Distinguishing among SN2/SN1/E1/E2 for a secondary bromide


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> **a**. 2-Bromo-3-phenylbutane reacts with $\ce{NaOMe}$ in methanol. What is the mechanism?
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> **b**. This compound then reacted with $\ce{CN-}$ in DMSO. What is the mechanism?
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### My Attempt


**a**. I am debating between $\mathrm{E1}$ and $\mathrm{S\_N1}$ since it is a polar protic solvent and the carbocation can be stabilized by the tertiary carbon and phenyl group. I am leaning towards $\mathrm{E1}$ since $\ce{OMe-}$ is a strong base. Am I correct?


**b**. If the answer to the first question was $\mathrm{E1}$, then 2-phenylbut-2-ene would be the product. I don't see how that would react with $\ce{CN-}$ in DMSO. So, would no reaction occur?




Reaction **a** proceeds via an $\mathrm{E2}$ mechanism. Firstly, elimination reactions are much more favorable in this molecule because of the conjugation of the resulting structure. If the reactant had been 2-bromo-2-phenylbutane, this would have proceeded via an $\mathrm{E1}$ mechanism, as this would delocalize the positive charge around the phenyl ring. There is no such delocalization in the ejection of the $\ce{Br-}$ from 3-bromo-2-phenylbutane though—it would simply be a 2º carbocation. Instead, the delocalization of the negative charge around the phenyl ring resulting form the extraction of the $\ce{H+}$ by $\ce{OMe-}$ is more stabilizing.


$\hspace{1.5cm}$[![E2](https://i.stack.imgur.com/TpjJz.png)](https://i.stack.imgur.com/TpjJz.png)


There are no possible reactions between an alkene and $\ce{CN-}$ in DMSO. If this were a haloalkene, further elimination could occur, and if it were in $\ce{H2O}$, [hydrocyanation](https://en.wikipedia.org/wiki/Hydrocyanation) could occur.


$\hspace{4cm}$[![DMSO](https://i.stack.imgur.com/d41GG.png)](https://i.stack.imgur.com/d41GG.png)


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