See image — Hydrocarbons Chemistry Question

💡 Solution & Explanation

Concept: In free radical bromination, bromine radical is highly selective due to its high activation energy for H-abstraction. It preferentially abstracts H from the most stable (highest substituted) carbon: 3° > 2° > 1°. Benzylic and allylic positions are also highly activated. We need to identify reactions where the major product is a secondary (2°) alkyl bromide. (P) Cyclohexane: All 12 hydrogens are equivalent and all are secondary (2°) hydrogens (each carbon in cyclohexane is bonded to 2 other carbons). Therefore, the only possible product is the 2° bromide. Major product = 2° halide. ✓ (Q) Methylcyclobutane: The ring carbons include one tertiary carbon (bearing the methyl group) and secondary carbons. The 3° H is preferentially abstracted by Br• due to selectivity of bromine. Major product = 3° bromide (bromination at tertiary carbon). ✗ (R) Ethylbenzene (PhCH2CH3): Benzylic position is the CH2 (secondary benzylic carbon, 2° benzylic). Bromine radical selectively abstracts from the benzylic position due to resonance stabilization of the benzylic radical. The benzylic carbon (CH2) is a secondary carbon attached to phenyl ring. Major product = 2° benzylic bromide (PhCHBrCH3). ✓ (S) 2-methylpentane: Structure is CH3CH(CH3)CH2CH2CH3. It has one 3° carbon (C2, bearing two methyl/ethyl groups). Bromine's selectivity strongly favors the 3° position. Major product = 3° bromide. ✗ (T) Decalin (bicyclo[4.4.0]decane, two fused cyclohexane rings): The ring junction carbons are tertiary (each bonded to 3 other carbons). Br• will preferentially abstract from these 3° positions. Major product = 3° bromide. ✗ (U) The structure shown appears to be a straight-chain or slightly branched alkane (drawn as a zig-zag with no branch shown clearly — appears to be n-pentane or similar straight chain). In a straight-chain alkane like n-pentane, there are no 3° carbons; the internal carbons are 2° and terminal carbons are 1°. Bromine selectively abstracts from 2° positions over 1° positions. Major product = 2° bromide. ✓ Thus, the reactions giving 2° halide as major product are P, R, and U. Options (a) P, Q, R, S — includes Q (3° product) and S (3° product), excludes U → incorrect Options (b) P, R, U — matches our analysis → correct Options (c) P, R, S, T — includes S and T (3° products) → incorrect Options (d) P, Q, R, S, T — includes Q, S, T (3° products) → incorrect Therefore, the correct answer is B.