See image — Isomerism and Stereochemistry Chemistry Question

💡 Solution & Explanation

# Solution: Optically Active Isomers of HOOC–CH(OH)–CH(OH)–CH(OH)–COOH **Step 1: Identify stereogenic centers** A stereogenic center requires a carbon with 4 different groups. Examine each carbon: - C1 (HOOC–): bonded to –COOH, –H, and one adjacent carbon → not stereogenic - C2, C3, C4 (middle carbons): each has –OH, –H, and two different carbon environments → **all are stereogenic centers** We have **3 stereogenic centers**. **Step 2: Apply the formula for maximum stereoisomers** Maximum stereoisomers = $2^n$ where $n$ =number of stereogenic centers $$2^3 = 8 \text{ stereoisomers}$$ **Step 3: Account for internal symmetry (meso compound)** The compound has a **plane of symmetry** through the central carbon (C3): - Left side (C1–C2–C3) is mirror image of right side (C3–C4–COOH) This creates a **meso form**: when C2 and C4 have opposite stereochemistry with C3 in the middle, the molecule becomes achiral (optically inactive). **Step 4: Count optically active isomers** - Total stereoisomers: 8 - Meso (optically inactive) isomers: Need opposite configurations at C2 and C4 - Only **2 enantiomeric pairs** have the same stereochemistry at both C2 and C4 → these are optically active **Maximum optically active isomers = 2** **The answer is (A) 2**