See image — Isomerism and Stereochemistry Chemistry Question

💡 Solution & Explanation

# Optical Activity Analysis **Key Concept:** A molecule is optically active if it contains a **chiral center** (stereogenic center) — a carbon atom bonded to four different groups. ## Analysis of Each Structure: **(1) Both Br atoms wedged (same side):** - Both Br groups point in the same direction - The cyclobutane ring has a plane of symmetry - **Not optically active** (achiral due to symmetry) **(2) One Br wedged, one Br dashed (opposite sides):** - Br atoms are on opposite faces of the ring - The molecule has a plane of symmetry through the ring - **Not optically active** (meso compound) **(3) Br atoms on adjacent carbons (one wedged, one dashed):** - Each Br-bearing carbon is bonded to: Br, H, and two different ring carbons - The molecule lacks a plane of symmetry - **This IS optically active** ✓ **(4) One Br wedged, one Br dashed (opposite sides, different positioning):** - Similar to (2), the configuration creates a plane of symmetry - **Not optically active** (meso compound) ## Answer: **Option (3)** is optically active because the two brominated carbons are on adjacent positions with opposite stereochemistry (one wedged, one dashed), creating **two distinct stereogenic centers without compensating symmetry**, making the molecule chiral.