See image — Hydrocarbons Chemistry Question

💡 Solution & Explanation

Step 1: Determine the molecular formula from vapour density (V.D.). Molar mass = 2 × V.D. = 2 × 36 = 72 g/mol. Step 2: Identify the hydrocarbon with molar mass 72. For alkanes CnH(2n+2): 12n + 2n + 2 = 72 → 14n = 70 → n = 5. So the formula is C5H12 (pentane isomers), MW = 72. ✓ For cycloalkanes CnH(2n): 12n + 2n = 72 → 14n = 72 → n = 5.14 (not integer). So cycloalkanes do not fit MW = 72. This eliminates options (c) cyclohexane (MW = 84) and (d) methyl-cyclohexane (MW = 98). Step 3: Among C5H12 isomers, identify which gives only ONE monochloro substitution product. - n-pentane has three types of hydrogen (C1, C2, C3), giving three monochloro products. - iso-pentane (2-methylbutane) has four types of hydrogen, giving four monochloro products. - neo-pentane (2,2-dimethylpropane) has only ONE type of hydrogen — all 12 hydrogens on the four equivalent methyl groups are identical. Therefore, chlorination gives only one monochloro product: 1-chloro-2,2-dimethylpropane (neopentyl chloride). Step 4: Why other options fail. - (a) iso-pentane: MW = 72 but gives 4 different monochloro products. - (c) cyclohexane: MW = 84 ≠ 72. - (d) methyl-cyclohexane: MW = 98 ≠ 72. Therefore, the correct answer is B.