Question

stereochemistry of the products of reduction depends on the reagent used,as you learned in Sections 20.5 and 20.6. With this in mind, how would you convert3,3-dimethylbutan-2-one $\left[{\mathrm{CH}}_3\mathrm{COC}\left({\mathrm{CH}}_3\right)\right]$ to: (a) racemic 3,3-dimethylbutan-2-ol $\left[{\mathrm{CH}}_3\mathrm{CH}\left(\mathrm{OH}\right)C{\left({\mathrm{CH}}_3\right)}_3\right]$;(b) only (R)-3,3-dimethylbutan-2-ol; (c) only (S)-3,3-dimethylbutan-2-ol?

Short answer

(a) You can convert 3,3-dimethylbutan-2-one to racemic 3,3-dimethylbutan-2-ol by the following method:

(b) You can convert 3,3-dimethylbutan-2-one to (R)-3,3-dimethylbutan-2-ol by the following method:

(c) You can convert 3,3-dimethylbutan-2-one to (S)-3,3-dimethylbutan-2-ol by the following method:

Step-by-step solution

Function of  NaBH4

${\mathbf{NaBH}}_{\mathbf{4}}$ is a strong reducing agent that reduces ketone to secondary alcohol and aldehyde to primary alcohol. Since chiral alcohol is formed after reduction, two kinds of alcohols are formed.

They are collectively called enantiomeric alcohols.

Function of CBS

CBS is a chiral reducing agent. It is a boronic compound that functions as an asymmetric reduction and forms enantiomeric alcohols.

(S)-CBS reagent provides hydrides that can attack the carbonyl carbon center position from the front and yields alcohol having an R-configuration where the alcohol group lies in the dash position.

The hydrides from (R)-CBS attack the carbonyl carbon center position from the backand yields alcohol having an S-configuration where the alcohol group lies in the wedge position.