# 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

## 1 Answer

The two commenters are collectively correct. But to clarify and augment their answers I would say that structure (d) cannot be correct because it is missing 2 carbon atoms relative to (c), and if you tried to cyclize (c) by forming esters, the result would be highly strained. Try making a molecular model of it and see how it looks.

Below is the product you would get if (c) cyclized. What is the difference between this product and (d)?

• Did not notice the missing methylene groups. So you say that even if the methylene groups were present, esterification wouldn't take place because of high ring strain? – user8244 Jun 29 '15 at 12:56
• That would be my back-of-the-envelope prediction. But I could be wrong. In any case it is not one of the choices presented in the question. Neither is the asymmetrical Canizzaro product. – iad22agp Jun 29 '15 at 13:54