See image — Hydrocarbons Chemistry Question

💡 Solution & Explanation

Step 1 - Determine the molecular formula. Mol. wt. = 86 for an alkane CnH(2n+2). 12n + 2n + 2 = 86 → 14n = 84 → n = 6. So the alkane is a hexane isomer (C6H14). Step 2 - Identify option (c). Option (c) is CH3-CH(CH3)-CH(CH3)-CH3, which is 2,3-dimethylbutane. MW = 6(12) + 14(1) = 72 + 14 = 86. Correct molecular weight. Step 3 - Count distinct hydrogen environments in 2,3-dimethylbutane. 2,3-dimethylbutane has the structure: (CH3)2CH-CH(CH3)2. - The four CH3 groups attached to the two CH carbons: all 12 hydrogens on these methyl groups are equivalent (by symmetry) → Type 1 hydrogens. - The two CH hydrogens (one on each central carbon): both are equivalent by symmetry → Type 2 hydrogens. So there are exactly 2 types of hydrogen environments, giving exactly 2 monobromo derivatives (excluding stereoisomers): 1-bromo-2,3-dimethylbutane (from CH3 bromination) and 2-bromo-2,3-dimethylbutane (from CH bromination). Step 4 - Check other options. (a) 2-methylpentane: has 4 distinct H environments → 4 monobromo derivatives. (b) 2,2-dimethylbutane (MW = 86): has 3 distinct H environments → 3 monobromo derivatives. (d) Neopentane (2,2-dimethylpropane): MW = 72, not 86. Eliminated by molecular weight. Step 5 - Conclusion. Only 2,3-dimethylbutane (option c) gives exactly two monobromo derivatives and has mol. wt. = 86. Therefore, the correct answer is C.