The nitrogen atom is not actually that bulky. The entire phthalimide molecule is flat and lies behind the nitrogen. It won’t result in steric hindrance between the nitrogen lone pair and the sigma-star orbital of the alkyl halide.

Good question! The advantage of using NaH is that it is a poor nucleophile and has no chance of reacting with our alkyl halide. (If we used NaOH, for instance, I might worry about performing an SN2 on that alkyl halide).

Another advantage is that it deprotonates the phthalimide irreversibly – there’s no equilibrium – and that the conjugate acid of H(-) is just H2, which bubbles away.

What happens is that the iodide ion (a good nucleophile!) acts as a nucleophile on the alkyl halide, creating an alkyl iodide, (which has a great leaving group), and then this is displaced by the nucleophile in the Gabriel.

For introductory organic chemistry purposes, it doesn’t really matter. The mechanism is a little easier to draw out with KOH.