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Basic Principles of Medicinal Chemistry Example Questions for EXAM #1 LECTURE

PHR 143M

Fill in the ScanTron sheet with the SINGLE BEST ANSWER to the following questions. ONLY the ScanTron sheet will be graded; you will receive no credit for answers marked on this question sheet. 1.

Below is a table listing the boiling points of a series of alkane hydrocarbons. What forces are responsible for the trend observed in this table?

Alkane n-hexane n-heptane n-octane A. B. C. D.

Boiling Point 69.0 °C 98.4 °C 126.0 °C Decreasing Inductive destabilization of the intermolecular attraction between alkanes with an increase in the number of bond dipoles (e.g., C–H bonds). Increasing Inductive stabilization of the intermolecular attraction between alkanes with a decrease in the number of bond dipoles (e.g., C–H bonds). Increasing van der Waals intermolecular attraction between alkanes with an increase in molecular size (e.g., number of electrons). Decreasing van der Waals intermolecular attraction between alkanes with an increase in molecular size (e.g., number of electrons).

2. Partition co-efficients are expressed as logP values. These logP values relate to the partitioning of the drug between an aqueous phase and an organic phase, typically water-saturated n-octanol. The logP value is defined as: A. The negative log of the ratio of the concentration of the drug in the aqueous phase over the concentration of the drug in the organic phase. B. The log of the negative of the ratio of the concentration of the drug in the aqueous phase over the concentration of the drug in the organic phase. C. The log of the ratio of the concentration of the drug in the organic phase over the concentration of the drug in the aqueous phase. D. The negative log of the ratio of the water solubility of the drug over the organic phase solubility of the drug.

3. Based upon the structures of the two barbiturates below, predict their relative partition co-efficients and water solubilities. O

NH

O

O

NH

NH

O NH

O

O

Butabarbital A. B. C. D. 4.

Butethal

Butabarbital is more water soluble and has a higher LogP than butethal Butabarbital is less water soluble and has a higher LogP than butethal. Butabarbital is more water soluble and has a lower LogP than butethal. Butabarbital is less water soluble and has a lower LogP than butethal. Two isomers of 2-butene exist due to:

A. The variation in the position of the carbon-carbon double bond. B. The variation in the disposition of the methyl groups attached to the two carbons that are connected with a double bond. C. The variation in the numbering system used to name this compound. D. The variation in the electronegativity of the sp3 and sp2 carbons in this compound. 5. Which of the following molecules is not aromatic (i.e., does not follow Huckel’s rule)? A.

OH

C. N

B.

D.

6. Which of the following lists is in increasing order of size (smallest molecule to largest molecule): A. B. C. D.

chloromethane, bromomethane, iodoform (CHI3 ), methanol methanol, bromomethane, iodoform (CHI3 ), chloromethane methanol, chloromethane, bromomethane, iodoform (CHI3 ) iodoform (CHI3 ), chloromethane, methanol, bromomethane

7. Which of the following molecules is most lipophilic? A.

C.

O

CH3

Cl Cl

B.

D.

CCl3

Cl

Cl

8. Which of the following has the larger molecular dipole: A. B.

CHCl3 C2 F6

9.

C. D.

CHF3 CH3 CH2 CH3

Which of the following alcohols is most water soluble?

A. B.

2-butanol 1-butanol

C. D.

1-hexanol 1-octanol

10. If the pKa of an acid is 4.5, at equilibrium at pH 7.0: A. B. C. D.

There is 4.5-times as much deprotonated form of the acid as the protonated form. There is 2.5-times as much protonated form of the acid as the deprotonated form. There is over 102 -times as much protonated form of the acid as the deprotonated form. There is over 102 -times as much deprotonated form of the acid as the protonated form.

