Courses
Courses for Kids
Free study material
Offline Centres
More
Store Icon
Store

Which of the following compounds is optically active?
A.${\text{C}}{{\text{H}}_{\text{3}}}{\text{C}}{{\text{H}}_{\text{2}}}{\text{COOH}}$
B.${\text{HOOC}} - {\text{C}}{{\text{H}}_{\text{2}}} - {\text{COOH}}$
C.\[{\text{C}}{{\text{H}}_{\text{3}}}{\text{CH(OH)COOH}}\]
D.\[{\text{C}}{{\text{l}}_{\text{2}}}{\text{CHCOOH}}\]

seo-qna
Last updated date: 13th Jun 2024
Total views: 394.2k
Views today: 10.94k
Answer
VerifiedVerified
394.2k+ views
Hint: For a compound to be optically active, the presence of a chiral centre is necessary. A carbon whose all valancy are satisfied by different substituents is known as a chiral centre. The compound having no chiral centre is known as optical inactive.

Complete Step by step answer:A compound having a chiral centre will be optically active. A carbon centre having four different substituents is known as a chiral centre.
The chiral molecule rotates the light in a clockwise direction are known as dextrorotatory and represented by \[\left( + \right)\] sign and the chiral molecule rotate the light in an anticlockwise direction are known as laevorotatory and represented by \[\left( - \right)\] sign.
Write the structure of all the compound to determine the chiral centre as follows:
seo images

The propionic acid has two hydrogens, one methyl and one carboxylic group. So, propionic acid has only three types of substituents. So, propionic acid is not optically active.
seo images

The malonic acid has two carboxylic, and two hydrogen groups. So, malonic acid has two types of substituents. So, malonic acid is not optically active.
seo images

The $2 - $hydroxypropanoic acid has one hydrogen, one methyl, one hydroxy and one carboxylic group. So, $2 - $hydroxypropanoic acid has four different substituents. So, $2 - $hydroxypropanoic acid is optically active.
seo images

The $2 - $chloroacetic acid has one hydrogen, two chlorine, and one carboxylic group. So, $2 - $ chloroacetic acid has three types of substituents. So, $2 - $chloroacetic acid is not optically active.
So, only $2 - $hydroxypropanoic acid is optically active.

Therefore, option (C) \[{\text{C}}{{\text{H}}_{\text{3}}}{\text{CH(OH)COOH}}\] is correct.

Note: A chiral molecule can be optical inactive in two conditions: first, if the molecule has symmetry such as tartaric acid having a plane of symmetry. Second, 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.