See image — Isomerism and Stereochemistry Chemistry Question

💡 Solution & Explanation

# Identifying the Optically Inactive Compound **Key Principle:** A compound is optically inactive if it lacks chiral centers or possesses a plane of symmetry (meso compound). ## Analysis of Each Option: **(A)** Biphenyl with $NH_2$ and $CH_3$ groups - Has **chiral centers** (quaternary carbons bearing different substituents) - **Optically active** ✗ **(B)** Biphenyl with $NHAc$ and $CH_3$ groups - The molecule has a **plane of symmetry** perpendicular to the biphenyl bond - The upper ring (with $CH_3$ and $NHAc$) is a mirror image of the lower ring (with $CH_3$ and $NHAc$) - Despite having structural asymmetry, the **overall molecular symmetry makes it a meso compound** - **Optically inactive** ✓ **(C)** Biphenyl with $COOH$, $NHAc$ groups - Has **multiple chiral centers** with distinct substituents - **Optically active** ✗ **(D)** Bicyclic compound with $NH$ and $CO$ groups - Has **chiral centers** in the fused ring system - **Optically active** ✗ ## Answer: **Option (B)** is optically inactive because the acetamido-methyl biphenyl possesses a plane of symmetry, making it a meso compound incapable of rotating plane-polarized light.