11. Which of the following statements concerning the relative acidities of trifluoroacetic acid and acetic acid is most correct? A. Trifluoroacetic acid is more acidic, because the deprotonated form (trifluoroacetate anion) is more inductively stabilized than is the deprotonated form of acetic acid (acetate anion). B. Trifluoroacetic acid is more acidic, because the deprotonated form of acetic acid (acetate anion)l is more inductively stabilized than is the deprotonated form of trifluoroacetic acid. C. Acetic acid is more acidic because the deprotonated form (acetate anion) is more inductively stabilized than is the deprotonated form of trifluoroacetic acid. D. Acetic acid is more acidic because the deprotonated form of trifluoroacetic acid is more inductively stabilized than is the deprotonated form of acetic acid. 12. Which of the functional groups shown in the compound is a tertiary alcohol? H OH OH 3

A. B. C. D. E.

CH2OH

1

2

Functional Group 1 Functional Group 2 Functional Group 3 Functional Groups 1 and 3 None of the functional groups are tertiary

13. Which of the following ethers would be expected to undergo a chemical reaction in the presence of a nucleophile? A.

C.

O

O

O

B.

O

D.

O

14. Which of the following two compounds is less basic (lower pKa for the conjugate acid) and why? NH2

NH2

O2N 4-Ethyl-phenylamine

4-Nitro-phenylamine

A. 4-Nitro-phenylamine is more basic because there is more resonance stabilization of the positive charge in the protonated form than in the case of 4-ethyl-phenylamine. B. 4-Ethyl-phenylamine is more basic because there is more resonance stabilization of the positive charge in the protonated form than in the case of 4-nitro-phenylamine. C. 4-Nitro-phenylamine is more basic because there is less resonance delocalization of the nitrogen lone pair in the conjugate base form than in the case of 4-ethyl-phenylamine. D. 4-Ethyl-phenylamine is more basic because there is less resonance delocalization of the nitrogen lone pair in the conjugate base form than in the case of 4-ethyl-phenylamine. 15. Which of the following chemical equations best represents the oxidation of a phenol? A

OH

O Air

B

OH

O Air

C

O CO2H

OH Air

D

OH Air

O

16. At neutral pH, which of the following amines would exist predominantly as the protonated, ionized form? A.

C.

NH2

H N

O2N

B.

D.

NH2

O NH

17. Which of the functional groups in the following compound is a tertiary amine? 2 3

HN

N

OH

NH2

A. B. C. D. E.

Functional group 1 Functional group 2 Functional group 3 Functional group 1 and 3 None of the functional groups indicated are tertiary amines.

1

18. Which of the following expressions best represents the predicted water solubility of the compound shown below?

HN

N

OH

NH2 A. Each of the three amine functional groups can solubilize approximately 34 carbon atoms (9–12 carbons for the three amine functional groups) and the hydroxyl functional group can solubilize 3–4 carbons. Because the number of carbon atoms in the molecule (14) is equivalent to the estimated number of carbons that can be solubilized (12–16), we can predict that the compound is water soluble. B. Each of the three amine functional groups can solubilize approximately 1 carbon atom (3 carbons for the three amine functional groups) and the hydroxyl functional group can solubilize 1 carbon. Because the number of carbon atoms in the molecule (14) is greater than the estimated number of carbons that can be solubilized (4), we can predict that the compound is insoluble in water. C. Each of the three amine functional groups can solubilize approximately 0.7 carbon atoms (2.1 carbons for the three amine functional groups) and the hydroxyl functional group can solubilize 1 carbon. Because the number of carbon atoms in the molecule (14) is greater than the estimated number of carbons that can be solubilized (3.1), we can predict that the compound is water soluble. D. Each of the three amine functional groups can solubilize approximately 67 carbon atoms (18–21 carbons for the three amine functional groups) and the hydroxyl functional group can solubilize 5–6 carbons. Because the number of carbon atoms in the molecule (14) is less the estimated number of carbons that can be solubilized (23–27), we can predict that the compound is insoluble in water.

