See image — Isomerism and Stereochemistry Chemistry Question

💡 Solution & Explanation

# Geometric Isomerism in Acyclic Compounds **Key Principle:** Geometric isomerism requires: 1. A C=C double bond (restricted rotation) 2. Two different groups on each carbon of the double bond 3. Acyclic structure only --- ## Analysis of Each Compound: **(i) $C_4H_8$** - Structure: $CH_2=CH-CH_2-CH_3$ (but-1-ene) - One carbon of the double bond has **two identical H atoms** → Cannot show geometric isomerism ✗ **(ii) $C_5H_{10}$** - Structure: $CH_3-CH=CH-CH_2-CH_3$ (pent-2-ene) - Both carbons of C=C have different groups: - Left C: $CH_3$ and $H$ - Right C: $CH_3$ and $CH_2CH_3$ - **Shows geometric isomerism:** cis and trans forms ✓ **(iii) $C_6H_{12}$** - Structure: $CH_3-CH=C(CH_3)_2$ (2-methylbut-2-ene) - Right carbon has **two identical $CH_3$ groups** → Cannot show geometric isomerism ✗ **(iv) $C_3H_5Cl$** - Structure: $CH_2=CH-CH_2Cl$ (allyl chloride) - Left carbon has **two identical H atoms** → Cannot show geometric isomerism ✗ **(v) $C_3H_4Cl_2$** - Structure: $CHCl=CH-CH_3$ (1,1-dichloropropene) - Left carbon has **two identical Cl atoms** → Cannot show geometric isomerism ✗ --- **Answer: Only (ii) $C_5H_{10}$ exhibits geometric isomerism** with cis-pent-2-ene and trans-pent-2-ene as isomers.