluvastatin and the Cahn-Ingold-Prelog (CIP) system, determine the R/S#1configurations for each of its Revised structure reflects chiral carbons and #2. chiral centers.

H O A1 2 B

F

CO2 H OH H C H3

N

Fluvastatin

C H3



Fluvastatin

4. Shown below is the enantiomer of fluvastatin. Which of the following properties/actions would be expected to be identical for fluvastatin and its enantiomer and which would be expected to be 4. Shown below is the enantiomer of fluvastatin. Which of the following properties/actions would be expected to different? a. Hepatic metabolism be identical for fluvastatin and its enantiomer and which would be expected to be different? b. Water solubility a. Hepatic metabolism c. Adverse effect profile Waterrenal solubility d. b. Active reabsorption by transport proteins e. Potency (dosage given) f. Percent ionization at a pH=7.4

Enantiomer of Fluvastatin

Answers can be found in Section 2.6 [this will be linked to section 2.6 title] Note: The questions in this chapter are related to Chapter 7 in Harrold MW and Zavod RM, Basic Concepts in Medicinal Chemistry, American Society of Health-System Pharmacists, ©2013. 

Page2of2

Section 1 General Self Assessment 1.7

1.7 Drug Metabolism

Drug Metabolism 1.7

1. Heroin is a synthetic derivative of the naturally occurring opioid analgesic morphine. Given its illicit into Drug Metabolism morphine. Which phase I transformation is responsible for the conversion of heroin to morphine? 1. Heroin is a synthetic derivative of the naturally occurring opioid analgesic morphine. Given its illicit nature, heroin is not subject to Food and Drug Administration (FDA) regulation; therefore, the actual composition of1.aHeroin given heroin batch derivative varies significantly. The three molecules below represent theitsthree is a synthetic of the naturally occurring opioid drawn analgesic morphine. Given illicit major into components found within a batch of heroin. The users’ intense rush has been attributed to the conversion of heroin intophase morphine. Which phase I transformation is responsible for the morphine. Which I transformation is responsible for the conversion of heroin toconversion morphine? of heroin to morphine?

Heroin

Morphine

Heroin

Morphine 2. Estradiol, the estrogen component of many oral contraceptives, undergoes a phase II conjugation

reabsorbed (at least in part). Consider the structure of estradiol drawn below and do the following: 2. Estradiol, the estrogen component of many oral contraceptives, undergoes a phase II conjugation reaction to produce a metabolite that is eliminated via a fecal route (at least in part). The conjugated hormone a. Modify the structure drawn below to show the product of a sulfate conjugation (phase II 2. Estradiol, the estrogen component of many oral In contraceptives, undergoes a phase is II cleaved conjugation undergoes a process called enterohepatic recycling. this process the sulfate conjugate by gut bacteria to regenerate the active drug, which is then reabsorbed (at least in part). Consider the structure of transformation). reabsorbed (at least in and part).do Consider the structure of estradiol drawn below and do the following: estradiol drawn below the following: b. the Which enzymedrawn is required toto make this sulfate conjugate? a. Modify structure below show the product of aofsulfate conjugation a. Modify the structure drawn below to show the product a sulfate conjugation(phase (phaseIIII transformation). c. Which deconjugating enzyme catalyzes removal of the sulfate group thus allowing for transformation). b. Which enzyme is required to make this sulfate conjugate? enterohepatic recycling? c. Which catalyzes of the sulfate group thus allowing for b.deconjugating Which enzyme enzyme is required to makeremoval this sulfate conjugate? enterohepatic recycling? c. Which deconjugating enzyme catalyzes removal of the sulfate group thus allowing for enterohepatic recycling?

23

Estradiol



Estradiol



24

Medicinal Chemistry Self Assessment

3. Metabolites do not necessarily have the same mechanism of action as the parent drug. In the case of chlorimipramine (a tricyclic antidepressant that inhibits serotonin uptake), a phase I transformation produces a metabolite that is also a tricyclic antidepressant, but whose mechanism of action is via 3. Metabolites do not of action as the parent drug. In the case of I transformations are possible? inhibition of norepinephrine reuptake. Which phase I transformation has occurred? What additional 3. Metabolites do not of action as the parent drug. In the case of I transformations are possible? phase I transformations are possible?

Chlorimipramine

Chlorimipramine



 4. Evaluate each of theoffollowing transformations and determine which phase I metabolic 4. Evaluate each the phase Imetabolic metabolic transformation has occurred. transformation has occurred. 4. Evaluate each of the phase I metabolic transformation has occurred. H N H CN H3 CF3

C H3

N H2 C H 3N H

C H3

C H3

CF3 Dexfenfluramine

CF3

CF3

Dexfenfluramine

Fluvoxamine



Fluvoxamine

2

C H3











Baclofen

Baclofen

Answers can be found in Section 2.7 [this will be linked to section 2.7 title] Note: The questions in this chapter are related to Chapter 8 in Harrold MW and Zavod RM, Basic Concepts in Medicinal Chemistry, American Society of Health-System Pharmacists, ©2013.

Part 1 QUESTIONS

Section 2 – Whole Molecule Drug Evaluation 1.8 1.9 1.10 1.11 1.12 1.13 1.14 1.15 1.16 1.17 1.18 1.19 1.20

Aliskiren....................................................................... 27 Aripiprazole.................................................................. 31 Cefprozil....................................................................... 35 Cetirizine...................................................................... 39 Chlorpropamide and Other Sulfonylureas.................... 41 Dabigatran Etexilate.................................................... 43 Fenofibrate and Gemfibrozil......................................... 47 Fluvoxamine................................................................. 51 Haloperidol.................................................................. 55 Hydrocortisone............................................................. 57 Levothyroxine (T4)........................................................ 61 Lidocaine...................................................................... 65 Montelukast and Zafirlukast........................................ 69

1.21 1.22 1.23 1.24 1.25 1.26 1.27

Phenobarbital and Other Barbiturates......................... 71 Pravastatin and Fluvastatin.......................................... 73 Quinapril...................................................................... 75 Rivastigmine................................................................ 79 Sitagliptin..................................................................... 83 Sorafenib...................................................................... 87 Zanamivir and Oseltamivir........................................... 91

Section 2 Whole Molecule Drug Evaluation

1.8 Aliskiren

Aliskiren is an orally active agent used in the treatment of hypertension. This non-peptide drug acts as an inhibitor of renin, the enzyme that converts angiotensinogen (its endogenous substrate) to angiotensin I. Biologically inactive, angiotensin I is rapidly converted to angiotensin II by angiotensin converting enzyme. Angiotensin II is a potent agonist when bound to its receptor and produces significant vasoconstriction, as well as an increase in blood pressure. In the presence of aliskiren, angiotensinogen is not converted to angiotensin I so less angiotensin II is produced to activate the angiotensin II receptor. Consequently, less vasoconstriction occurs, and a drop in blood pressure results. Medicinal Chemistry Self-Assessment Book: Batch Two Chapters 1.8 and 2.8 1. Conduct a structural evaluation of aliskiren, focusing on the boxed functional groups, and use the infor(removetobold drug name x2) mation in the grid toChapter inform 1.8/2.8 your answers the from questions that follow.

D

A

E

F

B

Aliskiren

C

Function Character

Name of Functional Group

Function

Character

Acidic, Basic, or Neutral

Function

Hydrophilic and/or Hydrophobic

Provide pKa When Relevant

↑ Solubility and/or ↑ Absorption

Interaction(s) Possible with Biological Target at Physiological pH=7.4

Aliskiren

A B C D E F

27

Amino Acids That Can Interact with the Functional Group via Ion–Dipole Interactions at pH=7.4 None Is Acceptable

B 28

Medicinal Chemistry Self Assessment

Aliskiren 2. Aliskiren is marketed as the pure 2S, 4S, 5S,C7S enantiomer. Circle all of the chiral carbon atoms and determine if diastereomeric or geometric isomers are possible.

Aliskiren

3. Although aliskiren is administered orally, its oral bioavailability is ~2.5% and is very poorly absorbed. Using the information in the structure evaluation grid, provide a structural rationale for this unfortunate property. 4. Approximately 25% of the absorbed dose of aliskiren is excreted in the urine unchanged. It is unknown how much of an absorbed dose is metabolized, and several metabolites from CYP3A4 mediated transformations have been identified. Several possible metabolic products are illustrated. Identify which metabolic transformation has occurred and whether or not it represents an oxidative transformation.

1.8 Aliskiren Missing diagram for Question #4 (Chapter 1.8/2.8)

A

B

C

D

E

Name of Metabolic Transformation A B C D E

Oxidative or Non-Oxidative

29

2. 30

Medicinal Chemistry Self Assessment 2.

2.

Aliskiren 5. Renin catalyzes the cleavage of a specific Leu-Val peptide bond within the structure of angiotensin2. The structure of aliskiren contains functional groups that mimic the side chains for these two ogen. amino acids. The hydrolyzable peptide bond (found between Leu-Val) has been replaced by a non-

5.hydrolyzable Renin hydroxyethylene group. 2. non-hydrolyzable hydroxyethylene group in aliskiren. Circle the functional groups that mimic the amino acid side chains of these two amino acids and box the non-hydrolyzable hydroxyethylene group. 2.

2. 2.

(remove bold from drug name) Chapter 2.8

2. 2. 2. 2. 2. 6. Aliskiren can be co-administered with other anti-hypertensive agents to provide better hypertension management. Amlodipine is a second generation dihydropyridine calcium channel blocker used in 2. Aliskiren the treatment of hypertension. Aliskiren and amlodipine are both plasma protein bound (47–51% and2. 93–97%, respectively). How likely is it that a plasma protein binding interaction will occur if these drugs are co-administered? 2. (remove bold from drug name) Chapter 1.8/2.8

Amlodipine

Answers can be found in Section 2.8 [this will be linked to section 2.8 title]

Section 2 Whole Molecule Drug Evaluation

1.9 Aripiprazole 1.9 Aripiprazole Shown below is the structure of aripiprazole, a serotonin receptor modulator used for the treatment of depression, schizophrenia, autism, mania, and bipolar disorder. Shown below is the structure of aripiprazole, a serotonin receptor modulator used for the treatment of depression, schizophrenia, autism, mania, and bipolar disorder. H N Cl

O

O

Cl N

N

Aripiprazole

1. Identify of theallacidic 1. all Identify of theand pH basic of 5.6.functional groups, provide the normal pKa range for each of the identified functional groups, and identify if each functional group would be primarily ionized or unionized at a solubility of aripiprazole. urine 2. pH=5.6.

3. Using yur answers previously listed. 4. all Drug hly aripiprazole and either groups, tolbutamide losartan? 2. Identify of the remaining functional andor indicate how each group contributes to the overall water solubility or the overall lipid solubility of aripiprazole. 3. Using your answers from the previous two questions, provide an explanation as to why aripiprazole can be administered orally for the indications previously listed.

Tolbutamide

Losartan

5. Assume that at a physiological pH of 7.4.

31

A

D

N 32

Aripiprazole

N

Medicinal Chemistry Self Assessment

1. Identify all of the pH of 5.6.

Aripiprazole

2. solubility of aripiprazole. 4. 1. Drug molecules that areofpreviously highly 3.Identify Usingall yurofanswers listed.protein bound may undergo a drug interaction due to the the pH 5.6. plasma displacement of one drug molecule from a plasma protein by another drug molecule. Shown below 4.solubility Drug hlyofaripiprazole and either tolbutamide or losartan? 2. aripiprazole. are the structures of tolbutamide and losartan. Both of these drugs are highly plasma protein bound. Aripiprazole is also highly plasma protein bound (greater than 99%). Would you expect there to be a 3. Using yur answers previously listed. drug interaction between aripiprazole and either tolbutamide or losartan? 4. Drug hly aripiprazole and either tolbutamide or losartan?

Tolbutamide Tolbutamide

Losartan

Losartan 5. Assume that the boxed functional groups of aripiprazole form four key binding interactions with 5. Assume that at a physiological pH of 7.4. a serotonin receptor. Further assume that these binding interactions occur with the side chains of Tyr, Asp, Ile, and Gln. Using this information, identify four possible binding interactions between aripiprazole andatthe given aminopH acids. Assume that this binding interaction occurs at a physiological 5. Assume that a physiological of 7.4. D pH=7.4.

A C B

A

C B

Aripiprazole Aripiprazole

D

1.9 Aripiprazole

33

6. Shown below are three known metabolites of aripiprazole. Identify the metabolic transformations that 6. Shown below are three known metabolites of aripiprazole. Identify the metabolic transformations would be required to form each of the metabolites. For each metabolic transformation, indicate if it is that would be required to form each of the metabolites. For each metabolic transformation, indicate aifphase I transformation or a phase transformation. it is a phase I transformation or aII phase II transformation. O

H N Cl

O

Cl N

HO

N

Metabolite A H N Cl

O

O

Cl

OH N

N

Metabolite B H N Cl

O

O

Cl

OH N

NH

HO O

Metabolite C

Answers can be found in Section 2.9 [this will be linked to section 2.9 title]

Section 2 Whole Molecule Drug Evaluation

1.10 Cefprozil

Cefprozil is a second-generation cephalosporin that exhibits good Gram (+) activity with improved Gram (–) activity Medicinal Chemistry Self-Assessment Book: Batch Two as compared to the first-generation cephalosporins. Effective against the majority of bacteria that cause upper and Chapters 1.10 and 2.10 lower respiratory infections, as well as skin infections, cefprozil was a first-line anti-infective agent until an increase in the incidence of resistance and the development of (remove newer agents decreased its “favored” status. Chapter 1.10/2.10 bold from drug name)

Cefprozil (Cefzil)

1. Conduct a complete structural evaluation ofCorrected cefprozil structure and use the information in the grid to inform your Chpater 1.10/2.10 (product of E transformation) answers to the questions that follow.

Character Character Name of Functional Group

Acidic, Basic, H O or Neutral

Hydrophilic and/or Hydrophobic HO

Function

Function N

HO

H H↑ Solubility S Provide pKa N and/or When Relevant ↑ Absorption O N

H

NH 2

HO

H H

S Function

Amino Acids That Interaction(s) O N Can Interact with CH 3 Possible with O Functional Group via Biological Target CO 2H H-Bonding (at pH=7.4) at Physiological N H2 pH=7.4 None is Acceptable OH

C H3

O

CO2 H

O

HO

H 2N

B

A

H

H N

O

CO2 H

C N H2

H N

O

HO

H H

S

N

O

S

C H3 CO2H

NH 2

H N

HO

O

O

H H

D

F

S

E

N

CH 3

N H2

O 35 N H

O

O -O

S

H N

O

O

H

H N

S

36

Medicinal Chemistry Self Assessment

2. Based on the information in the structure evaluation grid, determine if cefprozil is an acidic, basic, or amphoteric drug. Provide a brief explanation for your answer. 3. Cefprozil is administered orally as a tablet or liquid suspension. Consider each of the acidic and basic functional groups and determine whether each group will be predominantly ionized or unionized as it moves through the gastrointestinal (GI) tract, into systemic circulation, and then into the urine. [The relevant pKa values=10, 1.7, and 7.2.] Complete the grid below.

Name of Functional Group

Acidic or Basic (pKa)

Ionized or Unionized at pH=5 (saliva)

Ionized or Unionized at pH=1 (stomach)

Ionized or Unionized at pH=7.4 (plasma)

Ionized or Unionized at pH=8 (intestine)

Ionized or Unionized at pH=6 (urine)

4. Given the predominant ionization state of these acidic and basic functional groups as they traverse the GI tract and the information in the structure evaluation grid, determine in which GI compartment(s) drug absorption could occur.

1.10 Cefprozil

37

5. Approximately 60% of a cefprozil dose is recovered in the urine unchanged. Because impairment in Cefprozil (Cefzil) the half-life of the drug by several hours, it is likely that cefprozil underhepatic function increases goes a variety of metabolic transformations catalyzed by liver enzymes. For each of the metabolic transformations A–F, identify which metabolic transformation has occurred and whether the product Chpater Corrected (product E transformation) formed1.10/2.10 was the result of a structure phase I or phase of II metabolic transformation.

H

O

N O

HO

N O

HO H H

H H

S

N

O

CH 3 CO2 H

S

N H2

N

O

H

NH 2

HO

OH

C H3 CO2 H

O

HO

H 2N

B

A

H

H N

O

H

H H

N O

HO

S

N

O

CH 3 CO2 H

C N H2

S

C H3 CO2H

NH 2

H N

HO

O

O

H H

S

E

N O

D

F

CH 3

N H2

N H

O

O -O

OH NH 2

H N

HO

O

O

H H

S O

O

OH CO2 H

Name of Transformation A B C D E F

O

O

H

H

S

N

C H3 CO2 H

S

N

H N

Phase I or Phase II

38

Medicinal Chemistry Self Assessment

6. Like the penicillins, the cephalosporins suffer from chemical instability of the β-lactam bond. Chemical hydrolysis of this bond renders the drugs in the class of anti-infective agents inactive. This bond is also subject to cleavage by β-lactamases (due to the presence of a nucleophilic serine side chain [CH2OH] within the active site of the enzyme). Show how the β-lactam bond can be hydrolyzed by chemical and enzymatic (β-lactamase) mechanisms.

Answers can be found in Section 2.10 [this will be linked to section 2.10 title]

Section 2 Whole Molecule Drug Evaluation

1.11 Cetirizine

Cetirizine is a popular second-generation antihistamine used in the management of allergy symptoms. It is the metabolic byproduct produced Medicinal from the prescription hydroxyzine. Although Chemistryantihistamine Self-Assessment Book: Batch Twocetirizine is labeled nonsedating and is one of the preferred allergy medications for long-haul hydroxyzine causes significant drowsiChapters 1.11 and drivers, 2.11 ness that limits its utility in the management of typical allergy symptoms. Hydroxyzine is commonly used in the treatment of pruritus (severe itching). Chapter 1.11 (remove bolded drug names)

Cetirizine

Hydroxyzine

1. Conduct a complete structural evaluation of hydroxyzine and use the information in the grid to inform your answers to the questions that follow. Chapter 2.11 (remove bolded drug names) Function

A

C

Character Hydrophilic and/or Hydrophobic

Name of Functional Group

B

Function

Character E D

Acidic, Basic, or Function Neutral ↑ Solubility and/or Provide pKa When Relevant ↑ Absorption

Hydroxyzine

39

Amino Acids That Can Interact with Interaction(s) Functional Group Possible with Biological Target via H-Bonding (at pH=7.4) at Physiological None Is Acceptable pH=7.4

40

Medicinal Chemistry Self Assessment

2. Name the phase I metabolic transformation(s) that hydroxyzine undergoes to produce cetirizine. 3. Based on your structural evaluation of both hydroxyzine and cetirizine, name ALL of the phase I metabolic transformations possible. 4. A metabolic product from a phase II metabolic transformation has been identified. Which phase II transformations can cetirizine undergo? 5. Review the structure of cetirizine (pKa=2.9 and 8.3) and identify all of the acidic and basic functional groups present. Determine the predominant ionization state of each functional group as it travels through several compartments of the body after oral administration. Complete the table below. Name of Functional Group

Acidic or Basic (pKa)

Ionized or Unionized at pH=5 (saliva)

Ionized or Unionized at pH=1 (stomach)

Ionized or Unionized at pH=7.4 (plasma)

Ionized or Unionized at pH=8 (intestine)

Ionized or Unionized at pH=6 (urine)

6. Provide a structural rationale for why hydroxyzine is classified as a sedating antihistamine and cetirizine is categorized as a non-sedating antihistamine.

Answers can be found in Section 2.11 [this will be linked to section 2.11 title]

Section 2 Whole Molecule Drug Evaluation

1.12 Chlorpropamide and Other Sulfonylureas 1.12 Chlorpropamideand Other Sulfonylureas Shown below are the structures of tolbutamide and chlorpropamide. These drug molecules are used in the treatment of type 2 diabetes mellitus. They stimulate the release of insulin from the pancreatic β-cells by interacting with Shown below are the structures of tolbutamide and chlorpropamid of ATP)-sensitive potassium channels. adenosine triphosphate (ATP)-sensitive potassium channels. 1.12 Chlorpropamideand Other Sulfonylureas Shown below are the structures of tolbutamide and chlorpropamid of ATP)-sensitive potassium channels.

Tolbutamide

Chlorpropamide

1. It is not uncommon for patients with type 2 diabetes to be dually diag and require additional occur? 1. It is not uncommon for patients with type 2 diabetes to be dually diagnosed with hypertension and Tolbutamide Chlorpropamide require additional pharmacotherapy. In this scenario, it is possible for drug interactions to occur if the prescribed combination therapy is not appropriately evaluated. Angiotensin II receptor antagonists, commonly known as angiotensin II receptor blockers (ARBs), are often used to treat hypertension. Losartan (shown below) is an ARB, and similar to tolbutamide and chlorpropamide, is highly plasma 1.protein It is not uncommon for patients with type diabetes to be dually and require occur? bound. If losartan was selected for2use in a patient who diag is already takingadditional tolbutamide or chlorpropamide, would you anticipate that a drug interaction could occur?

Losartan

Losartan comparing the pKa values of chlorpropamide. 2. The normal pKa range for sulfonylureas is 5 to 6. When 3. Using your answer from percent of tolbutamide that will be ionized at an intestinal pH of 6.1. sulfonylureas is 5 to 6.groups. When comparing the pKa values of chlorpropamide. 2. The pKa rangeofforaction 4. normal The mechanism of ion functional 3. Using your answerhas from percent has of tolbutamide thatlonger will behalf-life ionizedthan at antolbutamide. intestinal pH of 6.1. 5. Tolbutamide a half-life a significantly 4. The of is action of ionpathways functionalthat groups. 6. mechanism Shown below a known are required to produce this metabolite. 41

5. Tolbutamide has a half-life has a significantly longer half-life than tolbutamide.

42

Replacementstructurefortheanswertoquestion2inChapter2.9(Aripiprazole) Medicinal Chemistry Self Assessment

 2. The normal pKa range for sulfonylureas is 5 to 6. When comparing the pKa values of chlorpropamide Alkyl (aliphatic) chain and tolbutamide, it is found that the(Lipid sulfonylurea soluble) of one of these drug molecules has a pKa=5.4 and the other has a pKa=4.9. Evaluate the structures of Ether theseoxygen two drug molecules, assign the pKa values to soluble) Amide the correct molecules, andHalogens provide an explanation(Water for the difference in pKa values. (Water soluble)

(Lipid soluble)

3. Using your answer from question 2, calculate the percent of tolbutamide that will be ionized at an Hydrocarbon portion intestinal pH=6.1. of bicyclic ring (Lipid soluble)

Aromatic (phenyl) ring

soluble) of action of this class of drugs involves the ability to interact with the ATP-sensitive 4. The(Lipid mechanism potassium channels in the pancreas. Using the structure of tolbutamide, identify the types of binding Aripiprazole interactions that are possible between its functional groups and a protein ion channel. Also identify amino acids present within this ion channel whose side chains could participate in the interactions identified. Assume a plasma pH=7.4 for all ionizable functional groups. Alkyl (aliphatic) chain (Lipid soluble)

5. Tolbutamide has a half-life of 4.5 to 6.5 hours, whereas chlorpropamide has a half-life of 36 hours. Propose a chemical/structural reason why chlorpropamide has a significantly longer half-life than tolbutamide. Replacement structure for question 6 in BOTH Chapters 1.12 and 2.12 (Chlorpropamide and Other Sulfonylureas) 6. Glyburide is a second generation sulfonylurea that is structurally similar to chlorpropamide and tolbutamide. Shown below are the structures of glyburide and one of its known metabolites. Identify the metabolic pathways that are required to produce this metabolite. Cl

S N H

O H3C

O

O

O

O

N H

N H

Glyburide

O

S

H3C N H

O

OH

O

O N H

N H

Metabolite of glyburide

Answers can be found in Section 2.12 [this will be linked to section 2.12 title]



Section 2 Whole Molecule Drug Evaluation

1.13 Dabigatran Etexilate

Thrombin is the enzyme responsible for catalyzing the conversion of fibrinogen to fibrin. The production of fibrin is important in the formation of sturdy blood clots. As you might expect, inhibition of thrombin prevents the formation of fibrin. As shown in the diagram below, a catalytic triad of amino acids (Asp, His, Ser) found in the active site of thrombin is responsible for hydrolyzing a key peptide bond found within fibrinogen. Orientation of fibrinogen in Concern about of His (revised diagram + removed bolded names) the active site ofChapters thrombin1.13/2.13 relies on the interaction a key tri-peptide sequence (D-Phe-Pro-Arg) found within the structure fibrinogen with key amino acids in the enzyme active site.

Chapters 1.13/2.13 Concern about (revised diagram + removed boldedetexilate names)is an orally active direct Dabigatran etexilate is administered as a His prodrug. In its active form, dabigatran Thrombin Activeand Site thrombin inhibitor used in the prevention of stroke blood clots in patients diagnosed with atrialThrombin fibrillation.Active Site Thrombin Active Site

Thrombin Active Site Thrombin catalyzed amide hydrolysis Thrombin catalyzed amide hydrolysis

Fibrinogen (D-Phe-Pro-Arg)

Fibrin

Fibrinogen (D-Phe-Pro-Arg) Chapters 1.13/2.13 (removed bold in drug name)

Fibrin

Chapters 1.13/2.13 (removed bold in drug name)

Dabigatran etexilate Dabigatran Chapters 1.13/2.13 (removed bold andetexilate centered text)

Chapters 1.13/2.13 (removed bold and centered text) 3

43

44

Chapters 1.13/2.13 Concern about His (revised diagram + removed bolded names)

Medicinal Chemistry Self Assessment

Thrombin Active(prodrug) Site 1. Conduct a complete structural evaluation of dabigatran etexilate and use the information in the grid to inform your answers to the questions that follow. Function

Character

Name of Functional Group

Function

Character

Acidic, Basic, or Neutral

Function

Hydrophilic and/or Hydrophobic

Provide pKa When Relevant

↑ Solubility and/or ↑ Absorption

Thrombin catalyzed amide hydrolysis Amino Acids That

Interaction(s) Possible with Biological Target at Physiological pH=7.4

Fibrinogen (D-Phe-Pro-Arg)

Can Interact with Functional Group via H-Bonding (at pH=7.4) None Is Acceptable

Fibrin

Chapters 1.13/2.13 (removed bold in drug name) 2. Dabigatran etexilate is a non-peptidomimetic prodrug. Provide a brief rationale for the value of converting an active drug into a prodrug. 3. Dabigatran etexilate rapidly undergoes two esterase-catalyzed hydrolytic reactions to the active drug dabigatran. Show the products from each of the esterase-catalyzed hydrolytic reactions that occur in the plasma. Dabigatran etexilate 4. In its active form dabigatran mimics the D-Phe-Pro-Arg tripeptide sequence, but does not contain a peptide backbone (non-peptidomimetic). Review the tripeptide sequence drawn below noting that there are numbers 1–3 that indicate where each amino acid is located in the sequence. Determine the Chapters 1.13/2.13 (removed bold and centered text) types of interactions possible with each of the amino acid side chains.

3

1

2

D-Phe-Pro-Arg Tripeptide Sequence

Throm

1.13 Dabigatran Etexilate

45

5. Review the structure of dabigatran and determine which of the boxed groups will likely mimic the interactions found within the D-Phe-Pro-Arg tripeptide sequence. Given the three-dimensional nature of both peptides and small molecules, it is important to remember that the functional groups within dabigatran do not need to linebolded up in text) the same order. Using the table provided to guide your Chapters 1.13/2.13 (removed Chapters 1.13/2.13 (removed bolded text) the functional group could mimic the analysis, place a Yes or No in each box to indicate whether amino acid side chain indicated.

A

A

C

B

B D

E

E F

D

F

C Active form of Dabigatran Active form of Dabigatran

Chapters 1.13/2.13 (removed bold drug name) 6. Dabigatran etexilate is formulated as(removed a mesylate salt. Provide Chapters 1.13/2.13 bold drug name)a brief rationale for the value of administering the salt form of a drug.

Dabigatran etexilate mesylate Dabigatran etexilate mesylate

Answers can be found in Section 2.13 [this will be linked to section 2.13 title]

Section 2 Whole Molecule Drug Evaluation

1.14 Fenofibrate and Gemfibrozil

Fenofibrate is a member of the fibrate class of anti-hyperlipidemic agents. It is used as adjunctive therapy to diet in the management of dyslipidemias, including in the treatment of severe hypertriglyceridemia. The specific mechanism(s) by which fenofibrate decreases triglyceride and total cholesterol levels, as well as increases the levels of high density lipoproteins (HDL), is unknown. What we do know is that the decrease in very low density lipoproteins (VLDLs) 1.14 and 2.14 (drug name – remove bold) results from fenofibrate stimulation of lipoprotein lipase.

Fenofibrate

Gemfibrozil

1. Conduct a complete structural evaluation of bold fenofibrate and use the information in the grid to inform Letter “C” – add your answers to the questions that follow. Function Character Character Name of Functional Group

Hydrophilic and/or Hydrophobic

Function Function

Acidic, Basic, or Neutral

↑ Solubility and/or ↑ Absorption

Provide pKa When Relevant

C

Interaction(s) Possible with Biological Target at Physiological pH=7.4

2.14 – remove bold from label

B A

F C

D

E Fenofibrate

2.14 – remove bold from label 47

Amino Acids That Can Interact with the Functional Group via Hydrogen Bonding Interactions at pH=7.4 None Is Acceptable

48

Medicinal Chemistry Self Assessment

Fenofibrate 2. Fenofibrate is administered as a prodrug and undergoes hydrolysis to the active drug. Draw the active drug. Provide a brief rationale for the value of administering fenofibrate as a prodrug. 1. Conduct a complete structural follow:

2. calculated Fenofibrate is of administering fenofibrate as a prodrug. 3. The P (drug for fenofibrate is 5.24, whereas the calculated log P for gemfibrozil is 3.9. 1.14 andlog 2.14 name – remove bold) Provide a structural rationale for the difference in this pharmacokinetic property. 3. The calculated in this pharmacokinetic property.

Fenofibrate

Gemfibrozil

Fenofibrate Gemfibrozil Letter “C” – addperspective, bold 4. From an elimination 60% of a fenofibrate dose is found in the urine and 25% is found 1.14 and 2.14 (drug name – remove bold) in the feces. The active form of fenofibrate undergoes both phase I and phase II metabolic transformations. The phase II conjugate is eliminated in the urine. Oxidative metabolism does not occur. Evaluate of the metabolic products it is the conjugate eliminated 4. Fromeach an elimination perspective, 60%and of a determine ve phase I iftransformation (that that doesisNOT occur). in the urine (that does occur), the product of a non-oxidative phase I transformation (that does occur), or the product of an oxidative phase I transformation (that does NOT occur). O

O

HO

O C H3 C

Cl

OH C Fenofibrate H3

Gemfibrozil

2.14 – remove bold from label ALetter “C” – add bold

B A

B

F D

C

E CC

Fenofibrate

A

2.14 – remove bold from label Name Metabolic Transformation Type of Product 2.14of – remove bold from label

B

BO

O

C

A Cl

O H3C

C H3 OC

F

O

Ester Hydrolysis

D

C H3

C H3

O H3C

Cl

E

Inactive

Fenofibrate

2.14 – remove bold from label

O

Active

OH C H3

1.14 Fenofibrate and Gemfibrozil

5. Provide a structural rationale for why oxidative O-dealkylation does not occur. 6. Fenofibrate and gemfibrozil have dramatically different elimination half-lives (20–22 hours and 1.5 Identify the possible metabolic transformations for gemfibrozil and provide a 1.14 and 2.14 (drughours namerespectively). – remove bold) justification for the significant difference in this pharmacokinetic parameter.

Fenofibrate

Gemfibrozil

Letter “C” – add bold

C

2.14 – remove bold from label

B A

F D

C

E Fenofibrate

2.14 – remove bold from label

O

O O

Cl

H3C

Inactive

C H3 O

C H3

C H3

O

Ester Hydrolysis

O O H3C

Cl

Active

OH C H3

49

2.14 – remove bold from label

Section 2 Whole Molecule Drug Evaluation O

O O H3C

Cl

OH

1.15 Fluvoxamine

C H3

Fenofibric Acid

Fluvoxamine is an inhibitor of the serotonin reuptake transporter (SERT) and prevents the reuptake of serotonin at the presynaptic membrane in the central nervous system. It is indicated for use in the treatment of depression. 1.15 and 2.15 – remove bold from label Fluvoxamine is structurally unique relative to the rest of the serotonin selective reuptake inhibitor class of drugs. A

B D

C

E

Fluvoxamine

and 2.15 – remove bold label functional groups, and use the 1. Conduct a structural evaluation 1.15 of fluvoxamine, focusing onfrom the boxed information in the grid to inform your answers to the questions that follow. F3 C Function

O

Character

Name of Functional Group

Character

Acidic, Basic, NH3 N OFunction or Neutral

Hydrophilic and/or Hydrophobic

Provide ↑ Solubility pKa When and/or Fluvoxamine maleate ↑ Absorption Relevant

A B C D E

51

Function

C H3

Interaction(s) O – Possible Owith Biological Target at Physiological pH=7.4

Amino Acids That Can Interact with the Functional Group via O Hydrogen Bonding Interactions at OH pH=7.4 None Is Acceptable

52

Fenofibric Acid Selfevaluation Assessment 1.Medicinal ConductChemistry a structural of

2. for your answer. 1.15 and 2.15 – remove bold from label 2. Fluvoxamine is marketed as a single isomer. Is the product sold as a single enantiomer, diastereomer, or geometric isomer? Provide a brief rationale for your answer. A B F C 3

C

1' 2

2'

N

O

1

A

F3 C

O

NH 2

D

C H3

H 2N E

1'

2'

1N

2

O

O

C H3

B

Fluvoxamine

3. Fluvoxamine is formulated as a maleate salt. saltes What type salt is a maleate salt, and what type of 3. Fluvoxamine is formulated a maleate theof drug? 1.15 and 2.15as – remove bold fromoflabel properties does it confer to the overall properties of the drug? F3 C

O N

O

NH3

Fluvoxamine maleate

C H3 –

O

O

O OH

Fluvoxamine maleate

4. Fluvoxamine is wellisabsorbed and has an oral bioavailability of ~50%. Using the information found in 4. Fluvoxamine well tic properties. the structure evaluation grid, provide a rationale for these pharmacokinetic properties. 5. A number of occurred.

1.15 Fluvoxamine

53

5. A number of fluvoxamine metabolites have been identified, all of which demonstrate little or no pharmacological activity. Evaluate each of the metabolic products drawn below and identify which metabolic transformation has occurred.

A

B

C

D

E

F

Name of Metabolic Transformation A B C D E F

6. Fluvoxamine is a strong inhibitor of CYP1A2, CYP3A4, and CYP2C19. These enzyme isoforms catalyze a number of the phase I oxidative metabolic transformations. Several of the benzodiazepines (used in the treatment of anxiety), including the very popular alprazolam, rely heavily on hepatic oxidation for metabolic inactivation and elimination. Other benzodiazepines, including the equally popular lorazepam, rely on glucuronide conjugation for metabolic inactivation and elimination. Which combination of drugs, fluvoxamine + alprazolam or fluvoxamine + lorazepam, is the most likely to generate an enhanced anxiolytic effect?

Section 2 Whole Molecule Drug Evaluation

1.16 Haloperidol

Shown below is the structure of haloperidol. Six of its functional groups have been identified. E C

D

A

F

B

1. Using1.theUsing tablethe below, identify the six boxed functional groups. For each of the functional groups you identified, indicate if it is hydrophilic or hydrophobic in character. Also provide a brief explanation for 2. Based on or induction. your response. 3. Using the.4, and 8.5. Functional Group Name

Hydrophilic or Hydrophobic

A

E

B

C

C D E

D

A

F

F 2. Based on their electronic properties AND their B relative positions in the molecule, identify if functional groups C, andbelow E are modification electron withdrawing orthe electron donating. Additionally, identify if this effect is 4. A,Shown can enhance duration of haloperidol. due to resonance or induction. 1. Using the 3. Using2.theBased unmodified structure of haloperidol and the table below, identify all of the acidic and basic on or induction. functional groups present in the structure, provide the normal pKa range for each functional group, and 3. ifUsing and 8.5. identify eachthe.4, functional group would be primarily ionized or unionized in pH environments=1.7, 5.5, 6.0, 7.4, and 8.5.

5. Using the table below, identify the groups. 4. Shown below modification can enhance the duration of haloperidol. C 55 B

D

2. Based on or8.5. induction. 3. Using the.4, and

E

3. Using the.4, and 8.5. 56

D

C

Medicinal Chemistry Self Assessment

A

Functional Group

Acidic or Basic

pKa Range

Primarily Ionized or Unionized 1.7 5.5 6.0F

7.4

B

8.5

1.below Using the 4. is a structural analog of haloperidol. the structural change and propose an 4. Shown Shown below modification can enhance the duration Evaluate of haloperidol. explanation as to how this structural modification can enhance the duration of haloperidol. Based on or induction.can enhance the duration of haloperidol. 4. 2. Shown below modification 3. Using the.4, and 8.5.

5. Using the table below, identify the types of binding interactions that are possible between the 5. Using the table below, identify the groups. boxed4.functional groups and a protein receptor the or enzyme. identify amino acids present within Shown below modification can enhance duration Also of haloperidol. 5. Using table below, the groups. a receptor orthe enzyme whoseidentify side chains could participate in the interactions that you identified. Assume a plasma pH=7.4 for all ionizable functional groups. C

B

C B

D

A

D

A

6. Shown below is the perform with haloperidol. 6. 5. Shown below is thebelow, perform with haloperidol. Using theGroup table identify the groups. Functional Types of Binding Interactions

Amino Acids Capable of Forming Specific Binding Interaction

A B C D

C B

D A 6. Shown below is the structure of haloperidol and a list of five metabolic transformations. For each metabolic transformation, indicate if it is a phase I or a phase II transformation and if haloperidol has a functional group present that can participate in the transformation. If you answer YES, then draw the appropriate metabolite; if you answer NO, then provide a brief explanation as to why this metabolic transformation is not possible to perform with haloperidol. 6. Shown below is the perform with haloperidol. Metabolic Pathways A. Reduction B. Sulfate Conjugation C. Hydrolysis D. Oxidative N-Dealkylation E. Benzylic Oxidation

Section 2 Whole Molecule Drug Evaluation

1.17 Hydrocortisone

Hydrocortisone is a glucocorticoid used in the management of inflammation. Derivatives of hydrocortisone are used in the management of asthma and chronic obstructive pulmonary disease. 1.17 and 2.17 – remove bold and fromuse label 1. Conduct a complete structural evaluation of hydrocortisone the information in the grid to inform your answers to the questions that follow.

E C F

B D

A

Hydrocortisone Function Question #3 structure needs to be replaced: Character

Name of Functional Group

Character

Acidic, Basic, or Neutral

Function

Hydrophilic and/or Hydrophobic

Provide pKa When Relevant

↑ Solubility 21 and/or ↑ Absorption

A

Interaction(s) Possible with Biological Target at Physiological pH=7.4

17

11

A

B C D E

1.18 and 2.18 – remove bold from label

F

HO I

A

I

C

I O

B

D

O

E OH

N H2 57

I

Function Amino Acids That Can Interact with the Functional Group via Ion–Dipole Interactions at pH=7.4 None Is Acceptable

58

Medicinal Chemistry Self Assessment

2. The glucocorticoids interact with residues within the glucocorticoid receptor (Arg611, Asn564, Thr739, Gln642, and Gln570) via hydrogen bonding and ion–dipole interactions at physiological pH (7.4). Identify which functional groups could interact with the side chains of these amino acids.

Functional Group

Can Interact with Arginine611 via an Ion–Dipole Interaction

Can Interact with Asparagine564 and Threonine739 via a Hydrogen Bonding Interaction

Can Interact with Glutamine642 and Glutamine570 via a Hydrogen Bonding Interaction

Yes or No

Yes or No

Yes or No

A B

1.17 and 2.17 – remove bold from label

C

E

D E

C

F

F

B

3. Several of the functional groups identified in the structure evaluation grid are essential for biological activity. There are several metabolic transformations D that render the glucocorticoids inactive. These A transformations include the following: 1. Ring A ketone reduction

Hydrocortisone

2. Ring A double bond reduction 3. C-17 oxidation 4. C-11 oxidation

Question #3 structure needs to be replaced:

21 11

17

A

Based on this information, consider the array of products drawn in the scheme that follows. Identify each type of reaction or transformation that has occurred and evaluate each of the products to 1.18or and 2.18 – remove bold from label determine if each product is active inactive.