19. Which of the following expressions best represents the predicted LogP of the compound shown below?

HN

N

OH

NH2 A. Each of the three amine functional groups has a π amine value of –1.0, and the hydroxyl functional group has a π hydroxyl value of 2.0, so the predicted logP is (3 x –1.0) + 2.0 = -1.0 for this compound. B. Each of the three amine functional groups has a π amine value of –1.0, and the hydroxyl functional group has a π hydroxyl value of –1.0. The phenyl ring has a πphenyl of 2.0 and each of the remaining 8 carbons have a π C (aliphatic) value of 0.5. There is an additional πIMHB value associated with the intramolecular hydrogen bond that can occur between the primary amine and hydroxyl functional groups (0.65) so the predicted logP is (3 x –1.0) + (-1.0) + (8 x 0.5) + 0.65 = 0.65 for this compound. C. Each of the three amine functional groups has a π amine value of 3–4, and the hydroxyl functional group has a πhydroxyl value of 3–4. The phenyl ring has a πphenyl of 2.0 and each of the remaining 8 carbons have a π C (aliphatic) value of 0.5. There is an additional π IMHB value associated with the intramolecular hydrogen bond that can occur between the primary amine and hydroxyl functional groups (0.65) so the predicted logP is (3 x 3.5) + (3.5) + (2.0) + (8 x 0.5) + 0.65 = 20.65 for this compound. D. The compound contains at least one amine functional group so there is a π amine value of –1.0, and it also contains one or more hydroxyl functional groups with a π hydroxyl value of –1.0. The compound contains one or more phenyl rings, so there is a π phenyl of 2.0. There are additional carbons, so there is a π C (aliphatic) value of 0.5. So the predicted logP is (–1.0) + (-1.0) + (2.0)+(0.5) = 0.5 for this compound.

20. Which of the functional groups in the following compound would be expected to have a pKa of around 10 (i.e., higher than 5, but lower than 15)? Me

Me N

2

NH2

1

OH OH

O

OH

O

O

3

A. B. C. D.

Functional group 1 Functional group 2 Functional group 3 Functional groups 1 and 3

21. Which of the chemical equilibria shown below is most likely to occur when chloral (Cl3 CCHO) is dissolved in water? O

O Cl

A.

Cl

H

+

H2 O

Cl

Cl

OH

Cl

Cl O

O Cl

B.

H

+

H2O

Cl Cl Cl

Cl Cl O Cl

C.

HO H

+

H2O

Cl

E.

+

HCl

OH H

O

O

D.

H3O +

Cl Cl

Cl Cl

Cl

+

H

+

H2O

Cl Cl

No chemical equilibrium is involved.

HO Cl Cl

H

22. Which of the following best explain the difference in pKa between benzoic acid and para-nitrobenzoic acid? CO2 H

O2 N

pKa = 4.21

CO2 H

pKa = 3.50

A. In the ionized form of benzoic acid there is complete delocalization of the carboxylate negative charge throughout the benzene ring; whereas, in the ionized form of para-nitrobenzoic acid, the is no delocalization of the carboxylate charge by the benzene ring. B. In the ionized form of para-nitrobenzoic acid there is complete delocalization of the carboxylate negative charge throughout the benzene ring; whereas, in the ionized form of benzoic acid, the is no delocalization of the carboxylate charge by the benzene ring. C. In the ionized form of para-nitrobenzoic acid there is a favorable inductive interaction; whereas, in the ionized form of benzoic acid, this inductive interaction is missing. D. In the ionized form of benzoic acid there is a favorable inductive interaction; whereas, in the ionized form of para-benzoic acid, this inductive interaction is missing. E. The ionized form of para-nitrobenzoic acid, being larger than the ionized form of benzoic acid, is better solvated and thus more stable in solution. 23.

The sodium salt of benzoic acid is: A. B. C. D.

More water soluble than benzoic acid. Less water soluble than benzoic acid. Has approximately that same water solubility as benzoic acid. Is unstable in water because is undergoes a fast reaction to form carbon dioxide.

23-25. For each of the structures presented below, match the circled functional group with the proper name from the list of names below. NOTE: Each lettered functional group name can be used once, more than once, or not at all in answering the following three questions. A. B. C. D.

Amide Alkene Phenol Amine

23.

24. H N

O H2N

S

O

CH2

N O CO2H

A

HO C 25.

CO2H

N Me

HN

B

CH C

OH

Basic Principles of Medicinal Chemistry Lab Example Questions for EXAM #1 ARROW PUSHING AND REVIEW LABS

PHR 143P

Fill in your answers to the questions below ON THIS SHEET. Keep your answers brief. ONLY ANSWERS WITHIN THE BOXED AREAS WILL BE GRADED. 1. Examine the aqueous equilibrium below. OH

OH

O

HO

OH

O

O HO

OH

O

OH

O

+

H+

-

(A) (5 points) Are there one or more reasonable resonance structures for the conjugate base that appears in the box? If so, draw them in the space below. \

(B) (5 points) Based upon your answer to part A above, predict the approximate pKa associated with the above equilibrium. HINT: would you predict the pKa is most similar to an alcohol (pKa ~15), a phenol (pKa ~ 10), or a carboxylic acid (pKa ~5). Explain how you made your prediction. \

2. (2 points) Draw the Lewis dot structure for formic acid (HCO2 H). \

3. (3 points) Draw the Lewis dot structure and any reasonable resonance forms for methylazide (CH3 N3 ).

