Exam 22008
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Chemistry 350 - Exam 2 Answers - April 3, 2008 Problem 1. (16 points) Fill in the blank with the correct organic product or reactant. If no reaction takes place, write NR. a)
H3C
CH2Br
NR - 2 points
DMSO + NaCN
CH2Cl
CH2OH
2 points
b)
+
SOCl2
+
I
Br
c)
NR - 2 points
O H3CS
OH
H3C
H
H + PPh3
d)
CH3
+ DEAD
2 points
+ CH3SH
1 pt if stereochem wrong or not shown
HO H
e)
+
f)
O H3C S Cl O
pyridine
O H 3C S O H O OR
2 pts 0 pts if stereochemistry is wrong
2 pts
C Cl
+
CH3OH (solvent)
Give 1 pt if they answer NR and indicate that the reaction would require heat.
C OMe
g)
Me Me
N Me
h)
O
MsO H
2 pts
! (Heat)
Me
CH3 N H3C CH3
I
+
OSiMe3 +
(CH3)3SiCl
1
CH3I
2 pts
Problem 2. (7 points) Assign the correct absolute configuration (R or S) in the spaces provided. If the indicated carbon is not a stereocenter, write “achiral” on the line.
achiral R
O O H R
S
OH
H O
NHMe
THC
MDMA 9
(Better known as Ecstasy)
(Δ -tetrahydrocannabinol, plant defense substance in Cannabis)
2 points for each R or S, 1 point for the achiral. No partial credit
2
Problem 3. (10 points) a) Each molecule has been assigned a letter. For those molecules that could be prepared using an SN2 reaction, write YES in the blank below. Otherwise write NO. Total score = number right minus number wrong to a minimum of zero. O
CH4
A NO
O
B
C
Yes or No give both credit
Yes
C
P CH3
D
SCH3
E
Yes
Yes
b) Each molecule has been assigned a letter. For those molecules that could be prepared using an SN1 reaction, write YES in the blank below. Otherwise write NO. Total score = number right minus number wrong to a minimum of zero. H
O C
OCH3
O
H
OCH2CH3
CH2OH H3C
CH3
O C
N H
(S) ONLY
A Yes
B No
C
D Yes
Yes
1 pt for each correct answer minus # incorrect, to a minimum of zero pts. 8 correct – 2 incorrect = 6 pts 4 correct – 6 incorrect = 0 pts No answer = wrong answer
3
E No
CH2CH3
Problem 4. (39 points) Mechanism Problems: Provide detailed, step-by-step mechanisms for the following reactions. In this question, we provide the reactants, products and reaction conditions. YOUR JOB IS TO: 1) Show the individual chemical steps and all intermediates formed in the process of converting the reactant to the product. 2) Use curved arrows to show all changes in bonding and lone pair electrons. 3) Show all formal charges and all contributing resonance structures. Remember, the sequence of steps you propose must convert the reactants into the products using only the materials provided. You may not use additional acids, bases or other reagents, since they are not present in the reaction mixture. 4) In some cases other products may form. Only show the mechanism that gives the indicated product. a) (12 points) O
CN Cl
NC
+ Cl
O
DMF solvent
3 points for each correct arrow. They do not need to redraw the product, or show Cl- as one of the products. O
O
NC
NC
O
Cl
Cl
CN
This answer also gets full credit. O
NC O
Cl CN
4
+ Cl
+ Cl
b) (12 points) O CH2Br
+
H
H
C
CH2 H
OH
O C
H
+ HBr
O 3 pts
2 points
2 points
CH2
Br
O H O H
O
H
2 points O
O OH
H
If the lone pairs on this oxygen are used as a nucleophile to attack the carbocation, rather than the lone pairs on the carbonyl oxygen atom, then a maximum of 8 points are given for this answer. 3 points
H
O
H
O
H
O
B H
O
This can be any reasonable base including :B, Br-, or they can simply write -H+
5
c) (15 points) In addition to the mechanism, give a one sentence explanation for why this reaction gives racemic products. OH Cl
O S
Cl Cl
+ SO2 + Racemic
N
OH
Cl
O S
Cl
N
3 points
N H
H
O
O S
Cl
O S
O
3 points
Cl
Cl
3 points
Cl Cl
- SO2 Even though the problem 3 points asked for all contributing resonance structures, this is they only one that is assigned points.
O
O S
They do not have to show pyridinium chloride or sulfur dioxide.
+ N H
Racemic
Cl O
O S
Cl
SO2 + Cl
Cl
OK to show this in place of the third step
N
H
O
O
Cl
O S
Cl
Total of 3 points for the first two steps if they show deprotonation of the alcohol by pyridine before attack on thionyl chloride, as long as all of it is correct, as shown above.
O
O S
Cl
Cl
Cl Total of 3 points for the last two steps if they show it as an SN2 reaction instead of SN1, as long as all of it is correct, as shown above, including inversion of stereochemistry.