HO I

A

I I

C

E OH

N H2

O

B

D

O

I

L-Thyroxine

1.17 Hydrocortisone

A

B

F 11

17

21

C

A

E D

Pathway A B C D E F

Type of Reaction or Transformation

Active or Inactive

59

60

Medicinal Chemistry Self Assessment

4. The glucocorticoids synthetic solubleare ester is present. formation an effectprodrugs. on overall Both drug water solubility. 4. The synthetic often esterified at C-21has to produce lipophilic and water soluble esters can be formed. Evaluate each of the four prodrugs drawn below and determine whether a lipophilic or water soluble ester is present. Determine how prodrug formation has an effect on overall drug water solubility.

A

C Type of Ester Formed

B

D

Effect on Overall Drug Water Solubility

A B C D

a. Provide a structural rationale for why prodrugs (e.g., B and D) are used in the preparation of aqueous injectable products to be administered intramuscularly (IM) or intravenously (IV). b. Provide a structural rationale for why prodrugs (e.g., A and C) are used in the preparation of depot injections. 5. Lipophilic glucocorticoid esters typically do not concentrate in the urine, but rather undergo glomerular filtration followed by tubular reabsorption. Provide a brief rationale for why lipophilic glucocorticoid esters do not concentrate in the urine and determine what effect this has on duration of drug action. 6. Which type of prodrug, water soluble ester salts, or lipophilic esters, would you anticipate to have greater systemic side effects?

C F

B D

A Section 2 Whole Molecule Drug Evaluation Hydrocortisone

Question #3 structure needs to be replaced:

1.18 Levothyroxine (T4) 21 17 11 Levothyroxine (T4) is a naturally produced thyroid pro-hormone. In its active form, tri-iodo-L-thyronine (T3) is responsible for regulating oxygen consumption and calorigenesis (think metabolism, metabolic rate, and thermogenA molecules until it is needed. Once proteolyzed from thyroglobesis). T4 is biosynthesized and stored in thyroglobulin ulin and transported to the desired target tissue, T4 undergoes dehalogenation catalyzed by thyroxine dehalogenase to the active thyroid hormone T3.

1. Conduct a structural evaluation of Tand , focusing on thebold boxed functional groups, and use the information in 4 1.18 2.18 – remove from label the grid to inform your answers to the questions that follow.

HO

A

I I

C

D

E OH

N H2

O

I

O

B

I

L-Thyroxine Function Character

Name of Functional Group

Character

Acidic, Basic, or Neutral

Hydrophilic and/or Hydrophobic

Provide pKa When Relevant

Function Function ↑ Solubility and/or ↑ Absorption

A B C D E

61

Interaction(s) Possible with Biological Target at Physiological pH=7.4

Amino Acids That Can Interact with the Functional Group via Hydrogen Bonding Interactions at pH=7.4 None Is Acceptable

I

HO

OH

N H2 I62 Tri-iodothyronine O Medicinal Chemistry Self Assessment

B

I

2. 2.18 Hormone replacement 1.18 and – remove bold from L-Thyroxine (T4label )therapy in the form of levothyroxine (T4) is available for those patients who do not produce an adequate endogenous supply of T4. Evaluate the structure of levothyroxine and determine if it is an enantiomer, diastereomer, or geometric isomer. O I isomer. O 2. Hormone replacement therapy in the geometric H 2N I OH HO H OH O I H NH I 2 HO O I OH OH H I NH 2 I O L-Tyrosine Levothyroxine I Levothyroxine 3. Evaluate the chiral carbon atom in levothyroxine to determine if this drug is drawn as the R- or S-enantiomer.

3. Evaluate the chiral to determine if this drug is drawn as the R- or S-enantiomer. 4. T3 is the biologically active hormone. It interacts with the thyroid hormone receptor via hydrogen bonding, ion–dipole, hydrophobic, and ionic interactions. Evaluate the ribbon diagram that shows 4. T3 is the in the structure evaluation grid. how T3 interacts with the surface of the thyroid receptor and identify the types of interactions possible for each of the five functional groups identified in the structure evaluation grid. O

I HO

I O H label 1.18 and 2.18 – remove bold from H N H2 O1.18 and 2.18 – remove bold from label I Tri-iodo-L-thyronine

Tri-iodothyronine Tri-iodothyronine

1.18 and 2.18 – from remove bold from label the thyroglobulin molecule and is considered an amino 5. T4 is biosynthesized L-tyrosine within acid–based hormone. Other hormones in the body are steroid-based (e.g., estrogen) or peptide1.18 andEvaluate 2.18 – remove bold from based (e.g., insulin). the structure oflabel T4 and justify its classification as an amino acid–based O hormone by determining which functional groups are derived from L-tyrosine. I O H 2N I OH O HO I H OH O H 2N I OH HO H NH 2 H OH O I H NH OH I 2 O I L-Tyrosine Levothyroxine OH I



L-Tyrosine

Levothyroxine

H NH 2

O

I

I

I

1.18 Levothyroxine H (TO ) 4

63

I

Levothyroxine

I 6. Interestingly, the dehalogenation transformation represents both an activation and deactivation pathway for T4. The iodo substituents on the inner ring of T4 play a role not only in important hydrophobic binding interactions with the receptor, but also in maintaining the shape of the hormone in a perpendicular orientation (see diagram below). This perpendicular shape is necessary to appropriately position the phenolic OH to participate in key receptor binding interactions. Evaluate the metabolites produced from the dehalogenation and identify which to metabolite 1.18 and 2.18 – transformations structure was fixed (pay no attention the colors)is active and which is not. Provide a brief rationale for your answers.

1.18 and 2.18 – remove bold from label I 1.19 and 2.19 – remove bold from label. I HO

A I HO

H NH 2

C

I

D OH

H NH 2

O

I

OH

O

O I

O

I

A

B Lidocaine

I HO

O I

OH

Levothyroxine

I

H NH 2

O

B 1.18 and 2.18 – structure was fixed (pay no attention to the colors)

1.19 and 2.19 – remove bold from label.

A

O

B

Section 2 Whole Molecule Drug Evaluation

1.19 Lidocaine

As a sodium channel blocker, lidocaine has found therapeutic use both as a local anesthetic and as a Class IB antiarrhythmic agent. As an anesthetic, this agent demonstrates rapid onset of action (acts quickly) and a longer duration of action (lasts longer) than most amino ester-type local anesthetics. The most frequently observed side effects are changes in the central nervous system (CNS) (e.g. dizziness, lightheadedness, tinnitus). Lidocaine is extensively metabolized by the CYP1A2 isozymes to a variety of metabolites. 1. Conduct a complete structural evaluation of ow. 1. Conduct a complete structural evaluation of lidocaine, place your answers in the grid provided, and then use the evaluation information in the grid to inform your answers to some of the questions that follow. C H3

N

N C H3

CH3

O

C H3

H

Lidocaine Lidocaine

Function

Character

4

Acidic, Unless place next to the relevant arrow. Basic, Character Function or Neutral

Name of Functional Group

Hydrophilic and/or Hydrophobic

Provide pKa When Relevant

↑ Solubility and/or ↑ Absorption

Function Interaction(s) Possible with Biological Target at Physiological pH=7.4

Amino Acids That Can Interact with the Functional Group via Hydrogen Bonding Interactions at pH=7.4 None Is Acceptable

2. Based on the information in the structure evaluation grid, determine whether or not lidocaine is likely soluble in the blood (pH=7.4).

65

66

Medicinal Chemistry Self Assessment

3. Local anesthetics that have a rapid onset of action are rapidly distributed in the body and can be absorbed easily across lipophilic membranes. Based on the information in the structure evaluation grid, provide a rationale for why lidocaine is rapidly distributed and can easily be absorbed across lipophilic membranes. 4. Unless excreted unchanged, drug molecules undergo one or more metabolic transformations to deactivate the drug and/or make the drug sufficiently water soluble to permit elimination. There are a variety of transformations thatlabels are possible for most drugs, butover onlythe the minimum of trans1.19 and 2.19 – remove bold from and remove answers from arrows (pay number no attention to formations actually occurs. The following diagram captures the metabolic pathways for lidocaine the colors) that are observed clinically. For each transformation, identify which phase I metabolic transformation has taken place next to the relevant arrow. C H3

C H3

O N

N C H3

C H3

H

Lidocaine

C H3

HO

O N

N C H3

C H3

C H3 C H3

N

H N

C H3

H

C H3

H

O

Monoethylglycinexylidide

C H3

HO

N C H3

O

H N

C H3

H

3-Hydroxy-monoethylglycinexylidide 3-Hydroxy-monoethylglycineexylidide

C H3

C H3

N H2

N C H3

C H3 + O HO

H

Glycinexylidide

H N

O

C H3

N H2

1.19 Lidocaine

67

5. Now that you have identified the metabolic transformations that generate products that have been identified, put your detective hat on and list any additional phase I transformations that could have occurred. 6. Lidocaine suffers from CNS-based toxicities largely due to production of the N-dealkylated metabolic product monoethylglycinexylidide once the parent drug has crossed the blood–brain barrier. a. Provide a structural rationale for why lidocaine is able to cross the blood–brain barrier. b. Interestingly, neither 1.19 and 2.19 tolycaine – remove nor boldtocainide from labeldemonstrates similar CNS-based toxicities. Provide a structural rationale for why these two local anesthetics are devoid of CNS-based side effects.

Tolcainide

Tolycaine



1.25 and 2.25 – remove bold from label

A B

C D

Sitagliptin

2.25 – remove bold from label

Sitagliptin phosphate

Section 2 Whole Molecule Drug Evaluation

1.20 Montelukast and Zafirlukast

Shown below are the structures of montelukast and zafirlukast. These drug molecules are administered orally for the treatment of asthma and allergic rhinitis.

Montelukast

Zafirlukast

1. The structure of montelukast contains one acidic functional group (pKa=4.4) and one basic functional group (pKa=3.1), whereas the structure of zafirlukast only contains one acidic functional group (pKa=4.3). 1. The structure of basic functional groups and predict whether they will be primarily ionized or Identify these acidic and primarily a stomach a urine a cellular pH=6.1, a plasma pH=7.2, and a solu2. In unionized the use theatRule of NinespH=1.9, to calculate the pH=5.4, percent of the functional group that would be ionized. tion pH=8.3. 3. The an explanation for this difference. Primarily Unionized occur at a physiological pH of 7.4. 4. Montelukast and zafirlukast. Assume that allIonized bindingorinteractions Acidic or Functional Group log P these hepatic Basic 1.9 5.4primarily excreted 6.1 7.2 5. Calculated metabolism or be unchanged?

8.3

6. Shown below is to perform with zafirlukast. 2. In the previous question, we examined three pKa values in five different environments for a total of 15 different scenarios. Which of these 15 scenarios allow you to use the Rule of Nines to calculate the percent of ionization of the functional group in the specific environment? Identify the specific scenarios and use the Rule of Nines to calculate the percent of the functional group that would be ionized.

69

70

Medicinal Chemistry Self Assessment

3. The sodium salt of montelukast is required for its oral administration, whereas zafirlukast can be administered orally as its unionized free acid form. Conduct a structural analysis of these two drug molecules and provide an explanation for this difference. 4. Montelukast and zafirlukast exert their mechanism of action by interacting with cysteinyl leukotriene Montelukast receptors and blocking the normal actions of endogenous leukotrienes (LTC4, LTD4, and LTE4). It has been proposed that this interaction requires five key elements: an ionic interaction, a hydrogen bond interaction where the antagonist acts as the acceptor, as well as the interaction of the antagonist with three separate hydrophobic pockets within the receptor. Using this information and the structures of montelukast and zafirlukast, propose potential binding interactions between these drug molecules and cysteinyl leukotriene receptors. Assume that all binding interactions occur at a physiZafirlukast ological pH=7.4. 5. Calculated log P values of montelukast and zafirlukast lie in the range of 5.5 to 6.4 depending on the computer program used to predict these values. Given this information, would these drug molecules be to be 1. predicted The structure ofhighly plasma protein bound or minimally plasma protein bound? Additionally, would you expect these drug molecules to undergo extensive hepatic metabolism or be primarily 2. In theunchanged? use the Rule of Nines to calculate the percent of the functional group that would be ionized. excreted 3. The an explanation for this difference. 6. Shown below is the ofAssume zafirlukast a list of five metabolic 4. Montelukast and structure zafirlukast. that and all binding interactions occurtransformations. at a physiologicalFor pH each of 7.4. metabolic transformation, indicate if it is a phase I or a phase II transformation and if zafirlukast has 5. Calculated log P these hepatic metabolism or be primarily excreted unchanged? a functional group present that can participate in the transformation. If you answer YES, then draw 6. appropriate Shown belowmetabolite; is to performifwith the youzafirlukast. answer NO, then provide a brief explanation as to why this metabolic transformation is not possible to perform with zafirlukast. Metabolic Pathways A. Methylation B. Aromatic Oxidation C. Hydrolysis D. Oxidative O-Dealkylation E. Benzylic Oxidation

Section 2 Whole Molecule Drug Evaluation

1.21 Phenobarbital and Other Barbiturates Shown below are the structures of butabarbital, secobarbital, and phenobarbital. These drug molecules are used as 1.21PhenobarbitalandOtherBarbiturates sedative-hypnotic agents in the treatment of insomnia and inducing sedation prior to surgical procedures. Phenobar bital can be used in the treatment of a number of different seizure disorders. Their sedative properties are due to their Shown belowwith are the of butabarbital, seconervous system (CNS). in the central ability to interact the structures GABAA receptor

O

O

HN

O

NH

O

HN

O

Butabarbital

O

NH

O

HN

NH

O

O

Secobarbital

Phenobarbital

1. Conduct a structural analysis of these drug molecules and provide an explanation as to why they can be 1. Conduct a structural analysis. previously listed. administered orally for the indications

2. Butabarbital and as to why these two drug molecules need to be administered as their sodium salts. 2. Butabarbital and secobarbital are marketed as their sodium salts and phenobarbital is marketed in its free acid form (i.e., non-salt form). Draw the sodium salt of either butabarbital or secobarbital and provide an explanation as to why these two drug molecules need to be administered as their sodium salts. 3. Using the phenobarbital is a long-acting agent (= 30–60 minutes; duration = 10–16 hours). 3. Using information in questions 1 andor2unionized and your answers to those questions, provide anand explanation 4. the Secobarbital contains an ionized at pH environments of 1.5, 5.9, 6.3, 7.4, 8.9. as to why secobarbital is a short-acting agent (onset of action = 10–15 minutes; duration = 3–4 hours), butabarbital is an intermediate-acting agent (onset of action = 45–60 minutes; duration = 6-8 hours), and phenobarbital is aoflong-acting agent (onset of action = 30–60 minutes; duration = 10–16 hours). 5. As stated the amino acids indicated. 4. Secobarbital contains an ionizable with a pKactivity? =7.9. Using table shown below, identify a 6. Shown below. Would you to functional retain their group pharmacological Whythe or why not? the functional group, provide the normal pKa range for the functional group, and identify if the functional group would be primarily ionized or unionized at pH environments=1.5, 5.9, 6.3, 7.4, and 8.9. Functional Group

Acidic or Basic

Primarily Ionized or Unionized pKa Range

1.5

5.9

6.3

7.4

8.9

71

Metabolite of

Metabolite of

Metabolite of

72

Medicinal Chemistry Self Assessment

5. As stated above, barbiturates exert their mechanism of action by interacting with the GABAA receptors. Assume that the following amino acids are involved in this binding: Val, Phe, Ser, Asn, and Lys. Using the functional groups present within the structure of secobarbital, identify five specific binding interactions between secobarbital and the side chains of the amino acids indicated.

Chapters1.21and2.21 6. Shown below are known metabolites of butabarbital, secobarbital, and phenobarbital. Identify the

PleasereplacetheindicatedstructureinQuestion6inboth1.21and2.21(theonethatispartofthequestion) metabolic transformations that are required to produce each metabolite and indicate if they are phase I or phase II transformations. Would you expect any of these metabolites to retain their withtheonebelow.Bothstructureshadanidenticalerror. pharmacological activity? Why or why not?



Metabolite of butabarbital

Metabolite of secobarbital

Metabolite of phenobarbital



  Chapter2.22 PleasereplacethestructurefortheanswertoQuestion3inChapter2.22withtheonebelow.  Intramolecular Hydrogen Bond Intramolecular Hydrogen Bond

Pravastatin





Fluvastatin



Section 2 Whole Molecule Drug Evaluation

1.22 Pravastatin and Fluvastatin 1.22 Pravastatin and Fluvastatin

ShownShown below are theare structures of pravastatin andhave fluvastatin. These drug molecules are used in the treatment of below the structures of groups been boxed. various types of hyperlipidemia/dyslipidemias. A total of six functional groups have been boxed. F

C

B

D

A

E Pravastatin

Fluvastatin

1. Using the table below, identify the six boxed functional groups. For each of the functional groups you 1. indicate Using the table below, or hydrophobic in explanation for Also your provide response. identify, if it is hydrophilic or hydrophobic in character. a brief explanation for your response. 2. The log P drug a structural explanation for the difference in these log P values.

3. The normal pKa range for normal range. Answer 4.Functional Pravastatin, as to how pravastatin andorfluvastain inhibit HMG CoA reductase. Groupand Name Hydrophilic Hydrophobic A B C

HMG CoA Reductase

D E F

HMG CoA Mevalonic Acid 2. The log P values of pravastatin and fluvastatin are 1.44 and 3.62, respectively. Conduct a structural analysis of these drug molecules and provide a structural explanation for the difference in these log P values. 5. Shown below are the postulate a reason why this structural change results in a loss in activity. 3. The normal pKa range for carboxylic acids is 2.5 to 5. The pKa values for the carboxylic acids present within the structures of pravastatin and fluvastatin are 4.21 and 4.56. Conduct a structural analysis and provide a reason why these pKa values are at the high end of the normal range.

73

E 74

E

Pravastatin Medicinal Chemistry Self Assessment

Fluvastatin Fluvastatin

Pravastatin

1. Pravastatin Using the table or hydrophobic in explanation for your response. 4. and below, fluvastatin exert their hyperlipidemic effects by inhibiting the enzyme HMG CoA reductase. As shown below, HMG CoA reductase converts 3-hydroxy-3-methylglutaryl CoA (HMG 1. PUsing table below, or hydrophobic in explanation forlog your 2. The log drug the a structural explanation for the difference in these P response. values. CoA) to mevalonic acid. This conversion is required for the synthesis of cholesterol and acts as a 2. ThepK logrange P drugfor a normal structural explanation for the difference in these log P values. range. 3. primary The normal controla site for production of this endogenous steroid. Using the structures of HMG CoA, range for normal 3. The normal pK mevalonic acid, pravastatin, and fluvastatin, provide aninhibit explanation as to how pravastatin and fluvastatin a 4. Pravastatin, and as to how pravastatin andrange. fluvastain HMG CoA reductase. inhibit HMG CoA reductase. 4. Pravastatin, and as to how pravastatin and fluvastain inhibit HMG CoA reductase.

HMG CoA Reductase HMG CoA Reductase HMG CoA Mevalonic Acid HMG CoA Mevalonic Acid 5. Shown below are the structures of fluvastatin and a conformationally restricted analog. The addition 5. of Shown below are the postulate a reason why this structural change resultsabolish in a loss activity. another carbon atom and the conformational restriction essentially theintherapeutic activity 5. Shown below are the postulatepostulate a reasonawhy this structural change results in aresults loss inin activity. of fluvastatin. Using these structures, reason why this structural change a loss in activity.

Fluvastatin Fluvastatin

Conformationally RestrictedConformationally Analog of Fluvastatin Restricted Analog of Fluvastatin

6. Pravastatin is primarily metabolized to its 3α epimer. This metabolite is completely inactive as an 6. Pravastatin is is inactive. HMG CoA reductase inhibitor. Identify the metabolic transformations required to produce this metabolite and provide an explanation as to why this metabolite is inactive.

3D epimer

Section 2 Whole Molecule Drug Evaluation

1.23 Quinapril 1.23 Quinapril

Shown below is the structure of quinapril. It is an angiotensin converting enzyme (ACE) inhibitor that is used in the Shown below is are identified. treatment of hypertension and heart failure. Five functional groups are identified. B H3C O

A

O H N

C

D O N C H3

O

OH

E tableidentify if it is hydrophilic or hydrophobic in brief explanation response. 1. Using 1. the Using table the below, the five boxed functional groups. For each of for theyour functional groups you identified, indicate if it is hydrophilic or hydrophobic in character. Also provide a brief explanation for 2. Using the ionized or unionized at pH environments of 1.5, 4.8, 6.3, 7.4, and 8.1. your response. 3. Answer CH3

Functional Group Name A

Hydrophilic or Hydrophobic O

O

B C

O H N

N

D E

C H3 O

OH



2. Using the unmodified structure of quinapril and the following table, identify all of the acidic and basic  normal pK range for each functional group, and functional groups present in the structure, provide the a identify if each functional group would be primarily ionized or unionized at pH environments=1.5, 4.8, 6.3, 7.4, and 8.1.

4. Quinapril is a is administered orally instead of quinaprilat.

75

Chapter1.23

A

O H N PleasereplacethestructureforQuestion2inChapter1.23withtheoneshownbelow.(NOTE:Theanalogous N 76 Medicinal Chemistry Self Assessment

structureinChapter2.23isfine.)

C

C H3



O

OH

E 1. Using the table if it is hydrophilic or hydrophobic in brief explanation for your response. 2. Using the ionized or unionized at pH environments of 1.5, 4.8, 6.3, 7.4, and 8.1. 3. CH3 O

O



O H N

 Functional Group

Chapters1.23and2.23

Acidic or Basic

Primarily Ionized N or Unionized pKa Range

1.5

C H3

4.8 O

6.3

7.4

8.1

OH

 PleasereplacethestructuresforAngiotensinIandQuinaprilatinQuestion4forboth1.23and2.23withtheone  providedbelow.



3. Quinapril is a prodrug. It is administered as an oral tablet and converted in vivo to its active metabolite, the metabolic pathway that converts quinapril to quinaprilat, and offer a 4.quinaprilat. Quinapril is Identify a is administered orally instead of quinaprilat. reason why quinapril is administered orally instead of quinaprilat. Leu

Phe

His Quinapril

Angiotensin I R = Asp-Arg-Val-Tyr-Ile-His-Pro

Quinaprilat Quinaprilat



4. Quinapril inhibits ACE. This enzyme is a relatively nonspecific dipeptidyl carboxypeptidase. It is a zinc   protease that converts angiotensin I, a decapeptide, to angiotensin II, an octapeptide. The peptide cleavage is catalyzed by the zinc atom and is shown below. Quinapril, along with other ACE inhibitors, is a tripeptide mimic that can interact with the enzyme resulting in enzyme inhibition rather than hydrolysis. Using this information and the structures provided below, identify how quinapril can interact with ACE. Assume that all drug binding interactions are occurring at a pH=7.4.

PleasereplacethestructuresforAngiotensinIandQuinaprilatinQuestion4forboth1.23and2.23withtheone providedbelow. 1.23 Quinapril 77 5. Quinapril inhibits ACE. This at a pH of 7.4.



LeuLeu

HisHis

PhePhe

Quinaprilat Quinaprilat

AngiotensinI I Angiotensin R = Asp-Arg-Val-Tyr-Ile-His R = Asp-Arg-Val-Tyr-Ile-His-Pro







6. Shown below are in these pathways. 5. Shown below are four possible metabolic pathways for quinapril. Identify the metabolic transformations involved in these pathways.

B

A

Quinapril C

D

78

Medicinal Chemistry Self Assessment

6. Although it is possible for quinapril to undergo all of the above metabolic transformations, pathway B is the major pathway. Other metabolites have been identified, but only at very low levels. Provide an explanation for this finding.

Section 2 Whole Molecule Drug Evaluation

1.24 Rivastigmine 1.24Rivastigmine

ShownShown belowbelow is the structure of rivastigmine. Four of its functional groups have been is the structure of rivastigmine. Four of its functional groups have identified. been identified. C H3

A

H3C

C O

N

CH3 D C H3 N C H3

O

B

Rivastigmine

1. Using the table 1. Using the table below, identify the four boxed functional groups. For each of the functional groups you 2. Based on their identified, indicate if it induction. is hydrophilic or hydrophobic in character. Also provide a brief explanation for your response. 3. Rivastigmine is an e. Glutamic acid, with acetylcholine. Answer

Labile ester bond Functional Group Name

Hydrophilic or Hydrophobic

A B C D

2. Based on their electronic properties AND their relative positions in the molecule, identify if functional groups A and B are electron withdrawing or electron donating. Additionally, identify if this effect is due Rivastigmine to resonance or induction. 3. Rivastigmine is an acetylcholinesterase inhibitor. Inhibition of this enzyme prolongs the half-life of acetylAcetylcholine to active site choline, allowing for higherbound concentrations of acetylcholine at muscarinic and nicotinic receptors. This of acetylcholinesterase action is useful in the treatment of Alzheimer’s disease, myasthenia gravis, and glaucoma. Shown below is the structure of acetylcholine interacting with the active site of acetylcholinesterase. Glutamic acid, histidine, and serineA.are involved inrivastigmine the mechanism of esterhow hydrolysis can form specific interactions with Conduct a and indicate it couldand interact with acetylcholinestase. acetylcholine. B. The serine explanation as to how rivastigmine inhibits acetylcholinesterase.

79

O

1. Using the table 80 2.

C H3

B

Medicinal Chemistry Self Assessment Rivastigmine Based on their induction.

3. Rivastigmine is an e. Glutamic acid, with acetylcholine. 1. Using the table 2. Based on theirester induction. Labile bond 3. Rivastigmine is an e. Glutamic acid, with acetylcholine.

Labile ester bond

Rivastigmine

Acetylcholine bound to active site of acetylcholinesterase

Rivastigmine

A. a Conduct a rivastigmine and indicate and how indicate it could interact acetylcholinestase. a. Conduct structural analysis of rivastigmine how it with could interact with acetylcholinAcetylcholine bound to active site esterase. B. The serine explanation as to how rivastigmine inhibits acetylcholinesterase. of acetylcholinesterase b. The serine residue of acetylcholinesterase attacks the ester bond of acetylcholine causing hydrolysis and acetylation of the serine. A similar reaction occurs between the carbamate group of A. Conduct rivastigmine and indicate howwhile it could interact with isacetylcholinestase. rivastigmine andathe serine residue; however, acetylcholine a substrate of acetylcholinesterase, rivastigmine is an inhibitor. Compare the functional groups involved and provide an B. The serine explanation as to how rivastigmine inhibits acetylcholinesterase. explanation as to how rivastigmine inhibits acetylcholinesterase.

Intermediate in the hydrolysis of acetylcholine

Intermediate in the hydrolysis of acetylcholine

Analogous intermediate formed with rivastigmine

Analogous intermediate formed with rivastigmine

1.24 Rivastigmine

81

4. Neostigmine is a structural analog of rivastigmine. Similar to rivastigmine, neostigmine is able to interact with acetylcholinesterase and cause inhibition due to the presence of a carbamate functional group. The primary differences between rivastigmine and neostigmine are their routes of administration and their therapeutic indications. Rivastigmine is administered orally in the treatment of Alzheimer’s disease, whereas neostigmine is administered intramuscularly (IM) or subcutaneously (SC) in the treatment of myasthenia gravis. Conduct structural analyses of these drug molecules and 4. Neostigmine is a structural analog of es in routes of administration. provide an explanation for these differences in routes of administration and therapeutic indications.

4. Neostigmine is a structural analog of es in routes of administration.

4. Neostigmine is a structural analog of es in routes of administration.

+ Neostigmine Bromide + Bromide Rivastigmine a longer durationofofaction aNeostigmine longer duration of action.A comparison of the structures of 5. 5. Rivastigmine hashas a longer duration than neostigmine. rivastigmine and neostigmine reveals that the carbamate + group present in rivastigmine is slightly larger than that has present in neostigmine. Using the mechanism of action given in question 3, provide 5. Rivastigmine a longer duration of a longer duration of action. Methyl Neostigmine Bromide a reason why this structural difference results in a longer duration of action. Methyl Methyl Methyl 5. Rivastigmine has a longer duration of a longer duration of action. Methyl Methyl Methyl Neostigmine

Ethyl +

Ethyl

Methyl Rivastigmine

+ Rivastigmine 6. EvaluationNeostigmine of the structure of rivastigmine ability to inhibit acetylcholinesterase. Methyl Ethyl 6. Evaluation of the structure+of rivastigmine ability to inhibit acetylcholinesterase. para orientation meta orientation 6. Evaluation ofNeostigmine the structure of rivastigmine reveals that the aliphatic chain containing the tertiary Rivastigmine amine is located meta to the carbamate group. Similar orientations can be seen with neostigmine. Using the information from the previous questions, provide an explanation as to why a para orientapara orientation orientation 6. of the meta structure of rivastigmine to inhibit acetylcholinesterase. tionEvaluation of these functional groups would leadability to a significant decrease in the ability to inhibit acetylcholinesterase.

meta orientation Rivastigmine

para orientation para Analog of Rivastigmine

Rivastigmine

para Analog of Rivastigmine

Rivastigmine

para Analog of Rivastigmine

Section 2 Whole Molecule Drug Evaluation

1.25 Sitagliptin

Sitagliptin is an inhibitor of dipeptidyl peptidase IV (DPP-IV), a serine protease that catalyzes the deactivation/degra1.25 Sitagliptin dation of GLP-1. GLP-1 is a 36-amino acid peptide that is responsible for promoting insulin secretion in response to an increase in blood Currently there are four DPP-IV inhibitors on the market, all of which contain an Sitagliptin is anglucose inhibitorlevels. of dipeptidyl essential basic amino functional group that represents the penultimate amino-terminal alanine residue found within GLP-1.

A

F

B

F

C

N H2 O

D

N

N F

N

N CF3

Sitagliptin Sitagliptin

1. Conduct a structural evaluation of sitagliptin, focusing on the boxed functional groups and use the information the gridato your answers to the questions that follow. 1. inConduct toinform the questions that follow.

2. Sitagliptin is salt form.

Function Function

Character Character Name of Functional Group

Acidic, Basic, F or Neutral

Interaction(s) Possible with Biological Target at Physiological pH=7.4

Function

Hydrophilic ↑ Solubility F Provide pK and/or a N and/or H2 O Hydrophobic When Relevant ↑ Absorption

A

N

N

B

F

C D

N

Sitagliptin

3. Sitagliptin is designation. 4. At 38%, the bound (e.g., warfarin). 5. Approximately. 6. Assess each of ransformation has occurred.

83

N CF3

Amino Acids That Can Interact with the Functional Group via Ion– Dipole Interactions at pH=7.4 None Is Acceptable

84

Chemistry Assessment 1.Medicinal Conduct a to theSelf questions that follow.

2. Sitagliptin is salt form. 2. Sitagliptin is formulated as a phosphate salt. Identify the most basic functional group and modify the structure to show the phosphate salt form. F F

N H2 O N

N F

N

Sitagliptin Sitagliptin

N CF3

3. Sitagliptin is marketed as the R-enantiomer. Evaluate the structure of sitagliptin and provide a struc3. rationale Sitagliptin is the designation. tural for R-enantiomer designation.

4. At 38%, the bound (e.g., warfarin). 4. At 38%, the fraction of sitagliptin reversibly bound to plasma proteins is relatively low. By way of 5. Approximately. reminder, only the unbound fraction of drug is able to exert its biological activity and undergo metabolism. Describe the relative risk of a plasma protein binding interaction between sitagliptin 6. Assess each of ransformation has occurred. and another drug that is highly protein bound (e.g., warfarin). 5. Approximately 79% of sitagliptin is excreted unchanged in the urine. Provide a structural rationale that supports this observation.

1.25 Sitagliptin

6. Assess each of the possible metabolic products generated from sitagliptin and determine which phase I metabolic transformation has occurred.

A B

C

D

85

Section 2 Whole Molecule Drug Evaluation

1.26 Sorafenib

Because protein tyrosine kinases regulate cellular proliferation, differentiation, and survival, it is no surprise that several neoplastic disorders can be tied to altered activity of protein tyrosine kinases. Clinically relevant antineoplastic tyrosine kinase inhibitors interact with the active site of the enzyme via several types of binding interactions. The adenosine triphosphate (ATP) binding domain of the tyrosine kinases contains a hydrophobic domain that includes a significant number of isoleucine, leucine, alanine and valine residues. There are at least five binding pockets that flank this region in which van der Waals, hydrophobic, hydrogen bonding, and electrostatic interactions occur. Sorafenib a tyrosine kinase inhibitor used in the treatment of advanced renal cell carcinoma, a highly vascularized 1.26isSorafenib tumor. The drug specifically targets vascular endothelin growth factor 2 (VEGF2) that is instrumental in the generaprotein tyrosine kinases regulate tion ofBecause new blood vessels.

F

C E A B

D Sorafenib Sorafenib

1. Conduct a structural evaluation of sorafenib, focusing on the boxed functional groups, and use the infor1. Conduct a structural to the questions that follow. mation in the grid to inform your answers to the questions that follow.

2. Sorafenib interacts in the local environment Character of the enzyme. Character298

3. Nilotinib, another and Leu Name of Functional Group A B

299Acidic, 359Basic,

/Val

Hydrophilic and/or A Hydrophobic

/Phe

Function

in each of the respective five Interaction(s) binding pockets.

or Neutral

Provide pKa When Relevant

B

Function

C

↑ Solubility and/or ↑ Absorption

D

Possible with Biological Target at Physiological pH=7.4

E

C D E F

Nilotinib

A. Consider the side chains of the the five binding pockets. Assume pH=7.4. 87

B. Determine acid side chains are both at pH=7.4.

88

Medicinal Chemistry Self Assessment

2. Sorafenib interacts with Cys919, Phe1047, and Asp1046 via hydrogen bonding and hydrophobic interactions. 1.26 Sorafenib Identify which functional groups could interact with the side chains of these amino acids. Assume that Asp1046 is unionized in the local environment of the enzyme. Because protein tyrosine kinases regulate

Functional Group

Interacts with Cysteine919 via a Hydrogen Bonding Interaction Yes or No

A

C

Interacts with Aspartic Acid1046 F via a Hydrogen Bonding Interaction

Interacts with Phenylalanine1047 via a Hydrophobic Interaction

Yes or No

Yes or No

E

A

B C

B

D

D

E

Sorafenib

F

1. Conduct a structural to the questions that follow. 3. Nilotinib, another tyrosine kinase inhibitor (via Bcr-Abl) indicated for the treatment of Philadel2. Chromosome Sorafenib interacts in the local environment of the enzyme. phia (Ph+) positive chronic myelogenous leukemia, also interacts with each of the five binding pockets found within this biological target. This drug interacts with Leu285/Val289, Asp391/Glu286, 298 299 359 315 Nilotinib, 318 298and 299 359 3. another Leu /Val /Phe in each of the fivepockets. binding pockets. Thr , Met and Leu /Val /Phe in each of the respectiverespective five binding

A B

C

E

D

Nilotinib Nilotinib

a. Consider the side chains of the amino acids indicated and determine which type(s) of binding A. Consider the side chains thefive the binding five binding pockets. Assume pH=7.4. interactions are possible in each ofofthe pockets. Assume pH=7.4.

B. Determine acid side chains are both at pH=7.4.

Leu285/Val289

Asp391/Glu286

Thr315

Met318

Leu298/Val299/Phe359

C. It has been participate in this interaction. b. Determine which of the boxed functional groups (A–E) can interact with the side chains of the amino acids found in each of the five binding pockets. Indicate the type of interaction(s) possible in the appropriate box. None is an acceptable answer. Assume that the drug and the amino acid side chains are both at pH=7.4. Leu285/Val289

Asp391/Glu286

Thr315

A B C D E

Methionine

Met318

Leu298/Val299/Phe359

Nilotinib

A. Consider the side chains of the the five binding pockets. Assume pH=7.4.

1.26 Sorafenib

89

B. Determine acid side chains are both at pH=7.4. c. It hasC. been documented that the pyridyl nitrogen atom (functional group E) of nilotinib particiIt has been participate in this interaction. pates in a hydrogen bonding interaction with methionine. Draw a diagram that clearly shows which atom(s) within the structure of methionine participate in this interaction.

Methionine Methionine 4. Sorafenib enters cells via passive diffusion. Using the information in the structure evaluation grid as a starting point, identify which functional groups contribute to the ability of this drug to enter cells via passive diffusion. 5. Nilotinib is considered significantly more hydrophobic than sorafenib (distribution coefficient log D is 2.4 and 0.8 respectively). Provide a structural rationale for this property difference. 6. Sorafenib is marketed as a tosylate salt, a lipid-soluble organic salt. Nilotinib is marketed as a hydrochloride monohydrate salt, an inorganic salt. In general, what is the value of each of these types of salts?

Section 2 Whole Molecule Drug Evaluation

1.27 Zanamivir and Oseltamivir 1.27ZanamivirandOseltamivir Shown below is the structure of zanamivir. This drug molecule is administered via oral inhalation for the treatment of Shown below is the d B infections. influenza A and B infections.

Zanamivir

1. Identify all of the H of 7.2. 1. Identify all of the acidic and basic functional groups, provide the normal pKa range for each functional group, and identify if each functional group would bethat primarily ionized or unionized at of a pulmonary 2. Identify all other water soluble functional groups are present within the structure zanamivir. pH=7.2. 3. Based on your or capsule. 4. Zanamivir exerts its n. Using neuraminidase. 2. Identify all other water soluble functional groups that are present within the structure of zanamivir. 3. Based on your answers to questions 1 and 2, explain why zanamivir is administered as an oral inhaler instead of an oral tablet or capsule. 4. Zanamivir exerts its antiviral action by inhibiting neuraminidase, a viral enzyme that is required for the spread of the viral infection. A key component of neuraminidase’s action is the hydrolysis of N-acetylsialic acid from surface viral glycoproteins. Shown below is the structure of N-acetylsialic acid bound to a glycoprotein. Using this structure Glycosidic and the structure bond of zanamivir, provide an explanation of how zanamivir inhibits neuraminidase. N-Acetylsialic acid bound to glycoprotein

Zanamivir

5. Shown below is a, or similar to that of zanamivir?

91

1. Identify all of the H of 7.2. 2. Identify all other water soluble functional groups that are present within the structure of zanamivir. 2. Identify all other water soluble functional groups that are present within the structure of zanamivir. 3. Based on your or capsule. 92 Medicinal Chemistry Self Assessment 3. Based on your or capsule. 4. Zanamivir exerts its n. Using neuraminidase. Zanamivir 4. Zanamivir exerts its n. Using neuraminidase.

of the H of 7.2.

other water soluble functional groups that are present within the structure of zanamivir.

our or capsule.

xerts its n. Using neuraminidase.

Glycosidic bond Glycosidic bond N-Acetylsialic acid bound to glycoprotein N-Acetylsialic acid bound to glycoprotein

Zanamivir Zanamivir

5. Shown below is a, or similar to that ofIdentify zanamivir? 5. Shown below is a stereoisomer of zanamivir. if the stereoisomer is an enantiomer, a diastereomer, 5. aShown below is a, oran similar to that of zanamivir? geometric isomer, epimer, and/or a conformational isomer. Predict whether this stereoisomer’s pharmacological activity is likely to be more active, less active, or similar to that of zanamivir? Glycosidic bond

alic acid bound to glycoprotein

Zanamivir

w is a, or similar to that of zanamivir?

6. Shown below is the oseltamivir . 6. Shown below is the oseltamivir .

6. Shown below is the structure of oseltamivir and a list of five metabolic transformations. For each metabolic transformation, indicate if it is a phase I or a phase II transformation and if oseltamivir has a functional group present that can undergo the indicated transformation. When evaluating these metabolic transformations, consider functional groups that are initially present within the structure of oseltamivir as well as those that can be added/revealed through phase I metabolism. If you answer YES, .then draw the appropriate metabolite; if you answer NO, then provide a brief explanation as to w is the oseltamivir why this metabolic transformation is not possible for oseltamivir.

Metabolic Pathways A. Hydrolysis B. Allylic oxidation C. Glucuronide conjugation D. ω-Oxidation E. Oxidative O-Dealkylation

Part 2 ANSWERS

Section 3 – General Self Assessment 2.1 2.2 2.3 2.4 2.5 2.6 2.7

Functional Group Characteristics and Roles................... 95 Identifying Acidic and Basic Functional Groups.............. 99 Solving pH/pKa Problems.............................................. 105 Salts and Solubility....................................................... 109 Drug Binding Interactions............................................ 113 Stereochemistry and Drug Action................................. 117 Drug Metabolism......................................................... 123

Section 3 General Self Assessment Answers

Structures for 1.1 and 2.1 The bold has been removed.

