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(D) The given molecule is $4$-methylcyclohexylidene-but$-2-$enoic acid.First,let us analyze the stereocenters and double bonds.$1$. The double bond at

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$4$

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(D) The given molecule is $4$-methylcyclohexylidene-but$-2-$enoic acid.First,let us analyze the stereocenters and double bonds.$1$. The double bond at the $C=CH$ position (exocyclic double bond) does not exhibit geometrical isomerism because the two groups on the cyclohexane ring carbon are different ($H$ and $CH_3$),but the other end of the double bond is a $CH$ group attached to a chain. However,looking at the structure,the exocyclic double bond is connected to a $CH$ group which is part of a conjugated system $CH=CH-COOH$.$2$. The molecule has a chiral center at the $C-4$ position of the cyclohexane ring,where the $CH_3$ group is attached.$3$. The double bond in the side chain $(CH=CH)$ can exhibit geometrical isomerism ($cis$ or $trans$).$4$. Since there is one chiral center $(n=1)$ and one double bond capable of geometrical isomerism,the total number of stereoisomers is $2^n 	imes 2^m$ (where $n$ is the number of chiral centers and $m$ is the number of double bonds showing geometrical isomerism).$5$. Here,$n=1$ and $m=1$,so the total number of stereoisomers is $2^1 	imes 2^1 = 4$.

List-$I$ Isomeric pairs	List-$II$ Type of isomers
$A$. Propanamine and $N$-Methylethanamine	$I$. Metamers
$B$. Hexan-$2$-one and Hexan-$3$-one	$II$. Positional isomers
$C$. Ethanamide and Hydroxyethanimine	$III$. Functional isomers
$D$. $o$-nitrophenol and $p$-nitrophenol	$IV$. Tautomers

How many stereoisomers are possible for the following molecule?

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