3 points for the correct explanation. This reaction gives racemic products because the substrate is benzylic and the intermediate carbocation is stabilized by resonance with two aromatic rings. The reaction proceeds through an SN1 mechanism, rather than SN2.
6
Problem 5. (8 points) For each series of compounds below, fill in the blanks with the letter that corresponds to the correct structure. a) (2 points) CO2H
CO2H
SH
F
A
CO2H
B
C A
B
strongest acid ________
weakest acid ________
b) (2 points) O O S CF3 O
O O S CH3 O
A
B
O O S CH3 O
C
C
A
strongest electrophile ________
weakest electrophile ________
c) (2 points) NH
NH
CN
N
A
B A
C C
strongest base ________
weakest base ________
d) (2 points) Me3Si
Me3C
A
B B
Me3P
C C
strongest base ________
weakest base ________
7
Problem 6. (20 points) Synthesis Problems: Propose a series of reactions (reagents and conditions, if necessary) that convert the Starting Material into the Final Product. You can add any other molecules to your reactions, as long as all of the carbon atoms in the starting material are present in the final product. To get credit, your starting material and product must contain the stereocenters exactly as specified in the problem. You do NOT need to draw mechanisms for each reaction. Structure your answer as follows: Starting Material(s)
Reagents-1
Product 1 Reagents-2
Product 2
etc.
Final Product (with correct stereochemistry)
a) (10 points) OH
O
starting material
final product
3 pts for correct alcohol with correct stereochemistry (S) OH
2 pts OH
TsCl - 2 pts pyridine - 1 pt
O
OTs NaH 2 pts for strong base. LDA also OK
starting material
final product
Also OK PBr3 (3 pts) to give BnBr (2 pts) SOCl2 (2 pts) and pyridine (1 pt) to give BnCl (2 pts) HBr (2 pts) and H2SO4 (1 pt) to give BnBr (2 pts) HI (3 pts) to give BnI (2 pts)
This answer gets NO CREDIT
or tosylate, mesylate, chloride, iodide, etc.
Br OH
O NaH
starting material
final product
8
b) (10 points) OH O
N3
O
starting material
final product PBr3 Br
OH O
O SOCl2, pyridine
Cl
OH O
O TsCl (or MsCl)
OR
Cl or Br OTs
OH pyridine
Cl
O
O
O
2 pts
3 pts
5 pts total for any of the above correct answers for the first step 0 pts for Mitsunobu with HCl or HBr 0 pts for OTs or OMs formation with inversion of stereochemistry
N3
OR Cl O
N3
Br O
O
5 pts for this correct answer to the second step 0 pts if this reaction is shown without inversion of configuration
9
Br O
H3C
CH2Br
NR - 2 points
DMSO + NaCN
CH2Cl
CH2OH
2 points
b)
+
SOCl2
+
I
Br
c)
NR - 2 points
O H3CS
OH
H3C
H
H + PPh3
d)
CH3
+ DEAD
2 points
+ CH3SH
1 pt if stereochem wrong or not shown
HO H
e)
+
f)
O H3C S Cl O
pyridine
O H 3C S O H O OR
2 pts 0 pts if stereochemistry is wrong
2 pts
C Cl
+
CH3OH (solvent)
Give 1 pt if they answer NR and indicate that the reaction would require heat.
C OMe
g)
Me Me
N Me
h)
O
MsO H
2 pts
! (Heat)
Me
CH3 N H3C CH3
I
+
OSiMe3 +
(CH3)3SiCl
1
CH3I
2 pts
Problem 2. (7 points) Assign the correct absolute configuration (R or S) in the spaces provided. If the indicated carbon is not a stereocenter, write “achiral” on the line.
