# Does esterification require reactants to be present in the same plane?

In the image that follows (7)

I believe that the first step is the Cannizzaro reaction which will give (c) as a product.

But since we are also adding $\ce{H+}$ in the second step, shouldn't an esterification reaction take place? That would give (d) as a final product.

However the answer given is (c). My only guess as to why esterification wouldn't take place is the fact that the phenyl groups are in perpendicular planes, therefore esterification is difficult? Is it necessary to have all the reactants in a plane for esterification?

• The nucleophile doesn't attack from the plane of the carbonyl group, it attacks below or above since this is where the $\pi^*$ orbital is (see chemtube3d.com/…). Personally I don't see why esterification would not occur though. – orthocresol Jun 28 '15 at 13:55
• structure d) lacks methylene groups so it can't be a product. – Mithoron Jun 29 '15 at 11:29