See image — Hydrocarbons Chemistry Question

💡 Solution & Explanation

Starting material (x) is a substituted cyclopentane bearing: a phenyl group on a wedge bond, a methyl group on a dash bond, a bromo substituent, and an exocyclic alkene (vinyl/allyl group). We analyze each reagent: (a) H2O, heat, pH 7: Neutral aqueous hydrolysis with heat. This is a mild SN1/solvolysis condition. The allylic bromide in (x) undergoes solvolysis (allylic system assists). The product would be the allylic alcohol with retention/racemization. This matches product C (cyclopentane with OH on a plain bond and vinyl group, allylic alcohol). (b) F3C-C(=O)-O-OH (trifluoroperacetic acid): This is a peracid, which epoxidizes the alkene. The exocyclic double bond in (x) is epoxidized to give an epoxide. Product D shows a cyclopentane ring with phenyl (wedge), methyl (dash), Br, and an epoxide ring (C=O with O bridge shown). So (b) -> D. (c) tBuOK, polar aprotic solvent: Strong, bulky base in polar aprotic solvent favors E2 elimination. The allylic Br undergoes E2 to give a diene or extended conjugation. Product A is the allylbenzene/diene product (no Br, extended alkene). So (c) -> A. (d) (1) O3, ether; (2) H2O, NaOH, H2O2: Ozonolysis followed by basic oxidative workup (Criegee with H2O2/NaOH gives carboxylic acid from aldehyde). The exocyclic double bond is cleaved oxidatively. One fragment becomes a carboxylic acid. Product F shows Br, carboxylic acid (O=C-OH), phenyl, methyl on the cyclopentane. So (d) -> F. (e) Br2, CCl4: Electrophilic addition of Br2 across the double bond (anti addition). The exocyclic alkene adds Br2 anti to give a vicinal dibromide. Product I shows three Br atoms total (the original Br plus two new ones from addition across the double bond, vicinal dibromide). So (e) -> I. (f) NBS, hv, CCl4: Radical allylic bromination (NBS with light). This introduces a new Br at the allylic position. The existing allylic system gains another Br. Product J shows a structure with two Br atoms on the allylic/vinyl portion and retained double bond. So (f) -> J. (g) (1) H3O(+); (2) NaOH, H2O: Acid-catalyzed opening followed by base. This could be hydration of the double bond (Markovnikov) via acid catalysis, then base workup. The alkene is hydrated to give the Markovnikov alcohol. Product E shows an aldehyde, which could arise from rearrangement or the acid/base sequence on an epoxide intermediate opening to give hemiacetal/aldehyde. Actually, (g) with H3O+ then NaOH/H2O on (x) with an allylic bromide: acid first could solvolyze the bromide to allylic alcohol or rearrange; NaOH workup then gives the aldehyde via oxidation level change. Product E (aldehyde on cyclopentane) fits. So (g) -> E. (h) (1) BH3, ether; (2) H2O2: Hydroboration-oxidation of the double bond. Anti-Markovnikov addition of OH. Product H shows the cyclopentane with phenyl (wedge), methyl (dash), Br, and an OH on a plain bond (anti-Markovnikov alcohol). So (h) -> H. (i) (1) OsO4; (2) NaOH, H2O: Osmium tetroxide dihydroxylation (syn addition of two OH groups). The exocyclic double bond gives a syn diol. Product B shows the cyclopentane with phenyl (wedge), methyl (dash), Br, and two OH groups (diol). So (i) -> B. (j) H2/Pd/C (EtOH): Catalytic hydrogenation removes the double bond and can also hydrogenolyze the allylic C-Br bond. Both the alkene and the C-Br bond are reduced. Product G shows the cyclopentane with phenyl (wedge), methyl (dash), and an ethyl/propyl group (no Br, no double bond - fully saturated). So (j) -> G. Summary of matches: a -> C (neutral hydrolysis, allylic alcohol) b -> D (peracid epoxidation) c -> A (E2 elimination to diene) d -> F (ozonolysis + oxidative workup to carboxylic acid) e -> I (Br2 addition, vicinal dibromide) f -> J (NBS radical allylic bromination) g -> E (acid then base, aldehyde product) h -> H (hydroboration-oxidation, anti-Markovnikov OH) i -> B (OsO4 syn dihydroxylation, diol) j -> G (hydrogenation, saturated product) Therefore, the correct answer is {"a": "C", "b": "D", "c": "A", "d": "F", "e": "I", "f": "J", "g": "E", "h": "H", "i": "B", "j": "G"}.