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(B) The given compound is $4-methylpent-3-enoic acid$ with the structure $(CH_3)_2C=CH-CH(CH_3)COOH$.$1$. Geometrical Isomerism: The double bond is be

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Stereoisomerism

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(B) The given compound is $4-methylpent-3-enoic acid$ with the structure $(CH_3)_2C=CH-CH(CH_3)COOH$.$1$. Geometrical Isomerism: The double bond is between $C_3$ and $C_4$. At $C_4$,there are two identical methyl groups attached,so it does not show geometrical isomerism.$2$. Optical Isomerism: There is a chiral carbon atom at the $C_2$ position (the carbon attached to $-H$,$-CH_3$,$-COOH$,and the $-CH=C(CH_3)_2$ group). Since it has a chiral center,it exhibits optical isomerism.$3$. Stereoisomerism: Since optical isomerism is a type of stereoisomerism,the compound exhibits stereoisomerism.

Which type of isomerism is exhibited by the given compound?
$(CH_3)_2C=CH-CH(CH_3)COOH$

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Which type of isomerism is exhibited by the given compound? $(CH_3)_2C=CH-CH(CH_3)COOH$