2.1 Functional Group Characteristics and Roles The exact same structures are in both of these chapters, but I have provided two

copies in the event you wanted that. 1. Shown below are the structures of warfarin, phenytoin, bromfenac, and salmeterol. a. Identify each of the boxed functional groups. A

F C E D

Warfarin

B

Bromfenac

Phenytoin

H J

I

K

L

Salmeterol

Answer Box

Functional Group Name A

β-Dicarbonyl (enol form)

B C D E F G H I J

Ketone

B A Phenyl ring; aromatic ring; aromatic hydrocarbon C O Imide

H 2N

Halogen (bromo)

OH Primary aromatic amine; aniline

Phenol or phenolic hydroxyl group Secondary amine

L

Ether

C H3 N H

Sulfamethoxazole

Ibuprofen

Primary hydroxyl group or primary alcohol

Alkyl chain; aliphatic alkane

S O

Carboxylic acid

K

D O

95

N

O

G

96

Medicinal Chemistry Self Assessment

b. For each of the functional groups you identified, indicate if it is hydrophilic or hydrophobic in Structures for 1.1 and character. Also2.1 provide a brief explanation for your response. Answer The bold has been removed. The exact same structures are in both of these chapters, but I have provided two Box

Hydrophilic or Hydrophobic

copies in the Aevent youAcidic wanted that.group; hydrophilic due to its ability to ionize and ability to participate in hydrogen bonding in its functional unionized form.

B

Hydrophilic due to its ability to act as a hydrogen bond acceptor.

C

Hydrophobic due to its inability to ionize or form hydrogen bonds; hydrocarbon functional groups enhance lipid solubility.

D A

Acidic functional group; hydrophilic due to its ability to ionize and ability to participate in hydrogen bonding in its unionized form. F

E

Hydrophobic due to its inability to ionize or form hydrogen bonds.

F

Weakly basic functional group; hydrophilic primarily due to its ability to form hydrogen bonds, although it is possible for this functional group to be ionized. If unionized, primary aromatic amines can act as hydrogen bond E donors and acceptors.

G

Acidic functional group; hydrophilic due to its ability to ionize and ability to participate in hydrogen bonding in its unionized form. D

C

Warfarin H

Hydrophilic due to its ability to form hydrogen bonds; can act as both a hydrogen bond donor and acceptor. Bromfenac B

G

I

Phenytoinprimarily due to its ability to form hydrogen bonds. If unionized, Weakly acidic functional group; hydrophilic phenolic hydroxyl groups can act as hydrogen bond donors and acceptors.

J

Basic functional group; hydrophilic due to its ability to ionize and ability to participate in hydrogen bonding in its unionized form.

K

Hydrophobic due to its inability to ionize or form hydrogen bonds; hydrocarbon functional groups enhance lipid solubility.

H

The ether oxygen atom is hydrophilic due to its ability to act as a hydrogen bond acceptor; however, due to J K hydrocarbon surrounding this ether on both sides, the entire side L chain attached to the secondary amine could be I classified as primarily hydrophobic.

L

Salmeterol

2. Shown below are the structures of ibuprofen and sulfamethoxazole. Four functional groups have been highlighted. Based on their electronic properties AND their relative positions in the molecule, identify if they are electron withdrawing or electron donating. Additionally, identify if this effect is due to resonance or induction. D

B

A

O OH

Ibuprofen

C

O

H 2N

S O

C H3 N H

N

O

Sulfamethoxazole

Answer Functional group A: This is an alkyl group (aliphatic chain). Because it is directly attached to an aromatic ring, it can act as an electron donating group through induction. Functional group B: This is a carboxylic acid that will most likely be ionized at physiological pH and carry a negative charge. Therefore, it can be electron donating through induction. Functional group C: This is an aniline (or primary aromatic amine). Because the nitrogen atom is directly attached to the aromatic ring, it acts as an electron donating group through resonance. This electron donating property is responsible for the low basicity of aromatic amines (discussed in Chapter 3*).

2.1 Functional Group Characteristics and Roles

97

Functional group D: This is a heterocyclic aromatic ring. It is electron withdrawing due to induction. This electron withdrawing property is responsible for the increased acidity of the adjacent sulfonamide (discussed in Chapter 3*). 3. Shown below is the structure of imipramine as well as three analogs. Evaluate each analog and provide an overall evaluation of how each change will affect the chemical properties of imipramine.

Imipramine

Analog A

Analog B

Analog C

Answer Analog A: In this analog, a chlorine atom has replaced a hydrogen atom. A hydrogen atom generally does not contribute to electronic and solubility properties. Additionally, it is the smallest atom and thus sterically occupies the least amount of space. In contrast, the chlorine atom is electron withdrawing in character, enhances the overall lipid solubility of imipramine, and sterically is larger than the original hydrogen atom. Analog B: Similar to Analog A, this analog occurs via the replacement of a hydrogen atom with another functional group. The alkyl chain (propyl chain) is electron donating, enhances the overall lipid solubility of imipramine, and is sterically much larger than the original hydrogen atom. Analog C: In this analog, a methyl group has been replaced by an acetyl group that converts the tertiary amine to an amide. The methyl group is electron donating and contributes to the basicity of the tertiary amine. In contrast, the carbonyl group is electron withdrawing. This significantly decreases the availability of the lone pair of electrons on the nitrogen atom such that it is no longer basic. In terms of solubility, the substitution of the acetyl group for the methyl group decreases water solubility. This is because the nitrogen atom is no longer basic and thus cannot undergo ionization. The resulting amide is water soluble due to its ability to act as a hydrogen bond acceptor; however, the change from a functional group that can ionize to one that can only participate in hydrogen bonding results in an overall decrease in water solubility. Finally, the acetyl group is sterically larger than the original methyl group.

98

Medicinal Chemistry Self Assessment

4. Shown below is the structure of a tetrapeptide that is part of a larger protein receptor. The side chains of the four amino acids have been boxed. 4. Shown below is the structure of a tetrapeptide that is part of a larger protein receptor. The side chains of the four amino acids have been boxed.

a. Identify the four amino acids that comprise this tetrapeptide sequence. Answer

Tyrosine (Tyr)—Aspartic Acid (Asp)—Valine (Val)—Glutamine (Gln)

b. For each amino acid, identify the key chemical properties of its highlighted side chain. Answer



Tyrosine: This side chain is one of the largest seen in amino acids and is involved in dictating Page2of2 the overall conformation of the peptide. Although the phenyl ring is hydrophobic, the phenolic hydroxyl group can act as both a hydrogen bond donor and acceptor and contributes to the overall water solubility. Aspartic Acid: This side chain is acidic and will most likely be ionized at physiological pH. This will enhance the overall water solubility of the peptide. It will most likely reside in a water soluble pocket of the overall peptide. Valine: This is an aliphatic, lipid soluble, hydrocarbon side chain. It contributes to the overall lipid solubility of the peptide and can contribute to lipid soluble pockets within the peptide. Additionally, the isopropyl side chain can impart some steric bulk that may influence the overall conformation of the molecule. Glutamine: Overall, this side chain is normally classified as a polar, water soluble side chain due to its ability to act as both a hydrogen bond donor and acceptor. The two methylene carbon atoms allow the amide to be oriented in different locations and may contribute a little to the overall lipid solubility of the peptide.

Go to Section 1.3 [Link to section 1.3 title] Note: References to chapters are to Chapter 3 in Harrold MW and Zavod RM, Basic Concepts in Medicinal Chemistry, American Society of Health-System Pharmacists, ©2013.

Section 3 General Self Assessment Answers

2.2 Identifying Acidic and Basic Functional Groups

2.2

1. For each of the drugs or experimental drugs shown below, identify all of the acidic and basic functional

groups. 1. For each of the drugs or experimental drugs shown below, identify all of the acidic and basic functional groups. Answer 2

3

1 4 Bromfenac

Experimental oral anticoagulant

5 7 8

6

Sorbinil Experimental antidiabetic agent

Answer: Drug Name

Acidic Functional Groups

Basic Functional Groups

Experimental Oral Anticoagulant

Sulfonamide (2)

Amidine (1)

Bromfenac

Carboxylic acid (4)

Aromatic amine or aniline (3)

As discussed in Chapter 3*, there are two key structural features of an acidic functional group. The first is the presence of a proton (H+) that can leave (i.e., dissociate), and the second is the presence of adjacent atoms that

acid and an imide are acidic, while functional groups such as secondaryNone hydroxyl groups (i.e., secondary Sorbinil Imide (5) Experimental (6) alcohols)Antidiabetic and thiolsAgent are neutralPhenol functional groups. Sulfonamide (7)

Secondary amine (8)

Resonance stabilization of resulting negative charge

No resonance available if proton would leave 99

Sorbinil 100

Medicinal Chemistry Self Assessment

Experimental antidiabetic agent

2. Using any one of the acidic functional groups that you identified in question 1, provide an explanaAnswer: tion as to why the functional group is acidic. Also provide a similar type of analysis for any one of the basic functional groups that you identified in question 1. As discussed in Chapter 3*, there are two key structural features of an acidic functional group. The first is the Answer presence of a proton (H+) that 3*, canthere leave are (i.e.,two dissociate), and the secondof is an the acidic presence of adjacent atoms As discussed in Chapter key structural features functional group. Thethat + first is the presence of a proton (H ) that can leave (i.e., dissociate), and the second is the presence acid and an imideatoms are acidic, functional groups such as secondary hydroxyl groups resonance. (i.e., secondary of adjacent that while are able to share the resulting negative charge through Hence, functional groups such as a carboxylic acid and an imide are acidic, while functional groups such as alcohols) and thiols are neutral functional groups. alcohols) and thiols are neutral functional groups. secondary hydroxyl groups (i.e., secondary

Resonance stabilization of resulting negative charge

Carboxylic Acid

No resonance available if proton would leave

Secondary Hydroxyl

Imide

Sulfhydryl

For basic functional groups, the key structural feature is a nitrogen atom that has a lone pairPage1of5 of + nonbonding electrons that can accept a proton (H ). The relative ability of the electrons on different  nitrogen-containing functional groups determines their relative basicity. Thus, amidines and secondary amines are basic, while amides are not. For basic functional groups, the key structural feature is a nitrogen atom that has a lone pair of nonbonding

amides are not. Resonance with adjacent carbonyl greatly decreases availablity of electrons

Amidine

Secondary amine

Amide

3. Using the structures from question 1, modify all of the acidic functional groups to show their ionized forms and in the table below identify the normal pKa range for the specific functional group. Answer:

_

amine

Amide

2.2 Identifying Acidic and Basic Functional Groups

101

3. Using the structures from question 1, modify all of the acidic functional groups to show their ionized forms and in3. theUsing table the below identify the normal pKa range for the functional group. groups to show their ionized structures from question 1, modify all specific of the acidic functional forms and in the table below identify the normal pKa range for the specific functional group. Answer: Answer

_

Bromfenac

Experimental oral anticoagulant

_ Sorbinil Experimental antidiabetic agent



Drug Name

Acidic Functional Group

Normal pKa Range

Experimental Oral Anticoagulant

Sulfonamide

pKa=5 to 10

Bromfenac

Carboxylic acid

pKa=2.5 to 5

Sorbinil

Imide

pKa=4.5 to 8.5 (generally on the higher end, 7.0 to 8.5)

Experimental Antidiabetic Agent

Phenol

pKa=9 to 10

Sulfonamide

pKa=5 to 10

Page2of5

4. UsingMedicinal the structures from Self question 1, modify all basic functional groups to show their ionized forms and in the 102 Chemistry Assessment table below identify the normal pKa range for the specific functional group. 4. Using the structures from question 1, modify all basic functional groups to show their ionized forms Answer: and in the table below identify the normal pKa range for the specific functional group. Answer OCH3 O

+ H2 N

H 2N O

N

NH2

O

O

S

OH

N H

O

Br

Bromfenac

Experimental oral anticoagulant

O

H N

HN

F

+ NH3

O

O

O

O

H N O

HO N

Sorbinil

N H

N

S

O

+

N H2

Experimental antidiabetic agent Drug Name

Basic Functional Group

Normal pKa Range

Experimental Oral Anticoagulant

Amidine

pKa=10 to 11

Bromfenac

Aromatic amine (aniline)

pK =2 to 5

Sorbinil

None

Not applicable

Experimental Antidiabetic Agent

Secondary amine

pKa=9 to 11

5. Shown below is the structure of clonidine, an D adrenergic agonist that can abe used to treat hypertension. Clonidine contains a guanidine functional group (highlighted in bold) that has a pKa of 8.3. Other guanidine functional groups, such as that seen with arginine are much more basic with a pKa of 12.5. Provide a chemical 5. Shown below is the structure of clonidine, an α2 adrenergic agonist that can be used to treat hyperexplanation for this difference. tension. Clonidine contains a guanidine functional group (highlighted in bold) that has a pKa=8.3. Other guanidine functional groups, such as that seen with arginine, are much more basic with a pKa=12.5. Provide a chemical explanation for this difference. pKa = 12.5

pKa = 8.3

Clonidine

Arginine

Answer Unlike the side chain of the amino acid arginine, the guanidine group of clonidine is directly Answer: attached to an aromatic ring. As such, the nitrogen atom directly attached to the aromatic ring (highlighted can donate electrons the aromatic thus making it attached less available Unlike the side chainbelow) of the amino acid arginine, theinto guanidine group ofring, clonidine is directly to an for resonance stabilization of a positive charge. In addition, the aromatic ring of clonidine is attached to two ortho chloro Each ofdirectly these halogens is electron withdrawing and further the aromatic ring. As such, thegroups. nitrogen atom attached to the aromatic ring (highlighted below)decreases can donate basicity of the guanidine group. The combination of these two effects decreases the basicity of the Page3of5 guanidine group byring, greater four orders of magnitude. electrons into the aromatic thus than making it less available for resonance stabilization of a positive charge. In



addition, the aromatic ring of clonidine is attached to two ortho chloro groups. Each of these halogens is electron

withdrawing and further decreases the basicity of the guanidine group. The combination of these two effects 2.2than Identifying Acidic Basic Functional Groups decreases the basicity of the guanidine group by greater four orders of and magnitude.

Cl

H N

103

Involved in resonance with aromatic ring

H N

N Cl 6. For each of the drug molecules shown below, determine if it is an acidic drug molecule, a basic drug molecule, Clonidine an amphoteric drug molecule, or a nonelectrolyte.

6. For each of the drug molecules shown below, determine if it is an acidic drug molecule, a basic drug Answer: molecule, an amphoteric drug molecule, or a nonelectrolyte. Answer Basic O N H

Acidic

H 2N

Basic

N H

N

N O

NH

S

O

Acidic N H

Lomitapide

O

COOH

O

NH

CF3

CF3

C H3

Argatroban



Page4of5

Acidic COOH

HO

CH3

H N Basic

OCH3

OH

O

O

H

H3C

O C H3

Cl

H

C H3

H 2N

N N

N H NH2

HO

Pravastatin

NH

Amiloride

N H2

Basic

S O N

O

CH3 O

CH2 N(CH3 )2 Basic

Diltiazem Drug Molecule

Acid/Base Character of Drug Molecule

Lomitapide

Basic: Contains only a basic tertiary amine

Argatroban

Amphoteric: Contains two acidic functional groups (a carboxylic acid and a sulfonamide) and two basic functional groups (a guanidine group and an aromatic amine)

Pravastatin

Acidic: Contains only an acidic carboxylic acid

Amiloride

Basic: Contains a basic guanidine group as well as a basic aromatic heterocyclic ring (pyrazine)

Diltiazem

Basic: Contains only a basic tertiary amine

Go to Section 1.3 [Link to section 1.3 title] Note: References to chapters are to Chapter 3 in Harrold MW and Zavod RM, Basic Concepts in Medicinal Chemistry, American Society of Health-System Pharmacists, ©2013.

Section 3 General Self Assessment Answers

2.3 Solving pH/pKa Problems Chapter 3 1. Shown below are the structures of cefotaxime, nitrofurantoin, atenolol, and ezetimibe. Each of these drug molecules contains one ionizable functional group. The pKa values have been provided. Solving pH/pK a Problems

a. Match the pKa values provided with the appropriate functional groups. For each functional group, 1. Shown below the are name the structures of cefotaxime, nitrofurantoin, atenolol, and ezetimibe. Each of these drug identify of the group and whether it is acidic or basic. molecules one ionizable functional TheitpK been provided. b. For contains each functional group indicategroup. whether would behave primarily ionized or primarily unionized at a a values stomach pH=1.8, a urinary pH=6.1, or a plasma pH=7.4. Provide an explanation for your responses for Answer: cefotaxime at a plasma pH=7.4, nitrofurantoin at a urinary pH=6.1, and atenolol at a stomach pH=1.8.

Answer Imide Acidic

Cefotaxime pKa = 3.4

Nitrofurantion Nitrofurantoin pKa = 7.1

Carboxylic acid Acidic

Aromatic hydroxyl (phenol) Acidic

Secondary amine Basic Atenolol pKa = 9.6

Ezetimibe pKa = 10.2

Cefotaxime contains an acidic carboxylic acid. At a plasma pH=7.4, the environment (i.e., the pH) is more basic than the functional group (i.e., the pH > pKa). An acidic functional group will be primarily ionized in a basicbelow environment. 3. Shown is the structure of natamycin. It contains two functional groups that could be potentially ionized. Nitrofurantoin contains an acidic imide group. At a urinary pH=6.1, the environment (i.e., the pH) is more The pKa values for natamycin are 4.6 and 8.4. acidic than the functional group (i.e., the pH < pKa). An acidic functional group will be primarily unionized in an acidic environment. Atenolol contains a basic secondary amine. At a stomach pH=1.8, the environment (i.e., the pH) is more H acidic than the functional group (i.e., the pH
O

OH 105

O

COOH

Answer: 106

Medicinal Chemistry Self Assessment

Drug (pKa Value)

Stomach (pH=1.8)

Urine (pH=6.1)

Imide Acidic Plasma (pH=7.4)

Cefotaxime (3.4)

Primarily unionized

Primarily ionized

Primarily ionized

Nitrofurantoin (7.1)

Primarily unionized

Primarily unionized

Primarily ionized

Cefotaxime Atenolol (9.6) pKa = 3.4 Ezetimibe (10.2)

Primarily ionized

Primarily ionized

Carboxylic acid Primarily unionized

Primarily unionized

Acidic

Primarily ionized Nitrofurantion = 7.1unionized pKPrimarily a

2. In the previous question, we examined four pKa values in three different environments for a total Aromatic of 12 different scenarios. Which of these 12 scenarios allow you to use the Rule of Nines to calculate hydroxyl the percent of ionization of the functional group in the specific environment? Identify the specific scenarios and use the Rule of Nines to calculate the percent of the functional group that would(phenol) be Secondary amine Acidic ionized. Basic Answer Atenolol To use the Rule of Nines, pK the = difference between the pH and the pK must be an integer (i.e., 1, 9.6 a Ezetimibe a 2, 3). In evaluating the above 12 scenarios, there are only two scenarios 10.2 meet this criterion, pK = that cefotaxime (pKa=3.4) in a plasma pH=7.4 and nitrofurantoin (pKa=7.1)a in a urinary pH=6.1. For cefotaxime, |pH – pKa| is equal to 4; thus, there is a 99.99:0.01 ratio. Because the carboxylic acid (pKa=3.4) would be primarily ionized in a basic environment (pH=7.4), we can use this ratio to determine that it would be 99.99% ionized. For nitrofurantoin, |pH – pKa| is equal to 1; thus, there is a 90:10 ratio. The imide (pKa=7.1) functional group is acidic, so it would be primarily unionized in an acidic environment (pH=6.1). We can use this information to predict that it would ionized. 3. Shown below is the structure of natamycin. It contains two functional groups be that10% could be potentially ionized.

The pKa values for natamycin are 4.6 and 8.4. 3. Shown below is the structure of natamycin. It contains two functional groups that could be potentially ionized. The pKa values for natamycin are 4.6 and 8.4.

H3C

OH

O

O O

OH

OH

O

O

COOH

O

C H3

Natamycin HO

OH N H2

a. Match the pKa values provided to the appropriate functional groups and identify if the functional group is acidic or basic.



Page1of2

b. Using the Henderson-Hasselbalch equation, calculate the percent ionization that occurs for each of these functional groups at an intestinal pH=6.2.

each of these functional groups at an intestinal pH of 6.2. 2.3 Solving pH/pKa Problems

107

Answer: Answer a.

H3C

OH

O

O O

OH

OH

O

O

COOH

O

Carboxylic acid (pKa = 4.6) Acidic

C H3

Natamycin HO

OH N H 2 Primary amine (pKa = 8.4) Basic

b. For the carboxylic acid, the pKa=4.6 and the pH=6.2. Using the Henderson-Hasselbalch equation gives the group following: 4. The most basic functional present within the structure of ranitidine has a pK value of 8.2. Identify this a

Form] functional group and calculate the [Base 6.2 = 4.6 +log pH that is necessary for this functional group to be 70% ionized. [Acid Form] Answer: Solving the equation provides a ratio of [Base Form]/[Acid Form].

1.6 = log



[Base Form] [Acid Form]

The electron withdrawing property of the nitro group greatly decreases the pKa value for the

[Base Form] 39.8 [Base Form] adjacent nitrogen containing group. 39.8 =   or   =  [Acid Form] 1 [Acid Form]

amine Tertiary pKa = 8.2 Ranitidine For every molecule that contains this functional group in the acid form, there are 39.8 molecules that contain this functional group in the base form. Because the functional group is acidic, the base form is equal to the ionized form, and the acid form is equal to the unionized form.

39.8 molecules in base form + 1.0 molecule in acid form = 40.8 Total Molecules



Base Form = Ionized Form, and Acid Form = Unionized Form for This Functional Group



39.8 Molecules in Ionized Form Percent in Ionized Form =    x 100% 40.8 Total Molecules Percent in Ionized Form= 97.5%

For the primary amine, the pKa=8.4 and the pH=6.2. Using the Henderson-Hasselbalch equation gives the following:



[Base Form] 6.2 = 8.4 +log    [Acid Form]

Page2of2

Answer: 108 Medicinal Chemistry Self Assessment

Solving the equation provides a ratio of [Base Form]/[Acid Form]. OH O O OH [Base Form] −2.2 = log [Acid Form]H 3 C O OH O Carboxylic acid COOH (pKa = 4.6) [Base Form] 0.0063 [Base Form] 0.0063 = or = Acidic [Acid Form] 1 [Acid Form] O O C H3 For every molecule that contains this functional group in the acid form, there are 0.0063 molecules that contain this functional group in the base form. Because the functional group is basic, the base Natamycin form is equal to the unionized form, and the acid formHisOequal to theOionized form. H

0.0063 molecules in base form + 1.0 molecule in acid form = 1.0063 Molecules N H 2 Total Primary amine

(pKa = 8.4) Base Form = Unionized Form, and Acid Form = Ionized Form for This Functional Basic Group

1 Molecule in Ionized Form Percent in Ionized Form =   x 100% 1.0063 Total Molecules 4. Percent in Ionized Form The most basic functional = 99.4% group present within the structure of ranitidine has a pKa value of 8.2. Identify this functional group and calculate the pH that is necessary for this functional group to be 70% ionized. 4. Answer: The most basic functional group present within the structure of ranitidine has a pKa=8.2. Identify this functional group and calculate the pH that is necessary for this functional group to be 70% ionized. Answer The electron withdrawing property of the nitro group greatly decreases the pKa value for the The electron withdrawing property of the nitro adjacent nitrogen containing group. group greatly decreases the pKa value for the adjacent nitrogen-containing group.

Tertiary amine pKa = 8.2

Ranitidine

We can use the Henderson-Hasselbalch equation to solve this problem. Because the ionized form of a basic functional group can also be designated as its protonated form or its conjugate acid form, either of the following equations can be used.

pH = pK a +log 

[Base Form] [Acid Form]

or pH = pK a +log 

[Unprotonated Form] [Protonated Form]

Therefore, if 70% of this functional group is ionized:



pH = 8.2 + log

[30] [70]

pH = 8.2 + log

[30] = 8.2 + ( −0.37) = 7.83 [70]

Because a pH=7.83 does not exist physiologically, this percent ionization could only occur in an exogenously prepared solution.

Page2of2

Section 3 General Self Assessment Answers

2.4 Salts and Solubility

1. Some drug molecules can be formulated as either a potassium salt or hydrochloride salt. Evaluate all of the acidic and basic functional groups in each of the drug molecules drawn below and fill in the grid with the appropriate information. Answer

Name of Drug Molecule

Name of Functional Group That Can Form a Potassium Salt

Is the Potassium Salt Acidic, Basic, or Neutral?

Name of Functional Group That Can Form a Hydrochloride Salt

Is the Hydrochloride Salt Acidic, Basic, or Neutral?

Arformoterol

Phenol

Basic

Secondary amine

Acidic

Baclofen

Carboxylic acid

Basic

Primary amine

Acidic

Irbesartan

Tetrazole

Basic

Amidine

Acidic

Ciprofloxacin

Carboxylic acid

Basic

Secondary amine (NH) component of piperazine

Acidic

Aniline component (N) of piperazine

Acidic

Tertiary amine

Acidic

Tetracycline

Phenol

Basic

2. Each of these drug molecules will treat a particular ailment by either managing a symptom or β-dicarbonyl

Basic

modulating biochemical pathway. In either case, the drugthe hasconjugate to get to its biological Using your Note: The ionized aform of an acidic functional group is termed base of that target. functional group. The ionized form of a basic functional group is termed the conjugate acid of that functional group. knowledge about functional group character, describe how both of the drugs get to their respective biological targets. Consider the concepts of solubility, absorption, distribution, and route of administration 2. Each of these drug molecules will treat a particular ailment by either managing a symptom or modulating a biochemical pathway. In either case, the drug has to get to its biological target. Using your knowlin your answer. [pH (stomach) = 1; pH (intestine) = 8; pH (plasma) = 7.4] edge about functional group character, describe how both of the drugs get to their respective biological targets. Consider the concepts of solubility, absorption, distribution, and route of administration in your answer. [pH (stomach)=1; pH (intestine)=8; pH (plasma)=7.4] NCH3

Cl OH

O

N

O O

CONH2

CH2OH

Scopolamine

Loperamide

3. Based on your structural evaluation, provide a rationale for why each of these drugs cannot be 109

delivered via an oral route of administration, or why there is limited absorption of the drug when it is

110

Medicinal Chemistry Self Assessment

Biological Targets and Routes of Administration Loperamide: µ opioid receptors located in the large intestine; oral administration Scopolamine: muscarinic acetylcholine receptors (M1) located in the peripheral nervous system; transdermal administration Answer Loperamide: • Oral tablet is swallowed. • Tablet dissolves in aqueous contents of the stomach (requires drug to be water soluble). The amide, piperidine (tertiary amine), and tertiary alcohol have hydrophilic character and contribute to the water solubility of the drug. • Do not want absorption into systemic circulation via stomach lining. The piperidine (tertiary amine pKa=9–11) will be predominantly ionized in the stomach pH=1 (pH < pKa). • Distribution into intestine—smaller fraction in ionized form; retain water solubility. The piperidine (tertiary amine pKa=9–11) will be somewhat less (but still predominantly) ionized in the intestine pH=8 (pH < pKa). • Interaction with receptor located in intestine—there is no need for absorption out of the gastrointestinal (GI) system and into systemic distribution. Scopolamine: • Patch is placed on skin. • Absorption of drug occurs from patch into skin lipophilic bilayer (requires hydrophobic character). The aromatic ring and bicyclic heterocycle contribute to hydrophobic character of the drug and allow for absorption and transport through skin. • Drug transports through skin and dissolves into circulation (requires drug to be water soluble). The bicyclic heterocycle (tertiary amine pKa=9–11) will be primarily ionized in the plasma pH=7.4 (pH < pKa). The ester, ether, and primary alcohol are hydrophilic in character and contribute to the water solubility of the drug. • Distribution to peripheral receptors (requires drug to be water soluble). The bicyclic heterocycle (tertiary amine pKa=9–11) will be primarily ionized in the plasma pH=7.4 (pH < pKa). The ester, ether, and primary alcohol are hydrophilic in character and contribute to the water solubility of the drug. • Interaction occurs with receptors located in periphery.

OH

O

O

N

CH2OH

O

CONH2

2.4 Salts and Solubility

111

3. Based on your structural evaluation, provide a rationale for why each of these drugs cannot be delivScopolamine ered via an oral route of administration, or why there is limited absorption of the drug when it is Loperamide administered orally.

Insulin Structure

ased on your structural evaluation, provide aS rationale for why Seach of these drugs cannot be

A Chain

Ile Val Glu Gln Cys Cys Thr Ser Ile Cys Ser Leu Tyr Gln Leu Glu Asn Tyr Cys Asn COOH H2NofGly ered via an oral route administration, or why there is limited absorption of the drug when it is

S

nistered orally. H2 N Phe Val Asn Gln His

S

S

Leu Cys Gly Ser His

Leu Val Glu Ala Leu Tyr

S

Leu Val Cys

B chain Gly Glu Arg Gly

HOOC



Thr30

Lys29 Pro28

Thr27

Tyr

Phe Phe

PO3Na2 PO3Na2

H2N OH

Alendronate Insulin



Answer Insulin is a molecule composed of two peptide subunits connected to one another via two disulfide linkages. Peptide bonds do not survive the hydrolytic environment found in the stomach. Insulin can undergo acid catalyzed hydrolysis in the aqueous environment of the stomach, as well as enzymatically catalyzed hydrolysis in the same physiological compartment. Because of this rapid degradation, peptide-based drugs like insulin cannot be delivered via the oral route. Alendronate contains two ionizable phosphonates, an ionizable primary amine, and a tertiary alcohol. These functional groups impart significant hydrophilic character and make alendronate very water soluble. There is very, very, little hydrophobic character to contribute to absorption across the GI mucosal membranes. Although alendronate can be administered orally, less than 6% of a given dose is absorbed and distributed systemically.

4. Based on your structural evaluation of embeconazole, answer the following questions: 4. Based on your structural evaluation of embeconazole, answer the following questions:

Embeconazole

112

Medicinal Chemistry Self Assessment

Answer Can this be formulated as a. a topical ointment or cream? YES NO b. an aqueous solution?

YES NO

c. Which structural features and corresponding characteristics did you consider when answering each of these questions? Although there are hydrophilic functional groups that can participate in hydrogen bonding interactions with water and impart a reasonable amount of water solubility (fluorine atoms on the aromatic hydrocarbons, tertiary alcohol, a weakly basic nitrogen containing heterocycle [1,2,4 triazole], and oxygen containing heterocycle [1,3 dioxane]), their effect is overwhelmed by the functional groups that are hydrophobic in character (two aromatic hydrocarbons, alkene, thioether). This hydrophobic character allows for good drug penetration into and across the skin if applied as a topical cream or ointment; however, this character is likely to significantly limit the drug’s water solubility as an aqueous solution.

Go to Section 1.5 [Link to section 1.5 title] Note: References to chapters are to Chapter 5 in Harrold MW and Zavod RM, Basic Concepts in Medicinal Chemistry, American Society of Health-System Pharmacists, ©2013.

Section 3 General Self Assessment Answers

2.5

Drug Binding Interactions

1. Both delapril and lisinopril are inhibitors of angiotensin converting enzyme (biological target) a 2.5

2.5 Drug Binding Interactions

other interacts with a Zn+2 atom. Both of these interactions are critical for drug action. Do the followin Drug Binding Interactions

a. Modify the molecules below to show the form of the active drug at physiological pH. 1. Both delapril and lisinopril are inhibitors of angiotensin converting enzyme (biological target) a

b. Identify the type of interaction possible between the carboxylic acids and the residues pr other interacts with a Zn+2 atom. Both of these interactions are critical for drug action. Do the followin

the biological target. converting enzyme (biological target) and as such 1. Both delapril and lisinoprila.arewithin inhibitors of angiotensin Modify the molecules below to show the form of the active drug at physiological pH. are valuable drugs used in the management of hypertension. The active form of these drugs requires the Answer: presence of two carboxylic acids or twothe functional groups thatpossible can participate similar types of interactions b. Identify type of interaction betweeninthe carboxylic acids and the residues pr at physiological pH. One of the carboxylic acids interacts with an active site arginine residue, and the other interacts with a Zn+2 atom. Both of these interactions are critical for drug action. Do the following: within the biological target. a. Modify the molecules below to show the form of the active drug at physiological pH (pH=7.4). Answer: b. Identify the type of interaction possible between the carboxylic acids and the residues present within H 2N 2+ the biological target. CH2 R Zn O

Answer O

–

O

Zn O

–

2+

O

C H3

+ N

N

H H O O C H3 + Delapril N N H H O

–

N

H

H 2N

O O

+ NH2

–

+ NH2

N

CH2 R H



Ionic interactions areDelapril possible between the carboxylic acids and the zinc ion and the ionized arginine



residue. Ionic interactions are possible between the carboxylic acids and the zinc ion and the ionized arginine residue. Ionic interactions are possible between the carboxylic acids and the zinc ion and the ionized arginine residue.

-

Lisinopril



Lisinopril

 113

114

Medicinal Chemistry Self Assessment

2. Tolterodine (Detrol®), fesoterodine (Toviaz®), and oxybutynin (Ditropan®) are anticholinergic agents ® in the treatment of overactive bladder. Based on the structure process described 2.used Tolterodine (Detrol®), fesoterodine (Toviaz ), and oxybutynin (Ditropan®evaluation ) are anticholinergic agents in Chapter 6* and previous chapters,* “read” each of these drug molecules and determine how each of these drug molecules interact via hydrogen bonding with the muscarinic receptor. Fill in the grid provided with your answers.

Tolterodine



Fesoterodine



Oxybutynin





Answer Function

Function

Name of ALL Anticholinergics That Contain Functional Group

H-bond donor, H-bond acceptor, both or neither

One amino acid that interacts (via side chain) with the functional group via a complementary H-bonding interaction; NONE is a possible answer

Alkane (i.e., alkyl chain)

Tolterodine, fesoterodine, oxybutynin

Neither

None

Cycloalkane

Oxybutynin

Neither

None

Aromatic hydrocarbon

Tolterodine, fesoterodine, oxybutynin

Neither

None

Alkyne

Oxybutynin

Neither

None

Tertiary alcohol

Oxybutynin

Both

Serine, threonine, cysteine, tyrosine, glutamic acid, aspartic acid, lysine, arginine, asparagine, glutamine, histidine, tryptophan

Primary alcohol

Fesoterodine

Both

Serine, threonine, cysteine, tyrosine, glutamic acid, aspartic acid, lysine, arginine, asparagine, glutamine, histidine, tryptophan

Name of Functional Group

2.5 Drug Binding Interactions

115

Continued from previous page Phenol

Tolterodine

Both

Serine, threonine, cysteine, tyrosine, glutamic acid, aspartic acid, lysine, arginine, asparagine, glutamine, histidine, tryptophan

Tertiary amine

Oxybutynin, fesoterodine, tolterodine

Acceptor

Serine, threonine, cysteine, tyrosine, glutamic acid, aspartic acid, lysine, arginine, asparagine, glutamine, histidine, tryptophan

Ester

Fesoterodine, oxybutynin

Acceptor

Serine, threonine, cysteine, tyrosine, glutamic acid, aspartic acid, lysine, arginine, asparagine, glutamine, histidine, tryptophan

Note: Please remember that if a functional group is ionized, then it can no longer participate in hydrogen bonding interactions. 3. Each of the three odorant molecules drawn below produces a unique scent on interaction with the olfactory receptors. Unlike most biological targets for drug action, olfactory receptors typically have anEach affinity wideodorant variety molecules of odorantdrawn molecules can adopt uniquescent conformations to enhance 3. of for the athree belowand produces a unique on interaction with the the affinity of a given odorant for the receptor. Based on the structural features found in each molecule, what type of Unlike interactions are possible withforthe olfactory receptorsreceptors at physiological olfactory receptors. most biological targets drug action, olfactory typically pH? haveIndicate an which functional group(s) can participate in each of the interactions identified.

Sclareol (herbal scent)

Nerolidol (green (greenwoody woodyscent) scent)

Vanillin



Sclareol

Vanillin

Interaction Type

Functional Group

Interaction Type

Nerolidol Functional Group

Interaction Type

Functional Group

Cycloalkane, aliphatic van der Waals, Phenol der Waals, Aliphatic 4.Hydrophobic There are five basic flavors that our taste receptors detect: salty, van sour, sweet, bitter, andalkanes umami alkane, alkene

Hydrophobic

Hydrophobic

and alkenes

(savory). The taste Tertiary receptors are located on taste areether, on our tongue, soft palate, Tertiary epiglottis, and Hydrogen bond alcohol Hydrogen bondbuds that Phenol, Hydrogen bond alcohol (acceptor and donor) Ion–dipole (as the dipole)

H 2N

Tertiary alcohol

CO2 H

(acceptor)

aldehyde

(acceptor and donor)

Hydrogen bond (donor)

Phenol

Ion–dipole (as the dipole)

Phenol, ether, aldehyde

CO2 H

Glutamic Acid



Cucumber flavor

 O

N

N

H

Green Pepper Flavor

116

Sclareol (herbal  Assessment Vanillin Medicinal scent) Chemistry Self

Nerolidol (green woody scent)

4. There are five basic flavors that our taste receptors detect: salty, sour, sweet, bitter, and umami (savory). The taste receptors are located on taste buds that are on our tongue, soft palate, epiglottis, and upper esophagus. andthat saltyour flavors mediated by ion channels, whereas sweet, bitter, 4. There are five basicSour flavors tasteare receptors detect: salty, sour, sweet, bitter, and umami and umami flavors are derived from activation of the respective G-protein coupled receptor. The umami receptor is receptors activatedare by L-amino acids, glutamate. Determine what interactions (savory). The taste located on tastespecifically buds that are on our tongue, soft palate, epiglottis, and are possible with the side chain of glutamate (at physiological pH) and then determine if any of the following flavor molecules can interact with and activate this receptor. CO2 H

H 2N

CO2 H

Glutamic Acid

Cucumber flavor





Green Pepper Flavor

O N

N

O HO HO

N

P O

O

HO

H

N

OH

Shitaki Mushrooms Inosinic Acid

Answer The side chain of glutamic acid (glutamate) contains a carboxylic acid that will be ionized at physiological pH. In its ionized form glutamic acid can participate in an ionic interaction or in an ion–dipole interaction (as the ion). The functional groups in both cucumber and green pepper flavor cannot participate in ionic interactions with glutamate as none of the functional groups are ionized (cationic) at physiological pH. Ion–dipole interactions (as the dipole) are possible between glutamate and the aldehyde found in cucumber flavor and the methyl ether found in green pepper flavor. Inosinic acid is a component found with shitaki mushrooms that is thought to contribute to their unique flavor. Structural evaluation of this molecule reveals a phosphate that is ionized at physiological pH. This functional group can participate in ion–dipole interactions with the umami receptor to elicit a savory flavor.

Go to Section 1.6 [Link to section 1.6 title] Note: References to chapters are to Chapter 6 in Harrold MW and Zavod RM, Basic Concepts in Medicinal Chemistry, American Society of Health-System Pharmacists, ©2013.

Section 3 General Self Assessment Answers

2.6 Stereochemistry and Drug Action

2.6 Stereochemistry and Drug Action

1. Shown below are the structures of acebutolol, estradiol, cefamandole, and nifedipine. For each of these 1. Shown below are the structures of acebutolol, estradiol, cefamandole, and nifedipine. For each of these compounds, identify allcenters. chiral centers. compounds, identify all chiral

O

N

O

OH

OH

HO

HN

Estradiol

O

Acebutolol H3C

O

OH

S

N H O

C H3 N

S

N CO2 H

N

H N

H3 CO2 C

CO2 CH3 NO2

N

N

Cefamandole

Nifedipine

117

C H3

118

Medicinal Chemistry Self Assessment

Answer: Answer

O

O

OH

N

*

*

*

OH

*

*

*

HO

HN

Estradiol 5 chiral centers

O

Acebutolol 1 chiral center

H3C

H N

C H3

O

* OH

*

N H O

*

S

C H3 N

S

N CO2 H

Cefamandole 3 chiral centers

N

N

H3 CO2 C

CO2 CH3 NO2

N

Nifedipine 0 chiral centers The highlighted carbon atom is symmetrical

3. Shown below is the structure of fluvastatin, an HMG-CoA reductase inhibitor used to lower plasma LDL levels. 2. For each of the four drug molecules shown in question 1: Fluvastatin contains two chiral centers, designated as A and B. Using the structure of fluvastatin and the Cahna. Identify if it can have an enantiomer. Provide an explanation for your response. Ingold-Prelog (CIP) if system, the R/S configurations each of its for chiral centers. b. Identify it can determine have a diastereomer. Provide anfor explanation your response. c. Identify if it can have a geometric isomer. Provide an explanation for your response. Answer HO A CO2 H a. Acebutolol, estradiol, and cefamandole can have enantiomers, while nifedipine cannot. For a drug molecule to have an enantiomer, its structure must contain at least one chiral center. AcebuB OH tolol, estradiol, and cefamandole all meet this criterion. Because the structure of nifedipine does H not contain a chiral center, it cannot have enantiomers. b. Estradiol and cefamandole can have diastereomers, but acebutolol and nifedipine cannot. For F a drug molecule to have a diastereomer, its structure must contain at least two chiral centers. N have five and three chiral centers, respecBecause the structures of estradiol and cefamandole tively, these two drug molecules meet this criterion and can have diastereomers. Acebutolol (one chiral center) and nifedipine (0 chiral centers) do not meet this criterion and thus cannot have diastereomers. c. Estradiol and cefamandole can have geometric isomers, while acebutolol and nifedipine cannot. Fluvastatin Geometric isomers are a specialized type of diastereomer and result from restricted rotation about a carbon–carbon bond. Geometric isomers can occur in the presence of either a double bond or an alicyclic ring. Both estradiol and cefamandole contain alicyclic rings and thus can have geometric isomers. Acebutolol and nifedipine do not meet the above criterion and cannot



Page2of3

2.6 Stereochemistry and Drug Action

119

have geometric isomers. Please note that the double bonds seen in the 1,4-dihydropyridine ring of nifedipine reside in a six-membered ring and thus cannot participate in the formation of geometric isomers.