achiral R
O O H R
S
OH
H O
NHMe
THC
MDMA 9
(Better known as Ecstasy)
(Δ -tetrahydrocannabinol, plant defense substance in Cannabis)
2 points for each R or S, 1 point for the achiral. No partial credit
2
Problem 3. (10 points) a) Each molecule has been assigned a letter. For those molecules that could be prepared using an SN2 reaction, write YES in the blank below. Otherwise write NO. Total score = number right minus number wrong to a minimum of zero. O
CH4
A NO
O
B
C
Yes or No give both credit
Yes
C
P CH3
D
SCH3
E
Yes
Yes
b) Each molecule has been assigned a letter. For those molecules that could be prepared using an SN1 reaction, write YES in the blank below. Otherwise write NO. Total score = number right minus number wrong to a minimum of zero. H
O C
OCH3
O
H
OCH2CH3
CH2OH H3C
CH3
O C
N H
(S) ONLY
A Yes
B No
C
D Yes
Yes
1 pt for each correct answer minus # incorrect, to a minimum of zero pts. 8 correct – 2 incorrect = 6 pts 4 correct – 6 incorrect = 0 pts No answer = wrong answer
3
E No
CH2CH3
Problem 4. (39 points) Mechanism Problems: Provide detailed, step-by-step mechanisms for the following reactions. In this question, we provide the reactants, products and reaction conditions. YOUR JOB IS TO: 1) Show the individual chemical steps and all intermediates formed in the process of converting the reactant to the product. 2) Use curved arrows to show all changes in bonding and lone pair electrons. 3) Show all formal charges and all contributing resonance structures. Remember, the sequence of steps you propose must convert the reactants into the products using only the materials provided. You may not use additional acids, bases or other reagents, since they are not present in the reaction mixture. 4) In some cases other products may form. Only show the mechanism that gives the indicated product. a) (12 points) O
CN Cl
NC
+ Cl
O
DMF solvent
3 points for each correct arrow. They do not need to redraw the product, or show Cl- as one of the products. O
O
NC
NC
O
Cl
Cl
CN
This answer also gets full credit. O
NC O
Cl CN
4
+ Cl
+ Cl
b) (12 points) O CH2Br
+
H
H
C
CH2 H
OH
O C
H
+ HBr
O 3 pts
2 points
2 points
CH2
Br
O H O H
O
H
2 points O
O OH
H
If the lone pairs on this oxygen are used as a nucleophile to attack the carbocation, rather than the lone pairs on the carbonyl oxygen atom, then a maximum of 8 points are given for this answer. 3 points
H
O
H
O
H
O
B H
O
This can be any reasonable base including :B, Br-, or they can simply write -H+
5
c) (15 points) In addition to the mechanism, give a one sentence explanation for why this reaction gives racemic products. OH Cl
O S
Cl Cl
+ SO2 + Racemic
N
OH
Cl
O S
Cl
N
3 points
N H
H
O
O S
Cl
O S
O
3 points
Cl
Cl
3 points
Cl Cl
- SO2 Even though the problem 3 points asked for all contributing resonance structures, this is they only one that is assigned points.
O
O S
They do not have to show pyridinium chloride or sulfur dioxide.
+ N H
Racemic
Cl O
O S
Cl
SO2 + Cl
Cl
OK to show this in place of the third step
N
H
O
O
Cl
O S
Cl
Total of 3 points for the first two steps if they show deprotonation of the alcohol by pyridine before attack on thionyl chloride, as long as all of it is correct, as shown above.
O
O S
Cl
Cl
Cl Total of 3 points for the last two steps if they show it as an SN2 reaction instead of SN1, as long as all of it is correct, as shown above, including inversion of stereochemistry.
3 points for the correct explanation. This reaction gives racemic products because the substrate is benzylic and the intermediate carbocation is stabilized by resonance with two aromatic rings. The reaction proceeds through an SN1 mechanism, rather than SN2.
6
Problem 5. (8 points) For each series of compounds below, fill in the blanks with the letter that corresponds to the correct structure. a) (2 points) CO2H
CO2H
SH
F
A
CO2H
B
C A
B
strongest acid ________
weakest acid ________
b) (2 points) O O S CF3 O
O O S CH3 O
A
B
O O S CH3 O
C
C
A
strongest electrophile ________
weakest electrophile ________
c) (2 points) NH
NH
CN
N
A
B A
C C
strongest base ________
weakest base ________
d) (2 points) Me3Si
Me3C
A
B B
Me3P
C C
strongest base ________
weakest base ________
7
Problem 6. (20 points) Synthesis Problems: Propose a series of reactions (reagents and conditions, if necessary) that convert the Starting Material into the Final Product. You can add any other molecules to your reactions, as long as all of the carbon atoms in the starting material are present in the final product. To get credit, your starting material and product must contain the stereocenters exactly as specified in the problem. You do NOT need to draw mechanisms for each reaction. Structure your answer as follows: Starting Material(s)
Reagents-1
Product 1 Reagents-2
Product 2
etc.
Final Product (with correct stereochemistry)
a) (10 points) OH
O
starting material
final product
3 pts for correct alcohol with correct stereochemistry (S) OH
2 pts OH
TsCl - 2 pts pyridine - 1 pt
O
OTs NaH 2 pts for strong base. LDA also OK
starting material
final product
Also OK PBr3 (3 pts) to give BnBr (2 pts) SOCl2 (2 pts) and pyridine (1 pt) to give BnCl (2 pts) HBr (2 pts) and H2SO4 (1 pt) to give BnBr (2 pts) HI (3 pts) to give BnI (2 pts)
This answer gets NO CREDIT
or tosylate, mesylate, chloride, iodide, etc.
Br OH
O NaH
starting material
final product
8
b) (10 points) OH O
N3
O
starting material
final product PBr3 Br
OH O
O SOCl2, pyridine
Cl
OH O
O TsCl (or MsCl)
OR
Cl or Br OTs
OH pyridine
Cl
O
O
O
2 pts
3 pts
5 pts total for any of the above correct answers for the first step 0 pts for Mitsunobu with HCl or HBr 0 pts for OTs or OMs formation with inversion of stereochemistry
N3
OR Cl O
N3
Br O
O
5 pts for this correct answer to the second step 0 pts if this reaction is shown without inversion of configuration
9
Br O
