Question

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: 

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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 (SN1 and E1) is supported by a weak base/poor nucleophile, and a bimolecular mechanism (SN2 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 SN2/E2 conditions. Since tertiary alkyl halides cannot undergo an SN2 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.

This image shows the three curved arrows for the E 2 elimination. The tertiary proton is removed by alkoxide, a new alkene is formed and the bromine is kicked off. The product is a tetra substituted alkene. This is the major product since it is the more substituted and more stable alkene. This image shows the three curved arrows for the E 2 limination. The primary proton is removed from one of the methyl substituents, a new alkene is formed and the bromine is kicked off. The disubstituted alkene is a minor product since it is less substituted and less stable.

However, when the base is bulky, as is 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. This image shows the analogous mechanism as with the methoxide, but with tert butoxide as the base. THe tertiary proton is removed by the base, an alkene is formed and the bromine is kicked off. This reaction is slower than the other possibility and will form the minor product.  This image shows the analogous mechanism as with the methoxide, but with tert butoxide as the base. The primary proton is removed by the base, an alkene is formed and the bromine is kicked off. This reaction is faster and the major product is the opposite of the major product with methoxide.

ans 1. c. ans2. E

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