Determine the number of optically active isom | vedclass

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

Question

(A) The given molecule is a derivative of a central cyclobutane ring substituted with two oxetane rings and two bromine atoms. To determine optical ac

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(A) The given molecule is a derivative of a central cyclobutane ring substituted with two oxetane rings and two bromine atoms. To determine optical activity,we must check for the presence of a plane of symmetry $(POS)$ or a center of inversion $(COS)$. If we examine the structure,the two bromine atoms can be oriented in different spatial arrangements relative to the central cyclobutane ring (cis or trans). However,in both the $cis$ and $trans$ configurations,the molecule possesses a plane of symmetry passing through the two carbon atoms bearing the bromine atoms or through the center of the cyclobutane ring. Since the molecule has a plane of symmetry in all its possible stereoisomeric forms,it is achiral. Therefore,the number of optically active isomers is $0$.

Determine the number of optically active isomers for the given structure.

Similar Questions

The Fischer projection formula of the given structure is:

How many stereoisomers does this molecule have?
$CH_3CH=CHCH_2CH(Br)CH_3$

Which one is the most stable cation in the following?

Which of the following is a meso compound?

Which of the following compounds is chiral?

Vedclass Test Series

Exam Paper Generator

Online Exam Module