CHEM 4345 EXAM 2

Dr. Bachrach - March 8, 2001

1) (20 points) Which is a stronger acid CH­3OH or CH3OD? Provide a justification.

Since D is heavier than H, the zero point energy associated with the stretch and cleavage of the O-H bond will be higher than the O-D bond. Therefore, the energy to cleave the O-H bond is less than for the cleavage of the O-D bond, making CH­­3OH a stronger acid than CH3OD.

2) (20 points) Pyridine reacts with CH­3I to form N-methyl pyridine. Explain the relative rates for substituted pyridines when reacted with CH3I.

compound

relative rate

1.0

2.27

0.47

0.042

The methyl group is electron donating group and would therefore make the nitrogen slightly more electron rich. Thus, the p-methylpyridine would be a better nucleophile and will react faster with CH3I and pyridine (in accord with the expt.). For the o-methyl and o,o-dimethyl pyridines, the methyl groups still act as electon donors, but now they can interfere with the ability of nitrogen to act as a nucleophile. They sterically block the nitrogen somewhat and will decrease the nucleophilicy, pretty dramatically in the case of the dimethylpyridine.

3) (20 points) Consider the quaternization of dimethylanilines by benzyl arenesulfonates:

A series of Hammett plots reveals that rX = -3.0, rY = +0.5, and rZ = +1.6. (These data are obtained by measuring the rates of a series of reactions where X is varied while keeping Y and Z = H to determine rX, where Y is varied while keeping X and Z = H to determine rY, etc.) In the space below, sketch the transition state for the reaction, showing the distribution of charges on the entering nucleophile, reaction center, and the leaving group.

4) (20 points) The orbital energy diagram for ethylene and butadiene is sketched below. Indicate the orbital interactions that dominate when these undergo a Diels-Alder reaction to form cyclohexene. What substituents might be placed on butadiene to increase the rate of this reaction?

5) (20 points) Consider the elimination reaction:

a) Draw a More O’Ferrall-Jenks diagram for this reaction. Label the axes and corners.

b) Draw the pathway for the E2 reaction, where the bond changes all happen simultaneously. Label any transition states on this path as TS1.

c) Draw the pathway for the reaction where the base removes the proton first, followed by a second step where the leaving group exits. Label any transition states on this path as TS2.