310N, Chem. Spring 2008 Professor Krische Midterm Exam II Average = 50 Grading Scale
90-100 = A+ 80-89 = A 70-79 = A65-69 = B+ 60-64 = B 55-59 = B50-54 = C+ 45-49 = C 40-44 = C37-39 = D+ 33-36 = D 30-32 = D0-29 = F
Name:
Problem 1. (16 points)
2. (24 points)
3. (12 points)
4. (18 points)
5. (20 points)
6. (10 points)
Total Points:
/100
1. (16 points) For the following pairs of compounds, circle the compound that reacts faster with the indicated nucleophile. For the faster reacting compound only, draw the product of its reaction with the indicated nucleophile.
A.
O versus Cl3C OCH3 H3C OCH3 O CH3MgBr (2 Equivalents) Then Dilute HCl (aq.) HO Cl3C CH3 OCH3
B.
O Cl O CH3 versus Cl CH3 O H3C H N CH3 H3C Pyridine N CH3 O CH3
C.
O O versus O O NaBH4 (2 Equivalents) Then Dilute HCl (aq.) HO OH
D.
O O O O versus O O HS S HO O
2. (24 points) Hydrolysis of benzonitrile produces benzoic acid. The hydrolysis occurs in two stages. The first stage involves hydrolysis of the nitrile to furnish an amide. The second stage involves hydrolysis of the amide to furnish an acid. Propose a mechanism to account for the benzonitrile hydrolysis. For each intermediate, draw all lone pair electrons, any formal charges, and all important resonance structures. Use the arrow pushing formalism to interconvert intermediates. A. First Stage: Hydrolysis of the nitrile to furnish an amide.
O N
HCl (1 eq.)
NH2
H2O,
H N H Cl H N N
H2O
H O Cl N H N H H Cl
H O
H
H O H N H
H O H N H
Cl
O
NH2
2 (continued) B. Second Stage: Hydrolysis of the amide to furnish an acid.
O O
NH2
HCl (1 eq.)
OH
+
NH4Cl
H2O,
H O O H NH2
Cl
H O
NH2
NH2
H O O
H O
Cl
H
NH2
NH2
H NH2 OH
Cl
OH2
H
O
H
H O NH3 NH3 OH O
H O NH3 OH OH
NH4
3. (12 points) Propose mechanisms to account for the following transformations. For each intermediate, draw all lone pair electrons, any formal charges, and all important resonance structures. Use the arrow pushing formalism to interconvert intermediates.
O H3C OCH3 H3C CH3 H3C NaOCH2CH3 HOCH2CH3
O
OCH2CH3 H3C CH3
Na O H3C OCH3 H3C CH3 Na OCH2CH3 H3C CH3 H3C O OCH3 OCH2CH3
O H3C OCH2CH3 H3C CH3
4. (18 points) Draw the major product expected when the indicated materials starting are subjected to the following multi-step transformations. A.
O 1) LiAlH4, Then H2O NH 2) CH3COCl, Pyridine N CH3 O
B.
O 1) HCl (aq.), Heat Ph CH3 2) OsO4, Then NaHSO3 Ph OH O OH CH3 Ph
C.
1) O N H Cat. HCl, D. S. Trap Benzene Solvent O 2) Cl Than H2O CH3 O O
CH3
4. (Continued) D.
OH OH 1) PBr 3 2) NaCN, DMSO 3) H2SO4(1 eq.), H2O O
E.
O 1) LDA, -78 C 2) O H CH3 Then H2O
o
O
OH
F.
O O
HO
1) SOCl2, Pyridine 2) (CH3)2CuLi
H3C
5. (20 points) The bromination of "acetophenone" under acidic conditions occurs by way of an enol intermediate. A. (15points) Propose a mechanism for this transformation. For each intermediate, draw all lone pair electrons, any formal charges, and all important resonance structures. Use the arrow pushing formalism to interconvert intermediates.
O OH O
H3C
Br2, HBr Br
H O H H3C Br H3C O
H O
H Br Br
H Br Br H2 C O Br
H O O
Br
Br
B. (05 points) Based on the mechanism you have proposed, clearly explain why products of "overbromination" are not observed in this transformation. Draw structures that illustrate key points. The first bromine atom removes electron density from the adjacent carbonyl through inductive effects causing the lone pairs on the carbonyl to be less basic. This hinders the reaction from proceeding because...