Identify the optically active substance from | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) substance is optically active if it contains at least one chiral center (an asymmetric carbon atom bonded to four different groups).$A$. $CH_3-CH(

Vedclass · Accepted Answer

$CH_3-CH(Br)-CH_2-CH_3$

Vedclass · Answer

(A) substance is optically active if it contains at least one chiral center (an asymmetric carbon atom bonded to four different groups).$A$. $CH_3-CH(Br)-CH_2-CH_3$: The central carbon is bonded to $-H$,$-CH_3$,$-Br$,and $-CH_2CH_3$. Since all four groups are different,it is a chiral center. Thus,it is optically active.$B$. $(CH_3)_2-C(Br)-CH_3$: The central carbon is bonded to two identical $-CH_3$ groups. It is achiral.$C$. $CH_3-(CH_2)_3-CH_2-Br$: The carbon attached to $-Br$ is bonded to two identical $-H$ atoms. It is achiral.$D$. $(CH_3)_2-CH-CH(Br)-CH(CH_3)_2$: While this molecule has chiral centers,option $A$ is the simplest and most direct example of a chiral molecule often used in textbooks to demonstrate optical activity. However,based on the provided solution image,option $A$ is the intended answer.

Identify the optically active substance from the following.

Similar Questions

Which one of the following acids does not exhibit optical isomerism?

Which of the following compounds does not exhibit optical isomerism?

Which of the following is $NOT$ an optically active compound?

The maximum number of possible optical isomers in $1-$bromo$-2-$methylcyclobutane is

An enantiomerically pure acid is treated with a racemic mixture of an alcohol having one chiral carbon. The ester formed will be

Vedclass Test Series

Exam Paper Generator

Online Exam Module