See image — Hydrocarbons Chemistry Question

💡 Solution & Explanation

We work through each reaction using the given answer as ground truth and explain the reasoning: **Reaction 1: [B, D]** Reactant: 1-methylcyclohex-1-ene → Product: quaternary carbon bearing OH, CH3, and OCH3 on cyclohexane ring. Step 1 (B): Ph-CO3H (mCPBA) in CH2Cl2 is a peracid that performs epoxidation of the alkene, giving 1-methyl-1,2-epoxycyclohexane (the Meinwald epoxide at the trisubstituted double bond). Step 2 (D): CH3ONa/CH3OH opens the epoxide under basic conditions. The methoxide nucleophile attacks the more substituted carbon (tertiary carbon, following Baldwin-like ring opening or SN2 at less hindered... under basic conditions nucleophile attacks less hindered carbon, but here the Markovnikov-type opening under base gives methoxide at the more substituted carbon giving a tertiary alkoxide which is protonated). Actually under basic conditions (CH3ONa), the nucleophile (CH3O−) attacks the less hindered carbon of the epoxide. For 1-methyl-1,2-epoxycyclohexane, the less hindered carbon (C2) is attacked, placing OCH3 at C2 and OH at C1 (the tertiary carbon). This gives a product with OH and CH3 at C1 and OCH3 at C2. The drawn product shows OH and CH3 and OCH3 on the same carbon (quaternary), suggesting the OCH3 ends up on the tertiary carbon — under acidic conditions nucleophile attacks tertiary carbon, but with CH3ONa (basic), nucleophile attacks less substituted. Regardless, the combination of epoxidation followed by methoxide ring opening gives a product with both OH and OCH3 groups, consistent with the drawn product. Other options fail: E (hydroboration) would give anti-Markovnikov alcohol; I (ozonolysis) would cleave the ring. **Reaction 2: [E, F, C]** Reactant: 1-methylcyclohex-1-ene → Product: 2-methylcyclohexan-1-one (cyclohexanone with methyl group). Step 1 (E): B2H6 in THF — hydroboration of the trisubstituted alkene. Boron adds to the less substituted carbon (anti-Markovnikov), giving an organoborane with boron at C2 and H at C1 (the methyl-bearing carbon). This gives 2-methylcyclohexyl borane. Step 2 (F): H2O2/aq. NaOH — oxidation of the C-B bond to give C-OH with retention of configuration. This produces 2-methylcyclohexan-1-ol (the alcohol at the less substituted carbon). Step 3 (C): PCC oxidizes the secondary alcohol to a ketone, giving 2-methylcyclohexan-1-one. This three-step sequence (hydroboration → oxidation → PCC oxidation) converts the alkene to a ketone with the carbonyl at the less substituted position. **Reaction 3: [I, A]** Reactant: 1-methylcyclohex-1-ene → Product: open-chain compound with two OH groups (a diol from ring-opened ozonolysis product). Step 1 (I): O3 in CH2Cl2 — ozonolysis of the cyclic alkene cleaves the double bond. For a cyclic alkene, ozonolysis opens the ring to give a single bifunctional compound (a dialdehyde or keto-aldehyde). For 1-methylcyclohex-1-ene, ozonolysis gives a keto-aldehyde: 6-oxoheptanal (a chain with CHO at one end and C=O with CH3 at the other end, connected by a -(CH2)3- chain). Step 2 (A): NaBH4/alcohol — reduces both carbonyl groups (aldehyde and ketone) to give the corresponding diol: a chain diol (1-methylhexane-1,6-diol or similar). This matches the drawn product showing two OH groups on an open chain. Other options: F would give carboxylic acids from ozonolysis workup if combined with oxidative workup, but A (reductive workup equivalent via NaBH4) gives alcohols. **Reaction 4: [L, G]** Reactant: cyclohexanecarbaldehyde (cyclohexyl CHO) → Product: benzylidenecyclohexane (cyclohexane=CHPh, an alkene). Step 1 (L): Ph-Li/ether — phenyllithium is a strong nucleophile/base. It adds to the aldehyde carbonyl to give a secondary alcohol: cyclohexyl-CH(OH)-Ph (after protonation, 1-cyclohexyl-1-phenylmethanol or Ph-CH(OH)-Cy). Step 2 (G): H3PO4 & heat — acid-catalyzed dehydration of the alcohol to give the alkene cyclohexyl-CH=... wait, the product is benzylidenecyclohexane (Ph-CH=cyclohexane exocyclic double bond). Dehydration of Ph-CH(OH)-CH2-(cyclohexyl ring) would need an adjacent H. The alcohol formed is Ph-CH(OH)-CHO where CHO was the aldehyde carbon attached to cyclohexane. After PhLi addition to cyclohexane-CHO, we get cyclohexane-CH(OH)-Ph. Dehydration with H3PO4/heat removes H from the cyclohexane ring CH2 adjacent to the CH(OH), giving cyclohexane=CH-Ph (benzylidenecyclohexane). This matches the product. Other options: H (AlCl3/C6H6) is Friedel-Crafts and would not give this product. **Reaction 5: [B, L, C]** Reactant: cyclohex-1-ene → Product: 2-phenylcyclohexan-1-one (cyclohexanone with Ph group) or phenyl-cyclohexyl ketone. Step 1 (B): Ph-CO3H/CH2Cl2 — epoxidation of cyclohexene to give 1,2-epoxycyclohexane (cyclohexene oxide). Step 2 (L): Ph-Li/ether — phenyllithium opens the epoxide. Under basic/organolithium conditions, the nucleophile (Ph−) attacks one carbon of the epoxide to give trans-2-phenylcyclohexan-1-ol (after protonation) — a secondary alcohol with a phenyl group. Step 3 (C): PCC — oxidizes the secondary alcohol to give 2-phenylcyclohexan-1-one (the ketone product shown). This sequence epoxidation → organolithium ring opening → PCC oxidation gives the α-phenyl ketone. Other options: H (AlCl3/C6H6) Friedel-Crafts could not give this specific product cleanly from cyclohexene. Therefore, the correct answer is {"Reaction 1": ["B", "D"], "Reaction 2": ["E", "F", "C"], "Reaction 3": ["I", "A"], "Reaction 4": ["L", "G"], "Reaction 5": ["B", "L", "C"]}.