See image — Hydrocarbons Chemistry Question

💡 Solution & Explanation

Concept: When HBr adds to an alkene, Markovnikov addition places Br on the more substituted carbon. For the product to be ONLY a pair of diastereomers (and not a mixture that includes enantiomers as the sole products, nor a single stereoisomer, nor constitutional isomers), the substrate must already possess one stereocenter before addition, and HBr addition must create exactly one new stereocenter. This gives two diastereomers (differing at the new center while the pre-existing center is fixed). Step 1 – Understand what 'only a pair of diastereomers' means: The product mixture contains exactly two stereoisomers that are diastereomers of each other. This requires: (i) the starting material already has one stereocenter (fixed configuration), and (ii) HBr addition creates one new stereocenter, giving two products that differ only at the new center — these are diastereomers. Step 2 – Analyze option (a): Vinylcyclohexane has no pre-existing stereocenter. HBr addition creates one new stereocenter, giving a pair of enantiomers (racemic mixture), not diastereomers. Eliminated. Step 3 – Analyze option (b): The structure shows a carbon bearing Cl (wedge) and H (dash) — this is a pre-existing stereocenter. The double bond is trisubstituted (two methyls). HBr addition (Markovnikov) places Br on the more substituted carbon of the double bond, creating a new stereocenter. With one fixed stereocenter and one new stereocenter, the products are a pair of diastereomers. However, the carbon receiving Br may be a tertiary carbon that becomes quaternary — need to check if it actually generates a new stereocenter. The double bond carbons in (b) are C(CH3)=C(CH3)— type; Markovnikov Br goes to the more substituted carbon which would be tertiary, potentially creating a new chiral center. This seems plausible, but option (c) is the answer. Step 4 – Analyze option (c): The structure shows a carbon with HO (wedge) and H (dash) bearing an ethyl group — a pre-existing stereocenter with fixed (R or S) configuration. This carbon is adjacent to a trisubstituted alkene (2-methylbut-2-ene-type). HBr adds: H goes to the less substituted end, Br goes to the more substituted carbon (Markovnikov), creating one new stereocenter. The pre-existing OH-bearing stereocenter remains intact. The two products differ only at the newly formed Br-bearing carbon while the OH-bearing carbon is fixed — these are diastereomers. No enantiomers are formed because the starting stereocenter is fixed (single enantiomer starting material as drawn). Thus ONLY a pair of diastereomers is formed. Step 5 – Analyze option (d): The quaternary carbon bearing H3C and CH3 is not a stereocenter (two identical methyl groups). HBr addition to the terminal alkene (CH2=CH-) by Markovnikov gives Br on the internal carbon, creating one new stereocenter. With no pre-existing stereocenter, only one new center forms, giving enantiomers, not diastereomers. Eliminated. Step 6 – Why (c) is uniquely correct: Option (c) has a fixed pre-existing stereocenter (the carbon bearing OH, H, ethyl, and the alkenyl chain) in a single enantiomeric form (as drawn with defined wedge/dash). HBr addition creates exactly one new stereocenter, producing exactly two products that are diastereomers of each other. No racemization of the existing center occurs under electrophilic addition conditions. Therefore, the correct answer is C.