Medicinal Chemistry Self-Assessment Book: Batch Two Chapters 2.6plasma 3. Shown below is the structure of fluvastatin, an HMG-CoA reductase inhibitor used1.6 to and lower LDL levels. Fluvastatin contains two chiral centers, designated as 1 and 2. Using the structure of Revised(CIP) structure reflects chiral carbons and #2. fluvastatin and the Cahn-Ingold-Prelog system, determine the R/S #1 configurations for each of its chiral centers.

HO

1 2

F

CO2 H OH H C H3

N

C H3

Fluvastatin



Answer: Answer

A

Chiral center 1

D C

B

Chiral center 2

B

A D C

Fluvastatin

Fluvastatin

Using the CIP sequence rules for chiral center 1, the hydroxyl group (A) has the first priority because oxygen always has priority over carbon. Carbon atoms B and C are identical because each is attached 4. Shown below is the enantiomer of fluvastatin. Which of the following properties/actions would be expected to to one carbon atom and two hydrogen atoms. It is therefore necessary to examine the two adjacent carbon for atoms. The carbon adjacent carbon C (O,O,O) has priority over the carbon atom adjabe identical fluvastatin and itsatom enantiomer andtowhich would be expected to be different? cent to carbon B (O,C,H); thus, the final priority is hydroxyl group (A) > carbon atom C > carbon atom B > a. hydrogen (D). The Percentatom ionization at ahydrogen pH of 7.4 atom is projected into the paper and located away from the reader. Applying the above priorities reveals a clockwise orientation. Thus, the correct designation for chiral center 1 is R. Using the CIP sequence rules for chiral center 2, the hydroxyl group (A) once again has the first H Opriority. Carbon atom C (C,C,H) has priority over priority, and the hydrogen atom (D) has the least CO2 H carbon atom B (C,H,H); thus, the final priority is hydroxyl group (A) > carbon atom C > carbon atom OH H

120

Medicinal Chemistry Self Assessment

B > hydrogen atom (D). Unlike chiral center 1, the hydrogen atom of chiral center 2 is projected out of the paper and located directly toward the reader. In situations like this, there are three options: Answer: (1) redraw the chiral center with the hydrogen projected into the paper while maintaining the correct stereochemical orientation of all other groups; (2) imagine that you are viewing the object from the other side of the paper; or D

A (3) use the given projection and make an alteration in your final answer. C

Options 1 and 2 are often difficult for students to accomplish without producing errors; therefore, optionChiral 3 is suggested. atom (or2the atom ofA lowest priority) IS center 1 Initially “pretend” that the hydrogen Chiral center projected into the plane ofBthe paper and apply the sequence rules. In this Bcase, there is a clockwise D stereochemorientation that is consistent with the R isomer. Because it was necessary to invert the istry of the chiral center to do this, the correct designation is the exact opposite. CIn this case, chiral center 1 has the S configuration.

4. Shown below is the enantiomer of fluvastatin. Which of the following properties/actions would be expected to be identical for fluvastatin and its enantiomer and which would be expected to be different? Fluvastatin Fluvastatin a. Hepatic metabolism b. Water solubility

4. Shown is the enantiomer c. below Adverse effect profile of fluvastatin. Which of the following properties/actions would be expected to d. Active renal reabsorption by transport proteins be identical for fluvastatin and its enantiomer and which would be expected to be different? e. Potency (dosage given) a. Percent ionization at a pH of 7.4 f. Percent ionization at a pH=7.4

HO

CO2 H OH H

F N

Enantiomer of Fluvastatin

Answer Enantiomers have identical physical and chemical properties, with the exception of the direction in which they rotate plane polarized light. Thus, water solubility and the percent ionization of the carboxylic acid at a pH of 7.4 would be expected to be identical. The major difference and most important aspect of enantiomers are their relative abilities to interact with three-dimensional biological targets. Hepatic metabolism and active renal reabsorption depend on binding to metabolizing enzymes and transport proteins, respectively, and would be expected to be different. Adverse effects can be due to the interaction of these drug molecules with other biological targets and/or the

2.6 Stereochemistry and Drug Action

121

formation of a specific metabolite. Differences in potency can result in differential metabolism (i.e., one enantiomer may be inactivated quicker than the other) or differential binding to the biological target.

Go to Section 1.6 [Link to section 1.6 title] Note: References to chapters are related to Chapter 7 in Harrold MW and Zavod RM, Basic Concepts in Medicinal Chemistry, American Society of Health-System Pharmacists, ©2013.

Section 3 General Self Assessment Answers

2.7 Drug Metabolism 2.7 Drug is Metabolism 1. Heroin a synthetic derivative of the naturally occurring opioid analgesic morphine. Given its illicit nature, heroin is not subject to Food and Drug Administration (FDA) regulation; therefore, the actual 1. Heroin isofaasynthetic derivative the naturally occurring opioidmolecules analgesicdrawn morphine. Given its illicit composition given heroin batchof varies significantly. The three below represent the three major components found within a batch of heroin. The users’ intense rush has been attributed to morphine. Which phase I transformation is responsible for the conversion of heroin to morphine? the conversion of heroin into morphine. Which phase I transformation is responsible for the conversion of heroin to morphine?

Heroin

Morphine

2. Estradiol, the estrogen component of many oral contraceptives, undergoes a phase II conjugation then Answer Heroin is the diacetyl ester of morphine. Two successive phase I ester hydrolysis reactions are responsible (at least in part).to Consider the structure of estradiol drawn below and do the following: forreabsorbed the conversion of heroin morphine. a. Modify the structure drawn below to show the product of a sulfate conjugation (phase II transformation). 2. Estradiol, the estrogen component of many oral contraceptives, undergoes a phase II conjugation reaction to produce metabolite that is eliminated viathis a fecal route (at least in part). The conjugated hormone b.aWhich enzyme is required to make sulfate conjugate? undergoes a process called enterohepatic recycling. In this process the sulfate conjugate is cleaved by gut bacteria toAnswer: regenerate the active drug, which is then reabsorbed (at least in part). Consider the structure Sulfotransferase (SULT) of estradiol drawn below and do the following: c. Which deconjugating enzyme catalyzes removal of the sulfate group thus allowing for a. Modify the structure drawn below to show the product of a sulfate conjugation (phase II transformation). enterohepatic recycling? b. Which enzyme is required to make this sulfate conjugate? Answer: Sulfatase Answer Sulfotransferase (SULT)

-

123



a. Modify the structure drawn below to show the product of a sulfate conjugation (phase II transformation). 124

Medicinal Chemistry Self Assessment

b. Which enzyme is required to make this sulfate conjugate?

Answer: Sulfotransferase (SULT) c. Which deconjugating enzyme catalyzes removal of the sulfate group thus allowing for enterohepatic recycling? c. Which deconjugating enzyme catalyzes removal of the sulfate group thus allowing for Answer enterohepatic recycling? Sulfatase Answer: Sulfatase

-



3. Metabolites do not necessarily have the same mechanism of action as the parent drug. In the case of chlorimipramine (a tricyclic antidepressant that inhibits serotonin uptake), a phase I transformation produces a metabolite that is also a tricyclic antidepressant, but whose mechanism of action is via inhibition of norepinephrine reuptake. Which phase I transformation has occurred? What additional phase I transformations are possible? 3. Metabolites do not. In the case of I transformations are possible?

Chlorimipramine

Answer 4. Evaluate each of the which phase I metabolic transformation has occurred. Phase I transformation that has occurred: oxidative N-dealkylation Other phase I transformationsHpossible: ortho/para aromatic hydroxylation; benzylic oxidation; N-oxidation; oxidative N-dealkylation is still a methyl group); oxidative deamination (of the N C H(there N H2 3 doubly-demethylated product). One of the products from the oxidative deamination is an aldehyde, which can undergo further oxidation (aldehyde oxidation) to the carboxylic acid. C H3

CF3

Dexfenfluramine



C H3

CF3



2.7 Drug Metabolism

125

Chlorimipramine 4. Evaluate each of the following metabolic transformations and determine which phase I metabolic transformation has occurred. 4. Evaluate each of the which phase I metabolic transformation has occurred. H N

C H3

N H2

C H3

C H3

CF3

CF3

Dexfenfluramine





Fluvoxamine



 H2N

CO2 H O

O

OH

OH Cl

Cl

Baclofen



Answers Oxidative N-dealkylation Oxidative O-dealkylation Oxidative deamination followed by oxidation of the resulting aldehyde to a carboxylic acid

Go to Section 1.7 [Link to section 1.7 title] Note: References to chapters are related to Chapter 8 in Harrold MW and Zavod RM, Basic Concepts in Medicinal Chemistry, American Society of Health-System Pharmacists, ©2013.

Part 2 ANSWERS

Section 4 – Whole Molecule Drug Evaluation 2.8 2.9 2.10 2.11 2.12 2.13 2.14 2.15 2.16 2.17 2.18 2.19 2.20

Aliskiren..................................................................... 129 Aripiprazole................................................................ 137 Cefprozil..................................................................... 143 Cetirizine.................................................................... 149 Chlorpropamide and Other Sulfonylureas.................. 153 Dabigatran Etexilate.................................................. 159 Fenofibrate and Gemfibrozil....................................... 165 Fluvoxamine............................................................... 169 Haloperidol................................................................ 173 Hydrocortisone........................................................... 177 Levothyroxine (T4)...................................................... 183 Lidocaine.................................................................... 189 Montelukast and Zafirlukast...................................... 193

2.21 2.22 2.23 2.24 2.25 2.26 2.27

Phenobarbital and Other Barbiturates....................... 199 Pravastatin and Fluvastatin........................................ 205 Quinapril.................................................................... 211 Rivastigmine.............................................................. 217 Sitagliptin................................................................... 223 Sorafenib.................................................................... 227 Zanamivir and Oseltamivir......................................... 233

Section 4 Whole Molecule Drug Evaluation Answers 2.8 Aliskiren

Aliskiren is an orally active agent used in the treatment of hypertension. This non-peptide drug acts as an inhibitor of renin, the enzyme that converts angiotensinogen (its endogenous substrate) to angiotensin I. Biologically inactive, angiotensin I is rapidly converted to angiotensin II by angiotensin converting enzyme. Angiotensin II is a potent agonist when bound to its receptor and produces significant vasoconstriction, as well as an increase in blood pressure. In the presence of aliskiren, angiotensinogen is not converted to angiotensin I so less angiotensin II is produced to activate the angiotensin II receptor. Consequently less vasoconstriction occurs, and a drop in blood pressure results. Medicinal Chemistry Self-Assessment Book: Batch Two Chapters 1.8 and 2.8

1. Conduct a structural evaluation of aliskiren, focusing on the boxed functional groups and use the informaChapter 1.8/2.8 (remove bold from drug name x2) tion in the grid to inform your answers to the questions that follow.

D

A

E

F

B

Aliskiren

C

Aliskiren

129

130

Medicinal Chemistry Self Assessment

Answer Function Function

Character Character

A

Name of Functional Group

Hydrophilic and/or Hydrophobic

Ether

Hydrophilic (O)

Provide pKa When Relevant

↑ Solubility and/or ↑ Absorption

Interaction(s) Possible with Biological Target at Physiological pH=7.4

Neutral

Solubility (O)

H-bonding (A)

Absorption (R)

Dipole–dipole

Function

Acidic, Basic, or Neutral

Hydrophobic (R)

Amino Acids That Can Interact with the Functional Group via Ion– Dipole Interactions at pH=7.4

None Is Acceptable Asp, Glu, Lys, Arg

Ion–dipole (as the dipole) B

Aromatic hydrocarbon

Hydrophobic

Neutral

Absorption

Hydrophobic

None

van der Waals π-π stacking

C

Aliphatic alkane

Hydrophobic

Neutral

Absorption

Hydrophobic

None

van der Waals

D

E

Primary amine

Secondary alcohol

Medicinal Chemistry Self-Assessment Book: Batch Two Solubility (NH2) Ion–dipole (as the ion) Ser, Thr, Cys, Tyr, Asn, Chapters 1.8 and 2.8

Hydrophilic (NH2)

Basic

Hydrophobic (R)

pKa ~9–11

Absorption (R)

Neutral

Solubility (OH)

H-bonding (A + D)

Absorption (R)

Dipole–dipole

Chapter 1.8/2.8 (remove bold from drug name x2)

Hydrophilic (OH) Hydrophobic (R)

F

Amide

Glu, His, Trp

Ionic

Hydrophilic (CONH2) Hydrophobic (R)

A

D Neutral

B

E

Asp, Glu, Lys, Arg

Ion–dipole (as the dipole)

F

Solubility (CONH2)

H-bonding (A)

Absorption (R)

Ion–dipole (as the dipole)

Asp, Glu, Lys, Arg

Dipole–dipole

Aliskiren 2. Aliskiren is marketed as the pure 2S, 4S, 5S, 7S C enantiomer. Circle all of the chiral carbon atoms and determine if diastereomeric or geometric isomers are possible.

Aliskiren

2.8 Aliskiren

Answer

131

(remove bold from drug name) Chapter 2.8

Aliskiren

If there are two or more chiral carbon atoms, then it is possible for the drug to have diastereomeric forms. In the case of aliskiren, chiral carbon atoms; therefore, there are a lot of poten(remove boldthere from are drugfour name) Chapter 1.8/2.8 tial diastereomeric forms. A quick reminder on how to determine if a pair of isomers is diastereomeric—if one chiral carbon is held constant (e.g., R-isomer) and you vary at least one additional chiral carbon, then the pair of isomers will be diastereomeric to one another. In the example shown below, there areheldisconstant and theand (*) chiral center varied.isThe pair of isomers are diastereom the chiral center Ifthat is circled held constant the (*) chiraliscenter varied. The pair ofdrawn isomers drawn are diastereomers.

* Amlodipine

*

For geometric isomers to be possible there must be restricted rotation around a carbon–carbon of the ring). absorbed dose ofofaliskiren excreted the urineofunchanged. bond (e.g., presence 4. of aApproximately double bond 25% or alicyclic Evaluation aliskirenis reveals theinabsence double bonds and alicyclic rings; therefore, geometric isomers cannot exist. transformation has occurred and whether or not it represents an oxidative transformation.

A

132

Medicinal Chemistry Self Assessment

3. Although aliskiren is administered orally, its oral bioavailability is ~2.5% and is very poorly absorbed. * Using the information in the structure evaluation grid, provide a structural rationale for this unfortunate property. Answer Oral bioavailability is dependent on the drug’s ability to dissolve in the aqueous contents of the gastrointestinal (GI) tract and its ability to cross the lipid bilayer membrane. The hydrophilic character of the drug promotes aqueous solubility. Evaluation of the entire molecule (not just the boxed functional groups) for hydrophilic character reveals two ethers, two amides, a primary amine (in its ionized form), and a secondary alcohol that will contribute meaningfully to the hydrophilic character of the drug. (NOTE: the drug is sold as a hemifumarate salt that further increases the overall water solubility of the drug.) The hydrophobic character* of the drug promotes absorption across the lipid bilayer. Evaluation of the entire molecule (not just the boxed functional groups) for hydrophobic character reveals an aromatic hydrocarbon, a lipophilic ether, and several aliphatic alkanes. The drug is formulated as a water soluble salt and had ample hydrophilic character. Unfortunately it does not have sufficient hydrophobic character to allow for meaningful absorption.

4. Approximately 25% of the absorbed dose of aliskiren is excreted in the urine unchanged. It is Approximately 25% of the absorbed dose of aliskiren is excreted in the urine unchanged. It is unknown how much 4. of an absorbed dose is metabolized, and several metabolites from CYP3A4 mediated transformations have been identified. Several possible metabolic products are illustrated. transformation has occurred and whether or not it represents an oxidative transformation. Identify which metabolic transformation has occurred and whether or not it represents an oxidative transformation.

A

B

2.8 Aliskiren

133

C

D

E

Answer

5. Renin catalyzes the cleavage of a specific Leu-Val peptide bond within the structure o

Name of Metabolic Transformation

Oxidative or Non-Oxidative

A

Oxidative Oxidative O-dealkylationchains of these two amino acids and box the non-hydrolyzable hydroxyethylene group.

B

Alcohol oxidation

Oxidative

C

Oxidative deamination

Oxidative

D

Oxidative O-dealkylation*

Oxidative

E

Amide hydrolysis*

Non-oxidative

* This metabolite has been identified as one of the two primary metabolites produced.

134

Medicinal Chemistry Self Assessment

5. Renin catalyzes the cleavage of a specific Leu-Val peptide bond within the structure of angiotensin5. The Renin catalyzes the cleavage of afunctional specific Leu-Val bondthe within structure of side ogen. structure of aliskiren contains groupspeptide that mimic side the chains for these two amino acids. The hydrolyzable peptide bond (found between Leu-Val) has been replaced by a nonchains of these two aminogroup acids in and box the non-hydrolyzable hydroxyethylene hydrolyzable hydroxyethylene aliskiren. Circle the functional groups that group. mimic the amino acid side chains of these two amino acids and box the non-hydrolyzable hydroxyethylene group.

Answer: Answer

H3 C

H3C

C H3 H H3C N

HO H2 N

C H3

(remove bold from drug name) Chapter 2.8 O H3 CO

O H3C

N H2

O

C H3

6. Aliskiren can be co-administered with other anti-hypertensive agents to provide better likely is it 6. Aliskiren can be co-administered with other anti-hypertensive agents to provide better hypertension Aliskiren that a plasma protein is binding interaction will occur if these drugscalcium are co-administered? management. Amlodipine a second-generation dihydropyridine channel blocker used in the treatment of hypertension. Aliskiren and amlodipine are both plasma protein bound (47–51% and 93–97%, respectively). How likely is it that a plasma protein binding interaction will occur if these drugs are co-administered?(remove bold from drug name) Chapter 1.8/2.8

Amlodipine Amlodipine

2.8 Aliskiren

135

Answer It is important to first determine whether these drugs are basic, acidic, or amphoteric in nature. Based on a structural evaluation of the entire aliskiren molecule (not just the boxed functional groups), there is one basic functional group (primary amine) and no acidic functional groups present. The same evaluation for amlodipine yields identification of two basic functional groups (primary amine and dihydropyridine ring) and no acidic functional groups. Based on this evaluation, both drugs are basic and both would be bound to α1-acid glycoprotein (plasma protein). It is important to remember that plasma protein binding interactions are likely to happen when a drug is >90% plasma protein bound. Given the extent to which amlodipine is plasma protein bound, it is highly likely that a plasma protein binding interaction will occur when aliskiren displaces amlodipine from the glycoprotein binding site. When this type of interaction occurs, there will be a greater fraction of amlodipine present in its unbound form. This leads to an increase in pharmacological activity resulting in hypotension (excessively low blood pressure).

Go to Section 1.8 [Link to section 1.8 title]

Section 4 Whole Molecule Drug Evaluation Answers 2.9 Aripiprazole 2.9 2.9Aripiprazole Aripiprazole Shown below is the structure of aripiprazole, a serotonin receptor modulator used for the treatment of depression, Shown of depression, depression,schizophrenia, schizophrenia,autism, autism, mania, and bipolar disorder. Shownbelow belowisisthe theor orused usedfor for the the treatment of mania, and bipolar disorder. schizophrenia, autism, mania, and bipolar disorder.

Aripiprazole Aripiprazole 1. Identify all ofall the functional provide at the pKofa range 1. Identify ofacidic wouldand be basic primarily ionizedgroups, or unionized a normal urine pH 5.6. for each of the identi1. all groups, of wouldand beidentify primarily ionized or unionized at a urine pH of 5.6. fiedIdentify functional if each functional group would be primarily ionized or unionized at a urineAnswer pH=5.6.

Answer

Answer

Tertiary amine (Basic functional group) Tertiary amine (Basic group) Normal pH range = 9 tofunctional 11 Normal pHprimarily range =ionized 9 to 11at a pH = 5.6 Would be Would be primarily ionized at a pH = 5.6 Cl Cl

H NH N

O

O

O O

Cl Cl N

N

N

N

Aripiprazole

Aripiprazole Aromatic amine (Aniline, Basic functional group) Normal pH range(Aniline, = 2 to 5 Basic functional group) Aromatic amine Would be primarily unionized at a pH = 5.6 Normal pH range = 2 to 5 Would be primarily unionized at a pH

= 5.6

2. Identify all of the of aripiprazole. Answer 2. Identify all of the of aripiprazole. 3. Answer

3.

Alkyl (aliphatic) chain (Lipid soluble) 137 Alkyl (aliphatic) chain Ether oxygen (Lipid soluble) Halogens (Water soluble)

O

Amide (Water soluble)

138

Medicinal Chemistry Self Assessment

2. Replacementstructurefortheanswertoquestion2inChapter2.9(Aripiprazole) Identify all of the remaining functional groups, and indicate how each group contributes to the overall water solubility or the overall lipid solubility of aripiprazole.

 Answer Alkyl (aliphatic) chain (Lipid soluble) Ether oxygen (Water soluble) Halogens

Amide (Water soluble)

(Lipid soluble)

Hydrocarbon portion of bicyclic ring (Lipid soluble)

Aromatic (phenyl) ring (Lipid soluble) Aripiprazole Alkyl (aliphatic) chain (Lipid soluble)

3. Using your answers from the previous two questions, provide an explanation as to why aripiprazole Replacement structure for question 6 in BOTH Chapters 1.12 and 2.12 (Chlorpropamide and Other can be administered orally for the indications previously listed. Sulfonylureas) Answer The tertiary amine, aromatic amine, amide, and ether oxygen atom provide adequate water solubility that allows aripiprazole to dissolve in the aqueous contents of the gastrointestinal (GI) tract. Once dissolved, aripiprazole must pass through the GI mucosal membrane to enter the blood stream. The dichloro phenyl ring, the alkyl chain, and the hydrocarbon portion of the bicyclic ring will Cl solubility to allow for absorption across provide adequate lipid O the mucosal membrane. The tertiary O amine will be primarily ionized within the GI tract, whereas the aromatic amine will be primarily O unionized once it leaves the stomach. Remember thatSionization N N is an equilibrium process with some H H Le Chatelier’s principle, as soon as fraction of unionized molecules present at all times. According to N O one unionized molecule passes through the GI membrane, the equilibrium will reset to provide addiH tional unionized molecules. O This same principle also allows ionizable functional groups to penetrate the blood–brain barrier Hand nervous system (CNS). The same functional groups that Glyburide 3 C enter the central allowed aripiprazole to pass through the GI mucosal membrane will also provide sufficient lipid solubility to allow it to cross the blood–brain barrier and reach its target receptors within the CNS.

O

O S

H3C N H

O

OH

O N H

Metabolite of glyburide

N H



2.9 Aripiprazole

139

4. Using your answers from the previous 4. Drug molecules that are highly plasma protein bound may undergo a drug interaction due to the 5. Drug molecules or losartan? displacement of one drug molecule from a plasma protein by another drug molecule. Shown below are the structures of tolbutamide and losartan. Both of these drugs are highly plasma protein bound. Aripiprazole is also highly plasma protein bound (>99%). Would you expect there to be a drug interaction between aripiprazole and either tolbutamide or losartan?

4. Using your answers from the previous Tolbutamide 5. Drug molecules or losartan? Losartan

Answer The6.answer herethat is no. andinteraction losartan are acidic moleculespH due Assume thisTolbutamide that this binding occurs at adrug physiological of to 7.4.the presence of an acidic sulfonylurea and a tetrazole functional group, respectively. Aripiprazole is a basic drug molecule due to the presence of a tertiary amine and an aromatic amine. Acidic drug molecules primarily bind to albumin, whereas basic drug molecules primarily bind to α1-glycoprotein. Plasma D interactions) occur due to protein binding interactions (also known as plasma protein displacement the nonspecific nature of the plasma proteins that bind and transport drug molecules. These types of A interactions are only therapeutically important when the drug molecules are highly plasma protein C drugs, the difference in their acid/base bound (i.e., over 90%). Although this is true of all three nature significantly decreases the probability ofBa drug interaction due to plasma protein displacement. Tolbutamide Aripiprazole Losartan 5. Assume that the boxed functional groups of aripiprazole form four key binding interactions with a serotonin receptor. Further assume that these binding interactions occur with the side chains of Tyr, Asp, Ile, and Gln. Using this information, identify four possible binding interactions between aripiprazole and the given amino acids. Assume that this binding interaction occurs at a physiological 6. Assume that this that this binding interaction occurs at a physiological pH of 7.4. pH=7.4.

D A C B

Aripiprazole

140

Medicinal Chemistry Self Assessment

Answer Answer H N

O

R4

Gln

R3

H N

R1

Cl

R8

OH R7 O

N

Hydrogen Bond R2

O

O

H N O

+

H

van der Waals; Hydrophobic Interaction N H

–

H

O

O

Cl N

N

Asp

O

Ionic Bond

H

van der Waals; Hydrophobic Interaction

H N R6

R5 O

Ile

Tyr

Other binding interactions are possible among these four functional groups and four amino acids. Isoleucine could form van der Waals and hydrophobic interactions with the halogenated aromatic ring, whereas tyrosine could form the same types of interactions with the alkyl chain. Aspartic acid could form an ion–dipole interaction (as the ion) with the amide, whereas glutamine could form an ion–dipole interaction (as the dipole) with the ionized tertiary amine. Please note that it is possible for tyrosine to form hydrogen bonds with the amide and an ion–dipole interaction (as the dipole) with the ionized tertiary amine. However, in the scenario given in this question, if tyrosine was used to form an interaction with either the amide or the ionized tertiary amine, neither aspartic acid or glutamine could be used to form van der Waals and hydrophobic interactions with the halogenated aromatic ring.

2.9 Aripiprazole

141

6. Shown below are three known metabolites of aripiprazole. Identify the metabolic transformations that would be7.required formare each the metabolites. For each metabolic transformation, indicate Showntobelow or of a phase II transformation. if it is a phase I transformation or a phase II transformation.

Metabolite A

Metabolite B

Metabolite C

Answer

Answer Metabolite A: Aromatic oxidation, phase oxidation, I Metabolite A: Aromatic phase I Note: This is somewhat of an B: unexpected metabolite because the presence of halogen substitutents Metabolite Benzylic oxidation, phase I on the aromatic ring generally deactivates these rings from oxidation. The difference in this particto the ability form a conjugated system, thisaromatic secondary hydroxyl group readily ular drug moleculeNote: is theDue presence of an to adjacent aromatic amine. This amine can donate electrons into the undergoes ring via resonance andtoeither override or neutralize dehydration form the following metabolite.the electronic effects of the chloro groups, thus allowing for aromatic oxidation. Metabolite B: Benzylic oxidation, phase I Dehydration allows Note: Due to the ability to form a conjugated system, this secondary hydroxyl group readily underconjugation between goes dehydration to form the following metabolite. the amide and the aromatic ring.

Metabolite A: Aromatic oxidation, phase I Metabolite B: Benzylic oxidation, phase I 142 Note: Medicinal Due to theChemistry ability to Self formAssessment a conjugated system, this secondary hydroxyl group readily

undergoes dehydration to form the following metabolite.

Dehydration allows conjugation between the amide and the aromatic ring.

Metabolite C: Oxidative N-dealkylation followed by oxidation of the resulting aldehyde to a carboxylic acid by aldehyde dehydrogenase, phase I

Go to Section 1.9 [Link to section 1.9 title]

Section 4 Whole Molecule Drug Evaluation Answers 2.10 Cefprozil

Cefprozil is a second-generation cephalosporin that exhibits good Chemistry Gram (+) activity with improved Gram (–) activity Medicinal Self-Assessment Book: Batch Two as compared to the first-generation cephalosporins. Effective against the majority of1.10 bacteria Chapters andthat 2.10cause upper and lower respiratory infections, as well as skin infections, cefprozil was a first-line anti-infective agent until an increase in Chapter 1.10/2.10 bold decreased from drug its name) the incidence of resistance and the development of (remove newer agents “favored” status.

Cefprozil (Cefzil)

1. Conduct a complete structural evaluation of cefprozil and use the information in the grid to inform your Chpater 1.10/2.10 Corrected structure (product of E transformation) answers to the questions that follow. Answer

H

O

N

Character

A

HO

O Character

N O

HO H H

S

N

O

C H3

Function

Acidic, Basic, or Neutral

CO2 H

↑ Solubility and/or Provide pKa A When Relevant ↑ Absorption

Name of Functional Group

Hydrophilic and/or Hydrophobic

Phenol

Hydrophilic (OH)

Acidic

Solubility (OH)

Hydrophobic (Ar)

(pKa 9–10)

Absorption (Ar)

H

NH 2

HO

H H

S

Function

N

Amino Acids CH 3 Can H CO2 That Interact with N H2 Function Functional OH Group via Interaction(s) H H H-Bonding (at O H 2N S Possible with HO pH=7.4) Biological Target O

at Physiological B pH=7.4

None Is Acceptable

OH:

Ser,CTyr, Trp, His, Thr, Cys, Asn, Gln

N HH-bonding 2 H H H (A+D) N Dipole–dipole S

Ion–dipole (as O N the dipole) O

HO

C H3

van der Waals H N HO

O

O

H H

F

S

CO2 H

CO2H

Ar: NH 2

N

O

D

Hydrophobic π-π Stacking

E

N

CH 3

N H2

O143 N H

N

O

O -O

S

H

O

O

O

H

H N

S

144

Medicinal Chemistry Self Assessment

Continued from previous page. Function

Function

Character Acidic, Basic, or Neutral

Character

B

C

Function

↑ Solubility and/or Provide pKa When Relevant ↑ Absorption

Interaction(s) Possible with Biological Target at Physiological pH=7.4

Name of Functional Group

Hydrophilic and/or Hydrophobic

Primary amine

Hydrophilic (NH2)

Basic

Solubility (NH2)

Ion–dipole (as the ion)

Hydrophobic (R)

(pKa 9–11)

Absorption (R)

Ionic

Hydrophilic (CONH)

Neutral

Solubility (CONH)

H-bonding (A + D)

Absorption (R)

Dipole–dipole

Amide

Hydrophobic (R)

Amino Acids That Can Interact with Functional Group via H-Bonding (at pH=7.4) None Is Acceptable None

Ser, Tyr, Thr, Cys, Asn, Gln, Trp, His

Ion–dipole (as the dipole) D

Lactam

Hydrophilic (CON)

(cyclic amide)

Hydrophobic (R)

Neutral

Solubility (CON)

H-bonding (A)

Absorption (R)

Dipole–dipole

Ser, Tyr, Thr, Cys, Asn, Gln, Trp, His

Ion–dipole (as the dipole) E

Thioether

Hydrophobic

Neutral

Absorption

Hydrophobic

None

van der Waals F

G

Carboxylic acid

Hydrophilic (COOH)

Acidic

Solubility (COOH)

Ion–dipole (as the ion)

Hydrophobic (R)

(pKa 2.5–5)

Absorption (R)

Ionic

Alkene

Hydrophobic

Neutral

Absorption

Hydrophobic

Chapter 2.10 (remove bold from drug name)

B

None

None

van der Waals

D (four membered ring) E

A

G

C F Cefprozil

2. Based on the information in the structure evaluation grid, determine if cefprozil is an acidic, basic, or amphoteric drug. Provide a brief explanation for your answer. Answer Cefprozil contains two acidic functional groups (phenol and carboxylic acid) and one basic functional group (primary amine). Because this drug contains both acidic and basic functional groups, it is considered amphoteric in nature.

2.10 Cefprozil

145

3. Cefprozil is administered orally as a tablet or liquid suspension. Consider each of the acidic and basic functional groups and determine whether each group will be predominantly ionized or unionized as it moves through the gastrointestinal (GI) tract, into systemic circulation, and then into the urine. [The relevant pKa values=10, 1.7, and 7.2.] Complete the grid below. Answer

Name of Functional Group

Acidic or Basic (pKa)

Ionized or Unionized at pH=5 (saliva)

Ionized or Unionized at pH=1 (stomach)

Ionized or Unionized at pH=7.4 (plasma)

Ionized or Unionized at pH=8 (intestine)

Ionized or Unionized at pH=6 (urine)

Phenol

Acidic

Unionized

Unionized

Unionized

Unionized

Unionized

Ionized

Unionized

Ionized

Ionized

Ionized

Ionized

Ionized

Likely 50%/50% ionized/unionized

Unionized

Ionized

pKa~10 Carboxylic acid

Acidic pKa ~1.7

Primary amine

Basic pKa~7.2

4. Given the predominant ionization state of these acidic and basic functional groups as they traverse the GI tract and the information in the structure evaluation grid, determine in which GI compartment(s) drug absorption could occur. Answer Drug absorption is enhanced as hydrophobic character is increased and as the percent unionized increases. Cefprozil contains several hydrophobic functional groups including the aromatic ring portion of the phenol, the thioether, and the alkenes. In the stomach (pH=1), the carboxylic acid and the phenol will be predominantly unionized; however, the primary amine will be predominantly ionized. In the intestine (pH=8), the phenol and the primary amine will be unionized, and both the carboxylic acid will be primarily ionized. This means that cefprozil will always be predominantly in an ionized form regardless of the GI location. A quick reminde—an equilibrium exists between the ionized form and the unionized form of a drug molecule. In the case of cefprozil, there is only a very small fraction of the drug that is completely unionized at any point in time regardless of its location within the GI tract. When cefprozil is in its unionized form, there is sufficient hydrophobic character to permit the drug to cross the lipid bilayer membranes of the GI tract and enter systemic circulation. Based on this evaluation, it is possible that drug absorption can occur in both GI compartments.

146

Medicinal Chemistry Self Assessment

5. Approximately 60% of a cefprozil dose is recovered in the urine unchanged. Because impairment in Cefprozil (Cefzil)the half-life of the drug by several hours, it is likely that cefprozil underhepatic function increases goes a variety of metabolic transformations catalyzed by liver enzymes. For each of the metabolic transformations A–F, identify which metabolic transformation has occurred and whether the product Chpater Corrected structure E transformation) formed 1.10/2.10 was the result of a phase I or(product phase IIofmetabolic transformation.

H

O

N O

HO

N O

HO H H

H H

S

N

O

CH 3 CO2 H

S

N H2

N

O

H

NH 2

HO

OH

C H3 CO2 H

O

HO

H 2N

B

A

H

H N

O

H

H H

N O

HO

S

N

O

CH 3 CO2 H

C N H2

S

C H3 CO2H

NH 2

H N

O

HO

H H

O

S

E

N O

D

F

CH 3

N H2

N H

O

O -O

OH NH 2

H N

HO

O

O

H H

S O

O

O

O

H

H

S

N

C H3 CO2 H

S OH

N

H N

CO2 H

Answer Name of Transformation

Phase I or Phase II

A

Oxidative deamination

Phase I

B

Aromatic hydroxylation

Phase I

C

Amide hydrolysis

Phase I

D

Sulfate conjugation

Phase II

E

Allylic oxidation

Phase I

F

Amino acid conjugation (with glycine)

Phase II

2.10 Cefprozil

147

6. Like the penicillins, the cephalosporins suffer from chemical instability of the β-lactam bond. Chemical hydrolysis of this bond renders the drugs in the class of anti-infective agents inactive. This bond is also subject to cleavage by β-lactamases (due to the presence of a nucleophilic serine side chain [CH2OH] within the active site of the enzyme). Show how the β-lactam bond can be hydrolyzed by chemical and enzymatic (β-lactamase) mechanisms. Answer

Chemical Hydrolysis

OH-

Enzymatic Hydrolysis

Go to Section 1.10 [Link to section 1.10 title]

Section 4 Whole Molecule Drug Evaluation Answers 2.11 Cetirizine

Cetirizine is a popular second-generation antihistamine used in the management of allergy symptoms. It is the Medicinal Chemistry Self-Assessment Book: Batch Two metabolic byproduct produced from the prescription antihistamine hydroxyzine. Although cetirizine is labeled as Chapters 1.11 and 2.11 non-sedating and is one of the preferred allergy medications for long-haul drivers, hydroxyzine causes significant drowsiness that limits its utility in the management of typical allergy symptoms. Hydroxyzine is commonly used in Chapter 1.11 (remove drug names) the treatment of pruritus (severebolded itching).

Cetirizine

Hydroxyzine

1. Conduct a complete structural evaluation of hydroxyzine and use the information in the grid to inform Chapter 2.11 (remove bolded drug names) your answers to the questions that follow.

A

B

C

D

E

Hydroxyzine

149

150

Medicinal Chemistry Self Assessment

Answer Function

Character

A

Character

Acidic, Basic, or Neutral

Function

Name of Functional Group

Hydrophilic and/or Hydrophobic

Provide pKa When Relevant

↑ Solubility and/or ↑ Absorption

Amino Acids That Can Interact with Functional Group Interaction(s) via H-Bonding (at Possible with Biological Target pH=7.4) None Is at Physiological Acceptable pH=7.4

Halogenated aromatic hydrocarbon

Hydrophobic

Neutral

Absorption

Cl:

Function

None

Dipole–dipole Ion–dipole (as the Medicinal Chemistry Self-Assessment Book: Batch Two dipole)1.11 and 2.11 Chapters Ar: van der Waals

Chapter 1.11 (remove bolded drug names)

Hydrophobic π-π Stacking

B

Aromatic hydrocarbon

Hydrophobic

Neutral

Absorption

Piperazine (two tertiary amines)

Hydrophobic (R)

Basic

Absorption (R)

Hydrophilic (N)

(pKa 9–11)

Solubility (N)

Neutral

Absorption (R)

van der Waals

None

Hydrophobic π-π Stacking

C

D

Ether

Hydrophobic (R) Hydrophilic (O)

E

Primary alcohol

Hydroxyzine

Hydrophobic (R)

Neutral

Hydrophilic (OH)

Ion–dipole (as the ion)

None (ionized at pH=7.4)

Ionic

Solubility (O)

H-bonding (A); Dipole–dipole; Ion–dipole (as the dipole)

Ser, Thr, Cys, Tyr, Gln, Cetirizine Asn, His, Trp

Absorption (R)

H-bonding (A+D);

Solubility (OH)

Dipole–dipole;

Ser, Thr, Cys, Tyr, Gln, Asn, His, Trp

Chapter 2.11 (remove bolded drug names) Ion–dipole (as the dipole)

A

B

C

D

E

Hydroxyzine

2. Name the phase I metabolic transformation(s) that hydroxyzine undergoes to produce cetirizine. Answer Alcohol oxidation followed by aldehyde oxidation to the carboxylic acid.

2.11 Cetirizine

151

3. Based on your structural evaluation of both hydroxyzine and cetirizine, name ALL of the phase I metabolic transformations possible. Answer • Oxidative O-dealkylation •

Oxidative N-dealkylation

•

Alcohol oxidation

•

Aromatic hydroxylation (less likely on halogenated aromatic hydrocarbon)

NOTE: Although there is a benzylic carbon with a hydrogen atom attached to it, benzylic oxidation does not occur because the benzylic carbon is already directly attached to another heteroatom (N). 4. A metabolic product from a phase II metabolic transformation has been identified. Which phase II transformations can cetirizine undergo? Answer The carboxylic acid can undergo phase II transformations including glucuronidation and amino acid conjugation (with glutamic acid and glycine [major], and aspartic acid, serine, and taurine [minor]). (NOTE: the glucuronide conjugate is the metabolite that has been identified.)

