Concept:For an alkene to exhibit geometric (cis-trans) isomerism,
both carbons of the \( \text{C=C} \) double bond must be attached to two
different groups.
Solution:- Let's analyze the double bonds:
- A: 3-Methyl-1-butene (\( \text{CH}_2=\text{CH-CH(CH}_3\text{)}_2 \)). Carbon-1 is attached to two identical Hydrogens. No geometric isomerism.
- B: 2,3-Dimethyl-2-butene. Both double-bonded carbons are attached to two identical methyl groups. No geometric isomerism.
- C: Methylcyclopentane has no double bonds (it's an alkane).
- D: 3-Methyl-2-pentene (\( \text{CH}_3\text{-CH=C(CH}_3\text{)-CH}_2\text{CH}_3 \)). Carbon-2 is attached to H and \( \text{CH}_3 \). Carbon-3 is attached to \( \text{CH}_3 \) and \( \text{CH}_2\text{CH}_3 \). Because both carbons have two different groups, it shows geometrical isomerism.
Why other options are incorrect:Options A and B violate the rule by having identical groups on at least one carbon of the double bond.
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