When a reactant that has an asymmetric center forms a product with a second asymmetric center, will the product contain diastereomers in unequal amounts?

1 Answer
Nov 23, 2015

Not necessarily.

This is a difficult question, because I would have to show a definitive counterexample. If I couldn't think of one, it wouldn't mean that the answer is yes. If I tried to find an example that affirmed the questioner, it would leave doubt.

So, suppose we want to prove that the answer is "not necessarily." That prompts us to find one example where one chiral compound reacts with another compound to form one product with two chiral centers, for which there is a racemic mixture. If one such example exists, then the answer is "not necessarily."

To do this, let's say we had one chiral reactant that reacts with something else in an #"S"_N1# reaction.


In an #\mathbf("S"_N1)# reaction, racemic mixtures are made due to the generation of a planar carbocation intermediate.

(This is because the nucleophile has an equal probability of attacking on either side of the plane.)

Thus, the products, which you can see are diastereomers (one or more, but not all, related stereocenters differ, and the two isomers are NOT mirror images of each other), by definition were made as a racemic mixture.

You can see that more clearly if we rotate the molecule on the right by #180^@# about the vertical axis.

(So we can see that they are not enantiomers, because they aren't mirror images, yet they are non-superimposable.)

Overall, the conclusion that I've come to is... "not necessarily."