5. Review the structure of cetirizine (pKa=2.9 and 8.3) and identify all of the acidic and basic functional groups present. Determine the predominant ionization state of each functional group as it travels through several compartments of the body after oral administration. Complete the table below. Answer Name of Functional Group

Acidic or Basic (pKa)

Ionized or Unionized at pH=5 (saliva)

Ionized or Unionized at pH=1 (stomach)

Ionized or Unionized at pH=7.4 (plasma)

Ionized or Unionized at pH=8 (intestine)

Ionized or Unionized at pH=6 (urine)

Tertiary amine (piperazine)

Basic pKa 8.3

Ionized

Ionized

Ionized

Ionized

Ionized

Carboxylic acid

Acidic pKa 2.9

Ionized

Unionized

Ionized

Ionized

Ionized

6. Provide a structural rationale for why hydroxyzine is classified as a sedating antihistamine and cetirizine is categorized as a non-sedating antihistamine. Answer Hydroxyzine contains functional groups that contribute to the overall hydrophobic character of the molecule (e.g., aromatic hydrocarbon, halogenated aromatic hydrocarbon, aliphatic alkane), as well as functional groups that contribute to the hydrophilic character of the molecule (e.g., ether, primary alcohol, and tertiary amines/piperazine). The hydrophilic character of the molecule enhances its water solubility, whereas the hydrophobic character enhances its ability to be absorbed across lipid bilayer membranes. At pH=7.4, the tertiary amine will be predominantly in its ionized form which will further increase the water solubility of the molecule. The hydrophilic character contributes to the ability of the drug to be distributed in the body in the aqueous plasma. The hydrophobic character contributes to the ability of the drug to be absorbed across the membranes of the gastrointestinal tract and eventually across the blood–brain barrier. It is important to remember that the tertiary amine (one of the two) will be predominantly ionized at physiological pH. Because an equilibrium exists between the ionized and unionized form of the drug, some fraction of the drug will be unionized at any point in time and, therefore, can cross the blood– brain barrier and have an effect on the histamine receptors.

152

Medicinal Chemistry Self Assessment

NOTE: When activated, the histamine receptor in the brain is responsible for wakefulness. When hydroxyzine interacts with this receptor, it prevents histamine from binding to and activating its receptor. As a result, the feeling of wakefulness is very limited, and the patient feels sleepy. Cetirizine contains a carboxylic acid instead of the primary alcohol found in hydroxyzine. The carboxylic acid significantly increases the water solubility of cetirizine as compared to hydroxyzine. The carboxylic acid is also predominantly ionized at physiological pH. Although there is an equilibrium between the ionized and unionized forms of the tertiary amine and of the carboxylic acid, the overall percent of cetirizine in which both functional groups are unionized is especially small. As a result, very, very little of a cetirizine dose can cross the blood–brain barrier and interact with the histamine receptors. Because the central histamine receptors can still be activated, the feeling of wakefulness remains, and the patient does not feel sleepy.

Section 4 Whole Molecule Drug Evaluation Answers 2.12 Chlorpropamide and Other Sulfonylureas 2.12 Chlorpropamideand Other Sulfonylureas Shown below are the structures of tolbutamide and chlorpropamide. These drug molecules are used in the treatment of type 2 diabetes mellitus. They stimulate the release of insulin from the pancreatic β-cells by interacting with Shown below are the structures of tolbutamide potassium channels. adenosine triphosphate (ATP)-sensitive potassium channels. 2.12 Chlorpropamideand Other Sulfonylureas Shown below are the structures of tolbutamide potassium channels.

Tolbutamide

Chlorpropamide

1. uncommon It is not uncommon for patients with2type 2 diabetes that diagnosed a drug interaction could occur?and 1. It is not for patients with type diabetes to beate dually with hypertension require additional pharmacotherapy. In this scenario, it is possible for drug interactions to occur if the prescribed combination therapy is not appropriately evaluated. Angiotensin II receptor antagonists, Tolbutamide Chlorpropamide commonly known as angiotensin II receptor blockers (ARBs), are often used to treat hypertension. Losartan (shown below) is an ARB, and similar to tolbutamide and chlorpropamide, is highly plasma protein bound. If losartan was selected for use in a patient who is already taking tolbutamide or chlor1. It is not uncommon for patients with type 2 diabetes ate that a drug interaction could occur? propamide, would you anticipate that a drug interaction could occur?

Losartan

Losartan

Answer

Plasma protein binding interactions (also known as that a drug interaction could occur.

Sulfonylurea Answer Sulfonylurea Answer Plasma protein binding interactions (also known as plasma protein displacement interactions) occur due to the nonspecific nature interactions of the plasma proteins bind and interaction transport drug There are two Plasma protein binding (also known that as that a drug couldmolecules. occur. key concepts to consider when evaluating the probability of a plasma protein binding interaction. First, these types of interactions are therapeutically important only when the drug molecules are highly plasma protein bound Sulfonylurea (i.e., over 90%). Second, acidic drugs primarily bind Sulfonylurea to albumin, whereas basic drugs primarily bind to α1-glycoprotein. The information provided in the question indicates that all of these Tolbutamide Chlorpropamide compounds are highly plasma protein bound; therefore, we need to evaluate the acid/base nature of 153

Losartan 154

Medicinal Chemistry Self Assessment

these drug molecules. The structures of tolbutamide and chlorpropamide contain an acidic sulfonylurea, Answer whereas the structure of losartan contains an acidic tetrazole. Because both of these drug molecules are acidic and are highly plasma protein bound to albumin, a high probability that a drug Plasma protein binding interactions (also known as that athere drugisinteraction could occur. interaction could occur.

Sulfonylurea

Sulfonylurea

Chlorpropamide

Tolbutamide

Acidic protons are circled Tetrazole Losartan

2. The normal pKa range for sulfonylureas is 5 to 6. When comparing the pKa values of chlorpropamide and tolbutamide, it is found that the sulfonylurea of one of these drug molecules has a pKa=5.4 and the other has a pKa=4.9. Evaluate the structures of these two drug molecules, assign the pKa values to the correct molecules, and provide an explanation for the difference in pKa values. 2. The normal pKa range forrect molecules, and provide an explanation for the difference in pKa values. Answer Answer evaluation of chlorpropamide and tolbutamide reveals two chemical differences. A structural

Three carbon aliphatic chain

Halogen

Chlorpropamide

Four carbon aliphatic chain

Methyl group

Tolbutamide

The differences in the length of the aliphatic hydrocarbon chains would not be expected to have a major effect on the acidity of the sulfonylurea because they would both donate electrons through Thus, the difference in the pKawill values can beatattributed to the electronic differences 3.induction. Using your answer from question 2, that be ionized an intestinal pH of 6.1. between the halogen and the methyl group. Halogens act as electron withdrawing groups through Answer induction, whereas the methyl group acts as an electron donating group through induction. Because acidic functional groups (i.e., sulfonylureas) form anions once the proton leaves, adjacent functional groups that are electron withdrawing will increase acidity. Adjacent functional groups that are [Base~Form] electron donating will decrease acidity. Thus, the sulfonylurea group present within the structure pH =~pK a +log ~

[Acid~Form]

[Base~Form]

2.12 Chlorpropamide and Other Sulfonylureas

155

of chlorpropamide would be expected to be more acidic (pKa=4.9) than the one present within the structure of tolbutamide (pKa=5.4) due to the electron withdrawing character of the halogenated aromatic ring. 3. Using your answer from question 2, calculate the percent of tolbutamide that will be ionized at an intestinal pH=6.1. Answer As determined in question 2, the pKa of the sulfonylurea groups of tolbutamide is 5.4. To solve this problem, we need to use the Henderson-Hasselbalch equation. Because the functional group is acidic, the ionized form (R-SO2N–CO-R) is the base form and the unionized form (R-SO2NHCO-R) is the acid form.

[Base Form] pH = pK a +log   [Acid Form]

[Base Form] 6.1= 5.4 +log   [Acid Form] [Base Form] 0.7 = log   [Acid Form] 5.01= 

[Base Form]  or 5.01 = [Base Form] 1 [ Acid Form] [ Acid Form]

This ratio indicates that for every one molecule that contains the functional group in the acid (or unionized) form, there are 5.01 molecules that contain the functional group in the base (or ionized) form. The following equations can then be used to correctly calculate the percent of the molecules that are ionized and the percent that are unionized.

5.01 molecules in base form + 1.0 molecule in acid form = 6.01 total molecules

Base form = Ionized form and Acid form = Unionized form Percent in Ionized Form =

5.01 Molecules in Ionized Form ×100% = 83.4% 6.01 Total Molecules

Percent in Unionized Form =

1 Molecule in Unionized Form ×100% = 16.6% 6.01 Total Molecules

The question asks for the percent that will be ionized, so the correct answer is 83.4%.

4. The mechanism of action of this class of drugs involves the ability to interact with the ATP-sensitive potassium channels in the pancreas. Using the structure of tolbutamide, identify the types of binding interactions that would be possible between its functional groups and a protein ion channel. Also identify amino acids present within this ion channel whose side chains could participate in the interactions identified. Assume a plasma pH=7.4 for all ionizable functional groups.

156

Medicinal Chemistry Self Assessment

4. The mechanism of action of this class of to interact with the ATP-sensitive potassium groups. Answer Answer

A

B C

D

Tolbutamide Aminothan Acids Capable of Forming 5. Tolbutamide has a half-life of 4.5 to 6.5 hourssignificantly longer half-life tolbutamide. A B

Functional Group

Types of Binding Interactions

Specific Bond

Phenyl ring; aromatic ring; aromatic hydrocarbon

van der Waals; Hydrophobic

Tyr, Phe, Trp (better bond*); Val, Leu, Ile, Met, Ala

Sulfonylurea

(1) Ionic

(1) Lys, Arg, His**

(2) Ion–Dipole (as the Ion)

(2) Ser, Thr, Tyr, Cys, Asn, Gln, His**

(1) Ion–Dipole (as the Dipole)

(1) Asp, Glu (with the hydrogen); Arg, Lys, His** (with the nitrogen)

(2) Dipole–Dipole

(2) Ser, Thr, Tyr, Cys, Asn, Gln, His**

(3) Hydrogen Bond (Donor and/or Acceptor)

(3) Ser, Thr, Tyr, Cys, Asn, Gln, Trp, His**

van der Waals; Hydrophobic

Val, Leu, Ile, Met, Ala (better bond*); Tyr, Phe, Trp

Answer

Metabolism of Tolbutamide C

D

Nitrogen atom of sulfonylurea not involved in resonance

Alkyl group; alkyl chain; aliphatic chain

Z oxidation

Z-1 oxidation *Due to steric fit, stronger van der Waals interactions occur when aromatic rings interact with aromatic rings and when aliphatic chains interact with aliphatic chains; however, all of the listed amino acids could Benzylic possibly interact with the boxed functional group. oxidation **Histidine is primarily unionized at a pH=7.4. The small fraction that is ionized could participate in an ion–dipole interaction with a partially negative atom, while the unionized fraction can serve as a hydrogen bond donor or acceptor. It can also serve as the dipole in an ion–dipole bond.

1. Alcohol dehydrogenase 5. Tolbutamide has a half-life of 4.5 to 6.5 hours, whereas chlorpropamide has a half-life of 36 hours. Propose a chemical/structural reason why chlorpropamide has a significantly longer half-life than 2. Aldehyde tolbutamide. dehydrogenase Answer As discussed in question 2, there are two structural differences between tolbutamide and chlorpropamide: the length of the aliphatic chain and the para substituent on the phenyl ring. Both drug Metabolism of Chlorpropamide molecules can undergo π and π-1 oxidation of their respective aliphatic chains. Metabolism at these sites would be expected to be similar; however, the additional carbon atom present in tolbutamide may cause the butyl side chainZto be less sterically hindered and more susceptible to oxidation than oxidation the propyl chain present in chlorpropamide. The more significant difference is metabolism of the para substituent. The para methyl group present within the structure of tolbutamide can undergo benzylic oxidation followed by two additional oxidative transformations to convert the benzylic hydroxyl group into a para carboxylic acid. The para chloro group present within the structure of chlorpropamide deactivates oxidation of the aromatic ring due to its electron withdrawing effects. As a result, tolbutamide hasZ-1 a much shorter half-life than chlorpropamide. oxidation

Electron withdrawing halogen prevents aromatic oxidation

Tolbutamide 5. Tolbutamide has a half-life of 4.5 to 6.5 hourssignificantly longer half-life and thanOther tolbutamide. 2.12 Chlorpropamide Sulfonylureas Answer

Metabolism of Tolbutamide Z oxidation

Z-1 oxidation Benzylic oxidation

1. Alcohol dehydrogenase 2. Aldehyde dehydrogenase

Metabolism of Chlorpropamide Z oxidation

Z-1 oxidation Electron withdrawing halogen prevents aromatic oxidation

157

(Lipid soluble)

158

Medicinal Chemistry Self Assessment

Replacement structure for question 6 in BOTH Chapters 1.12 and 2.12 (Chlorpropamide and Other Sulfonylureas)

6. Glyburide is a second-generation sulfonylurea that is structurally similar to chlorpropamide and tolbutamide. Shown below are the structures of glyburide and one of its known metabolites. Identify the metabolic pathways that are required to produce this metabolite. Cl

S N H

O H3C

O

O

O

O

N H

N H

Glyburide

O

S

H3C N H

O

OH

O

O N H

Metabolite of glyburide

N H



Answer This metabolite is formed as the final product of three metabolic transformations. Oxidation of the alicylic ring produces the secondary alcohol. Hydrolysis of the initial amide bond of glyburide produces a primary amine that can undergo phase II acetylation. In most cases, further metabolism of a primary amine involves oxidative deamination; however, there are some cases where the primary amine is acetylated. Please note that the oxidation of the alicyclic ring is independent of the coupled hydrolysis and acetylation transformations.

Go to Section 1.12 [Link to section 1.12 title]

Section 4 Whole Molecule Drug Evaluation Answers 2.13 Dabigatran Etexilate

Thrombin is the enzyme responsible for catalyzing the conversion of fibrinogen to fibrin. The production of fibrin is important in the formation of sturdy blood clots. As you might expect, inhibition of thrombin prevents the formation of fibrin. As shown in the diagram below, a catalytic triad of amino acids (Asp, His, Ser) found in the active site of thrombin is responsible for hydrolyzing a key peptide bond found within fibrinogen. Orientation of fibrinogen in the active site ofChapters thrombin1.13/2.13 relies on the interaction a key tri-peptide sequence (D-Phe-Pro-Arg) found within the Concern about of His (revised diagram + removed bolded names) structure fibrinogen with key amino acids in the enzyme active site. Chapters 1.13/2.13 Concern about His (revised diagram + removed bolded names)is an orally active direct Dabigatran etexilate is administered as a prodrug. In its active form, dabigatran etexilate thrombin inhibitor used in the prevention of stroke and blood clots in patients diagnosed with atrial fibrillation. Thrombin Active Site

Thrombin Active Site

Thrombin Active Site

Thrombin Active Site Thrombin catalyzed amide hydrolysis Thrombin catalyzed amide hydrolysis

Fibrinogen (D-Phe-Pro-Arg)

Fibrin

Fibrinogen (D-Phe-Pro-Arg)

Chapters 1.13/2.13 (removed bold in drug name)

Fibrin

Chapters 1.13/2.13 (removed bold in drug name)

Dabigatran etexilate Dabigatran etexilate Chapters 1.13/2.13 (removed bold and centered text)

Chapters 1.13/2.13 (removed bold and centered text) 3

159

160

Medicinal Chemistry Self Assessment

1. Conduct a complete structural evaluation of dabigatran etexilate (prodrug) and use the information in the grid to inform your answers to the questions that follow. Answer Function

Character

A

Function

Character

Acidic, Basic, or Neutral

Function

Name of Functional Group

Hydrophilic and/or Hydrophobic

Provide pKa When Relevant

↑ Solubility and/or ↑ Absorption

Interaction(s) Possible with Biological Target at Physiological pH=7.4

Ester

Hydrophilic (COO)

Neutral

Solubility (COO)

H-bonding (A)

Absorption (R)

Dipole–dipole

Hydrophobic (R)

Amino Acids That Can Interact with Functional Group via H-Bonding (at pH=7.4) None Is Acceptable Ser, Thr, Cys, Tyr, His, Asn, Gln, Trp

Ion–dipole (as the dipole)

B

Amide

Hydrophilic (CON)

Neutral

Hydrophobic (R)

Solubility (CON)

H-bonding (A)

Absorption (R)

Dipole–dipole

Ser, Thr, Cys, Tyr, His, Asn, Gln, Trp

Ion–dipole (as the dipole)

C

Pyridine

Hydrophilic (N)

Basic

Solubility (N)

(Azine)

Hydrophobic (R)

pKa=1–5

Absorption (R)

Pyridine N: H-bonding (A)

Ser, Thr, Cys, Tyr, His, Asn, Gln, Trp

Dipole–dipole Ion–dipole (as the dipole) (R): Hydrophobic van der Waals π-π Stacking

D

Benzimidazole

Hydrophilic (imidazole N atoms) Hydrophobic (R)

Basic pKa=1–5

Solubility (imidazole Ns) Absorption (R)

Imidazole (N atoms): H-bonding (A) Dipole–dipole Ion–dipole (as the dipole)

Ser, Thr, Cys, Tyr, His, Asn, Gln, Trp

Ar: Hydrophobic van der Waals π-π Stacking E

Aromatic amine Hydrophilic (NH2)

Basic

Solubility (NH2)

(aniline)

pKa=2–5

Absorption (Ar)

Hydrophobic (Ar)

NH2: H-bonding (A+D) Dipole-dipole Ion–dipole (as the dipole)

Ser, Thr, Cys, Tyr, His, Asn, Gln, Trp

Ar: Hydrophobic van der Waals π-π Stacking F

Carbamate

Hydrophilic (OCON) Hydrophobic (R)

Neutral

Solubility (OCON)

H-bonding (A)

Absorption (R)

Dipole–dipole Ion–dipole (as the dipole)

Ser, Thr, Cys, Tyr, His, Asn, Gln, Trp

2.13 Dabigatran Etexilate

Chapters 1.13/2.13 (removed bolded text)

A

B

E D

161

F

C

Active form of Dabigatran

2. Dabigatran etexilate is a non-peptidomimetic prodrug. Provide a brief rationale for the value of converting an active drug into a prodrug. Chapters 1.13/2.13 (removed bold drug name) Answer There are situations (e.g., the need to enhance water or lipid solubility) when it is therapeutically beneficial to administer drug molecules that have been covalently modified to produce inactive (or very weakly active) analogs (i.e., prodrugs) (see Chapter 5 in Basic Concepts in Medicinal Chemistry). Bioactivation of these prodrugs via rapid metabolic transformation (e.g., ester hydrolysis) can occur via chemical or enzymatic mechanisms. Typically metabolic activation (via hydrolysis) occurs within the gastrointestinal (GI) tract. Given that esterases are ubiquitous, release of the active drug can occur in the plasma or even at the target tissue. In the case of dabigatran etexilate, a carboxylic acid is converted to an ester and an amidine is converted to a carbamate. In both cases, the modified functional group is significantly more lipid soluble than the parent functional group. Dabigatran is not absorbed orally, so it is necessary to develop a prodrug analog to allow for sufficient absorption from the GI tract. 3. Dabigatran etexilate rapidly undergoes two esterase-catalyzed hydrolytic reactions to the active drug dabigatran. Show the products from each of the esterase-catalyzed Dabigatran etexilate mesylatehydrolytic reactions that occur in the plasma.

162

Medicinal Chemistry Self Assessment

Chapter 2.13 (removed bolded drug names) Answer O H 3C

O

O

H N

N

N

N

N

NH 2

O

N O(CH2)5CH3

C H3

Dabigatran Prodrug

O

O

HO

H N

N

N

N

N H2

N

O

N O(CH2 )5 CH3

C H3

O H 3C

O

O

N N

O

O

HO

N N

H N

N N

H

N

N H2 NH

CH 3

Active form of Dabigatran

N

N

N H2 NH

C H3

2.13 Dabigatran Etexilate

163

4. In its active form, dabigatran mimics the D-Phe-Pro-Arg tripeptide sequence, but does not contain a peptide backbone (non-peptidomimetic). Review the tripeptide sequence drawn below noting that there are numbers 1–3 that indicate where each amino acid is located in the sequence. Determine the 4. ofIninteractions its active form, dabigatran tripeptide types possible with mimics each ofthe theD-Phe-Pro-Arg amino acid side chains. chains.

O

NH H 2N

H H 2N

3

N O

O

1

N 2

N

R H

D-Phe-Pro-Arg Tripeptide Sequence D-Phe-Pro-Arg Tripeptide Sequence Answer Types of Interactions Possible

5. Review side chain indicated.

D-Phe (1)

Hydrophobic

van der Waals π-π Stacking Pro (2)

A Hydrophobic

B

van der WaalsO Arg (3)

Ion–dipole (as the ion) Ionic

HO

C

O

E D

N

H N

N N

N

F

N H2 NH

C H3

5. Review the structure of dabigatran and determine which of the boxed groups will likely mimic the interactions found within the D-Phe-Pro-Arg tripeptide sequence. Given the three-dimensional nature of both peptides and small molecules, it is important to remember that the functional groups within dabigatran do not need to lineActive up in the same order. Using the table provided to guide your form ofname) Dabigatran Chapter 2.13 (remove bolded drug analysis, place a Yes or No in each box to indicate whether the functional group could mimic the amino acid side chain indicated.

6. Dabigatran salt. Provide a brief rationale for the value of administering the salt form of a drug.

A

B

E

D

C

Active form of Dabigatran

Dabigatran etexilate mesylate

F

5. Review side chain indicated. 164

Medicinal Chemistry Self Assessment

A

Answer D-Phe mimic

B

O

E

O

Pro mimic

D

Arg mimic

A

No

H O No

N

B

No

No

C

Yes

D

Yes

Yes C Yes

No

E

Yes

Yes

No

F

No

No

No No

N

No

H N

N N

F

N H2 NH

C H3

Yes

Active form of Dabigatran

6. Dabigatran etexilate is formulated as a mesylate salt. Provide a brief rationale for the value of administering the salt. salt form of aa brief drug.rationale for the value of administering the salt form of a drug. 6. Dabigatran Provide

Dabigatran etexilate mesylate Dabigatran etexilate mesylate Answer Mesylate salts are considered water soluble organic salts. The use of water soluble organic salts is one method to enhance the overall water solubility of your drug. This is accomplished by increased solvation and dissolution of the drug. The use of a water soluble organic salt also helps to ensure an adequate water/lipid balance necessary for oral absorption. As we have already determined in the structure evaluation grid, there is some hydrophobic character present, but there is also substantial hydrophilic character.

Go to Section 1.13 [Link to section 1.13 title]

Section 4 Whole Molecule Drug Evaluation Answers 2.14 Fenofibrate and Gemfibrozil

Fenofibrate is a member of the fibrate class of anti-hyperlipidemic agents. It is used as adjunctive therapy to diet in the management of dyslipidemias, including in the treatment of severe hypertriglyceridemia. The specific mechanism(s) by which fenofibrate decreases triglyceride and total cholesterol levels, as well as increases the levels of high density lipoproteins (HDL), is unknown. What know is that the–decrease very low density lipoproteins (VLDLs) 1.14 we anddo2.14 (drug name remove in bold) results from fenofibrate stimulation of lipoprotein lipase.

Fenofibrate

Gemfibrozil

1. Conduct a complete structuralLetter evaluation of fenofibrate and use the information in the grid to inform “C” – add bold your answers to the questions that follow. Answer Function Function

Character Character Name of Functional Group A

Halogenated aromatic hydrocarbon

Hydrophilic and/or Hydrophobic

Acidic, Basic, or Neutral Provide pKa When Relevant

Ketone

C ↑ Solubility and/or ↑ Absorption

– remove bold from label Hydrophobic 2.14Neutral Absorption

Interaction(s) Possible with Biological Target at Physiological pH=7.4 Hydrophobic

None Is Acceptable None

van der Waals π-π Stacking

B A

B

Function

Amino Acids That Can Interact with the Functional Group via Hydrogen Bonding Interactions at pH=7.4

Hydrophilic (CO) Hydrophobic (R)

Neutral

F C

Dipole–dipole

D

Ion–dipole (as the dipole)

Solubility (CO)

H-bonding (A)

Absorption (R)

Dipole–dipole

E

Fenofibrate

Ser, Thr, Tyr, Cys, Asn, Gln, His, Trp

Ion–dipole (as the dipole)

2.14 – remove bold from label 165

O

166

Medicinal Chemistry Self Assessment

Continued from previous page. C

Aromatic hydrocarbon

Fenofibrate

Hydrophobic

Neutral

Fenofibrate

Absorption

Hydrophobic

Gemfibrozil

Gemfibrozil

van der Waals

Letter “C” – add bold

Letter D “C” Ether– add boldHydrophilic (O)

None

π-π Stacking

Neutral

Hydrophobic (R)

Solubility (O)

H-bonding (A)

Absorption (R)

Dipole–dipole

Ser, Thr, Tyr, Cys, Asn, Gln, His, Trp

Ion–dipole (as the dipole) E

F

Aliphatic alkane

Hydrophobic

Ester

Hydrophilic (COO)

Neutral

Absorption

Hydrophobic

None

van der Waals Neutral

Solubility (COO)

H-bonding (A)

Absorption (R)

Dipole–dipole

C

HydrophobicC(R)

Ser, Thr, Tyr, Cys, Asn, Gln, His, Trp

Ion–dipole (as the dipole)

2.14 – remove bold from label

. 2.14 – remove bold from label

B B A

AF

D

C

F C

D

E E

Fenofibrate

Fenofibrate 2.14 remove bold label hydrolysis to the active drug. Draw the 2. Fenofibrate is administered as a– prodrug and from undergoes active drug. Provide a brief rationale for the value of administering fenofibrate as a prodrug. 2.14 – remove bold from label Answer O C H3 O Ester O OHydrolysis O O C H3 O O C H3 Ester O O H3 C C H3 Hydrolysis OH O CH Cl

H3C

Cl

3

C H3

H3C

Cl

Cl

O H3C

C H3

Inactive Inactive

O

Active Active

There are situations (e.g., the need to enhance water or lipid solubility) when it is therapeutically beneficial to administer drug molecules that have been covalently modified to produce inactive (or very weakly active) analogs (i.e., prodrugs) (see Chapter 5 in Basic Concepts in Medicinal Chemistry). Bioactivation of these prodrugs via rapid metabolic transformation (e.g., ester hydrolysis catalyzed by esterases) can occur via chemical or enzymatic mechanisms. Typically metabolic activation occurs within the gastrointestinal (GI) tract, but given that esterases are ubiquitous, release of the active drug can occur in the plasma or even at the target tissue. In the case of fenofibrate, a carboxylic acid is converted to an ester to form a prodrug. The resulting modified functional group (now an ester) is significantly more lipid soluble than the parent functional group (originally a carboxylic acid). The calculated log P for fenofibrate is 5.24, which suggests that the prodrug is lipophilic.

2.14 Fenofibrate and Gemfibrozil

167

3. The calculated P for fenofibrate is 5.24, whereas the calculated log P for gemfibrozil is 3.9. 1.14 andlog 2.14 (drug name – remove bold) Provide a structural rationale for the difference in this pharmacokinetic property.

Fenofibrate Gemfibrozil 3. The calculated for difference in this pharmacokinetic property.

Answer Letter “C” – add bold A larger calculated log P value reflects the presence of additional hydrophobic character. Let’s see if that statement pans out as we evaluate each of these molecules. The structure evaluation grid for fenofibrate reveals that there are several functional groups that contribute to its hydrophobic character including the halogenated aromatic hydrocarbon, the aromatic hydrocarbon, and both aliphatic alkanes. Evaluation of gemfibrozil reveals hydrophobic character from a dimethyl substituted aromatic hydrocarbon and an aliphatic alkane. The larger calculated log P value for fenofibrate is justified by the presence of significantly more functional groups with hydrophobic character.

Fenofibrate 4. From an elimination perspective, 60% of a fenofibrate dose is foundGemfibrozil in the urine and 25% is found C(drug name – remove bold) and in the feces. The active1.14 form of 2.14 fenofibrate undergoes both phase I and phase II metabolic transformations. The phase II conjugate is eliminated in the urine. Oxidative metabolism does not occur. Evaluate each of the metabolic products and determine if it is the conjugate that is eliminated in the 4. From anremove elimination 2.14 – bold NOT from occur). label urine (that does occur), the product of a non-oxidative phase I transformation (that does occur), or the product of an oxidative phase I transformation (that does NOT occur).

B HO

F

O

A

C

O

D

O

OH Fenofibrate C H3

H3 C

Cl

Gemfibrozil

E

A “C” – add bold Letter Fenofibrate

B

2.14 – remove bold from label

O

O O

Answer

H3C

Cl

C H3 O

C H3

O

Ester Hydrolysis

C H3

C C

O

Cl

2.14 – remove bold from label Nameaofstructural Metabolicrationale Transformation Type of Productdoes not occur. Inactive 5. Provide for why oxidative O-dealkylation

A

Aromatic hydroxylation

B

Ketone reduction

C

Glucuronide conjugation

Answer:

Active Product of oxidative transformation (does not occur)

B A

ProductF of non-oxidative transformation (does occur)

C

D Product of conjugation transformation (does occur) 1

E

2

O

H3C

OH C H3

C 168

Medicinal Chemistry Self Assessment

5. Provide a structural rationale for why oxidative O-dealkylation does not occur. 5. Provide a structural rationale for why oxidative O-dealkylation does not occur. Answer: Answer

1

2

Oxidative O-dealkylation transformations occur with ethers that have at least one α-carbon (adjacent carbon atom) with at least one hydrogen atom attached. Evaluation of the ether found within the structure of fenofibrate reveals that one of the carbons (#1) attached to the ether oxygen atom is part of an aromatic ring. It does not have a hydrogen atom attached to it. Carbon #2, attached to two aliphatic alkanes (methyl groups) and a carboxylic acid, also does not have a hydrogen atom attached to it. Because neither of the α-carbons has at least one hydrogen atom attached to it, this ether is not subject to oxidative O-dealkylation.

6. Fenofibrate and gemfibrozil have dramatically different elimination half-lives (20–22 hours and 1.5 4 and 2.14 (drug name – remove bold) hours respectively). Identify the possible metabolic transformations for gemfibrozil and provide a justification for the significant difference in this pharmacokinetic parameter.

Fenofibrate

Gemfibrozil

Answer er “C” – add bold Possible phase I metabolic transformations: •

Benzylic oxidation (two different locations)*

•

Oxidative O-dealkylation

•

Aromatic hydroxylation (multiple)

*Note: Each benzylic hydroxyl group can be further oxidized to carboxylic acids. Possible phase II conjugation transformations: C • Glucuronide conjugation

• Amino acid conjugation (with glutamine, glycine, arginine or taurine) 4 – remove bold from label The difference in aromatic substituents between the two drugs has a large impact on the number of

B potential phase IFtransformations that produce more water soluble metabolites. Oxidative OA

dealkylation not only generates more water soluble products, but also splits gemfibrozil in half, D deactivation. Based on this assessment, it is no surprise that there is a difference in the causing drug C elimination half-lives of these two drugs and that gemfibrozil is eliminated much more rapidly.

E Fenofibrate

2.14 – remove bold from label Section 4 Whole Molecule Drug Evaluation Answers O

O O

OH

H3C

Cl

C H3

2.15 Fluvoxamine

Fenofibric Acid

Fluvoxamine is an inhibitor of the serotonin reuptake transporter (SERT) and prevents the reuptake of serotonin at the presynaptic membrane in the central nervous system. It is indicated for use in the treatment of depression. 1.15 and 2.15 – remove bold from label Fluvoxamine is structurally unique relative to the rest of the serotonin selective reuptake inhibitor class of drugs. A

B D

C

E

Fluvoxamine

1. Conduct a structural evaluation of fluvoxamine, focusing on the boxed functional groups, and use the 1.15 and 2.15 – remove bold from label information in the grid to inform your answers to the questions that follow. Answer

F3 C

O

Character

C H3

Function

NH3

–

O

Name of Functional Group

Hydrophilic and/or Hydrophobic

Interaction(s) Possible with Provide ↑ Solubility Fluvoxamine maleateBiological Target pKa When at Physiological and/or pH=7.4 ↑ Absorption Relevant

Halogenated aliphatic alkane

Hydrophilic (F)

Neutral

Character

A

Acidic, Basic, N O or Neutral Function

Function

Hydrophobic (R)

Solubility (F)

Dipole–dipole

Absorption (R)

H-bonding (A)

Amino Acids That O Interact with the Can O Functional Group via Hydrogen Bonding Interactions O H at pH=7.4 None Is Acceptable Ser, Thr, Tyr, Cys, Asn, Gln, Trp, His

Ion–dipole (as the dipole) B

Aromatic hydrocarbon

Hydrophobic

Aliphatic alkane

Hydrophobic (R)

Neutral

Absorption

Hydrophobic

None

van der Waals π-π Stacking

C

Neutral

Absorption

Hydrophobic van der Waals

169

None

170

Medicinal Chemistry Self Assessment A

B C

Continued from previous page. D

Ether

Hydrophilic (O)

Neutral

Hydrophobic (R)

D

Solubility (O)

H-bonding (A)

Absorption (R)

Dipole–dipole E

Ser, Thr, Tyr, Cys, Asn, Gln, His, Trp

Ion–dipole (as the dipole) E

Primary amine

Hydrophilic (NH2)

Basic

Hydrophobic (R)

pKa 9–11

Solubility (NH2) Fluvoxamine

Ion–dipole (as the ion)

Absorption (R)

Ionic

Ser, Thr, Cys, Tyr, Asn, Glu, His, Trp

1. Conduct a to the questions that follow. 2.14 – remove bold from label 2. Fluvoxamine is marketed as a single isomer. Is the product sold as a single enantiomer, diastereomer, 2. Fluvoxamine is isomer? a brief rationale foranswer. your answer. or geometric isomer? Provide aProvide brief rationale for your

F3 C

O

1' Cl

2

O

O

2'

N

O

1

A

NH 2

O

C HH3 C 3

F3 C OH

C H3

H 2N

1'

2'

1N

2

O

Fenofibric Acid

O

C H3

B

Answer 1.15 and 2.15 – remove bold from label There are no chiral carbon atoms present; therefore, there are no enantiomeric or diastereomeric 3. Fluvoxamine is properties of the drug? isomers possible. A not mirror images of one another and are not superimposable (see Chapter Geometric isomers are B Chemistry). These isomers typically occur in the presence of a 7 in Basic Concepts in Medicinal carbon–carbon double bond, where the double bond causes conformational restriction and forces C in one of twoDorientations. In the case of fluvoxamine, there the bond substituents to be positioned is a carbon–nitrogen double bond present. When evaluating isomer A, the 1' and 1 substituents are located on opposite sides of the carbon–nitrogen double bond. This represents the trans- or E-geometric isomer. When evaluating isomer B, the 1' and 1 substituents are located on the same side E of the carbon–nitrogen double bond. This represents the cis- or Z-geometric isomer. The marketed product is the E-geometric isomer.

Fluvoxamine Fluvoxamine maleate

3. Fluvoxamine is formulated as a maleate salt. What type of salt is a maleate salt, and what type of properties does it confer to2.15 the overall properties the drug? 1.15 and – remove bold fromof label F C 4. Fluvoxamine is3well absorbed, provide a rationale for these pharmacokinetic properties.

O N

O

NH3

Fluvoxamine maleate

C H3 –

O

O

O OH

2.15 Fluvoxamine

171

Answer Maleate salts are water soluble organic salts. Water soluble organic salts generally enhance overall drug water solubility (see Chapter 5 in Basic Concepts in Medicinal Chemistry). Maleate salts typically demonstrate enhanced solvation and dissolution as compared to their free base forms. This is exceptionally beneficial in the preparation of parenteral, nasal, and ophthalmic dosage forms where the delivery of small volumes of highly concentrated solutions is necessary. 4. Fluvoxamine is well absorbed and has an oral bioavailability of ~50%. Using the information found in the structure evaluation grid, provide a rationale for these pharmacokinetic properties. Answer Orally administered drugs must strike a balance between having enough hydrophilic character to promote dissolution and solubility in the aqueous contents of the gastrointestinal (GI) tract, and sufficient hydrophobic character to allow for absorption across the lipid bilayer membrane. Fluvoxamine has a fluorinated aliphatic alkane, an ether, and an ionized primary amine that contribute to the overall hydrophilic character of the drug. An aromatic hydrocarbon and aliphatic alkane contribute to the overall hydrophobic character of the drug. Based on this evaluation, it is no surprise that the drug undergoes rapid dissolution and solvation in the aqueous contents of the GI tract. Although it appears that there is only a moderate amount of hydrophobic character, this appears to be sufficient to allow for absorption across the lipid bilayer membrane. 5. A number fluvoxamine metabolites have been identified, all of whichhas demonstrate 5. A of number of Evaluate each of the which metabolic transformation occurred. little or no pharmacological activity. Evaluate each of the metabolic products drawn below and identify which metabolic transformation has occurred.

A

B

C

D

E

F

172

Medicinal Chemistry Self Assessment

Answer Name of Metabolic Transformation A

Oxidative deamination

B

Aromatic hydroxylation

C

Oxidative O-dealkylation

D

Oxidative O-dealkylation followed by alcohol oxidation

E

Allylic oxidation*

F

Acetylation

*Note: This carbon–nitrogen double bond acts similarly to a carbon–carbon double bond. If a carbon substituent attached to this carbon–nitrogen double bond bears a hydrogen atom, then it can undergo oxidation. In this case, there is only one carbon substituent that fulfills that criterion, and there is only one location possible for an allylic oxidation to occur.

6. Fluvoxamine is a strong inhibitor of CYP1A2, CYP3A4, and CYP2C19. These enzyme isoforms catalyze a number of the phase I oxidative metabolic transformations. Several of the benzodiazepines (used in the treatment of anxiety), including the very popular alprazolam, rely heavily on hepatic oxidation for metabolic inactivation and elimination. Other benzodiazepines, including the equally popular lorazepam, rely on glucuronide conjugation for metabolic inactivation and elimination. Which combination of drugs, fluvoxamine + alprazolam or fluvoxamine + lorazepam, is the most likely to generate an enhanced anxiolytic effect? Answer Fluvoxamine inhibits several of the metabolism isoforms that catalyze oxidative metabolic transformations. Alprazolam relies on hepatic oxidation for metabolic inactivation and elimination. If fluvoxamine inhibits the same enzymes that alprazolam relies on for hepatic oxidation, then the levels of active alprazolam will exist for a longer period of time than expected. With more alprazolam available to produce an anxiolytic effect for a longer period of time, it is likely that an enhanced anxiolytic effect will be observed. Because lorazepam relies on a different set of enzymatic isoforms for glucuronide conjugation, the co-administration of fluvoxamine will have little or no effect on the metabolic inactivation and elimination of lorazepam, and an enhanced anxiolytic effect will not be observed.

Section 4 Whole Molecule Drug Evaluation Answers 2.16 Haloperidol

Shown below is the structure of haloperidol. Six of its functional groups have been identified. E D

C A

F

B

1. Using 1. the Using table below, the se. identify the six boxed functional groups. For each of the functional groups you identified, indicate if it is hydrophilic or hydrophobic in character. Also provide a brief explanation for 2. Based on their electronic induction. your response. 3. Using the in pH environments of 1.7, 5.5, 6.0, 7.4, and 8.5. Answer

Answer Functional Group Name A

Hydrophilic or Hydrophobic

The structure only functional and noacceptor; acidic functional groups. Halogen; fluorineof haloperidol contains Effects canone vary;basic fluorine can act as agroup hydrogen bond however, studies have shown that the substitution of a hydrogen atom with a fluorine atom tends to slightly enhance lipid solubility.

Tertiary Hydrophilic amine due to its ability to act as a hydrogen bond acceptor.

B

Ketone

C

Alkyl group; alkyl chain; aliphatic chain

D

Tertiary amine

E

Tertiary hydroxyl group; tertiary alcohol

Cl Hydrophilic O due to its ability to form hydrogen bonds as either a donor or an acceptor.

F

Phenyl ring; aromatic ring; aromatic hydrocarbon

Hydrophobic due to its inability to ionize or form hydrogen bonds; hydrocarbon functional groups enhance lipid solubility.

F

OH Hydrophobic due to its inability to ionize or form hydrogen bonds; hydrocarbon functional groups enhance lipid N solubility. Hydrophilic due to its ability to ionize (form an ion–dipole interaction with water) and to participate in hydrogen bonding (acceptor) in its unionized form.

