The reacting \b[<Homo> is the tertiary-amine <nitrogen> <lone pair>]. The relevant \b[<LUMO> is the C-I <antibonding orbital>, $\sigma^*_{\mathrm{C-I}}$, of MeI]. Backside lone-pair donation into that orbital forms N-C while weakening C-I. The neutral OH group is not first converted to an alkoxide in the stated reaction.

Solution

Past exam of the chemistry course of the year 1 of the natural sciences course of the University of Cambridge / 2021 / Paper 1 / 2 / b / i / Solution

 Solution (source code)