a. 2-Bromo-3-phenylbutane reacts with $\ce{NaOMe}$ in methanol. What is the mechanism?

b. This compound then reacted with $\ce{CN-}$ in DMSO. What is the mechanism?

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?

  • $\begingroup$ I read part b as 2-broom-3-phenylbutane is reacted with cyanide. Otherwise, it doesn't make sense. $\endgroup$
    – jerepierre
    May 7, 2016 at 3:58

1 Answer 1


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.


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 could occur.



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