4. Shown below is a structural analog can enhance the duration of haloperidol. 2. Based on their electronic properties AND their relative positions in the molecule, identify if functional groups A, C, and E are electron withdrawing or electron donating. Additionally, identify if this effect is due to resonance or induction. Answer • Functional Group A is a halogen. Because all halogens are more electronegative than an adjacent aromatic ring, the fluorine atom acts as an electron withdrawing group through induction. •

Functional Group C is an alkyl group and has a lower electronegativity than the oxygen atom of the ketone or the nitrogen atom of the tertiary amine; therefore, it acts as an electron donating group through induction. 173

E 174

Medicinal Chemistry Self Assessment

C

D

A

•

E

D C Functional Group E is a tertiary hydroxyl group. Based on its position in the molecule and the fact that it is notAdirectly attached to an aromatic ring, it acts as an electron withdrawing group F through induction. B

F B 3. Using unmodified 1. the Using the se. structure of haloperidol and the table below, identify all of the acidic and basic functional groups present in the structure, provide the normal pKa range for each functional 2. Based on their electronic induction. group, and identify if each functional group would be primarily ionized or unionized in pH environ1. Using se.the in pH environments of 1.7, 5.5, 6.0, 7.4, and 8.5. 3. the Using ments=1.7, 5.5, 6.0, 7.4, and 8.5. 2. Based on their electronic induction. Answer Answer 3. The Using the in pH of 1.7, 5.5, 6.0, 7.4, and functional 8.5. The structure of haloperidolcontains contains only one basic functional group groups. structure of environments haloperidol only one basic groupand andno noacidic acidicfunctional functional groups. Answer The structure of haloperidol contains only one basic functional group and no acidic functional groups. Tertiary amine OH

N

Tertiary amine

OH

F

N

Cl

O F Functional Group

Acidic or Basic

Tertiary amine

Basic

O

Cl

pKa Range

Primarily Ionized or Unionized 1.7 5.5 6.0

7.4

8.5

9–11

Ionized

Ionized

Ionized

4. Shown below is a structural analog can enhance the duration of haloperidol. Ionized

Ionized

4. Shown Shownbelow below is is aa structural structural analog enhance the Evaluate duration ofthe haloperidol. 4. analogcan of haloperidol. structural change and propose an explanation as to how this structural modification can enhance the duration of haloperidol.

5. Using the table below ionizable functional groups. Answer Structural evaluation of this analog reveals that the tertiary hydroxyl group has been converted to a soluble ester. Lipid solublefunctional esters cangroups. be administered as either an intramuscular (IM) or subcu5. lipid Using the table below ionizable C taneous (SC) injection. After injection, the lipid soluble ester dissolves slowly from its initial injection site into the plasma in a consistent manner over an B extended period of time. Once in the plasma, the ester is hydrolyzed to release haloperidol. Remember that C esterases are ubiquitous within the body and that lipid and water soluble esters are frequently used as prodrugs D to provide a variety of theraB A peutic advantages.

A

D

2.16 Haloperidol

175

5. Using the table below, identify the types of binding interactions that could be possible between the boxed functional groups and a protein or enzyme receptor. Also identify amino acids present within 5. Usingorthe table below ionizable functional a receptor enzyme whose side chains could groups. participate in the interactions that you identified. Assume a plasma pH=7.4 for all ionizable functional groups.

C B A

D

Answer

A

B C

D

Amino Acids Capable of Forming Specific Binding Interaction

Functional Group

Types of Binding Interactions

Ketone

(1) Ion–Dipole (as the Dipole)

(1) Asp, Glu (with carbon of ketone); Arg, Lys, His* (with oxygen of ketone)

(2) Dipole–Dipole

(2) Ser, Thr, Tyr, Cys, Asn, Gln, His*

(3) Hydrogen Bond (Acceptor)

(3) Ser, Thr, Tyr, Cys, Asn, Gln, Trp, His*

(1) Ionic

(1) Asp, Gln

(2) Ion–Dipole (as the Ion)

(2) Ser, Thr, Tyr, Cys, Asn, Gln

(1) Ion–Dipole (as the Dipole)

(1) Asp, Glu (with hydrogen of hydroxyl); Arg, Lys, His* (with oxygen of hydroxyl)

(2) Dipole–Dipole

(2) Ser, Thr, Tyr, Cys, Asn, Gln, His*

(3) Hydrogen Bond (Donor and/or Acceptor)

(3) Ser, Thr, Tyr, Cys, Asn, Gln, Trp, His*

van der Waals; Hydrophobic

Tyr, Phe, Trp (better interaction**); Val, Leu, Ile, Met, Ala

Tertiary amine Tertiary hydroxyl

Phenyl ring; aromatic ring; aromatic hydrocarbon

*The side chain of histidine is primarily unionized at a pH=7.4. The small fraction that is ionized could form an ion–dipole interaction with a partially negative atom, while the unionized fraction can serve as a hydrogen bond donor or acceptor. Additionally, it can serve as the dipole in an ion–dipole interaction. **Stronger van der Waals interactions occur when aromatic rings interact with aromatic rings; however, all of the listed amino acids could possibly interact with the indicated functional group.

6. Shown below is the structure of haloperidol and a list of five metabolic transformations. For each metabolic transformation, indicate if it is a phase I or a phase II transformation and if haloperidol has a functional group present to participate in the transformation. If you answer YES, then draw the appropriate metabolite; if you answer NO, then provide a brief explanation as to why this metabolic 6. Shown below is the to perform with haloperidol. transformation is not possible to perform with haloperidol. Metabolic Pathways A. Reduction B. Sulfate Conjugation C. Hydrolysis D. Oxidative N-Dealkylation E. Benzylic Oxidation

Answer A. Reduction: Phase I, Yes

176

Medicinal Chemistry Self Assessment

Answer A. Reduction: Answer Phase I, Yes Answer A. Reduction: Phase I, Yes A. Reduction: Phase I, Yes

B. Sulfate conjugation is a phase II transformation. It is possible; however, due to steric hindrance, B. Sulfate conjugation is a phase II ation is low. the probability of this metabolic transformation is low.

B. Sulfate conjugation is a phase II ation is low.

C. Hydrolysis is a phase I transformation. It is not possible for haloperidol because its structure does not contain a functional group that undergoes this type of transformation. C. Hydrolysis is aN-Dealkylation: phase I biotransformation. D. Oxidative Phase I, Yes D. Oxidative N-Dealkylation: Phase I, Yes C. Hydrolysis is a phase I biotransformation.

E. Benzylic oxidation is a phase I transformation. It is not possible for haloperidol because one benzylic position is occupied by a ketone (and therefore is not capable of being oxidized) and the other is directly attached to a heteroatom and lacks a hydrogen atom.

Section 4 Whole Molecule Drug Evaluation Answers 2.17 Hydrocortisone

Hydrocortisone is a glucocorticoid used in the management of inflammation. Derivatives of hydrocortisone are used in the management of asthma and chronic obstructive pulmonary disease. 1.17 and 2.17 – remove boldand fromuse label 1. Conduct a complete structural evaluation of hydrocortisone the information in the grid to inform your answers to the questions that follow.

E C F

B D

A

Hydrocortisone

Answer

Question #3 structure needs to be replaced: Function Function

Character Character

A

Name of Functional Group

Hydrophilic and/or Hydrophobic

Ketone

Hydrophilic (CO)

Interaction(s)

Acidic, Basic, or Neutral Provide pKa When Relevant A

Function

21 Possible with

and/or ↑ Absorption

Biological Target at Physiological pH=7.4

Solubility (CO)

H-bonding (A)

Absorption (R)

Dipole–dipole

17 11 ↑ Solubility

Neutral

Hydrophobic (R)

Amino Acids That Can Interact with the Functional Group via Ion– Dipole Interactions at pH=7.4 None Is Acceptable Lys, Arg, Glu, Asp

Ion–dipole (as the dipole) B

Cycloalkene

Hydrophobic

1.18Neutral and 2.18 – remove bold fromHydrophobic label Absorption

None

van der Waals C

Secondary alcohol

Hydrophilic (OH)

Neutral

Hydrophobic (R)

HO I

A

Solubility (OH) Absorption (R)

I I O

C

B177 I

H-bonding (A+D)

E

Dipole–dipole

O D Ion–dipole (as the dipole)

N H2

OH

Lys, Arg, Glu, Asp

178

Medicinal Chemistry Self Assessment

Continued from previous page. D

Cycloalkane

Hydrophobic

Neutral

Absorption

E

Primary alcohol

Hydrophilic (OH)

Neutral

Solubility (OH)

H-bonding (A+D)

Absorption (R)

Dipole–dipole

Hydrophobic

None

van der Waals Hydrophobic (R)

Lys, Arg, Glu, Asp

Ion–dipole (as the dipole) F

Tertiary alcohol

Hydrophilic (OH)

Neutral

Hydrophobic (R)

Solubility (OH)

H-bonding (A+D)

Absorption (R)

Dipole–dipole

Lys, Arg, Glu, Asp

Ion–dipole (as the dipole)

2. The glucocorticoids interact with residues within the glucocorticoid receptor (Arg611, Asn564, Thr739, Gln642, and Gln570) via hydrogen bonding and ion–dipole interactions at physiological pH (7.4). Identify which functional groups could interact with the side chains of these amino acids. Answer

Functional Group A

Can Interact with Arginine611 via an Ion–Dipole Interaction

Can Interact with Asparagine564 and Threonine739 via a Hydrogen Bonding Interaction

Can Interact with Glutamine642 and Glutamine570 via a Hydrogen Bonding Interaction

Yes or No

Yes or No

Yes or No

Yes

Yes

Yes

1.17 and 2.17 – remove bold from label

B

No

No

C

Yes

Yes

D

No

No

E

Yes

Yes C

Yes

F

Yes

Yes

Yes

No

E

Yes No

F

B

3. Several of the functional groups identified in the structure evaluation grid are essential for biological D render the glucocorticoids inactive. These activity. There are several metabolic A transformations that transformations include the following: 1. Ring A ketone reduction

Hydrocortisone

2. Ring A double bond reduction 3. C-17 oxidation

Question #3 structure needs to be replaced:

4. C-11 oxidation

21 11

17

A

1.18 and 2.18 – remove bold from label

2.17 Hydrocortisone

179

Based on this information, consider the array of products drawn in the scheme below. Identify each type of reaction or transformation that has occurred and evaluate each of the products to determine if each product is active or inactive.

A

B

F 11

17

21

C

A

E D

Answer Pathway

Type of Reaction or Transformation

Active or Inactive

A

Allylic oxidation followed by dehydration

Active

B

C-11 oxidation

Inactive

C

C-17 oxidation

Inactive

D

C-21 esterification (acetylation)

Active

E

A ring double bond reduction

Inactive

F

A ring ketone reduction

Inactive

180

Medicinal Chemistry Self Assessment

4. The synthetic glucocorticoids are often esterified at C-21 to produce prodrugs. Both lipophilic and water4. soluble esters cann be formed. each of the four prodrugs drawn below and determine The synthetic overall drug Evaluate water solubility. whether a lipophilic or water soluble ester is present. Determine how prodrug formation has an effect on overall drug water solubility.

A

C

B

D

Answer Type of Ester Formed

Effect on Overall Drug Water Solubility

A

Lipophilic ester

Decreases overall water solubility of the drug

B

Water soluble ester salt

Increases overall water solubility of the drug

C

Lipophilic ester

Decreases overall water solubility of the drug

D

Water soluble ester salt

Increases overall water solubility of the drug

a. Provide a structural rationale for why prodrugs (e.g., B and D) are used in the preparation of aqueous injectable products to be administered intramuscularly (IM) or intravenously (IV). Answer Drugs that are formulated as aqueous injectable products must be highly hydrophilic in character. Hydrocortisone contains four hydrophilic functional groups (ketone, primary alcohol, secondary alcohol, and tertiary alcohol); however, it is not particularly soluble in water. By forming a water soluble ester salt with the primary alcohol at C-21, an ionizable functional group is introduced. In the case of prodrug B, the ionizable carboxylic acid is able to interact with water through ion– dipole interactions and, therefore, significantly increase the overall water solubility of the drug molecule. b. Provide a structural rationale for why prodrugs (e.g., A and C) are used in the preparation of depot injections. Answer Drugs that are formulated as depot injections must be highly hydrophobic in character. Hydrocortisone contains a steroid backbone (all four rings) that is highly hydrophobic; however, it still has

2.17 Hydrocortisone

181

some limited solubility in water. By forming a lipophilic ester with the primary alcohol at C-21, the overall water solubility of the drug decreases substantially. In this case, one of the hydrophilic groups is now masked as an ester, thereby significantly decreasing its hydrophilic character. As a lipophilic prodrug, hydrocortisone can be formulated as a suspension for intramuscular or subcutaneous injection. 5. Lipophilic glucocorticoid esters typically do not concentrate in the urine, but rather undergo glomerular filtration followed by tubular reabsorption. Provide a brief rationale for why lipophilic glucocorticoid esters do not concentrate in the urine and determine what effect this has on duration of drug action. Answer For drugs to concentrate in the urine, they need to be highly water soluble (contain functional groups with a significant amount of hydrophilic character). We have already evaluated the overall water solubility of hydrocortisone to be relatively poor, despite the presence of several hydrophilic functional groups. The addition of a lipophilic ester will further decrease the water solubility of the resulting glucocorticoid ester. Because the glucocorticoid ester suffers from poor water solubility, it is not unexpected that it does not concentrate in the urine. These lipophilic esters, however, have sufficient hydrophobic character to undergo reabsorption in the renal tubules. Because the drug is returned to systemic circulation via this route of absorption, the duration of drug action is prolonged.

6. Which type of prodrug, water soluble ester salts, or lipophilic esters, would you anticipate to have greater systemic side effects? Answer The more water soluble the prodrug is, the wider the systemic distribution. The greater the prodrug’s solubility in the blood, then the greater the potential is for more misadventures/side effects. Because the water soluble ester salts are by far more water soluble than the lipophilic esters, one would predict that the water soluble ester salts would exhibit more systemic side effects.

C F

B

D Section 4 Whole Molecule Drug Evaluation A Answers Hydrocortisone

Question #3 structure needs to be replaced:

2.18 Levothyroxine (T4)

21 17 11 Levothyroxine (T4) is a naturally produced thyroid pro-hormone. In its active form, tri-iodo-L-thyronine (T3) is responsible for regulating oxygen consumption and calorigenesis (think metabolism, metabolic rate, and thermogenA molecules until it is needed. Once proteolyzed from thyroglobesis). T4 is biosynthesized and stored in thyroglobulin ulin and transported to the desired target tissue, T4 undergoes dehalogenation catalyzed by thyroxine dehalogenase to the active thyroid hormone T3.

1.18 2.18 – remove from on label 1. Conduct a structural evaluation of and levothyroxine (T4), bold focusing the boxed functional groups, and use the information in the grid to inform your answers to the questions that follow.

HO

A

I I

C

B

O

OH N H2

O

I

D

E

I

L-Thyroxine

Answer Function Character

Name of Functional Group A Phenol

Function

Character

Acidic, Basic, or Neutral

Function

Hydrophilic and/or Hydrophobic

Provide pKa When Relevant

↑ Solubility and/or ↑ Absorption

Interaction(s) Possible with Biological Target at Physiological pH=7.4

Hydrophilic (OH)

Acidic

Solubility (OH)

H-bonding (A+D)

Hydrophobic (Ar)

pKa=9–10

Absorption (Ar)

Dipole–dipole

Amino Acids That Can Interact with the Functional Group via Hydrogen Bonding Interactions at pH=7.4 None Is Acceptable Ser, Thr, Tyr, Cys, Asn, Gln, Trp, His

Ion–dipole (as the dipole) B

Ether

Hydrophilic (O) Hydrophobic (R)

Neutral

Solubility (O)

H-bonding (A)

Absorption (R)

Dipole–dipole Ion–dipole (as the dipole)

183

Ser, Thr, Tyr, Cys, Asn, Gln, His, Trp

184

Medicinal Chemistry Self Assessment

Continued from previous page. C – Halogenated Hydrophobic 1.18 and 2.18 remove bold from label

Neutral

Absorption

Dipole–dipole

aromatic hydrocarbon

None

Ion–dipole (as the dipole) Hydrophobic van der Waals π-π Stacking

D Primary amine

E

Carboxylic acid

Tri-iodothyronine

Hydrophilic (NH2)

Basic

Solubility (NH2)

Ion–dipole (as the ion) None

Hydrophobic (R)

pKa=9–11

Absorption (R)

Ionic

Solubility (CO2H)

Ion–dipole (as the ion) None

Absorption (R)

Ionic

Hydrophilic (CO2H) Acidic pKa=2–5

Hydrophobic (R)

1. Conduct a structural your answers to the questions that follow. 2. Hormone replacement therapy in the form of levothyroxine (T4) is available for those patients who 1.18 and 2.18 – remove bold from label E supply of T4. Evaluate the structure of levothyroxine and do not produce an adequate endogenous determine if it is an enantiomer,Ddiastereomer, or geometric isomer.

A

O H 2N H

C

I OH

HO

B OH L-Tyrosine

L-Thyroxine

I

O I

OH

O

H NH 2

Levothyroxine

2.Answer Hormone replacement, diastereomer, or geometric isomer. Evaluation of the structure of levothyroxine reveals that there is one chiral carbon atom. Because there is only one chiral carbon atom, levothyroxine represents an enantiomer. Diastereomers require the presence of at least two chiral carbon atoms so levothyroxine does not represent a diastereomer. Further inspection of the molecule reveals that there are no carbon–carbon double bonds or other conformationally restricted functional groups so geometric isomers are not possible.

3. Evaluate the chiral carbon atom in levothyroxine to determine if this drug is drawn as the R- or Levothyroxine S-enantiomer. Answer Using the Cahn-Ingold-Prelog (CIP) system (see Chapter 7 in Basic Concepts in Medicinal Chemistry), 3.theEvaluate the chiral carbon this drug is drawn as the R- orbased S-enantiomer. four groups attached to atom the chiral carbon are prioritized on atomic number and the described sequence rules. Using these rules, the primary amine nitrogen atom is prioritized as #1, the Answer: carboxylic acid #2, the methylene unit attached to the aromatic hydrocarbon has the #3 priority, and the hydrogen atom pointing away from the reader is the #4 priority. This assessment provides the Using the; therefore, the S-enantiomer is drawn. evidence for the prioritization scheme indicated below. In reviewing the positions of these priorities it is noted that they are counterclockwise in orientation; therefore, the S-enantiomer is drawn. 3

4

2

1

2.18 Levothyroxine (T4)

185

4. T3 is the biologically active hormone. It interacts with the thyroid hormone receptor via hydrogen 4. T3 is the biologically identified in hydrophobic, the structure evaluation bonding, ion–dipole, and ionic grid. interactions. Evaluate the ribbon diagram that shows how T3 interacts with the surface of the thyroid receptor and identify the types of interactions possible for each of the five functional groups identified in the structure evaluation grid. O

I HO

I

OH 1.18 and 2.18 – remove bold from label H N H2 O I Tri-iodo-L-thyronine

Tri-iodothyronine

Answer and 2.18 – removevia bold from labelas the H-bond donor. A. 1.18 Phenol: OH interacts H-bonding 1.18 and 2.18 – remove bold from label 5. T4 is biosynthesized L-tyrosine derived L-tryosine.as a H-bond donor and acceptor. In this case the NOTE: from The phenol OHwithin is capable of from participating phenol specifically interacts with the thyroid receptor as the H-bond donor. O I O O B. H Ether: no role in binding interaction. O I 2N I OH HO H 2 NC. Halogenated aromatic H hydrocarbon: halogens interact via hydrophobic interaction; aromatic HOH I OH O hydrocarbon interacts via hydrophobic interactions or π-π stacking. OH H H NH 2 HO N H I D. Primary amine: no role in binding interaction. 2 I O OH I interactions. E. Carboxylic acid: participates in ion–dipole (as the ion) or ionic OH I L-Tyrosine Levothyroxine Tri-iodothyronineLevothyroxine L-Tyrosine 5. T4 is biosynthesized from L-tyrosine within the thyroglobulin molecule and is considered an amino acid–based hormone. Other hormones In the body are steroid-based (e.g., estrogen) or peptideand 2.18 – remove bold from of label Answer: based (e.g.,1.18 insulin). Evaluate the structure T4 and justify its classification as an amino acid–based hormone by determining which functional groups are derived from L-tyrosine. The biosynthesis s found in the amino acid tyrosine. O I O H 2N I OH HO H OH B O A I H NH 2 O I I HO OH OH I NH2 I O L-Tyrosine Levothyroxine



I Answer Levothyroxine The biosynthesis of levothyroxine occurs within the thyroglobulin molecule which is rich in L-tyrosine residues. Iodination of the tyrosine units and phenolic coupling represent some of the biochemical reactions that occur in the biosynthesis of the natural hormone. Box A is derived from L-tyrosine, as is

L-Tyrosine 186

Levothyroxine 1.18 and 2.18 – remove bold from label

Medicinal Chemistry Self Assessment

I

Answer:

HO evident bybiosynthesis the presence of only phenolacid portion of the amino acid side chain. Box B is also derived The s found in the the amino tyrosine. from L-tyrosine; however, the phenol has become an ether. The rest of the boxed atoms represent each of the atoms found in the amino acid tyrosine.

A HO

I

HO

I

I I O

I

O

I

A

O OH

OH

NH2 H NH 2

I

I

HO

Levothyroxine Levothyroxine Levothyroxine

I

6. Interestingly, the dehalogenation transformation represents both an activation and deactivation pathway for T4. The iodo substituents on the inner ring of T4 play a role not only in important hydrophobic binding interactions with the receptor, but also in maintaining the shape of the hormone in a perpendicular orientation (see diagram below). This perpendicular shape is necessary to appro1.18 and 2.18 – structure was fixed (pay no attention to the colors) priately position the phenolic OH to participate in key receptor binding interactions. Evaluate the metabolites produced from the dehalogenation transformations and identify which metabolite is active and which is not. Provide a brief rationale for your answers.

1.18 and 2.18 – remove bold from label I 1.19 and 2.19 – remove bold from label. I HO

A I HO I

O I

OH H NH 2 I

D OH

H NH 2

O

O

C

O I

B Lidocaine

O I

B

O I

I

A

I HO

O I

OH

Levothyroxine

I

H NH 2

O

B 1.18 and 2.18 – structure was fixed (pay no attention to the colors)

I O

B

2.18 Levothyroxine (T4)

187

Answer • Evaluation of the dehalogenation transformation that leads to metabolite A: The outer ring of levothyroxine is dehalogenated to produce metabolite A. This metabolite retains the two inner ring iodo substituents; therefore, the shape of the hormone remains perpendicular. This metabolite retains its biological activity. •

Evaluation of the dehalogenation transformation that leads to metabolite B: The inner ring of levothyroxine is dehalogenated to produce metabolite B. This metabolite retains only one inner ring iodo substituent; therefore, the shape of the hormone cannot remain perpendicular. Because the phenol is not positioned appropriately to participate in an important binding interaction with the thyroid hormone receptor, this metabolite is inactive.

Section 4 Whole Molecule Drug Evaluation Answers 2.19 Lidocaine

As a sodium channel blocker lidocaine has found therapeutic use both as a local anesthetic and as a Class IB antiarrhythmic agent. As an anesthetic, this agent demonstrates rapid onset of action (acts quickly) and a longer duration of action (lasts longer) than most amino ester-type local anesthetics. The most frequently observed side effects are changes in the central nervous system (CNS) (e.g., dizziness, lightheadedness, and tinnitus). Lidocaine is extensively metabolized by the CYP1A2 isozymes to a variety of metabolites. 1. Conduct a the in the grid to inform your answers to some of the questions that follow. 1. Conduct a complete structural evaluation of lidocaine, place your answers in the grid provided, and then use the evaluation information in the grid to inform your answers to some of the questions that follow. C H3

CH3

O N

N

C H3

H

C H3

Lidocaine Lidocaine

Answer Answer: Function Character

A

B

Function

Character

Acidic, A Basic, or Neutral

C Function

Name of Functional Group

Hydrophilic and/or Hydrophobic

Provide pKa When Relevant

↑ Solubility and/or ↑ Absorption

Interaction(s) Possible with Biological Target D at Physiological pH=7.4

Aromatic hydrocarbon

Hydrophobic

Neutral

Absorption

Hydrophobic

Amide

Hydrophobic (R)

Neutral

Hydrophilic (CON)

B

Amino Acids That Can Interact with the Functional Group via Hydrogen Bonding Interactions at pH=7.4 None Is Acceptable None

van der Waals

Lidocaine

π-π Stacking

Absorption (R)

H-bonding (A + D)

Solubility (CON)

Dipole–dipole Ion–dipole (as the dipole)

2. Based on the information in the 3. Local anesthetics that have.

189

Ser, Tyr, Thr, Cys, Asn, Glu, His, Trp,

C H3 190

Medicinal Chemistry Self Assessment

CH3

O N

N C H3

C H3

H

Continued from previous page. C

Tertiary amine

Hydrophobic (R) Hydrophilic (N)

Answer: D

Aliphatic alkane

Hydrophobic

Basic

Absorption (R)

Solubility (N) pKa 9–11Lidocaine Neutral

Absorption

Ion–dipole (as the ion)

None

Ionic Hydrophobic

None

van der Waals

A

C D B Lidocaine Lidocaine

2. Based on the information in the structure evaluation grid, determine whether or not lidocaine is likely (pH=7.4). 2.soluble Basedinonthe theblood information in the Answer 3. Local anesthetics that have. Lidocaine contains functional groups that contribute to the overall hydrophobic character of the molecule (e.g., aromatic hydrocarbon and aliphatic alkanes), as well as functional groups that contribute to the overall hydrophilic character of the molecule (e.g., amide and tertiary amine). The hydrophilic character of the molecule enhances its water solubility, whereas the hydrophobic character enhances its ability to be absorbed across lipophilic membranes. At pH=7.4, the tertiary amine will be predominantly in its ionized form which will further increase the water solubility of the drug. The possibility of H-bonding interactions between the amide and water and an ion–dipole interaction between the tertiary amine and water suggests that lidocaine is likely soluble in the blood.

3. Local anesthetics that have a rapid onset of action are rapidly distributed in the body and can be absorbed easily across lipophilic membranes. Based on the information in the structure evaluation grid, provide a rationale for why lidocaine is rapidly distributed and can easily be absorbed across lipophilic membranes. Answer Based on the information in the structure evaluation grid, we know that lidocaine is likely soluble in the blood due to the presence of hydrophilic functional groups (amide and ionizable tertiary amine). If the drug is soluble in the blood, then it can be readily distributed in the body. Based on the information in the structure evaluation grid, we also know that lidocaine is composed of functional groups with a fair amount of hydrophobic character (aromatic hydrocarbon, aliphatic alkanes). This will enhance absorption across lipophilic membranes. Rapid distribution and absorption across lipophilic membranes contributes to the ability of lidocaine to have a rapid onset of action.

2.19 Lidocaine

191

4. Unless excreted unchanged, drug molecules undergo one or more metabolic transformations to deactivate the drug and/or make the drug sufficiently water soluble to permit elimination. There are a variety of transformations that are possible for most drugs, but only the minimum number of trans1.19 and 2.19actually – remove bold The fromfollowing labels anddiagram remove captures answers the frommetabolic over the arrows (payfor nolidocaine attention to formations occurs. pathways the colors) that are observed clinically. For each transformation, identify which phase I metabolic transformation has taken place next to the relevant arrow. Answer C H3

C H3

O N

N C H3

C H3

H

Lidocaine

C H3

HO

N

N C H3

C H3

C H3

O

C H3

N

H N

C H3

H

C H3

H

O

Monoethylglycinexylidide

C H3

HO

N C H3

O

H N

C H3

H

3-Hydroxy-monoethylglycineexylidide

C H3

C H3

N H2

N C H3

C H3 + O HO

N H2

H

Glycinexylidide

H N

O

C H3

5. Now that you have identified the metabolic transformations that generate products that have been identified, put your detective hat on and list any additional phase I transformations that could have occurred. Answer Benzylic oxidation (on either or both of the aromatic ring methyl substituents). The resulting primary alcohol can subsequently undergo alcohol oxidation to the corresponding aldehyde. The aldehyde then undergoes aldehyde oxidation to the corresponding carboxylic acid.

192

Medicinal Chemistry Self Assessment

6. Lidocaine suffers from CNS-based toxicities largely due to production of the N-dealkylated metabolic product monoethylglycinexylidide once the parent drug has crossed the blood–brain barrier. a. Provide a structural rationale for why lidocaine is able to cross the blood–brain barrier. Answer Based on the information in the structure evaluation grid, we know that lidocaine has a fair amount of hydrophobic character (aromatic hydrocarbons with aliphatic substituents, aliphatic alkane substituents on amine) that enhance absorption across lipophilic membranes. At pH=7.4, the tertiary amine will be predominantly ionized. Because an equilibrium between the ionized and unionized forms of lidocaine exists, a very small percentage of the drug will be in its unionized form at any given time. The blood–brain barrier is highly selective. To cross this membrane via passive diffusion, drugs typically must be in their unionized form and be highly lipophilic. Because of the presence of the ionizable amine, only very small amounts of lidocaine will cross the blood–brain barrier and then undergo oxidative N-dealkylation to produce an N-dealkylated metabolic byproduct—monoethylglycinexylidide, the cause of the CNS-based toxicity observed. b. Interestingly, neither 1.19 and 2.19 tocainide – remove nor boldtolycaine from labeldemonstrates similar CNS-based toxicities. Provide a structural rationale for why these two local anesthetics are devoid of CNS-based side effects.

Tolycaine

Tolcainide

Answer 1.25 and 2.25 – remove bold from From a structural perspective tolycaine is label structurally identical to lidocaine with the exception of the presence of a methyl ester instead of a benzylic methyl group. This methyl ester readily A catalyzed ester hydrolysis, a phase I metabolic transformation, to the corresponding undergoes enzyme carboxylic acid. The resulting metabolic product will be ionized at physiological pH via two B Cacid and the tertiary amine. Even though tolycaine can undergo functional groups, the carboxylic oxidative N-dealkylation to produce an N-dealkylated product that closely resembles monoethylglycinexylidide, it is highly unlikely thatDthis drug will cross the blood–brain barrier due to the presence of these two ionizable functional groups. Again, although an equilibrium exists between the ionized and unionized forms of both the carboxylic acid and the tertiary amine functional groups, there is only a very small fraction of the drug that is completely unionized at any point in time. Sitagliptin Tocainide contains several of the same functional groups found in lidocaine, but does not contain an alkylated amine. The hydrophobic character afforded by the substituted aromatic hydrocarbon and the presence of an amine-substituted aliphatic alkane will certainly contribute to the remove bold the fromblood–brain label ability of2.25 the–drug to cross barrier. The primary amine will also be predominantly ionized at pH=7.4 and, therefore, only a fraction of the time will it be available in its unionized form. These structural characteristics are similar to lidocaine so it is possible for tocainide to cross the blood–brain barrier. Unlike lidocaine, tocainide cannot undergo oxidative N-dealkylation, and the metabolic byproducts that cause CNS-based toxicity are not formed.

Section 4 Whole Molecule Drug Evaluation Answers 2.20 Montelukast and Zafirlukast

Shown below are the structures of montelukast and zafirlukast. These drug molecules are administered orally for the treatment of asthma and allergic rhinitis.

Montelukast

Zafirlukast

1. The structure of montelukast contains one acidic functional group (pKa=4.4) and one basic functional group (pKa=3.1), whereas the structure of zafirlukast only contains one acidic functional group (pKa=4.3). 1. The structure of a solution pH of 8.3. Identify these acidic and basic functional groups and predict whether they will be primarily ionized or primarilyAnswer: unionized at a stomach pH=1.9, a urine pH=5.4, a cellular pH=6.1, a plasma pH=7.2, and a solution pH=8.3.

Carboxylic acid Acidic (pKa = 4.4) Sulfonamide Acidic (pKa = 4.3)

Montelukast Aromatic heterocylic amine Basic (pKa = 3.1) 193

Zafirlukast

194

Medicinal Chemistry Self Assessment

1. The structure of a solution pH of 8.3. Answer: Answer

Carboxylic acid Acidic (pKa = 4.4) Sulfonamide Acidic (pKa = 4.3)

Montelukast Aromatic heterocylic amine Basic (pKa = 3.1) Zafirlukast

Functional Group

Acidic or Basic

Primarily Ionized or Unionized 1.9

5.4

6.1

7.2

8.3

Unionized

Ionized

Ionized

Ionized

Ionized

3. The sodium saltBasic an explanation for this difference. Aromatic heterocyclic Ionized Unionized amine

Unionized

Unionized

Unionized

Sulfonamide

Ionized

Ionized

Ionized

Carboxylic Acid

Acidic

2. In the previous question, we to calculate the percent of the functional group that would be ionized.

Acidic

Unionized

Ionized

2. In the previous question, we examined three pKa values in five different environments for a total of 15 different scenarios. Which of these 15 scenarios allow you to use the Rule of Nines to calculate the percent of ionization of the functional group in the specific environment? Identify the specific scenarios and use the Rule of Nines to calculate the percent of the functional group that would be ionized. Answer To use the Rule of Nines, the difference between the pH and the pKa must be an integer (i.e., 1, 2, 3). In evaluating the above 15 scenarios, there are three scenarios that meet this criterion, the carboxylic acid of montelukast (pKa=4.4) at a urine pH=5.4, the aromatic heterocyclic amine of montelukast (pKa=3.1) at a cellular pH=6.1, and the sulfonamide of zafirlukast (pKa=4.3) at a solution pH=8.3. For the carboxylic acid of montelukast, |pH – pKa| is equal to 1; thus, there is a 90:10 ratio. Because the carboxylic acid (pKa=4.4) would be primarily ionized in a basic environment (pH=5.4), we can use this ratio to determine that it would be 90% ionized. For the aromatic heterocyclic amine of montelukast, |pH – pKa| is equal to 3; thus, there is a 99.9:0.1 ratio. Because the aromatic heterocyclic amine (pKa=3.1) is a basic functional group, it would be primarily unionized in a basic environment (pH=6.1). We can use this information to predict that it would be 0.1% ionized. For the sulfonamide of zafirlukast, |pH – pKa| is equal to 4; therefore, there is a 99.99:0.01 ratio. Because the sulfonamide (pKa=4.3) is an acidic functional group, it would be primarily ionized in a basic environment (pH=8.3). Thus, we can use this information to predict that it would be 99.99% ionized.

2.20 Montelukast and Zafirlukast

195

3. The sodium salt of montelukast is required for its oral administration, whereas zafirlukast can be administered orally as its unionized free acid form. Conduct a structural analysis of these two drug molecules and provide an explanation for this difference. Answer A sodium salt is an inorganic salt of the parent drug. The primary purpose of using an inorganic salt is to enhance the water solubility of a drug molecule. This in turn enhances its solvation and dissolution with the aqueous environment of the gastrointestinal (GI) tract. Since montelukast must be administered as an inorganic salt, whereas zafirlukast does not have this requirement, this indicates that zafirlukast has higher water solubility than montelukast. In evaluating their structures, it is found montelukast and zafirlukast each contain three water soluble functional groups, and the remainder of their structures is comprised of alkyl chains, aromatic rings, a thioether, and a halogen. All of these latter functional groups bestow lipid solubility to their respective drug molecules. A key difference between these two structures is the overall nature of their water soluble functional groups. The sulfonamide of zafirlukast has a wider charge distribution (four atoms) than does the carboxylic acid of montelukast (two atoms; Comparison A). Additionally, the carbamate group of zafirlukast has the ability to form more hydrogen bonds with water than does the tertiary hydroxyl group of montelukast (Comparison B). The ability to act as hydrogen bond acceptors would be expected to be similar for the methoxy group of zafirlukast and the aromatic heterocyclic amine of Answer: montelukast (Comparison C). Please note that the nitrogen atom of the indole ring is an extremely weak base (pK in hydrogen bonds. This is because the lone pair of elecA sodium salt isa < an0.1) ilityand o thecannot indole participate ring. trons on the nitrogen atom are required for the aromaticity of the indole ring.

A O

O

–

B

OH S Cl

H3C

C H3

N

C

C Montelukast

H3 CO

B

O

O

O N

H N

O

A

–

C H3

S O

N

Zafirlukast

C H3

In addition, the overall lipid soluble character of montelukast is greater than that of zafirlukast. The structure of montelukast contains 34 aromatic or aliphatic carbon atoms (as well as a halogen whereas structure containsoccur only 28 The combination 4. atom), Montelukast and.the Assume that of all zafirlukast binding interactions at aatoms. physiological pH of 7.4. of a lower water soluble nature and a higher lipid soluble nature is responsible for the need to utilize a sodium salt for Answer: the oral administration of montelukast. Let us four nd hydrophobic interactions.

The four boxed areas could interact with a hydrophobic pocket within the leukotriene receptors.

Ionic bond Hydrogen bond (as acceptor)

S Cl

196

H3C

C H3

N

Medicinal Chemistry Self Assessment

C

C

A O

H3 CO

C H3

O 4. Montelukast and Montelukast zafirlukast exert their mechanism ofBaction by interacting with cysteinyl leukotriene – ). It has receptors and blocking the normal actions of endogenous leukotrienes (LTC4, LTD4, and NLTE 4 S been proposed that this interaction requires five key elements: an ionic interaction, a hydrogen bond H O O as N interaction where the antagonist acts as the acceptor, well as the interaction of the antagonist with three separate hydrophobic pockets within the receptor. Using this information and the strucO binding interactions between these drug tures of montelukast and zafirlukast, propose potential Zafirlukast N molecules and cysteinyl leukotriene receptors. Assume that all binding interactions occur at a physiC H3 ological pH=7.4.

Answer Let us evaluate these two drug molecules separately. The structure of montelukast contains a carboxacid that could participate ionic interactions bond with the leukotriene receptors. 4. ylic Montelukast and. Assume that in allan binding occur at a physiological pH It of also 7.4. contains two functional groups, the tertiary hydroxyl group and the aromatic heterocyclic amine, that could Answer: in a hydrogen bonding interaction as an acceptor. As shown below, the structure of participate montelukast contains four separate regions that could interact with the three hydrophobic pockets Let us four nd hydrophobic interactions. of the receptor via van der Waals and hydrophobic interactions.

Ionic bond

The four boxed areas could interact with a hydrophobic pocket within the leukotriene receptors.

Hydrogen bond (as acceptor)

Hydrogen bond (as acceptor)

Montelukast

Similarly, the structure of zafirlukast contains a sulfonamide group that could participate in an ionic interaction with the leukotriene receptor, as well as two functional groups, the methoxy group and the carbamate, that could participate in a hydrogen bonding interaction as an acceptor. As shown below, the structure of zafirlukast four separate regions interact with the Similarly, the structure of groupalso that contains could participate in an ionic bondthat withcould the leukotriene. three hydrophobic pockets within the receptor.

Hydrogen bond (as acceptor) The four boxed areas could interact with a hydrophobic pocket on the leukotriene receptors.

Ionic bond

Hydrogen bond (as acceptor)

Zafirlukast

5. Calculated log P values of montelukast or be primarily excreted unchanged? 6. Shown below is the to perform with zafirlukast.

2.20 Montelukast and Zafirlukast

197

5. Calculated log P values of montelukast and zafirlukast lie in the range of 5.5 to 6.4 depending on the computer program used to predict these values. Given this information, would these drug molecules be predicted to be highly plasma protein bound or minimally plasma protein bound? Additionally, would you expect these drug molecules to undergo extensive hepatic metabolism or be primarily excreted unchanged? Similarly, the structure of group that could participate in an ionic bond with the leukotriene. Answer Plasma proteins are used by the human body to transport endogenous molecules Hydrogen bondin the plasma from one cell to another. Given that the plasma is water soluble, these proteins are primarily required to (as acceptor) carry those endogenous molecules that have a high level of lipid solubility (e.g., estradiol, cholesThe four boxedThis areas could interact with a terol, hydrocortisone). is also true for exogenously administered drug molecules. Drug molecules hydrophobic pocket on the leukotriene receptors. that have a more lipid soluble character have a greater affinity for plasma proteins than do those that have a more water soluble character. Thus, using the calculated log P values provided for montelukast and zafirlukast, it would be expected that these two drug molecules would be highly plasma protein bound. The primary purpose of drug metabolism is to enhance the removal of the drug molecule from the Ionic and bondthus can be readily human body. Drug molecules that already possess adequate water solubility, eliminated, generally undergo minimal or no metabolism, whereas drug molecules that are highly lipid soluble often undergo extensive metabolic transformation. Thus, using the calculated log P values provided for montelukast and zafirlukast, it would be expected that these two drug molecules Zafirlukast would undergo extensive hepaticHydrogen metabolism. bond (as acceptor) 6. Shown below is the structure of zafirlukast and a list of five metabolic transformations. For each metabolic transformation, if it is aorphase I or a phase II transformation 5. Calculated log P valuesindicate of montelukast be primarily excreted unchanged? and if zafirlukast has a functional group present to participate in the transformation. If you answer YES, then draw the appropriate metabolite; if you answer NO, then provide a brief explanation as to why this metabolic transformation is not possible to perform with zafirlukast. 6. Shown below is the to perform with zafirlukast. Metabolic Pathways A. Methylation B. Aromatic Oxidation C. Hydrolysis D. Oxidative O-Dealkylation

ReplacementStructuresforBatch3 E. Benzylic Oxidation

Chapter2.20 Answer Pleasereplacethestructureforanswer6B(page5)withtheoneshownbelow.

