Question

Why –(-) butan- 2 -ol is optically inactive ?

Answer 1

Hint: For a compound to be optically active, the presence of a chiral center is necessary. A compound having a chiral center can be optically inactive due to the presence of symmetry or some factor that can cancel the optical activity.

Complete answer:
A compound having a chiral center will be optically active. A center having four different substituents is known as a chiral center.
The chiral molecule rotates the light in a clockwise direction are known as dextrorotatory and represented by $$(+)$$ sign and the chiral molecule rotate the light in an anticlockwise direction are known as laevorotatory and represented by $$(-)$$ sign.
A chiral molecule can be optical inactive in two conditions:
- If the molecule has symmetry such as tartaric acid having a plane of symmetry.
- If the molecule exists in two enantiomers and both are present in equal amounts and both rotate the light in opposite directions.
This type of mixture when two enantiomers of a chiral molecule present in the same amount and both rotate the light in opposite directions, is known as a racemic mixture.
Write the structure of –(–) butan- 2 -ol as follows:

The carbon in red colour is the chiral center because it has four different substituents.
So, butan-2-ol should be optically active.
But the (-) and (+) enantiomers of butan-2-ol, both exist in equal amounts, so both rotate the light in opposite directions in equal amounts. So, rotation of both enantiomers cancel out by each other thus the –(–) butan-2-ol is optically inactive.
Therefore, –(–) butan-2-ol is optically inactive due to racemic mixture.

Note: The racemic mixture is optical active but the optical activity of the racemic mixture is not observed due to the cancelation of optical activity of both enantiomers by each other.