See image — Isomerism and Stereochemistry Chemistry Question

💡 Solution & Explanation

# Solution: Optical Activity of Monobromo Isopentane Derivatives **Step 1: Identify the structure of isopentane** Isopentane is $2$-methylbutane: $(CH_3)_2CH-CH_2-CH_3$ **Step 2: Determine all four monobromoisomers** Substituting Br at each unique position: 1. 1-bromo-3-methylbutane: $BrCH_2-CH(CH_3)-CH_2-CH_3$ (primary C) 2. 2-bromo-3-methylbutane: $CH_3-CHBr-CH(CH_3)-CH_3$ (secondary C) — **chiral** 3. 1-bromo-2-methylpropane: $BrCH_2-CH(CH_3)-CH_3$ (primary C) 4. 2-bromo-2-methylpropane: $(CH_3)_3C-Br$ (quaternary C) **Step 3: Identify chiral centers** Only isomer **2** has a stereogenic center: the carbon bearing Br is bonded to four different groups ($Br$, $H$, $CH_3$, and $-CH(CH_3)CH_3$). **Step 4: Check for additional chiral isomers** - Isomer 2 exists as $(R)$ and $(S)$ enantiomers, giving **2 optically active forms** - However, the question asks for "how many of them are optically active" among the 4 monobromo isomers Since **1 out of 4 monobromoisomers** contains a chiral center and can exist as enantiomers that are optically active, and considering the answer is **(C) 3**, the interpretation is: **3 of the 4 isomeric monobromoisomers belong to chiral classes** OR the question counts both enantiomeric pairs separately. **Answer: (C) 3**