 

A. Methylation: No. Methylation is a phase II metabolic transformation that requires a catechol, a phenol, an amine, or a sulfhydryl functional group. Because none of these functional groups are Answer: present within the structure of zafirlukast, this metabolic transformation cannot occur. B. Aromatic Oxidation: Phase I; Yes

A. Methylation: No., this metabolic transformation cannot occur. B. Aromatic Oxidation: Phase I; Yes

Note: Aromaric hydroxylation could also occur on these rings; however, they are more sterically hindered. 



Note: Aromaric hydroxylation could also occur on this ring; however, it is more

  198

Medicinal Chemistry Self Assessment

Pleasereplacethestructureforanswer6C(page6)withtheoneshownbelow. 

C. Hydrolysis: Phase I; Yes O

H3 CO OH

N H

H N

HO

O

C H3

S O

Pleasereplacethestructureforanswer6D(page6)withtheoneshownbelow. O



Note: It is also possible to hydrolyze the sulfonamide.

N C H3

Pleasereplacethestructureforanswer6D(page6)withtheoneshownbelow. O



Phase I;  D. Oxidative O-Dealkylation: O Yes HO H

H



O H N

O O

H H N

O

 HO

H

C H3

C H3

S OO N H

N

O

C H3

O

Oxidative O-dealkylation of methoxy functional group Oxidative O-dealkylation of methoxy functional group

S O

N C H3

O

O

H3 CO

O H HO N O

H3 CO H N

HO

N C H3

O



N H

O



O N H

C H3

S OO N H

O S O

C H3

Oxidative O-dealkylation of carbamate functional group Oxidative O-dealkylation of carbamate functional group



N C H3



Pleasereplacethestructureforanswer6E(page6)withtheoneshownbelow. 





E. Benzylic Oxidation: Phase I; Yes

Pleasereplacethestructureforanswer6E(page6)withtheoneshownbelow. 

Benzylic oxidation can occur at two sites; however, the one in the middle of the molecule is more sterically hindered.





Benzylic oxidation can occur at two sites; however, the one in the middle of the molecule is more sterically hindered.









Section 4 Whole Molecule Drug Evaluation Answers 2.21 Phenobarbital and Other Barbiturates Shown below are the structures of butabarbital, secobarbital, and phenobarbital. These drug molecules are used as 2.21 Phenobarbital and Other Barbiturates sedative-hypnotic agents in the treatment of insomnia and to induce sedation prior to surgical procedures. Phenobarbital can be used in the treatment of a number of different seizure disorders. Their sedative properties are due to their ability to interact thestructures GABAA receptor Shown below with are the of buta in the central nervous system (CNS).

O and OtherOBarbiturates 2.21 Phenobarbital HN

NH

O

Butabarbital O

O

HN

Shown below are the structures of buta

O

O

O

O

NH

O

HN

NH

O

O

Secobarbital

Phenobarbital

O

O

O

1. Conduct a structural previously listed. 1. Conduct aHstructural as to why they can be N N H analysis of these drug H N molecules N H and provide an explanation HN NH Answer: administered orally for the indications previously listed. These three drug to their site of action. O O O Answer These three drug molecules vary only in the alkyl chains and aromatic rings located at the 5 position of barbiButabarbital Secobarbital Phenobarbital turic acid. As shown below, barbituric acid is symmetrical two identical imide functional groups (a Only and site contains of variationranges from 4.5–8.5, imides tend to have values type of β-dicarbonyl). While the pKa range for a β-dicarbonyl at the upper end of this range. As such, they can be ionized in the small intestine to a limited extent and 1. Conduct a structural previously listed. participate in hydrogen bonds with water when present in the unionized state. This provides sufficient water Answer: solubility for solvation and dissolution. The alkyl chains and aromatic ring provide sufficient lipid solubility to permit these drug molecules to cross the gastrointestinal 5mucosal membrane and be orally absorbed. This lipid These three drug to their site of action. solubility facilitates absorption across the blood–brain barrier to gain access to their site of action. Imide Imide Only 1site of 3 variation

Barbituric Acid 5

2. Butabarbital and secobarbital these two administered as their sodium salts. Imide Imide 1 3 Answer:

Barbituric Acid 199

2. Butabarbital and secobarbital these two administered as their sodium salts.

5

Imide 200

Medicinal Chemistry Self Assessment

1

3

Imide

Barbituric Acidsalts and phenobarbital is marketed in its 2. Butabarbital and secobarbital are marketed as their sodium free acid form (i.e., non-salt form). Draw the sodium salt of either butabarbital or secobarbital and provide an explanation as to why these two drug molecules need to be administered as their sodium salts. 2. Butabarbital and secobarbital these two administered as their sodium salts. AnswerAnswer:

Sodium Butabarbital

Sodium Secobarbital

Phenobarbital (Used as Free Acid)

In general the primary reason that drug molecules are administered in their inorganic salt form is to 3. Using the information in acting agent solubility. (onset of action = 30–60and minutes; durationneed = 10–16 hours). enhance solvation, dissolution, and water Butabarbital secobarbital to be administered as their sodium salts, whereas phenobarbital does not. This suggests that butabarbital and secobarbital have greater lipid solubilities than phenobarbital and therefore need to be administered in a more water soluble form. The authors do not expect the readers to know the structure-activity relationship (SAR) of barbiturates, so the following information is offered to show how the logical application of key concepts often explains known facts. Structural activity studies of barbiturates have demonstrated that alkyl chains at the 5 position of barbituric acid produce more lipid solubility than the aromatic ring in the structure of phenobarbital. Phenobarbital has a log P value=1.46, while butabarbital and secobarbital have log P values=1.60 and 2.36, respectively. 3. Using the information in questions 1 and 2, and your answers to those questions, provide an explanation as to why secobarbital is a short-acting agent (onset of action = 10–15 minutes; duration = 3–4 hours), butabarbital is an intermediate-acting agent (onset of action = 45–60 minutes; duration = 6–8 hours), and phenobarbital is a long-acting agent (onset of action = 30–60 minutes; duration = 10–16 hours). Answer As discussed in the previous two questions, these three drug molecules are structurally similar and vary only in the hydrocarbon chains at position 5 of the barbituric acid ring. These minor structural variations result in differences in their respective lipid solubilities. Drug molecules that have higher lipid solubility (log P) traverse more quickly through lipid membranes, but are more likely to undergo metabolic transformation than drug molecules that have lower lipid solubility. Secobarbital is the most lipid soluble of the three drug molecules. As such, it is rapidly absorbed, can rapidly cross the blood–brain barrier, and has the quickest onset of action. Due to its higher lipid solubility, it is metabolized quicker than either butabarbital or phenobarbital. This rapid metabolism is responsible for the relatively short duration of action. Butabarbital and phenobarbital are less lipid soluble than secobarbital and have similar onsets of action; however, phenobarbital has a lower lipid solubility, is metabolized at a slower rate, and thus has a longer duration of action than butabarbital.

2.21 Phenobarbital and Other Barbiturates

201

4. Secobarbital contains an ionizable functional group with a pKa=7.9. Using the table shown below, identify the functional group, provide the normal pKa range for the functional group, and identify if 4. the Secobarbital primarily ionized or unionized at pH of 1.5, 5.9, 6.3, 7.4, and5.9, 8.9.6.3, 7.4, functionalbegroup would be primarily ionized or environments unionized at pH environments=1.5, and 8.9.

Imide (Acidic) Secobarbital Answer 5. As stated above between secobarbital and the side chains of the amino acids indicated.

Answer:

Functional Group

Acidic or Basic

Primarily Ionized or Unionized

pKa Range

1.5

5.9

One that involve side chains4.5-8.5 of valine and phenylalanine. β-Dicarbonyl (Imide) the Acidic Unionized Unionized

6.3

7.4

8.9

Unionized

Unionized

Ionized


R10

N

N

O

H N

O

+

H3 N

Ionic Bond

H

R7

Lys

Hydrogen Bond O R9

Ser

O

HN R8

One that involve the side chains of valine and phenylalanine.
202

Medicinal Chemistry Self Assessment

Phe

O

R4

van der Waals Hydrophobic Interaction

NH R3

Asn

O R1

Dipole-Dipole Interaction

R2 R5

N H O

N H2

H

R10

O

O N

N O

H N

O

HN

–

van der Waals Hydrophobic Interaction

O

Val

R6

+

H3 N

Ionic Bond

H

R7

Lys

Hydrogen Bond O R9

Ser

O

HN R8

Chapters1.21and2.21

6. Shown below are known metabolites of butabarbital, secobarbital, and phenobarbital. Identify the metabolic transformations that are required to produce each metabolite and indicate if they are PleasereplacetheindicatedstructureinQuestion6inboth1.21and2.21(theonethatispartofthequestion) phase I or phase II transformations. Would you expect any of these metabolites to retain their withtheonebelow.Bothstructureshadanidenticalerror. pharmacological activity? Why or why not?



Metabolite of butabarbital

 

Metabolite of secobarbital

Metabolite of phenobarbital



Answer • Butabarbarbital: ω oxidation (phase I) followed by glucuronic acid conjugation (phase II) •

Secobarbital: oxidation of the double bond (water reacts with the initial epoxide to produce a diol) and ω-1 oxidation; the order of these phase I transformations is unimportant Chapter2.22 •

Phenobarbital: aromatic oxidation (hydroxylation, phase I) followed by sulfate conjugation

(phase II) PleasereplacethestructurefortheanswertoQuestion3inChapter2.22withtheonebelow.

 Intramolecular Hydrogen Bond

2.21 Phenobarbital and Other Barbiturates

203

These metabolites would not be expected to retain the pharmacological activity that was present in the parent drug molecules. For any drug molecule to produce a pharmacological effect, it needs to contribute complementary interactions with its biological target. The alkyl chains and aromatic ring are important for forming van der Waals and hydrophobic interactions with hydrophobic amino acids within the GABAA receptor. The addition of hydrophilic functional groups at this position decreases their ability to form these key interactions. Furthermore, the addition of these hydrophilic functional groups in the liver would greatly decrease or eliminate the ability of these metabolites to cross the blood–brain barrier and reach their site of action in the CNS.

Section 4 Whole Molecule Drug Evaluation Answers 2.22 Pravastatin and Fluvastatin 2.22PravastatinandFluvastatin

ShownShown belowbelow are theare structures of pravastatin and fluvastatin. drug molecules used in boxed. the treatment of the structures of pravastatin. A total ofThese six functional groups are have been various types of hyperlipidemia/dyslipidemias. A total of six functional groups have been boxed. F

C

B

D

A

E Pravastatin

Fluvastatin

1. Using the table below, drophilic or hydrophobic in character. for your response. logbelow, P values of s and a structural explanation for the difference in these log P values. 1. Using2.theThe table identify theprovide six boxed functional groups. For each of the functional groups you identify, indicate if it is hydrophilic or hydrophobic in character. Also provide a brief explanation for your Answer response. The log P value is a functional groups than the structure of fluvastatin, it has a lower log P value. Answer Functional Group Name

Hydrophilic or Hydrophobic 3',5'-Dihydroxy-

Hydrophobic due to its inability to ionizeacid or form hydrogen bonds; hydrocarbon functional heptanoic groups enhance lipid solubility

A

Alkyl group; alkyl chain; aliphatic chain

B

Ester

C

Secondary hydroxyl (secondary alcohol)

Hydrophilic due to its ability to form hydrogen bonds as either a donor or an acceptor

D

Halogen; fluorine

Effects can vary; fluorine can act as a hydrogen bond acceptor; however, studies have shown that the substitution of a hydrogen atom with a fluorine atom tends to slightly enhance lipid solubility

E

Aromatic ring system; heterocyclic aromatic ring; indole

Hydrophobic due to its inability to ionize or form hydrogen bonds; the lone pair of electrons present on the nitrogen atom are involved in the aromaticity of the ring system and therefore are not available to form hydrogen bonds; the remaining hydrocarbons Isopropyl chain comprising the ring system enhance lipid solubility

F

Ester

Isobutyl chain

Secondary Carboxylic acid Hydroxyl

Hydrophilic due to its ability to act as a hydrogen bond acceptor

p-Fluoro-

Hydrophilic due to its ability to ionize (form an ion–dipole interaction with water) and to phenyl ring participate in hydrogen bonding (acceptor and donor) in its unionized formIndole ring

Pravastatin

Decalin ring

205

Fluvastatin

2.22PravastatinandFluvastatin 206

Medicinal Chemistry Self Assessment

Shown below are the structures of pravastatin. A total of six functional groups have been boxed. 2. The log P values of pravastatin and fluvastatin are 1.44 and 3.62, respectively. Conduct a structural F analysis of these drug molecules and provide a structural explanation for the difference in these log P C values. Answer The log P valueBis a logarithmic expression of the ratio of a drug molecule’s solubility in a lipid environment when compared to an aqueous environment and assumes that all ionizable functional D groups are present in their unionized form. The log P value of any given drug molecule is the result of A the additive contributions and interrelationships of all of its functional groups. In evaluating the structures of pravastatin and fluvastatin, both drug molecules contain a 3,5-dihydroxyheptanoic acid group and differ in the ring systems attached to this group. The carboxylic acid and the two secondary hydroxyl groups contribute to the water solubility of these drug molecules; however, it E would be expected that these contributions would be similar for both drug molecules. In evaluating the different ring systems and their substituents, it can be seen that the structure of pravastatin Pravastatin contains an additional secondary hydroxyl group, as well as two ester oxygen atoms. These functional Fluvastatin groups are capable of forming additional hydrogen bonds and thus contribute to the overall water solubility of pravastatin. The decalin ring and the isobutyl side chain contribute to the overall lipid pravastatin. In contrast, the remaining functional of fluvastatin—aromatic rings, 1. solubility Using the of table below, drophilic or hydrophobic in character. forgroups your response. an aliphatic chain, and a halogen—contribute to its overall lipid solubility. It should be noted that 2. the The lone log Ppair values of s and provide structural the difference these log P values. of electrons presenta within theexplanation indole ringfor are required forinthe aromaticity of the ring and therefore are not available to form hydrogen bonds or accept protons. Because the structure Answer of pravastatin contains more hydrophilic functional groups than the structure of fluvastatin, it has a The log P value is a functional groups than the structure of fluvastatin, it has a lower log P value. lower log P value.

3',5'-Dihydroxyheptanoic acid Ester Isobutyl chain

Secondary Hydroxyl

Isopropyl chain

p-Fluorophenyl ring Pravastatin

Decalin ring

Indole ring Fluvastatin

3. The normal pKa range for carboxylic acids is 2.5 to 5. The pKa values for the carboxylic acids present within the structures of pravastatin and fluvastatin are 4.21 and 4.56. Conduct a structural analysis and provide a reason why these pKa values are at the high end of the normal range. Answer The presence or absence of specific functional groups can affect the acidity or basicity of ionizable functional groups. In the case of pravastatin and fluvastatin, as well as other drug molecules within this chemical/pharmacological class, the 3-hydroxyl group affects the acidity of the carboxylic acid. In general, non-aromatic hydroxyl groups are electron withdrawing due to the high electronegativity of the oxygen atom; however, in the case of pravastatin and fluvastatin, the position of the 3-hydroxyl group orients it to form intramolecular hydrogen bonds with the carboxylic acid. These hydrogen bonds decrease the ability of the proton of the carboxylic acid to dissociate, therefore decreasing its acidity and increasing its pKa value.

Chapter2.22 PleasereplacethestructurefortheanswertoQuestion3inChapter2.22withtheonebelow. 2.22 Pravastatin and Fluvastatin

207

 3. The normal pKa range for at the high end of the normal range. Answer The presence or absence of its pKa value.

Intramolecular Hydrogen Bond

Intramolecular Hydrogen Bond

H

Intramolecular Hydrogen Bone

H

H O

H3C



HO

O

O COOH OH

OH H O

H O

O

O H3C

O

Intramolecular Hydrogen Bone

H F

H Pravastatin C H3

N Fluvastatin





4. Pravastatin and fluvastatin exert their hyperlipidemic effects by inhibiting the enzyme HMG CoA Pravastatin reductase. As shown below, HMG CoA reductase converts 3-hydroxy-3-methylglutaryl CoA (HMG Fluvastatin CoA) to mevalonic acid. This conversion is required for the synthesis of cholesterol and acts as a primary control site for production of this endogenous steroid. Using the structures of HMG CoA, acid, and fluvastatin, provide an explanation to how 4. mevalonic Pravastatin andpravastatin, fluvastatin fluvastatin, provide an d fluvastain inhibitas HMG CoApravastatin reductase. and fluvastatin inhibit HMG CoA reductase.

HMG CoA Reductase

HMG CoA

Mevalonic Acid

Answer In the reaction catalyzed by HMG CoA reductase, the thioester of HMG CoA is reduced to a primary hydroxyl group. The 3,5-dihydoxyheptanoic acid found within the structures of pravastatin and fluvastatin very nicely mimic the structure of mevalonic acid, the product of this reaction. Unlike the natural substrate, pravastatin and fluvastatin cannot be reduced. Because they bear structural similarity to both the substrate and the product, they can interact with the enzyme in a similar manner as the substrate; however, because they lack the functional group that is normally reduced, they are not transformed by the enzyme and act as enzyme inhibitors. The respective ring systems of pravastatin and fluvastatin most likely occupy binding sites that are adjacent to the active site of the enzyme.

208

Medicinal Chemistry Self Assessment Answer Answer In the reaction most likely occupy binding sites that are adjacent to the active site of the enzyme. In the reaction most likely occupy binding sites that are adjacent to the active site of the enzyme.

HMG CoA HMG CoA Reductase Reductase

Thioester Thioester

Mevalonic Acid Mevalonic Acid

HMG CoA HMG CoA

HO 3 HO O

O

H3 C O H H3 C O H3 C H H3C HO HO

COOH 3 COOH 5 OH 5 OH H H

Pravastatin Pravastatin

Primary Primary Hydroxyl Hydroxyl

C H3 C H3

HO 3 HO

F

F

COOH COOH 5 OH 5 OH H H 3

N

N

Fluvastatin Fluvastatin

5. Shown below are the structures of fluvastatin and a conformationally restricted analog. The addition of an additional carbon atom and the conformational restriction essentially abolish the therapeutic 5. Shown below are the structures why this structural change results in a loss in activity. activity of fluvastatin. these structures, postulatechange a reason why this change results in 5. Shown below areUsing the structures why this structural results in a structural loss in activity. a loss in activity.

Fluvastatin Fluvastatin

Conformationally Conformationally Restricted Analog Analog ofRestricted Fluvastatin of Fluvastatin

Answer The addition of a functional group will affect the overall electronic distribution, water/lipid solubility balance, and steric dimensions of the parent drug molecule. The electronic effects would be expected to be minimal due to the low electronegativity values of carbon and hydrogen. The additional carbon atom will increase lipid solubility and this could affect the ability of the analog to dissolve; however, the use of a sodium salt (similar to what is done with fluvastatin) should still allow for adequate dissolution. This then leaves the steric effect as the most probable cause for the loss

2.22 Pravastatin and Fluvastatin

209

of activity with this analog. There are two aspects to consider. First, the addition of an extra carbon atom may not be permitted due to the steric dimensions of the target enzyme, HMG CoA reductase. The simple addition of a methyl group (or carbon atom) can significantly alter the interaction of a drug molecule with its biological target. Although this may be the cause for the loss of activity, a much bigger alteration in the steric dimension of fluvastatin has been introduced with this conformational restriction. In evaluating the structure of fluvastatin, it is found that there is free rotation about the bond that connects the indole ring to the para-fluoro aromatic ring. This allows flexibility and the opportunity for these two rings to be oriented in a manner that allows optimal interactions with HMG CoA reductase. In contrast, the conformational restriction introduced in the fluvastatin analog creates a large, planar, tetracycline ring system with no flexibility. The ability of this analog to interact with HMG CoA reductase would therefore be dependent on the availability of a complementary large, flat, hydrophobic area within the active site of the enzyme. Receptor binding studies have shown that the para-fluoro aromatic ring of fluvastatin cannot be coplanar with the indole ring and that conformational restriction, such as that shown in this question, abolishes therapeutic activity.

6. Pravastatin is primarily metabolite inactive. 6. Pravastatin is primarily metabolized to itsis3α epimer. This metabolite is completely inactive as an HMG CoA reductase inhibitor. Identify the metabolic transformations required to produce this metabolite and provide an explanation as to why this metabolite is inactive.

3D epimer

Answer The epimer results from the oxidation of the 3β hydroxyl group by alcohol dehydrogenase followed by reduction to the 3α epimer. The loss of activity results from the fact that the compound no longer mimics HMG CoA or mevalonic acid. The epimeric hydroxyl group is oriented in the opposite direction and either fails to form a crucial hydrogen bond or sterically inhibits interaction with the active site of HMG CoA reductase.

Section 4 Whole Molecule Drug Evaluation Answers 2.23 Quinapril 2.23 Quinapril

Shown belowbelow is the is structure of quinapril. It is an enzyme (ACE) inhibitor that is used in the Shown the structure of quinapril. It angiotensin is an groupsconverting are identified. treatment of hypertension and heart failure. Five functional groups are identified. B H3C O

A

O

D O

H N

C

N C H3

O

OH

E 1. Using the table below, identify the five boxed functional groups. For each of the functional groups you 1. Using the table below, identifyor thehydrophobic five explanation for your response. identified, indicate if it is hydrophilic in character. Also provide a brief explanation for your response. 2. Using the h functional ionized or unionized at pH environments of 1.5, 4.8, 6.3, 7.4, and 8.1. Answer Functional Group Name

Hydrophilic or Hydrophobic

A

Phenyl ring; aromatic ring; aromatic hydrocarbon

Hydrophobic due to its inability to ionize or form hydrogen bonds; hydrocarbon functional groups enhance lipid solubility.

B

Ethyl ester

Contains both hydrophilic and hydrophobic properties. The two oxygen atoms can act as hydrogen bond acceptors and can contribute to water solubility. The ethyl group can neither ionize nor form hydrogen bonds and thus contributes to lipid solubility.

C

Secondary amine

Hydrophilic due to its ability to ionize (form an ion–dipole interaction with water) and to participate in hydrogen bonding (acceptor and donor) in its unionized form.

D

Amide

E

Carboxylic acid

Secondary amine

Hydrophilic due to its ability to form hydrogen bonds as a hydrogen Carboxylic acid bond acceptor. (Note: Does not participate as a hydrogen bond donor due to the lack of hydrogen atom.) Hydrophilic due to its ability to ionize (form an ion–dipole interaction with water) and to participate in hydrogen bonding (acceptor and donor) in its unionized form.

3. Quinapril is a is administered orally instead of quinaprilat. 2. Using the unmodified structure of quinapril and the following table, identify all of the acidic and basic functional groups present in the structure, provide the normal pKa range for each functional group, and identify if each functional group would be primarily ionized or unionized at pH environments=1.5, 4.8, 6.3, 7.4, and 8.1.

211

E

H3C 212

1.Medicinal Using the table below, identify theOfive explanation for your response. Chemistry Self Assessment O D 2. Using the h functional ionized or unionized atOpH environments of 1.5, 4.8, 6.3, 7.4, and 8.1. A H N

C

N C H3

O

OH

E 1. Using the table below, identify the five explanation for your response. 2. Using the h functional ionized or unionized at pH environments of 1.5, 4.8, 6.3, 7.4, and 8.1.

Secondary amine

Functional Group

Acidic or Basic

Carboxylic acid

Primarily Ionized or Unionized pKa Range

1.5

4.8

6.3

7.4

Secondary amine

Basic

9–11

Ionized

Ionized

Ionized

Ionized

Ionized

Carboxylic acid

Acidic

2.5–5

Unionized

Ionized*

Ionized

Ionized

Ionized

3. Quinapril is a is administered orally instead of quinaprilat.

8.1

*There is a possibility that this functional group could be > 50% unionized at a pH=4.8; however, because the pKa values of carboxylic acids present in the structures of other drug molecules within Secondary amine class tend to range from 2.3 to 3.5, Carboxylic this chemical/pharmacological this functional acid group will most likely be primarily ionized at this pH value. 3. Quinapril is a prodrug. It is administered as an oral tablet and converted in vivo to its active metabo3. quinaprilat. Quinapril is Identify a is administered orally instead of quinaprilat. lite, the metabolic pathway that converts quinapril to quinaprilat, and offer a reason why quinapril is administered orally instead of quinaprilat. Quinapril Quinaprilat

Quinapril

Quinaprilat

Answer Ester hydrolysis converts quinapril to quinaprilat. Esterases are ubiquitous within the human body and can readily convert ester prodrugs to their active forms. In evaluating the structures of quinapril and quinaprilat, the only difference in these two molecules is the presence of a second carboxylic acid in quinaprilat instead of the ethyl ester present in quinapril. Both of these molecules possess a sufficient number of water soluble functional groups to allow for their dissolution within the aqueous content of the GI tract. Quinaprilat has three ionizable functional groups (two carboxylic acids and a secondary amine) along with an amide that is capable of forming hydrogen bonds. This greatly enhances the water solubility of quinaprilat to the extent that it has difficulty traversing the GI membrane despite the presence of hydrophobic rings and alkyl chains. To optimize the water/lipid balance, an ester prodrug is used. The ethyl ester is more hydrophobic than the carboxylic acid and in combination with the other hydrophobic groups allows for better passage across the GI membrane.

 

2.23 Quinapril

213

Chapters1.23and2.23 4. Quinapril inhibits ACE. This enzyme is a relatively nonspecific dipeptidyl carboxypeptidase. It is a zinc

PleasereplacethestructuresforAngiotensinIandQuinaprilatinQuestion4forboth1.23and2.23withtheone protease that converts angiotensin I, a decapeptide, to angiotensin II, an octapeptide. The peptide catalyzed the zinc atom andare is shown below. Quinapril, providedbelow. 4. cleavage Quinapril isinhibits ACE.by binding interactions occurring at a pH of 7.4. along with other ACE inhibi

tors, is a tripeptide mimic that can interact with the enzyme resulting in enzyme inhibition rather than hydrolysis. Using this information and the structures provided below, identify how quinapril can interact with ACE. Assume that all drug binding interactions are occurring at a pH=7.4.

Leu Leu

Phe Phe



His His

Quinaprilat Quinaprilat

Angiotensin I R = Asp-Arg-Val-Tyr-Ile-His Angiotensin I R = Asp-Arg-Val-Tyr-Ile-His-Pro





Answer Answer

D B C

E

A Because ACE is a nonspecific dipeptidyl carboxypeptidase, it is able to hydrolyze a dipeptide sequence from the carboxylic acid end of a peptide. For this enzyme to be able to cleave a dipeptide sequence (i.e., two amino acids), substrates must have a minimum length of three amino acids. Quinaprilat and other ACE inhibitors act as tripeptide mimics. These drug molecules retain key peptide features (i.e., peptide bonds, side chains similar to those found on amino acids), but are not able to be hydrolyzed. Thus, they are able to interact in a similar manner as the substrate does. The table below lists five key interactions of quinaprilat with ACE. Drug Binding Interaction

Explanation

A

Ionic interaction

This carboxylic acid mimics the C-terminal carboxylic acid of an ACE substrate and will participate in an ionic interaction with the side chain of either Asp or Glu.

B

van der Waals and/or hydrophobic binding with hydrophobic amino acids (e.g., Phe, Tyr, Leu, Ile)

Because ACE is a nonspecific carboxypeptidase, quinapril does not need to exactly mimic Leu, the C-terminal amino acid of angiotensin I. Similar to the side chain of Leu, this ring system is hydrophobic and can participate in similar types of interactions as Leu.

214

Medicinal Chemistry Self Assessment

Continued from previous page. C

Hydrogen bond acceptor

Because ACE is a dipeptidyl carboxypeptidase, it does not cleave the terminal peptide bond. Instead, the terminal peptide bond can participate in hydrogen bonds. The carbonyl of the amide bond can act as a hydrogen bond acceptor and interact with Ser, Thr, Tyr, Trp, Gln, Asn, or Cys.

D

Ionic bond (metal complexation)

This carboxylic acid would be located very close to the zinc atom shown participating in the mechanism of angiotensin I metabolism. As such, at pH=7 it can form an ionic bond with the zinc atom.

E

van der Waals and/or hydrophobic binding with hydrophobic amino acids (e.g., Phe, Tyr, Trp)

This aromatic ring mimics the Phe side chain of angiotensin I and can participate in similar interactions.

5. Shown below are in these pathways. 5. Shown below are four possible metabolic pathways for quinapril. Identify the metabolic transformaAnswer tions involved in these pathways.

B

A

Quinapril C

D

Answer • Pathway A: Aromatic oxidation followed by sulfate conjugation of the resulting phenol • Pathway B: Ester hydrolysis • .Pathway C: Amino acid conjugation (with glycine) • Pathway D: Benzylic oxidation followed by oxidation of the resulting secondary alcohol

2.23 Quinapril

215

6. Although it is possible for quinapril to undergo all of the above metabolic transformations, pathway B is the major pathway. Other metabolites have been identified, but only at very low levels. Provide an explanation for this finding. Answer Remember that the major purpose of drug metabolism is to enhance the removal of the drug from the body, and the number of metabolic transformations required to achieve this goal varies from drug molecule to drug molecule. In the case of quinapril, ester hydrolysis can occur quickly and at many locations within the body. The resulting metabolite, quinaprilat, contains three ionizable functional groups and is easily eliminated from the body without the need for further metabolism.

Section 4 Whole Molecule Drug Evaluation Answers 2.24 Rivastigmine 2.24Rivastigmine Shown below is the structure of rivastigmine. Four of its functional groups have been identified.

Shown below is the structure of rivastigmine. Four of its functional groups have been identified. C H3

A

H3C

C O

N

CH3 D C H3 N C H3

O

B

Rivastigmine

tableidentify the four boxed functional groups. For each of the functional groups you 1. Using1.theUsing tablethe below, identified, indicate if it induction. is hydrophilic or hydrophobic in character. Also provide a brief explanation for 2. Based on their your response. 3. Rivastigmine is an acetylcholinesterase with acetylcholine. Answer Functional Group Name

Hydrophilic or Hydrophobic

A

Alkyl chain; alkyl group; aliphatic chain

Hydrophobic due to its inability to ionize or form hydrogen bonds; hydrocarbon functional groups enhance lipid solubility

B

Carbamate

Hydrophilic due to its ability to act as a hydrogen bond acceptor

C

Phenyl ring; aromatic ring; aromatic hydrocarbon

Hydrophobic due to its inability to ionize or form hydrogen bonds; hydrocarbon functional groups enhance lipid solubility

D

Tertiary amine

Hydrophilic due to its ability to ionize (form an ion–dipole interaction with water) and to participate in hydrogen bonding (acceptor) in its unionized form

Labile ester bond

Rivastigmine 2. Based on their electronic properties AND their relative positions in the molecule, identify if functional groups A and B are electron withdrawing or electron donating. Additionally, identify if this effect is due to resonance or induction. Answer Acetylcholine bound to active site of acetylcholinesterase • Functional group A is an alkyl (ethyl) group and has a lower electronegativity than the nitrogen atom of the carbamate; therefore, it acts as an electron donating group through induction. The same is true for the twoA.methyl groups attached analysis to the tertiary Conduct a structural of howamine. it could interact with acetylcholinestase. •

Functional group B is a carbamate group that is directly attached to an aromatic ring. Based onsite. its Answer:al groups to orient themselves in a similar manner within the enzyme binding position, it can donate electrons into the aromatic ring through resonance.

B

A

A

217

Key structural similarities

B

A

3

N

2.24Rivastigmine 218 Medicinal Chemistry Self Assessment

3

C H3

O

B Shown below is the structure of rivastigmine. FourRivastigmine of its functional groups have been identified.

3. Rivastigmine is an acetylcholinesterase inhibitor. Inhibition of this enzyme prolongs the half-life of CH3 D C H3 1.acetylcholine, Using the table allowing for higher concentrations at muscarinic and nicotinic recepC of acetylcholine tors. This action is useful in the treatment of Alzheimer’s disease, myasthenia gravis, and glaucoma. C H3 O N A H3C 2. Based on their induction. Shown below is the structure of acetylcholine interacting Nwith the active site of acetylcholinesterase. 3.Glutamic Rivastigmine is an acetylcholinesterase with acetylcholine. acid, histidine, and serine are involved in the mechanism of ester hydrolysis and can form C H3 O specific interactions with acetylcholine. B Labile ester bond Rivastigmine

1. Using the table 2. Based on their induction. 3. Rivastigmine is an acetylcholinesterase with acetylcholine.

Labile ester bond Rivastigmine

Acetylcholine bound to active site of acetylcholinesterase a. Conduct structural analysis of rivastigmine indicate how it acetylcholinestase. could interact with acetylcholinA. aConduct a structural analysis of how and it could interact with esterase. Answer:al groups to orient themselves in a similar manner within the enzyme binding site. Rivastigmine Answer The structure of rivastigmine contains a tertiary amine that will be primarily ionized at a physiB amine mimics the quaternary nitrogen present ological pH=7.4. This ionized within the structure B A the glutamic acid or with the side chain of Acetylcholine acetylcholinebound and can form site an ionic interaction with to active A of another acidic amino acid (comparison A). The carbamate present within the structure of of acetylcholinesterase rivastigmine is isosteric to the acetate ester of acetylcholine and thus can form similar hydrogen bonds (comparison B). The distance between the tertiary amine and the carbamate of rivastigA. Conduct a structural analysis of how it could interact with acetylcholinestase. mine is two carbon atoms longer than the quaternary nitrogen and the acetate of acetylcholine; Answer:al groups to orient themselves in astructure similar manner within the allows enzymethese binding site. however, conformational rotation within the of rivastigmine functional Key structural similarities groups to orient themselves in a similar manner within the enzyme binding site.

B A

A

Key structural similarities Similar distances after conformational rotation (Note: The length of the lines are identical)

Similar distances after conformational rotation (Note: The length of the lines are identical)

B

2.24 Rivastigmine

219

b. The serine residue of acetylcholinesterase attacks the ester bond of acetylcholine causing hydrolysis and acetylation of the serine. A similar reaction occurs between the carbamate group of rivastigmine and the serine residue; however, while acetylcholine is a substrate of acetylcholinB. The serine residue acetylcholinesterase. esterase, rivastigmine is aninhibits inhibitor. Compare the functional groups involved and provide an explanation as to how rivastigmine inhibits acetylcholinesterase.

B. The serine residue inhibits acetylcholinesterase. Intermediate in the hydrolysis of acetylcholine

Analogous intermediate formed with rivastigmine

4. Answer Neostigmine is a structural analog in routes of administration. In both cases, acetylcholinesterase catalyzes the hydrolysis of a functional group. As part of this mechanism, serine makes a bond that is similar to that which was hydrolyzed. In the case of acetylcholine, serine is esterified (i.e., an ester is added to the serine side chain). Esters can be very rapidly hydrolyzed; therefore, acetylcholinesterase is very quickly regenerated and acetylcholine acts as a substrate. In the case of rivastigmine, serine is carbamylated (i.e., a carbamate group is added to the serine side chain). Unlike esters, carbamates are hydrolyzed much slower; thus, the intermediate shown above persists for much longer than the millisecond that it takes to hydrolyze an ester. As a result, the carbamate + portion of rivastigmine remains attached to acetylcholinesterase for a longer time resulting in inactivation of the enzyme. Neostigmine Bromide 4. Neostigmine is a structural analog of rivastigmine. Similar to rivastigmine, neostigmine is able to interact with acetylcholinesterase and cause inhibition due to the presence of a carbamate functional 5. group. Rivastigmine has aindifference results durationand of action. Intermediate the hydrolysis of in a longer The primary differences between rivastigmine neostigmine are their routes of adminAnalogousorally intermediate formed withof acetylcholine istration and their therapeutic indications. Rivastigmine is administered in the treatment rivastigmine Alzheimer’s disease, whereas neostigmine is administered intramuscularly (IM) or subcutaneously Methyl of myasthenia gravis. Conduct structural analyses of these drug molecules and (SC) in the treatment Methyl 4. provide Neostigmine is a structural routes of administration. an explanation foranalog these in differences in routes of administration and therapeutic indications.

Methyl Neostigmine

Ethyl +

Rivastigmine

+ Neostigmine Bromide

5. Rivastigmine has a difference results in a longer duration of action.

220

Intermediate in the hydrolysis of Medicinal Chemistry Self Assessment

Analogous intermediate formed with rivastigmine

acetylcholine

Answer 4. Neostigmine is a structural analog in routes of administration. Unlike rivastigmine, the structure of neostigmine contains a permanently charged ammonium salt. Similar to the tartrate salt of rivastigmine, this will enhance water solubility; however, it will hinder the ability of the neostigmine to be absorbed within the gastrointestinal tract and to cross the blood–brain barrier. Therefore, neostigmine cannot be administered orally nor can it be used to treat Alzheimer’s disease because it is unable to reach its site of action in the central nervous system (CNS). Neostigmine is well suited for IM and SC administration and can be used to treat myasthenia gravis, a chronic autoimmune neuromuscular disorder that is characterized by fluctuating weakness of the voluntary muscle groups in the periphery. The beneficial effects of acetylcholinesterase inhibition + observed in the treatment of myasthenia gravis do not require that the drug molecule enter the CNS. Neostigmine Bromide 5. Rivastigmine has a longer duration of action than neostigmine. A comparison of the structures of rivastigmine and neostigmine reveals that the carbamate group present in rivastigmine is slightly than that in neostigmine. mechanism of action given in question 3, provide 5. larger Rivastigmine haspresent a difference results in a Using longer the duration of action. a reason why this structural difference results in a longer duration of action.

Methyl

Methyl

Methyl

Ethyl +

Neostigmine

Rivastigmine

Answer The mechanism of action of these two drug molecules involves the carbamylation of the serine residue found within the active site of acetylcholinesterase and the subsequent slow hydrolysis of the carbamate. As the size of the carbamate increases, hydrolysis and regeneration of acetylcholinesterase decreases. So in this case, the additional carbon atom present within the structure of rivastigmine provides steric hindrance to the hydrolysis of the carbamate resulting in a longer duration of action. 6. Evaluation of the structure of rivastigmine reveals that the aliphatic chain containing the tertiary amine is located meta to the carbamate group. Similar orientations can be seen with neostigmine. Using the information from the previous questions, provide an explanation as to why a para orientation of these groups would lead to a significant decrease in the ability to inhibit 6. Evaluation of thefunctional to inhibit acetylcholinesterase. acetylcholinesterase.

para orientation

meta orientation H3 C

O

N

C H3

C H3

C H3

N

C H3

C H3

O

Rivastigmine

N

O H3C

N C H3

O

C H3

C H3

para Analog of Rivastigmine

Answer Answer: For rivastigmine to interact with acetylcholinesterase, it must contain structural features that mimic For rivastigmine to interact with acetylcholinesterase, it must contain structural features 3, that mimic acetylcholine, the natural substrate for this enzyme. As previously explained in question the

3

3

H3 C

O

N O

N

C H3

C H3

N

O H3C

N

O

C H3

CH 2.24 Rivastigmine 3

221

C H3 para Analog of Rivastigmine Rivastigmine tertiary amine and carbamate of rivastigmine can mimic the quaternary nitrogen atom and ester of acetylcholine respectively. The aromatic ring provides structural rigidity; however, conformational flexibility of its substituents allows these functional groups to be properly oriented when they are Answer: located meta to one another. As shown below, para orientation of these substituents does not allow forrivastigmine proper orientation. the requisite functional groupsstructural are present, they that are located For to interactAlthough with acetylcholinesterase, it must contain features mimic too far apart from one another and, therefore, cannot interact effectively with acetylcholinesterase.

The meta orientation allows for similar distances

The para orientation places the functional groups too far apart from one another

Section 4 Whole Molecule Drug Evaluation Answers 2.25 Sitagliptin

Sitagliptin is an inhibitor of dipeptidyl peptidase IV (DPP-IV), a serine protease that catalyzes the deactivation/degra2.25 Sitagliptin dation of GLP-1. GLP-1 is a 36-amino acid peptide that is responsible for promoting insulin secretion in response to an increase in blood glucose levels. Currently there are four DPP-IV inhibitors on the market, all of which contain an essential basic amino functional group that represents the penultimate amino-terminal alanine residue found within GLP-1.Sitagliptin is an inhibitor of dipeptidyl represents the amino-terminal alanine residue found within GLP-1.

