Question

Which of the following compound is optically active?

• (A) ${CH}_3{CH}_{2}COOH$
• (B) ${CH}_{3}CHOHCOOH$
• (C) $HOOC\cdot {CH}_2\cdot COOH$
• (D) ${CH}_{3}CO.COOH$

Answer 1

The Correct Option is B

Explanation:
${CH}_{3}CHOHCOOH$ compound is optically active.Propanoic acid has the following structure:

On evaluating the structure, we see that there is no chiral carbon atom present. Therefore, propanoic acid is not optically active.When we observe the structure of lactic acid, we see that it is attached to four different functional groups and has a chiral carbon.Also, there are no elements of symmetry, so Lactic acid shows optical activity. It is present as D and L isomers.

Malonic acid is symmetrical and thus not chiral.

${CH}_{3}CO.COOH$ also does not have any chiral carbon atom, so it is not chiral.Hence, the correct option is (B).

Answer 2

Ans: An optically active compound is the molecule that has the ability to rotate the plane of the polarised light. In the optically active compound they are not superimposable on their mirror images. They are the chiral compound which means that they have all the four different groups attached to the carbon atom. In the question, CH₃CHOHCOOH is optically active. Only it has a carbon atom that have all four different atoms attached to it.