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

Compare the basic strengths of the given compounds:

Which carbocation is the most stable?

Determine the configuration at $C-2$ and $C-3$ of the given compound.

Which statement regarding the following structures is true?

The number of possible isomers of butene $(C_4H_8)$ are

Vedclass Test Series

Exam Paper Generator

Online Exam Module