Question

Which of the following will give Walden inversion?
(a) ${ C }_{ 4 }{ H }_{ 9 }-CHD-Br$
(b)
(c)
(d) All

Answer 1

Walden inversion refers to the inversion of configuration of an optically active compound such that the product is either an enantiomer of the original compound (if the original compound has only one stereo-center) or is a diastereoisomer of the original compound (if the original compound has more than one stereo-center).

Complete step by step answer:
In order to solve this question we first need to know the meaning of Walden inversion. It basically refers to the inversion of configuration of an optically active compound such that the product is either an enantiomer of the original compound (if the original compound has only one stereo-center) or is a diastereoisomer of the original compound (if the original compound has more than one stereo-center).
Walden inversion is generally seen in ${ S }_{ N }2$ reactions. ${ S }_{ N }2$ reactions follow second order kinetics i.e. the rate of the reaction depends on both the concentration of the substrate and the nucleophile. When the nucleophile approaches the substrate, the bond between the carbon atom and the leaving group present in the substrate starts breaking and a new carbon-nucleophile bond starts forming. Both of these processes take place simultaneously in a single step and no intermediate is formed. Thus ${ S }_{ N }2$ reactions are concerted reactions. These reactions occur through a transition state; the attack of the nucleophile occurs from the back side and the leaving group leaves from the front side. This results in the Walden inversion.
For Walden inversion to take place, the reactant has to be optically active or more specifically the carbon atom to which the leaving group (-Br) is attached should be a stereo-center. ${ C }_{ 4 }{ H }_{ 9 }-CHD-Br$ is an optically active compound and hence will give Walden inversion.

Both and are optically inactive compounds and therefore will not give Walden inversion.
So, the correct answer is “Option A”.

Note: In order to determine whether a compound is optically active or not we can use the trick. Generally compounds containing stereo-centers/chiral centers are optically active. However that is not the case with Meso compounds.