 Solution

The reacting Homo is the tertiary-amine nitrogen lone pair. The relevant LUMO is the C-I antibonding orbital, ${\sigma }_{\mathrm{C}-\mathrm{I}}^{\ast }$, 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.