See image — Hydrocarbons Chemistry Question

💡 Solution & Explanation

Step 1 - Concept: Dihydroxylation stereochemistry. Syn dihydroxylation (cold KMnO4 or OsO4) adds two OH groups from the same face, giving a syn addition product. Anti dihydroxylation (peracid epoxidation followed by acid-catalyzed ring opening, RCO3H/H3O+) adds two OH groups from opposite faces, giving an anti addition product. Step 2 - Determine what stereochemistry is needed for each starting material to give meso 3,4-hexanediol. Meso 3,4-hexanediol has an internal plane of symmetry: the two stereocenters at C3 and C4 have opposite configurations (R and S), making the molecule achiral (meso). Step 3 - cis-3-hexene to meso 3,4-hexanediol (reagent a): cis-3-hexene has the two ethyl groups on the same side of the double bond. Syn dihydroxylation (cold KMnO4) adds both OH groups to the same face. When both OHs add syn to cis-3-hexene, the product has C3 and C4 stereocenters with opposite configurations (one R, one S), which is the meso compound. Therefore reagent a = cold KMnO4. Step 4 - trans-3-hexene to meso 3,4-hexanediol (reagent b): trans-3-hexene has the two ethyl groups on opposite sides of the double bond. Syn dihydroxylation of trans-3-hexene would give the (R,R) and (S,S) enantiomers (racemic mixture), not the meso compound. Anti dihydroxylation (peracid epoxidation then acid-catalyzed ring opening, RCO3H/H3O+) adds OH groups to opposite faces. For trans-3-hexene, anti addition gives C3 and C4 with opposite configurations, yielding the meso compound. Therefore reagent b = RCO3H/H3O+. Step 5 - Match with answer choices: Reagent a = cold KMnO4, reagent b = RCO3H/H3O+ corresponds to option (b): Cold KMnO4, RCO3H/H3O+. Step 6 - Why other options fail: Option (a) Cold KMnO4, OsO4 - both are syn dihydroxylation agents; OsO4 on trans-3-hexene gives racemic diol, not meso. Option (c) RCO3H/H3O+, cold KMnO4 - reverses the reagents; anti addition on cis-3-hexene gives racemic diol, not meso. Option (d) None of these - incorrect since option (b) is valid. Therefore, the correct answer is B.