Determine the number of optically active isom | vedclass

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(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

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(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 $0$.

$A$. Acetaldehyde,vinyl alcohol	$1$. Enantiomers
$B$. Eclipsed and staggered ethane	$2$. Tautomers
$C$. $(+)2$-butanol,$(-)2$-butanol	$3$. Chain isomers
$D$. Methyl-$n$-propylamine and diethylamine	$4$. Conformational isomers
	$5$. Metamers

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

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Match the following:
$A$. Acetaldehyde,vinyl alcohol $1$. Enantiomers
$B$. Eclipsed and staggered ethane $2$. Tautomers
$C$. $(+)2$-butanol,$(-)2$-butanol $3$. Chain isomers
$D$. Methyl-$n$-propylamine and diethylamine $4$. Conformational isomers
$5$. Metamers

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