See image — Isomerism and Stereochemistry Chemistry Question

💡 Solution & Explanation

# Optical Isomers in 2,3-Dichlorobutane **Step 1: Draw the structure of 2,3-dichlorobutane** $$CH_3-CHCl-CHCl-CH_3$$ Chlorine atoms are attached at carbons 2 and 3. **Step 2: Identify chiral centers** - Carbon 2: bonded to $CH_3$, $H$, $Cl$, and $CHClCH_3$ → **chiral** - Carbon 3: bonded to $CH_3$, $H$, $Cl$, and $CHClCH_3$ → **chiral** Both carbons have four different groups, so there are 2 chiral centers. **Step 3: Determine stereoisomers** With 2 chiral centers: $2^n = 2^2 = 4$ possible stereoisomers - RR configuration - SS configuration - RS configuration - SR configuration **Step 4: Check for meso compounds** The RS and SR configurations are actually the **same compound** (meso-2,3-dichlorobutane) because the molecule has an internal plane of symmetry. This meso form is **optically inactive**. **Step 5: Count optically active isomers** Only the **RR and SS** enantiomers are optically active (they are non-superimposable mirror images). **Answer: (B) 2** The two optically active isomers are the RR and SS enantiomers of 2,3-dichlorobutane.