2-chlorobutane + Cl2 C4H8Cl2(isomers) Total number of optically active isomers shown by C4H8Cl2, obt — JEE Mains Chemistry Past Papers Chemistry Question

💡 Solution & Explanation

**Step 1: Identify the reaction** 2-chlorobutane undergoes free radical chlorination with Cl₂. The hydrogen atoms are replaced by chlorine atoms, producing C₄H₈Cl₂ isomers. **Step 2: Determine possible C₄H₈Cl₂ isomers** Starting with 2-chlorobutane (CH₃CHClCH₂CH₃), chlorination can occur at different positions: - Position 1: 1,2-dichlorobutane - Position 2: 2,2-dichlorobutane (no new chiral center) - Position 3: 2,3-dichlorobutane - Position 4: 2,4-dichlorobutane **Step 3: Identify chiral centers** - **1,2-dichlorobutane**: Has one chiral center (at C2). Gives 2 enantiomers (optically active). - **2,2-dichlorobutane**: No chiral center. Not optically active. - **2,3-dichlorobutane**: Has two chiral centers (at C2 and C3). Gives 2² = 4 stereoisomers: 2 enantiomers + 1 meso compound. Only 2 are optically active (the enantiomeric pair). - **2,4-dichlorobutane**: Has one chiral center (at C2). Gives 2 enantiomers (optically active). **Step 4: Count total optically active isomers** - 1,2-dichlorobutane: 2 enantiomers - 2,3-dichlorobutane: 2 enantiomers (meso form is not optically active) - 2,4-dichlorobutane: 2 enantiomers Total = 2 + 2 + 2 = **6 optically active isomers** Therefore, the answer is **6**.