Question

Is the reduction of benzoin to hydrobenzoin using NaBH4 and EtOH stereoselective: will only meso (cis) product be formed or a mixture of meso/racemic products?

Answer 1

In this reaction the use of Sodium Borohydride (NaBH4) will cause the reduction of the Benzoin so that the Sodium cation will chelate with the oxygen atoms of both of the central carbons (the carbon linkage between the 2 phenyl groups), that way it will in a sense ‘freeze’ the stereo isomer into a meso fashion. Now the carbonyl will be reduced and the resulting molecule will be a meso product. Even though there will be some of the other isomers made but the favoured product will be the meso compound.

Another way to look at the product is through what is known as Cram’s Rule. Cram’s Rule states that:

"If the molecule is observed along it’s axis… …the oxygen of the carbonyl orients itself between the small and the medium sized groups. The rule is that the incoming group preferentially attacks on the side of the plane containing the small group"

So, when we look at the projection of the benzoin compound we see that the carbonyl oxygen is oriented (for steric purposes) between the hydrogen and hydroxyl of the next carbon:

and that the NaBH4 will attack from the plane of the hydrogen causing the phenyl group to orient between the hydrogen and phenyl of the other carbon thus leaving the hydroxyl groups in the meso isomer.

The meso isomer. A meso isomer is a molecule with 2 chiral centres in which the attached functional groups on each are the same.

A meso compound is one whose molecules are superimposable on their mirror images even though they contain chiral centres.

So, now you know , meso (cis) is favoured but overall both meso/racemic product will form.