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# From the list of structures on the right, select the major product formed when the following alkyl bromide:

From the list of structures on the right, select the major product formed when the following alkyl bromide:

1) is treated with sodium methoxide in DMSO.

2) is treated with sodium t-butoxide in DMSO.

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When considering the reactions of alkyl halides, we need to remember that a unimolecular mechanism (${\text{S}}_{\text{N}}\text{1}$ and E1) is supported by a weak base/poor nucleophile, and a bimolecular mechanism (${\text{S}}_{\text{N}}\text{2}$ and E2) is supported by a strong base/good nucleophile.

The combination of an alkoxide (a strong base) and DMSO (a polar aprotic solvent) creates ideal ${\text{S}}_{\text{N}}\text{2}$/E2 conditions. Since tertiary alkyl halides cannot undergo an ${\text{S}}_{\text{N}}\text{2}$ reaction, we can definitely conclude that both reactions are E2 eliminations.

In an E2 reaction, the base removes a proton from a carbon in a β‑position to the halide. There are three β‑carbons relative to bromine. Which one is going to lose its proton to the methoxide base? Since a more substituted alkene is more stable, the reaction typically follows Zaitsev's rule, and the proton is removed from a β‑carbon that is bonded to the fewest hydrogens − in this case, the tertiary carbon.

However, when the base is bulky, as is ${\text{t}}$‑butoxide, its approach to the tertiary carbon is sterically hindered. Consequently, the bulky base will remove a more accessible hydrogen from one of the β‑methyl groups, thereby generating the less substituted alkene as the major product. The more substituted alkene is only a minor product in the second reaction.

ans 1. c. ans2. E

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