Q.$\text{tert}$-Butyl bromide solvolyses in aqueous ethanol far faster than methyl bromide under the same conditions. Explain this using carbocation stability.

Solvolysis proceeds by the $S_N1$ mechanism, whose rate-determining step is ionisation of the substrate to form a carbocation. $\text{tert}$-Butyl bromide ionises to the tertiary $\text{tert}$-butyl cation, which is substantially stabilised by hyperconjugation and inductive electron donation from its three attached methyl groups, giving this ionisation a comparatively low activation energy. Methyl bromide, by contrast, would have to ionise to the bare methyl cation, which has no adjacent carbon to offer any stabilisation at all and is exceptionally high in energy -- so high that this ionisation essentially does not occur under ordinary conditions. Since the rate of $S_N1$ tracks the ease of this ionisation step direc …

Trace the rate difference back to the stability of the specific carbocation each substrate would have to form, since that stability direc …

Explaining the rate difference in terms of steric hindrance (relevant to SN2, not SN1) rather than carbocation sta …