\

4. (3 points) Draw the Lewis dot structure and any reasonable resonance forms for dimethylsulfoxide (CH3 SOCH3 ). \

5. (2 points) Circle the atoms that bear a partial positive charge in the following compound: O

6. (2 points) Circle the atoms that bear a negative charge or a partial negative charge in the following compound: O– NH2 OH

OH OH O

O

O

7. (8 points) Show the electron flow (push arrows) for the following reaction: Show the movement of electrons by curved arrow notation for the following transformation:

O H3 C

O S

–

CH3

O

O

O

O S

CH3 O

O CH3

H

O

H

O

N

N

N

N O

CH3

O

H

CH3

H

8. (10 points) Show the- electron flow (push arrows) for the following reaction: Cl

Me Me

Cl

N

N+

Cl MeOH

Me

Cl

OMe

OMe

N

Me + HCl

MeOH

Me

N

OMe OMe

Cl Cl

-

N+ + HCl

9. (10 points) Using the pKa values given, decide whether the equilibrium will lie toward the right or the left. Give a justification for your choice. O O2N

pKa

C

O O– + CH3NO2

10.2

O2N

C

OH + –CH2NO2

3.44

10. (5 points) A solution of formic acid (0.01 M) is 13% ionized. What is the pH of this solution?

11. Below are two isomers of nitrophenol with their corresponding pKa’s. (A) (5 points) Write out the chemical equations for the ionization of these two phenols (the chemical equation to which the pKa’s refer). OH OH

NO2 pKa = 7.2

NO2 pKa = 8.3

(B) (10 points) With regards to the ionized form of each phenol, indicate the difference, if any in the contribution of the following in terms of stabilizing or destabilizing effects on one isomer versus the other: -

Inductive Effects Resonance

(C) (5 points) Based upon your answer to part B, explain the observed difference in pKa as a consequences of differences in one or more of these effects.

12. Lidocaine is a weak organic base that exist in aqueous solution as a mixture of Lidocaine and the protonated form, LidocaineH +. Me NH Me O

N

(A) (5 points) Draw out the equilibrium involving the protonated and non-protonated froms of Lidocaine in water. Draw out the complete structure of the protonated form, showing the site of protonation.

(C) (5 points) What is the pK a for Lidocaine?

13. (2 points) Show the electron flow (push arrows) for the following reaction: Show the movement of electrons by curved arrow notation for the following transformation:

Br

+

–

Br

14. (A)(4 points) You need to prepare 30 mL of a 10 mM solution of KCl.. How much of a 2.5M solution of KCl would you need in order to make your 10 mM solution?

(B) (4 points) The 10 mM KCl solution that you prepared was further diluted to a total volume of 250 mL. What is the final chloride ion concentration of this 250 mL of solution (assuming that the water you used has no chloride ions).

15. (A) (5 points) You are in charge of a biopharmaceutical laboratory and you are extracting drug metabolites from a patient’s blood. You extracted 3 mL of blood plasma with with 20 mL of dichloromethane. A 2 mL aliquot of the bottom, dichloromethane layer was removed and diluted with 2 mL of acetonitrile. A 1 mL aliquot of the resulting solution was diluted to 25 mL total volume with 1:1 dichloromethane/acetonitrile in a volumetric flask. Approximately 0.5 mL of this solution was removed and placed in a cuvette. By measuring the absorbance of this solution, it was determined that the concentration of drug metabolite in this solution was 0.04 nM. What was the concentration of drug metabolite in the original dichloromethane solution (underlined above)?

15(B) (5 points) Assuming that the extraction step was quantitative, how much drug metabolite was present in the 3 mL of blood plasma? What was the concentration of metabolite in the blood plasma?