See image — Isomerism and Stereochemistry Chemistry Question

💡 Solution & Explanation

# Solution: Optically Active Isomers of $C_4H_7Br$ **Step 1: Determine the degree of unsaturation** $$\text{DBE} = \frac{2(4) + 2 - 7 + 1}{2} = \frac{4}{2} = 2$$ Two degrees of unsaturation indicates either two double bonds, one triple bond, or one ring + one double bond. **Step 2: Identify possible structures** For $C_4H_7Br$ with optical activity, we need a chiral center (carbon with 4 different groups). - **But-3-yn-1-bromide**: $Br-CH_2-CH_2-C\equiv CH$ — No chiral center (carbons 2 and 3 lack 4 different groups) - **Bromocyclopropane derivatives**: $C_3H_5Br$ ring + one more carbon — Cyclopropane without substituents creates no chiral center in simple bromocyclopropane - **1-Bromo-2-methylcyclopropane**: The carbon bearing the methyl group and bromine has 4 different groups (Br, H, CH₃, and the cyclopropane ring carbons are different) — **This is chiral** **Step 3: Count stereoisomers** 1-Bromo-2-methylcyclopropane has **one chiral center**, producing **one pair of enantiomers** (R and S forms). Each enantiomer is optically active, but they represent **1 optically active compound type** (or 2 enantiomers of a single compound). By the standard interpretation in JEE/NEET chemistry, the answer counts **distinct optically active structures** rather than counting both enantiomers separately. **The answer is (A) 1** — only one structure of $C_4H_7Br$ possesses a chiral center and exhibits optical activity.