See image — Hydrocarbons Chemistry Question

💡 Solution & Explanation

Step 1: Draw n-butane structure. n-Butane is CH3-CH2-CH2-CH3. It has two types of hydrogen atoms: the 6 equivalent primary H atoms on C1 and C4, and the 4 equivalent secondary H atoms on C2 and C3. Step 2: Identify monochloro products from each type of hydrogen abstraction. Product 1: Chlorination at C1 (primary carbon) gives 1-chlorobutane (CH3CH2CH2CH2Cl). This is a simple achiral molecule — only one structure, no stereoisomers. Product 2: Chlorination at C2 (secondary carbon) gives 2-chlorobutane (CH3CHClCH2CH3). C2 becomes a chiral center (four different substituents: H, Cl, CH3, CH2CH3). This produces two enantiomers: (R)-2-chlorobutane and (S)-2-chlorobutane. Step 3: Count all distinct structures including stereoisomers. - 1-chlorobutane: 1 compound (no chiral center) - (R)-2-chlorobutane: 1 compound - (S)-2-chlorobutane: 1 compound Total = 1 + 2 = 3 monochloro products Step 4: Evaluate why other options fail. - Option (a) 2: Incorrect — ignores the stereoisomers of 2-chlorobutane. - Option (c) 4 and (d) 5: Incorrect — there are only two distinct types of C-H bonds in n-butane, giving only two constitutional isomers, one of which has two stereoisomers, totaling 3. Therefore, the correct answer is B.