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My teacher stated that allylic carbocations are comparable in stability to tertiary carbocations. $\def\SN#1{\mathrm{S_N}#1}$

With this in mind I am confused why question 13 of this MIT practice test states that these two molecules are only expected to undergo $\SN 2$ reactions:

benzylic reaction

allylic reaction

An $\SN 1$ reaction on the first molecule with the phenyl substituent would yield a primary carbocation, yes, but this primary carbocation is also benzylic, and can be stabilized through resonance.

An $\SN 1$ on the second molecule similarly yields a carbocation, but this is an allylic carbocation, and can easily be stabilized through resonance and delocalization of pi-electrons.

So, would $\SN 1$ be valid pathways of reactions for both of these molecules, or is $\SN 2$ simply the only pathway for these molecules to react?

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1 Answers1

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I planned to add this as a comment but I don't have enough reputation ... A definite answer can only come by referring to the literature and/or from the lab. In the context of an undergraduate exam it is reasonable to accept both possible answers ($\mathrm{S_N1}$ and $\mathrm{S_N2}$ ) for allylic and benzylic halides. $\ce{EtOH}$ solvolysis of benzyl bromide is a known reaction and the hypothesis that it goes via $\mathrm{S_N1}$ is the most reasonable one. Is it possible that the solvolysis acually goes via $\mathrm{S_N2}$ instead? The answer lies on performing the appropriate mechanistic studies.

However, in this case, having checked the actual exam question, it specifies acetone as the solvent. This polar aprotic solvent favours $\mathrm{S_N2}$ conditions, which is aprotic (see this question also). Incidentally, aside from the use of $\ce{KI}$ over $\ce{NaI}$, this halide exchange is basically the Finkelstein reaction, which occurs via $\mathrm{S_N2}$. (Credit to Greg E. who wrote this in a now deleted comment.)

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