Why is the rate of solvolysis of t-butyl chloride lesser than that of 1-chloro-1-methylcyclohexane?

Question

I came across a question regarding relative rate of solvolysis. According to the answer key:

My analysis:

As in all the three cases, the Cl atom is bounded to a tertiary carbon, hence, all of them are ${S_N1}$ reactions. So, the rate of solvolysis will depend on the stability of the carbocation formed. Upon considering hyperconjugation, t-Bu cation is coming out to be more stable than the tertiary cycloalkyl carbocations.

But, according to the given answer, t-BuCl has the least solvolysis rate as compared to the others. Could anyone please explain why is it so?

Answer 1

According to the definition given by The IUPAC Compendium of Chemical Terminology: The Gold Book^[1], solvolysis is defined as:

Generally, reaction with a solvent, or with a lyonium ion or lyate ion involving the rupture of one or more bonds in the reacting solute. More specifically the term is used for substitution elimination and fragmentation reactions in which a solvent species is the nucleophile.

It clearly states that elimination occurring due to a solvent is also considered solvolysis and thus elimination in case of t-butyl is also considered.

This means you are right in your reasoning of t-butyl being the most susceptible to solvolysis among the given choices.

I believe this was an error from the question setter. They must have been referring to the substitution reaction due to the solvent but ended up creating a poorly worded question.

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Reference:

^{(1) The IUPAC Compendium of Chemical Terminology: The Gold Book; Gold, V., Ed.; International Union of Pure and Applied Chemistry (IUPAC): Research Triangle Park, NC, 2019.}