See image — GOC and Organic Chemistry Basics Chemistry Question

💡 Solution & Explanation

Concept: Acidity of C-H bonds in hydrocarbons depends on the stability of the conjugate base (carbanion) formed upon deprotonation. Greater stabilization of the carbanion means higher acidity. Step 1 - Identify the structures: (a) 1-Methylindane: fully saturated five-membered ring fused to benzene; the acidic proton would be at a benzylic sp3 carbon. Benzylic carbanions are stabilized by resonance with the aromatic ring. (b) The structure in (b) appears to be 3-methyl-1H-indene or a related compound where the methyl-bearing carbon is allylic AND benzylic with conjugation into a cyclopentadienyl-type system. Specifically, (b) shows a partially saturated six-membered ring fused to a cyclopentadiene ring with a methyl group; the methyl C-H is at a position where deprotonation gives a carbanion that is both allylic and benzylic, and importantly can delocalize into an extended conjugated system approaching cyclopentadienyl anion character. (c) Exocyclic methylene compound: the =CH2 protons are vinylic, which are less acidic than benzylic. (d) 1-Methylindene type with aromatic ring: benzylic/allylic stabilization similar to (a) but slightly different connectivity. Step 2 - Key reasoning for (b) being most acidic: In structure (b), the methyl group is attached to a carbon that is part of a conjugated diene system within the five-membered ring that is fused to a partially unsaturated six-membered ring. Deprotonation of the methyl group yields a carbanion that is stabilized by extensive conjugation - the resulting anion has cyclopentadienyl-like character with delocalization across multiple double bonds in both rings. This provides significantly greater stabilization than simple benzylic stabilization seen in (a) or (d). Step 3 - Why other options fail: (a) Only benzylic stabilization from one aromatic ring; sp3 CH at ring junction is benzylic but less stabilized than (b). (c) Vinylic/exo-methylene protons are generally less acidic than benzylic positions; no special carbanion stabilization. (d) The methyl-bearing carbon in (d) is at a non-conjugated position relative to the double bond in the five-membered ring, giving less carbanion stabilization than (b). Step 4 - Conclusion: Structure (b) provides the most stabilized conjugate base upon deprotonation of the methyl group due to extended conjugation and near-aromatic (cyclopentadienyl anion-like) stabilization, making it the most acidic hydrocarbon among the four isomers. Therefore, the correct answer is B.