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(A) To exhibit geometrical isomerism,the molecule must have a double bond with different groups attached to each carbon atom of the double bond. To ex

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$4-$Bromopent$-2-$ene

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(A) To exhibit geometrical isomerism,the molecule must have a double bond with different groups attached to each carbon atom of the double bond. To exhibit optical isomerism,the molecule must contain at least one chiral center (a carbon atom bonded to four different groups). In $4-$Bromopent$-2-$ene $(CH_3-CH(Br)-CH=CH-CH_3)$: $1$. The double bond between $C_2$ and $C_3$ allows for geometrical isomerism ($cis$ and $trans$ forms). $2$. The $C_4$ atom is a chiral center because it is bonded to a hydrogen atom,a methyl group,a bromine atom,and a propenyl group. Therefore,$4-$Bromopent$-2-$ene exhibits both geometrical and optical isomerism.

Column-$I$	Column-$II$
$A$. 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

The compound which exhibits both geometrical and optical isomerism is

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