See image — Isomerism and Stereochemistry Chemistry Question

💡 Solution & Explanation

Geometrical (cis-trans) isomerism about a C=C double bond requires that each carbon of the double bond bears two different substituents. If either carbon of the double bond carries two identical groups, geometrical isomerism is not possible. Concept: For a C=C to show geometrical isomerism, neither sp2 carbon should have two identical groups attached to it. Option (a): CH3 - C(CH3) = CH - CH2 - CH3 The left carbon of the double bond is attached to CH3 and CH3 (two identical methyl groups) as well as being part of the double bond. Because this carbon bears two identical substituents (both CH3), it cannot show geometrical isomerism. This is the compound that will NOT show geometrical isomerism. Option (b): CH3 - CH(CH3) - CH = CH - CH2 - CH3 The double bond carbons are: left C bears H and CH(CH3)(CH3) chain substituent (i.e., H and -CH(CH3)(CH3) chain — actually H and a carbon chain group, both different), right C bears H and -CH2CH3 (both different). So both double bond carbons have two different groups → shows geometrical isomerism. Option (c): CH3 - CH = CH - CH3 Left C of double bond: CH3 and H (different). Right C: CH3 and H (different). Both carbons have two different substituents → shows geometrical isomerism (cis and trans 2-butene). Option (d): CH3 - CH2 - CH = CH - CH2 - CH3 Left C: H and CH2CH3 (different). Right C: H and CH2CH3 (different). Shows geometrical isomerism. Only option (a) has a double bond carbon bearing two identical groups (two CH3 groups on the same sp2 carbon), so it cannot exhibit geometrical isomerism. Therefore, the correct answer is A.