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?


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.)


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.