Concept:Geometric isomerism is exhibited by alkenes where
both carbons of the double bond are attached to two different groups.
Solution:- Let's test each molecule:
- A. 1-bromo-2-chloropropene: C1 has H and Br; C2 has Cl and \( \text{CH}_3 \). Since each carbon has two different groups, it shows geometric isomerism.
- B. 2,3-dimethylpropene: (Wait, propene only has 3 carbons. A substituent at C3 of propene makes it butene. Assuming it means 2-methylpropene, C1 has two H's). It does not show geometric isomerism.
- C. 2-pentene: (\( \text{CH}_3\text{CH}=\text{CHCH}_2\text{CH}_3 \)). C2 has H and \( \text{CH}_3 \); C3 has H and \( \text{CH}_2\text{CH}_3 \). It shows geometric isomerism.
- Therefore, both A and C exhibit the property.
Why other options are incorrect:Selecting only A or C is incomplete. Option B is incorrect due to identical groups on one end of the double bond.
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