See image — Hydrocarbons Chemistry Question

💡 Solution & Explanation

Step 1: Identify compound A and B. Compound A is CH3CH2CH=CH2 (but-1-ene). Under HBr with peroxide (R2O2), the reaction proceeds via anti-Markovnikov (free radical) addition, giving CH3CH2CH2CH2Br (1-bromobutane), which is compound B. Without peroxide (normal Markovnikov), HBr would give CH3CH2CHBrCH3 (2-bromobutane). So A = CH3CH2CH=CH2 (but-1-ene) and B = CH3CH2CH2CH2Br (1-bromobutane). Compare A and B: A is but-1-ene and B is 1-bromobutane. These have different functional groups (alkene vs alkyl bromide), so they are not isomers of each other in the traditional sense. However, the question asks about the relation between A and B as a pair, and C and D as a pair. Step 2: Identify compound C and D. Compound C is CH3CH2C≡CH (but-1-yne). Under hydroboration-oxidation (B2H6 then H2O2/OH^-), alkynes undergo anti-Markovnikov addition of water. For a terminal alkyne, this gives an aldehyde: CH3CH2CH2CHO (butanal), which is compound D. Step 3: Determine the relation between A and B. A = CH3CH2CH=CH2 (but-1-ene, molecular formula C4H8) B = CH3CH2CH2CH2Br (1-bromobutane, molecular formula C4H9Br) These are not isomers. But in context, the question likely intends comparison within each reaction as structural isomer pairs or asking about what type of isomerism exists between the reactant and product, or comparing A with another compound that would give B under Markovnikov conditions. Actually, re-reading: the question says 'Relation between A and B' - A is but-1-ene and the Markovnikov product would be 2-bromobutane, while B (anti-Markovnikov) is 1-bromobutane. So 1-bromobutane and 2-bromobutane are position isomers (same molecular formula C4H9Br, same functional group, different position of Br). Thus A and B here likely refers to the two possible products: 1-bromobutane vs 2-bromobutane, which are position isomers. Step 4: Determine the relation between C and D. C = CH3CH2C≡CH (but-1-yne, C4H6) D = CH3CH2CH2CHO (butanal, C4H8O) These have different molecular formulas and different functional groups. But comparing the Markovnikov product of hydration of but-1-yne (which would give CH3CH2COCH3, methyl ethyl ketone, a ketone) vs the anti-Markovnikov product D (butanal, an aldehyde): both have formula C4H8O. Methyl ethyl ketone (butanone) and butanal are functional group isomers (same molecular formula C4H8O, different functional groups: ketone vs aldehyde). Step 5: Conclude. Relation between A and B (the two possible bromination products of but-1-ene): Position isomers. Relation between C and D (the two possible hydration products of but-1-yne): Functional isomers. This matches option (b): Position, Functional. Why other options fail: (a) Position, chain - chain isomerism would require different carbon skeletons, not applicable here. (c) Chain, Identical - incorrect on both counts. (d) Metamer, Functional - metamerism involves different alkyl groups on same functional group; not applicable to the A/B pair. Therefore, the correct answer is B.