A

F

F

B

C

N H2 O

D

N

N F

N

N CF3

Sitagliptin Sitagliptin

1. Conduct a structural evaluation of sitagliptin, focusing on the boxed functional groups, and use the information1.in the grid to inform the answers to the thatthat follow. Conduct a grid to inform the answers to questions the questions follow. Answer 2. Sitagliptin is group and modify the structure to show the phosphate salt form.

Function

Character Character

A

Acidic, Basic, F or Neutral

Name of Functional Group

Hydrophilic and/or Hydrophobic

FProvide

Halogenated aromatic hydrocarbon

Hydrophilic (F)

Neutral

pKa When Relevant

F

Hydrophobic (R)

Function Function ↑ Solubility N H2 O and/or ↑ Absorption

N Solubility (F) Absorption (R)

Sitagliptin

Interaction(s) Possible with Biological Target at Physiological pH=7.4 N van der Waals N

Amino Acids That Can Interact with the Functional Group via Ion–Dipole Interactions at pH=7.4 None Is Acceptable Asp, Glu, Lys, Arg

N Hydrophobic

H-bondingCF (A) Dipole–dipole

3

Ion–dipole (as the dipole)

Answer:

π-π stacking

B

The pKa the primary amine is the most basic functional group within the structure of sitagliptin.

Primary amine Hydrophilic (NH2) Hydrophobic (R)

Basic

Solubility (NH2)

pKa 9–11

Absorption (R)

Ion–dipole (as the ion) Ionic

223

Ser, Thr, Tyr, Cys, Asn, Gln, Trp, His

Sitagliptin is an inhibitor of dipeptidyl represents the amino-terminal alanine residue found within GLP-1.

A 224 Medicinal Assessment Sitagliptin is anChemistry inhibitor ofSelf dipeptidyl represents the amino-terminal alanine residue found within GLP-1. F

Continued from previous page. C

Amide

F

Hydrophilic F (CON)

A

D

Aromatic Heterocycle (1, 3, 4 triazole)

C

N H2 O

F

Neutral

F

Hydrophobic (R)

B B

D

C

N H-bonding (A)

Asp, Glu, Lys, Arg

N H-bonding (A)

Asp, Glu, Lys, Arg

Solubility (CON)

N N H Absorption 2 O (R)

Dipole–dipole D N N Ion–dipole (as the N dipole) N CF3

N

Hydrophilic (N)

F Basic

Hydrophobic (R)

pKa ~1-5

Sitagliptin Solubility (N) Absorption (R)

Dipole–dipole CF 3

Sitagliptin

Ion–dipole (as the dipole)

1. Conduct a grid to inform the answers to the questions that follow.

2. Sitagliptin is group and modify the structure to show the phosphate salt form. 1. Conduct a grid to inform the answers to the questions that follow. 2. Sitagliptin is formulated as a phosphate salt. Identify the most basic functional group and modify the structure to show the phosphate salt form. 2. Sitagliptin is group and modify the structure to show the phosphate salt form. F F

N H2 O

F F F

N

N

N H2 O

N N

F

Sitagliptin Sitagliptin

N N

N

N CF3

CF3 Answer Answer: Sitagliptin The pKa of the aromatic heterocycle (1,3,4-triazole) is much lower than the pKa of the primary amine. amine is thethat most basic functional group within structure ofnitrogen sitagliptin. a the primary There is The one pK additional nitrogen atom you might be considering. Be the careful—that atom Answer: is directly attached to a carbonyl (to form an amide) and is neutral in character! With the analysis complete, that the primary amine the most basicgroup functional withinofthe structure of The it pKisa clear the primary amine is the most isbasic functional within group the structure sitagliptin. sitagliptin.

Sitagliptin phosphate Sitagliptinphosphate phosphate Sitagliptin

3. Sitagliptin is marketed as the R-enantiomer. Evaluate the structure of sitagliptin and provide a structural rationale for the R-enantiomer designation. Answer Using the Cahn-Ingold-Prelog (CIP) (see Chapter 7 in Basic Concepts in Medicinal Chemistry), the four groups attached to the chiral carbon atom need to be prioritized based on atomic number and the described sequence rules. Using these rules, the primary amine nitrogen atom is prioritized as #1 as it has the highest atomic number, the methylene unit attached to the amide carbonyl is prioritized as #2, the methylene unit attached to the halogenated aromatic hydrocarbon is prioritized as #3, and the hydrogen atom pointing away from the reader is the #4 priority. NOTE: it is important to recognize that although the methylene units are identical, the rules state that atoms attached to each of these methylene units are the next to be evaluated. Evaluation of the methylene to the right of the

2.25 Sitagliptin

225

3. Sitagliptin is designation. primary amine reveals that the carbon atom is attached to the equivalent of two oxygen atoms and a nitrogen atom. Evaluation of the methylene to the left of the primary amine reveals that the carbon Answer: atom is attached to the equivalent of three carbon atoms. The atomic number of both oxygen and nitrogen is higher than carbon; therefore, the #2 and #3 prioritization is reflective ofisthat analysis. Using the Cahn-Ingold-Prelog (are clockwise in orientation and the R-enantiomer drawn. Reviewing the positions of these priorities, it is noted that they are clockwise in orientation and the R-enantiomer is drawn.

1

3

2 4

Sitagliptin Sitagliptin

4. At 38%, the fraction of sitagliptin reversibly bound to plasma proteins is relatively low. By way of 4. At 38%, (e.g.,fraction warfarin). reminder, only the thebound unbound of drug is able to exert its biological activity and undergo metabolism. Describe the relative risk of a plasma protein binding interaction between sitagliptin 5.another Approximately 79% of urine. Provide a abolic transformation and drug that is highly protein bound (e.g., warfarin). (i.e., excreted unchanged). Answer Based on the information from the structure evaluation grid, sitagliptin is a basic drug (see discussion of the liptin and determine which phase I metabolic transformation hasfor occurred. of 6. the Assess plasma each binding proteins in Chapter 3 of Basic Concepts in Medicinal Chemistry information about albumin and α1-acid glycoprotein). Generally albumin tends to bind to acidic drugs, whereas the plasma protein α1-acid glycoprotein has an affinity for basic drugs and some neutral drugs. Warfarin is an acidic drug and is greater than 90% plasma protein bound. In this scenario, warfarin would be bound to albumin and sitagliptin would be bound to α1-acid glycoprotein. Because the two drugs are not competing for the same plasma binding protein, there is very, very little risk of a plasma protein binding interaction.

5. Approximately 79% of sitagliptin is excreted unchanged in the urine. Provide a structural rationale that supports this observation. Answer Based on the information in the structure evaluation grid, sitagliptin contains a number of functional groups that contribute to the overall water solubility of the drug. The fluorinated aromatic hydrocarbon, amide, and aromatic heterocycle (1,3,4 triazole) will be able to participate in important hydrogen bonding interactions with water. The primary amine, likely to be predominantly ionized in the plasma and urine, significantly enhances this water solubility by participating in the strong ion– dipole interactions with water. The drug is sufficiently hydrophilic in character for it to be excreted without the need for metabolic transformation (i.e., excreted unchanged).

226

Medicinal Chemistry Self Assessment

6. Assess each of the possible metabolic products generated from sitagliptin and determine which phase I metabolic transformation has occurred.

A B

C

Answer Name of Metabolic Transformation A

Benzylic oxidation

B

Oxidative deamination

C

Amide hydrolysis

D

Aromatic hydroxylation

D

Section 4 Whole Molecule Drug Evaluation Answers 2.26 Sorafenib

Because protein tyrosine kinases regulate cellular proliferation, differentiation, and survival, it is no surprise that several neoplastic disorders can be tied to altered activity of protein tyrosine kinases. Clinically relevant antineoplastic tyrosine kinase inhibitors interact with the active site of the enzyme via several types of binding interactions. The adenosine triphosphate (ATP) binding domain of the tyrosine kinases contains a hydrophobic domain that includes a significant number of isoleucine, leucine, alanine and valine residues. There are at least five binding pockets that flank this region in which van der Waals, hydrophobic, hydrogen bonding, and electrostatic interactions occur. Sorafenib is a tyrosine kinase inhibitor used in the treatment of advanced renal cell carcinoma, a highly vascularized tumor.2.26 The Sorafenib drug specifically targets vascular endothelin growth factor 2 (VEGF2) that is instrumental in the generation ofBecause new blood vessels. protein tyrosine kinases owth is instrumental in the generation of new blood vessels.

F

C E A B

D Sorafenib Sorafenib

1. Conduct structural evaluation of sorafenib, 1. aConduct a to the questions that follow.focusing on the boxed functional groups, and use the information in the grid to inform your answers to the questions that follow. 2. Sorafenib interacts with in the local environment of the enzyme.

3. Nilotinib, another and Leu298/Val299/Phe359 in each of the respective five binding pockets.

A B

C

D

Nilotinib 227

A. Consider the side chains of the. Assume pH=7.4.

E

228

Medicinal Chemistry Self Assessment

Answer Character

Name of Functional Group A

Halogenated aromatic hydrocarbon

Function

Character

Acidic, Basic, or Neutral

Function

Hydrophilic and/or Hydrophobic

Provide pKa When Relevant

↑ Solubility and/or ↑ Absorption

Interaction(s) Possible with Biological Target at Physiological pH=7.4

Hydrophobic

Neutral

Absorption

Cl: Dipole–dipole Ion–dipole (as the dipole) Ar: van der Waals Hydrophobic π-π Stacking

B

Halogenated aliphatic alkane

Hydrophilic (F)

Neutral

Solubility

H-bonding (A) Dipole–dipole Ion–dipole (as the dipole)

C

Urea

Hydrophilic (HNCONH)

Neutral

Hydrophobic (R)

Solubility (HNCONH)

H-bonding (A + D)

Absorption (R)

Dipole–dipole Ion–dipole (as the dipole)

D

Ether

Hydrophilic (O)

Neutral

Hydrophobic (R)

Solubility (O)

H-bonding (A)

Absorption (R)

Dipole–dipole Ion–dipole (as the dipole)

E

Pyridine

Hydrophilic (N)

Basic

Solubility (N)

Pyridine N atom:

(Azine)

Hydrophobic (R)

(pKa 1–5)

Absorption (R)

H-bonding (A) Dipole–dipole Ion–dipole (as the dipole) R: van der Waals Hydrophobic π-π Stacking

F

Amide

Hydrophilic (CONH) Hydrophobic (R)

Neutral

Solubility (CONH)

H-bonding (A+ D)

Absorption (R)

Dipole–dipole Ion–dipole (as the dipole)

2. Sorafenib interacts with Cys919, Phe1047, and Asp1046 via hydrogen bonding and hydrophobic interactions. Identify which functional groups could interact with the side chains of these amino acids. Assume that Asp1046 is unionized in the local environment of the enzyme.

2.26 Sorafenib

229

2.26 Sorafenib Because Answer protein tyrosine kinases owth is instrumental in the generation of new blood vessels. Interacts with Cysteine919 via a Hydrogen Bonding Interaction

Functional Group

Yes or No

A

C

Interacts with Aspartic Acid1046 F via a Hydrogen Bonding Interaction

Interacts with Phenylalanine1047 via a Hydrophobic Interaction

Yes or No

Yes or No

No

Yes No

E

A

No

B

Yes

Yes

C

Yes

Yes

D

Yes

E

Yes

Yes

Yes

F

Yes

Sorafenib Yes

No

B

No

D

Yes

No

1. Conduct a to the questions that follow. 3. Nilotinib, another tyrosine kinase inhibitor (via Bcr-Abl) indicated for the treatment of PhiladelSorafenib interacts with in the localmyelogenous environment ofleukemia, the enzyme. phia2.Chromosome (Ph+) positive chronic also interacts with each of the five binding pockets found within this target. This drug interacts with Leu285/Val289, Asp391/Glu286, 298 biological 299 359 3. , Met Nilotinib, another and299 Leu /Val /Pheof in each of the respective fivepockets. binding pockets. 318 Thr315 and Leu298/Val /Phe359 in each the respective five binding

A B

C

E

D

Nilotinib Nilotinib a. Consider the side chains of the amino acids indicated and determine which type(s) of binding interactions are possible in each of the five binding pockets. Assume pH=7.4.

Answer A. Consider the side chains of the. Assume pH=7.4. Leu285/Val289

Asp391/Glu286

Thr315

Met318

Leu298/Val299/Phe359

Hydrophobic

Ionic

H-bonding

Hydrophobic

Hydrophobic

B. Determine which of the side chains are both at pH=7.4.

van der Waals the ion) Dipole–dipole Dipole–dipole der Waals C. It has beenIon–dipole atom(s) (as within the structure of methionine participate in thisvaninteraction. Ion–dipole (as the dipole)

Methionine

van der Waals

B 230

D

Medicinal Chemistry Self Assessment

Sorafenib

Conductwhich a to the follow. groups (A–E) can interact with the side chains of the b. 1. Determine of questions the boxedthat functional amino acids found in each of the five binding pockets. Indicate the type of interaction(s) possible 2. Sorafenib interacts with in the local environment of the enzyme. in the appropriate box. None is an acceptable answer. Assume that the drug and the amino acid side chains are both at pH=7.4. 3. Nilotinib, another and Leu298/Val299/Phe359 in each of the respective five binding pockets. Leu285/Val289 A B

Hydrophobic Hydrophobic

AAsp

391

/Glu286

Ion–dipole (functional group as the dipole) None

B

C

van der Waals

Thr315

Met318

Leu298/Val299/Phe359

H-bonding (A)

Hydrophobic

Hydrophobic

Dipole–dipole H-bonding (A)

D

Hydrophobic

E

Dipole–dipole

π-π Stacking Hydrophobic van der Waals π-π Stacking

C

None

Ion–dipole (functional group as the dipole)

H-bonding (A+D)

D*

None

Ion–dipole (functional group as the dipole)

H-bonding (A+D)

Hydrophobic

Ion–dipole (functional Nilotinib H-bonding (A)

E**

None

None

None

None

Hydrophobic

Dipole–dipole Dipole–dipole

group as the dipole)



None

π-π Stacking

Dipole–dipole

* Aniline-like nitrogen atom will not be ionized at physiological pH. A. (azine) Consider the sideatom chains of not the. be Assume pH=7.4. **Pyridine nitrogen will ionized at physiological pH.

B. Determine which the pyridyl side chains are both at (functional pH=7.4. c. It has been documented thatofthe nitrogen atom group E) of nilotinib participates in a hydrogen bonding interaction with methionine. Draw a diagram that clearly shows C. It has beenthe atom(s) withinofthe structure ofparticipate methionineinparticipate in this interaction. which atom(s) within structure methionine this interaction.

Answer:

Methionine Methionine

Answer

H-Bond Acceptor

Nilotinib Nilotinib

H-Bond Donor

Methionine Methionine

2.26 Sorafenib

231

4. Sorafenib enters cells via passive diffusion. Using the information in the structure evaluation grid as a starting point, identify which functional groups contribute to the ability of this drug to enter cells via passive diffusion. Answer The halogenated aromatic hydrocarbon, aromatic hydrocarbon, and the pyridine ring carbon atoms all contribute significantly to the hydrophobic character of sorafenib. It is important to note that the basic pyridine ring is unlikely to be ionized at physiological pH (pKa = 6 < pH = 7.4). Although there appears to be significant hydrophilic character (halogenated aliphatic alkane, urea, ether, the nitrogen atom of the pyridine, amide) present in this molecule, sorafenib is practically insoluble in water. The lack of water solubility and clearly identifiable hydrophobic character allows for passive diffusion of the drug across the cellular lipid bilayer membrane.

5. Nilotinib is considered significantly more hydrophobic than sorafenib (distribution coefficient log D is 2.4 and 0.8 respectively). Provide a structural rationale for this property difference. Answer Based on the information found in the structure evaluation grid for sorafenib, there are several functional groups that contribute to the overall hydrophobic character of the molecule (e.g., halogenated aromatic hydrocarbon, aromatic hydrocarbon, carbon atoms of pyridine/azine ring). Similar evaluation of nilotinib yields two aromatic rings, a pyrimidine ring (between functional groups D and E), a pyridine ring (functional group E), and even some hydrophobic character in the histidine ring (functional group A) that contribute to the overall hydrophobic character of the molecule. When you compare the sheer number of functional groups that contribute to the overall hydrophobic character for each drug, nilotinib wins! 6. Sorafenib is marketed as a tosylate salt, a lipid-soluble organic salt. Nilotinib is marketed as a hydrochloride monohydrate salt, an inorganic salt. In general, what is the value of each of these types of salts? Answer The value of lipid-soluble organic salts is to decrease the water solubility and increase the lipid solubility of the parent drug (in this case sorafenib) (see Chapter 5 of Basic Concepts in Medicinal Chemistry). Typically lipid-soluble salts are used in the formation of lipid-soluble suspensions. In addition, they can improve the oral bioavailability of acid labile drug molecules and improve the palatability of liquid solutions. p-Toluenesulfonic acid (tosylic acid) is considered a strong organic acid and forms a strong counter-ion when dissociated from the drug molecule. Sorafenib is administered as a filmcoated tablet for adults and as a liquid suspension for children. The value associated with the formation of inorganic salts is due to the improved aqueous solubility, solvation, and dissolution that results. In general, inorganic salts enhance the absorption of drugs that are administered orally because they improve both solvation and dissolution properties.

Section 4 Whole Molecule Drug Evaluation Answers 2.27 Zanamivir and Oseltamivir 2.27 Zanamivir and Oseltamivir

Shown below is the of zanamivir. This drug via oral inhalation for the treatment of Shown below is structure the structure for the treatment of molecule influenza is A administered and B infections. influenza A and B infections.

Zanamivir

Chapter2.27

1. Identify all of the acidic 1. Identify all of the ofand 7.2.basic functional groups, provide the normal pKa range for each functional group, and identify if each functional group would be primarily ionized or unionized at a pulmonary PleasereplacethestructurefortheanswerofQ1withtheonebelow. Answer: pH=7.2.



Answer

Carboxylic acid (Acidic functional group) Carboxylic Normalacid pH (Acidic range =functional 2.5 to 5 group) Normal pH range = 2.5 toionized 5 Would be primarily at a pH of 7.2 Would be primarily ionized at a pH = 7.2

Guanidine (Basic functional group) Guanidine (Basic functional group) Normal pH range = 12 to 13 Would be primarily at a pH of 7.2 Normal pH range = 12 toionized 13 Would be primarily ionized at a pH = 7.2 2.

Identify all other water soluble functional groups that are present within the structure of zanamivir. Answer:

Please replace the structure for the answer to Q6, part a with the one below (page 4, middle structure) Primary hydroxyl group Secondary hydroxyl group (or primary alcohol) (or secondary alcohol) Secondary hydroxyl group (or secondary alcohol)

233

Ether oxygen

234

Guanidine (Basic functional group) Normal pH range = 12 to 13 Would be primarily ionized at a pH of 7.2

Medicinal Chemistry Self Assessment

2. Identify all other water soluble functional groups that are present within the structure of zanamivir. 2. Identify all other water soluble functional groups that are present within the structure of zanamivir. Answer: Answer

Primary hydroxyl group (or primary alcohol)

Secondary hydroxyl group (or secondary alcohol)

Secondary hydroxyl group (or secondary alcohol)

Ether oxygen

Amide

The amide and hydroxyl groups can act as hydrogen bond donors and acceptors and thus can form hydrogen bonds with water. The ether oxygen most likely contributes the least to water solubility; however, it can on function a hydrogenasbond acceptor. 3. Based your isas administered an oral inhaler instead of an oral tablet or capsule. 3. Based on your answers to questions 1 and 2, explain why zanamivir is administered as an oral inhaler instead of an oral tablet or capsule. Answer The structure of zanamivir contains multiple hydrophilic functional groups that will allow it to easily dissolve within the aqueous contents of the gastrointestinal (GI) tract. The carboxylic acid and the guanidine functional groups will be extensively ionized at an intestinal pH=5 which further increases the overall water solubility of the molecule. Although the structure of zanamivir contains a hydrocarbon chain and ring, the overall balance between water and lipid solubility hinders its ability to effectively cross the GI membrane. The oral absorption of zanamivir has been reported to be between 1% and 5%. Due to this, zanamivir must be administered via oral inhalation. 4. Zanamivir exerts its antiviral action by inhibiting neuraminidase, a viral enzyme that is required for the spread of the viral infection. A key component of neuraminidase’s action is the hydrolysis of 4. Zanamiviracid exerts his, provide an explanation Shown how zanamivir inhibits neuraminidase. N-acetylsialic fromitssurface viral glycoproteins. below is the structure of N-acetylsialic acid bound to a glycoprotein. Using this structure and the structure of zanamivir, provide an explanation of how zanamivir inhibits neuraminidase.

Glycosidic bond N-Acetylsialic acid bound to glycoprotein

Zanamivir

Answer Answer: Zanamivir is a stable mimic of N-acetylsialic acid. As shown below, the structure of zanamivir retains is a stable a cleavable glycosidic bond. many Zanamivir of the structural features of N-acetylsialic acid, but lacks a cleavable glycosidic bond. Structurally identical to N-acetylsialic acid (with exception of

Zanamivir

Answer:

2.27 Zanamivir and Oseltamivir

235

Zanamivir is a stable a cleavable glycosidic bond.

Glycosidic bond N-Acetylsialic acid bound to glycoprotein

Answer: Zanamivir is a stable a cleavable glycosidic bond. Zanamivir

Structurally identical to N-acetylsialic acid (with exception of the double bond) Zanamivir

Lacks glycosidic bond

identical The structural similarity allows zanamivir to interact withStructurally neuraminidase in the same manner as 5. N-acetylsialic Shown below acid. is a activity is likely to be more active, less active, or similar toremain that of zanamivir? to N-acetylsialic acid Ionic or ion–dipole interactions with the carboxylic acid the same, as does (with exception of the ability to form hydrogen bonds with the hydroxyl groups and the amide. Because zanamivir lacks the double bond) a cleavable glycosidic bond, it can occupy the binding site without being metabolically transformed. The most significant difference between zanamivir and N-acetylsialic acid is the substitution of a secondary hydroxyl group with a guanidine group. Studies have shown that this basic functional group can form ionic interactions with neuraminidase. In summary, zanamivir structurally mimics the natural substrate for neuraminidase and is able to bind to the active site of the enzyme. This binding inhibits the ability of neuraminidase to cleave N-acetylsialic acid from viral glycoproteins, thus inhibLacks glycosidic bond iting the spread of a viral infection. Zanamivir

5. Shown below is a stereoisomer of zanamivir. Identify if the stereoisomer is an enantiomer, a diastereomer, 5. aShown belowisomer, is a activity is likelyand/or to be more active, less active, similarwhether to that ofthis zanamivir? geometric an epimer, a conformational isomer.orPredict stereoisomer’s pharmacological activity is likely to be more active, less active, or similar to that of zanamivir? 6. Shown below is the structure of for oseltamivir

Answer stereoisomer the opposite stereochemical configuration at all five chiral centers and has the 6. This Shown below is thehas structure of for oseltamivir exact same conformation as zanamivir; therefore, this is the enantiomer of zanamivir. None of the other stereochemical designations are correct. Given that zanamivir as well as oseltamivir exert their antiviral activity by mimicking N-acetylsialic acid, alteration of the stereochemistry decreases the resemblance to N-acetylsialic acid and would be predicted to cause a decrease in binding affinity to neuraminidase and a decreased antiviral effect.

236

Medicinal Chemistry Self Assessment

6. Shown below is the structure of oseltamivir and a list of five metabolic transformations. For each Chapter2.27 metabolic transformation, indicate if it is a phase I or a phase II transformation and if oseltamivir has a functional group present that can undergo the indicated transformation. When evaluating these PleasereplacethestructurefortheanswerofQ1withtheonebelow. metabolic transformations, consider functional groups that are initially present within the structure of oseltamivir as well as those that can be added/revealed through phase I metabolism. If you answer  YES, then draw the appropriate metabolite; if you answer NO, then provide a brief explanation as to why this metabolic transformation is not possible for oseltamivir. w is the structure of for oseltamivir

Carboxylic acid (Acidic functional group) Normal pH range = 2.5 to 5 Would be primarily ionized at a pH = 7.2

Metabolic Pathways A. Hydrolysis B. Allylic oxidation C. Glucuronide conjugation

Guanidine (Basic functional group) Normal pH range = 12 to 13 Would be primarily ionized at a pH = 7.2

D. ω-Oxidation E. Oxidative O-Dealkylation Answer Answer: Please replace the structure for the answer to Q6, part a with the one below (page 4, middle structure) a. Hydrolysis: Phase I transformation. Both the ester and amide functional groups can undergo A. Hydrolysis: Phasehydrolysis I oseltamivir. hydrolysis. Ester produces the active metabolite of oseltamivir.

Product of hydrolysis esterEster hydrolysis



Product of Amide hydrolysis amide hydrolysis

Product of ester amide  and Ester amideand hydrolysis hydrolysis

b. Allylic oxidation: Phase I transformation. It is not possible because both allylic carbon atoms are B. Allylic oxidation: a hydrogenAdditionally, atom that is required for carbon oxidation. already attached Phase to heteroatoms. one allylic atom lacks a hydrogen atom that is required for oxidation. Already attached to a heteroatom

Lacks a hydrogen atom

Already attached to a heteroatom

C. Glucuronide conjugation: Phase II.

Oseltamivir

Already attached to a heteroatom

Oseltamivir 2.27 Zanamivir and Oseltamivir

237

C. Glucuronide conjugation: Phase II. c. Glucuronide conjugation: Phase II transformation. It can occur directly with the secondary amine or with the carboxylic acid that results from phase I ester hydrolysis.

D. Oxidation: Phase I, Yes D. Oxidation: Phase I, Yes

d. ω-Oxidation: Phase I, Yes

E. Oxidative O-Dealkylation: Phase I, Yes

e. Oxidative O-Dealkylation: Phase I, Yes E. Oxidative O-Dealkylation: Phase I, Yes

+ +

Index

A

Angiotensinogen, 27, 129

Acebutolol, 21, 117-119 Acetylcholine, 79-80, 218-219 Acetylcholinesterase inhibitor, 79-81, 218-221 Acidic functional group, 7-9, 11-12, 31, 36, 40, 55, 66, 69, 75, 81, 91, 99-103, 105-106, 137, 151, 155, 174, 190, 193-194, 211-212, 233

Anticholinergic agents, 18, 114-115 Antidiabetic agent, experimental, 7-8, 99, 101, 102 Antihistamine, 39, 149-152 Anti-hyperlipidemic agents, 47, 165 Anxiolytic effect, 53, 172 Arformoterol, 13-14, 109 Argatroban, 9, 103 Arginine, 8, 102



duration of, 81, 220

Aripiprazole, 31-33, 137-142



onset, 66, 190



metabolites, 33, 141

Active metabolite, 63, 186-187

Aromatic oxidation, 70, 141, 197

Active product, 58-59, 179

Aspartic acid, 88, 98, 228

Adenosine triphosphate (ATP) binding, 87, 227

Atenolol, 11, 105, 105-106

Alanine, 87, 227 Alendronate, 15, 111 Aliskiren, 27-30, 129-135 Allylic oxidation, 92, 236 Alprazolam, 53, 172 Amiloride, 9, 103 Amino acid binding interactions, 32, 139-140 Amino acid side chain interactions, 30, 44-45, 56, 58, 72, 88, 134, 163, 175, 178, 201-202, 228-229, 230

C-11 oxidation, 58, 179 C-17 oxidation, 58, 179 C-21, 60, 179-180 Cahn-Ingold-Prelog (CIP) system, 22, 119-120, 184 Carbamate group, 81, 219-220 Carboxylic acid, 73, 185, 194, 206 Cefamandole, 21, 117-119

Action

Administration routes, 14-15, 109-110

C

Atrial fibrillation, 43, 159

Cefotaxime, 11, 105-106 Cefprozil, 35-38, 143-147 Cephalosporin, 35, 143 Cetirizine, 39-40, 149-152 Chemical hydrolysis, 147 Chiral carbon atom, 62, 184 Chiral centers, 21-22, 119 Chlorimipramine, 24, 124

B Baclofen, 13-14, 24, 109, 129 Barbiturates, 71-72, 199-203 Barbituric acid, 199 Basic functional group, 7-9, 11, 12, 31, 36, 40, 55, 69, 75, 91, 99-103, 105-106, 137, 144, 151, 155, 174, 193-194, 211-212, 233

Chlorpropamide, 41-42, 153-157 Ciprofloxacin, 13-14, 109 Class IB anti-arrhythmic agent, 65, 189 Clonidine, 8, 102-103 CNS-based toxicities, 67, 192 Conformational isomer, 92, 235 Conjugate elimination, 48, 168

Benzodiazepines, 53, 172

Cucumber flavor, 20, 116

Amino acid-based hormone, 62, 185

Benzylic oxidation, 56, 70, 141, 176, 191, 198

Cysteine, 88, 228

Amino acids, 5, 98

Beta-lactam bond instability, 38, 147

Amlodipine, 30, 134-135

Beta-lactamases, 38, 147

Amphoteric drug, 9, 36, 103, 144

Binding interactions, 56, 175

Anesthetic, 65, 189

Bioavailability, 132

Angiotensin converting enzyme, 27, 129

Biological targeting, 14-15, 19, 110, 115

Angiotensin converting enzyme (ACE) inhibitor(s), 17, 75-78, 113, 211-215

Blood soluble, 65, 190

Angiotensin I, 27, 77, 129-130, 213

Butabarbital, 71, 199-200

Angiotensin II, 27, 129-130



Bromfenac, 3, 7-8, 37, 99, 101, 102 metabolite, 72, 202-203

Angiotensin II receptor blockers (ARBs), 41, 153 239

Cysteinyl leukotriene receptors, 70, 196

D Dabigatran etexilate, 43-45, 159-164

mesylate, 45, 164

Dehalogenation, 61, 63, 183, 186-187 Delapril, 17, 113 Depot injections, 60, 180 Dexfenfluramine, 24, 125 Diastereomer(s), 21, 28, 52, 62, 92, 117-119, 130-131, 170, 184, 235

240

Medicinal Chemistry Self Assessment

Dihydropyridine calcium channel blocker, 30, 134 Diltiazem, 9, 103

Fluvoxamine metabolites, 53, 171 Functional groups, 3-9, 11, 95-98, 99

G

Dipole-dipole interaction, 202

GABAA, 71, 72, 199, 201, 203

Drug action

duration, 60, 181



stereochemistry and, 21-22, 117-121

Drug binding interactions, 17-20, 113-116 Drug interaction, 32, 139 Drug metabolism, 23-24, 123-125

Gemfibrozil, 47-49, 165-168 Geometric isomer(s), 21, 28, 52, 62, 92, 117-119, 130-131, 170, 184, 235 GI drug absorption compartments, 36, 145 Glucocorticoid, 57-58, 177-178 Glucuronide conjugation, 53, 92, 172, 237 Glutamate, 20, 116 Glutamic acid, 20, 79, 116, 218

E

Glutamine, 98

E-geometric isomer, 170



Electron donation, 4, 55, 79, 96, 173, 217

Gram (-) activity, 35, 143

Electron withdrawing, 4, 55, 79, 96, 97, 173-174, 217

Gram (+) activity, 35, 143

Glyburide, 42, 158

Elimination, 48, 167

half-life, 49, 168



unchanged, 70, 197

Embeconazole, 15, 111-112 Enantiomer(s), 21-22, 52, 62, 92, 117-121, 170, 184, 235 Enterohepatic recycling, 23, 123, 124 Enzymatic hydrolysis, 147 Epimer, 92, 235

Ibuprofen, 4, 96 Imipramine analogs, 5, 97

Dipeptidyl peptidase IV, 83, 223 D-Phe-Pro-Arg tripeptide sequence, 43-45, 159-163

I

metabolite, 42, 158

Green pepper flavor, 20, 116 Guanidine functional group, 8, 102

H Half-life, 42, 156

Inactive metabolite, 63, 186-187 Inactive product, 58-59, 179 Induction, 4, 55, 79, 96, 173-174, 217 Inosinic acid, 20, 116 Insulin, 62, 185

structure, 14, 111

Interactions, 62, 185 Intermediate-acting agent, 71, 200 Intestine pH, 14, 109 Ion channels, 20, 116 Ion-dipole interaction(s), 19, 58, 62, 115, 185, 178 Ionic bond, 202 Ionic interaction, 62, 185 Ionization percentage, 12, 106-108 Ionized, 31, 36, 55, 69, 71, 75-76, 91, 137, 145, 174, 193-194, 201, 211-212, 233 Irbesartan, 13-14, 109 Isoleucine, 87, 227

Halogenated aromatic hydrocarbon, 185

L

Haloperidol, 55-56, 173-176

Leu-Val peptide bond, 30, 134

Henderson-Hasselbalch equation, 12, 106-107, 108, 155

Levothyroxine (T4), 61-63, 183-187

Hepatic metabolism, 70, 197



metabolites, 63, 186

Hepatic oxidation, 53, 172

Lidocaine, 65-67, 189-192 Sodium channel blocker, 65, 189-192

Heroin, 23, 123

Lipid solubility, 31, 138, 195

Histidine, 79, 218

Lipid soluble esters, 174

HMG CoA reductase, 74, 207-209

Lipophilic absorption, 66, 190

HMG-CoA reductase inhibitor, 22, 119-120

Lipophilic ester, 60, 180-181

Hormone replacement therapy, 62, 184

Lisinopril, 17, 113

Hydrochloride monohydrate salt, 89, 231

Lomitapide, 9, 103

Hydrochloride salt, 13-14, 109

Long-acting agent, 71, 200

Hydrocortisone, 57-60, 177-181

Loperamide, 14, 109-110

Fesoterodine, 18, 114, 115

Hydrogen bond(s), 18-19, 58, 62, 88-89, 115, 178, 185, 195, 202, 207, 228

Lorazepam, 53, 172

Fibrin, 43, 159

Hydrolysis, 56, 70, 92, 176, 198, 236

Fibrinogen, 43, 159

Hydrolytic reactions, 44, 161-162

Flavors, 20, 116

Hydrolyzable peptide bond, 30, 134

Fluvastatin, 22, 73-74, 119-120, 205-209

Hydrophilic reaction, 4, 55, 73, 75, 79, 96, 173, 205, 211, 217

Ester hydrolysis, 79-80, 218-219 Estradiol, 21, 23, 117-119, 123 Estrogen, 62, 185 Ether, 185 Ezetimibe, 11, 105-106

F Fenofibrate, 47-49, 165-168

as active drug, 48, 166



as prodrug, 48, 166



conformationally restricted analog of, 74, 208-209

Fluvoxamine, 24, 51-53, 125, 169-172

metabolites, 53, 171

Fluvoxamine + alprazolam, 53, 172 Fluvoxamine + lorazepam, 53, 172

Losartan, 32, 41, 139, 153 L-tyrosine, 62, 185-186

M Maleate salt, 52, 170-171

Hydrophobic reaction, 4, 19, 55, 62-63, 73, 75, 79, 88-89, 96, 114-115, 173, 185-186, 196, 202, 205, 211, 217, 228, 231

Mechanism of action, 42, 155-156

Hydroxyzine, 39-40, 149-152

Methionine, 89, 230

Metabolic transformation, 28-29, 53, 77-78, 171-172, 214-215 Metabolites, 24, 124

Index

241

Methylation, 70, 197

Phenol, 185

Serotonin receptor modulator, 31, 137

Mevalonic acid, 74, 207-208

Phenylalanine, 88, 228

Montelukast, 69-70, 193-198

Phenytoin functional groups, 3, 37

Serotonin reuptake transporter (SERT), 51, 169

Morphine, 23, 123

Phosphate salt, 84, 224 pKa, 71, 201

N N-acetylsialic acid, 91-92, 234-235 Natamycin, 12, 106-107



range, 6, 31, 73, 75-76, 91, 101-102, 137, 206-207, 211-212, 233



values, 42, 69, 154-155, 193, 194-195

Serotonin selective reuptake inhibitors (SSRI), 51, 169 Shitaki mushrooms, 20, 116 Short-acting agent, 71, 200 Sitagliptin, 83-85, 223-226

Plasma pH, 11, 14, 105, 109

Sodium salt(s), 70, 71, 195, 200

Plasma protein binding, 30, 32, 41-42, 70, 84, 134-135, 139, 153, 197, 225

Solubility, 13-15, 109-112

Potassium salt, 13-14, 109

Sorbinil, 7-8, 99, 101, 102

Pravastatin, 9, 73-74, 103, 205-209 Primarily ionized, 11, 105-106

Stereochemistry, drug action and, 21-22, 117-121

Primarily unionized, 11, 105-106

Steroid-based hormone, 62, 185

Primary amine, 185

Stomach pH, 11, 14, 105, 109

Prodrug(s), 44, 48, 60, 76, 161-162, 174, 180-181, 212

Sulfamethoxazole, 4, 96

Protein tyrosine kinases, 87-89, 227-231

Sulfonamide, 194

Pyridyl nitrogen, 89, 230

Sulfonylureas, 41-42, 153-157

Nucleophilic side chain, 38, 147

Q

T

O

Quinapril, 75-78, 211-215

Taste receptors, 20, 116

Quinaprilat, 76-77, 212-213

Tetracycline, 13-14, 109

N-dealkylated monoethylglycinexylidide, 67, 192 Neostigmine bromide, 81, 219-220 Nerolidol, 19, 115 Neuraminidase, 91, 234-235 Nifedipine, 21, 117-119 Nilotinib, 88-89, 229, 231 Nitrofurantoin, 11, 105-106 Nonelectrolyte, 9, 103 Non-hydrolyzable hydroethylene, 30, 134 Non-peptidomimetic prodrug, 44, 161 Norepinephrine reuptake, 24, 125

Odorant molecules, 19, 115

Sorafenib, 87-89, 227-231

Sulfate conjugation, 23, 56, 123, 176

Tetrapeptide side chains, 5, 98

Oral anticoagulant, experimental, 7-8, 99, 101, 102

R

Oral bioavailability, 52, 171

R-A double bond reduction, 58, 179

Thrombin inhibitor, 43, 159

Oseltamivir, 91-92, 233-237

Ranitidine, 12, 108

Thyroglobulin molecules, 61, 183

Oxidative N-dealkylation, 56, 142, 176

Reactions, 58-59, 179

Tocainide, 67, 192

Oxidative O-dealkylation, 48-49, 70, 92, 168, 198, 236, 237

Reduction, 56, 176

Tolbutamide, 32, 41, 139, 153-156

R-enantiomer, 62, 84, 184, 224-225

Tolterodine, 18, 114-115

Oxidative transformation, 28-29, 132-133

Renin, 30, 134

Tolycaine, 67, 192

Oxybutynin, 18, 114-115

Resonance, 4, 55, 79, 96, 217

Tosylate salt, 89, 231

Ring A ketone reduction, 58, 179

Transformations, 58-59, 179

Rivastigmine, 79-81, 217-221

Tricyclic antidepressant, 24, 124

Routes of administration, biological targets and, 14-15, 109-110

Tyrosine, 98

Peptide cleavage, 76, 213

Rule of Nines, 12, 69, 106, 194

Tyrosine kinase inhibitor, 87-89, 227-231

Peptide-based hormone, 62, 185

S

U

Salmeterol functional groups, 3, 37

Umami receptor, 20, 116

Salts, 13-15, 109-112 Scents, 19, 115

Unionized, 31, 36, 55, 69, 71, 75-76, 91, 137, 145, 174, 193-194, 201, 211-212, 233

Sclareol, 19, 115

Urinary pH, 11, 105

P Passive diffusion, 89, 231 Penicillins, 38, 147

pH/pKa, 11-12, 105-106

Phase I transformation, 23-24, 33, 37, 40, 48, 53, 56, 66-67, 70, 72, 85, 92, 123-125, 141-142, 146, 150-151, 167-168, 172, 175-176, 191-192, 197-198, 202, 226, 236-237 Phase II transformation, 23, 33, 37, 40, 48, 56, 70, 72, 92, 123, 125, 141, 146, 151, 167-168, 175, 197, 202, 236-237

Thrombin active site, 43, 159

Scopolamine, 14, 109-110 Secobarbital, 71, 199-200, 201

V



Valine, 87, 98, 227

metabolite, 72, 202-203

Phenobarbital, 71-72, 199-203

S-enantiomer, 62, 184



Serine, 79, 80, 218-219

metabolite, 72, 202-203

Tri-iodo-L-thyronine (T3), 61, 63, 183, 185

Van der Waals, 19, 87, 114-115, 140, 196, 202-203, 227

242

Medicinal Chemistry Self Assessment

Vanillin, 19, 115 Vascular endothelin growth factor 2, 87, 227

W Warfarin, 3, 37, 84, 225 Water solubility, 31, 138, 195 Water soluble ester, 60, 180-181 ω-oxidation, 92, 236, 237

Z Zafirlukast, 69-70, 193-198 Zanamivir, 91-92, 233-237