Question

Why $\pm$ butan $- 2 -$ ol is optically inactive. Give reasons.

Answer 1

The existence of dextro and laevo isomers of a molecule is possible only if it has an asymmetric carbon in the molecule. Since, here it is given that the molecule has both dextro and laevo isomers in it indicated by $\pm$ , it should be an enantiomer.

Complete step by step answer:
Any molecule having one asymmetric carbon atom exists in two configurational isomeric forms which are non-superimposable mirror images. Such molecules which have the same molecular formula, same structure, but different configurations are called optical isomers.
Configurational isomers that are non-superimposable mirror images are called enantiomers; a pair of such compounds is called an enantiomeric pair. The molecules of an enantiomeric pair have the ability to rotate plane polarised light into two different directions, i.e., left and right. The compound which rotates the plane polarised light to the left side is called laevo and the compound which rotates the plane polarised light into the right side is called dextro. The magnitude of rotation will be the same in both the cases.
An equimolar mixture of the two enantiomers of an enantiomeric pair is called racemic mixture or racemate, often represented as $\pm$ form or $({{dl}})$ form. Such a mixture is an optically inactive mixture because the two enantiomers will be rotating the plane polarised light in equally opposite directions and hence cancel each other’s rotation. This phenomenon is known as external compensation.

Hence, in this question, $\pm$ butan $- 2 -$ ol represents that butan $- 2 -$ ol exists as a racemic mixture, indicating that it is optically inactive.

Note:
Racemic mixture can be separated into dextro and laevo forms by the method called resolution. The conversion of dextro and laevo compounds into a racemic mixture is called racemisation. It can be caused by heat, light, or by exposure to chemical reagents.