See image — Isomerism and Stereochemistry Chemistry Question

💡 Solution & Explanation

# Geometric Isomerism in Alkenes **Requirement for geometric isomerism:** Each carbon of the $C=C$ double bond must have **two different groups** attached to it. **Analysis of each option:** **(1)** $H$ and $H$ on left carbon; $CH_3$ and $H$ on right carbon - Left carbon has two identical $H$ atoms → **No geometric isomerism possible** **(2)** $I$ and $CH_3$ on left carbon; $CH_3$ and $H$ on right carbon - Left carbon: $I \neq CH_3$ ✓ - Right carbon: $CH_3 \neq H$ ✓ - All four groups are different → **Cis-trans isomerism IS possible** - This can exist as cis (both bulky groups on same side) and trans (on opposite sides) **(3)** $CH_3$ and $CH_3$ on left carbon; $CH_3$ and $H$ on right carbon - Left carbon has two identical $CH_3$ groups → **No geometric isomerism possible** **(4)** $Br$ and $Br$ on left carbon; $CH_3$ and $H$ on right carbon - Left carbon has two identical $Br$ atoms → **No geometric isomerism possible** **Answer: Option (2)** shows geometric isomerism because both sp² carbons have two different substituents, allowing distinct cis and trans configurations around the rigid double bond.