See image — Hydrocarbons Chemistry Question

💡 Solution & Explanation

The starting material in all four reactions is 4-tert-butylcyclohexene (the triple-line substituent represents a tert-butyl or similar bulky group locked equatorial at C4, and the double bond is between C1 and C2). Step 1 - Identify each reaction type and its stereochemical outcome: (a) OsO4 followed by NaOH/H2O: OsO4 dihydroxylation is a syn addition. Both OH groups are added to the same face of the double bond. For 4-tert-butylcyclohexene, syn addition gives the cis-diol (both OH groups on the same face). This corresponds to product (r) - the cis-1,2-diol with both OH groups shown on the same face (both wedge or both dash relative to each other). (b) BH3/ether followed by H2O2/NaOH/H2O (hydroboration-oxidation): This is also an overall syn addition, giving anti-Markovnikov alcohol. The boron adds to the less substituted carbon (C1) with syn selectivity, and oxidation retains configuration. The product is the alcohol where OH is added syn to the ring face, giving a trans relationship between the OH at C1 and the tert-butyl at C4 in the chair. This corresponds to product (p) - the monoalcohol with trans stereochemistry (OH on wedge at C1 with the bulky group on dash at C4, or the trans isomer). (c) Cl2, H2O (chlorohydrin formation): Cl2 in water proceeds via a bromonium/chloronium ion intermediate, followed by anti addition of water (nucleophile attacks from the opposite face). This gives anti addition of Cl and OH. The product is the trans-chlorohydrin. This corresponds to product (q) - Cl and OH on adjacent carbons in a trans (anti) relationship. (d) Cl2/CCl4 (halogenation without water): Cl2 in an inert solvent (CCl4) proceeds via a chloronium ion with anti addition of both chlorides. This gives the trans-1,2-dichloride. This corresponds to product (s) - the trans-dichloride with Cl atoms on adjacent carbons in anti relationship (one wedge, one dash). Step 2 - Verify why other options fail: - (a) cannot be (p) because OsO4 gives a diol, not monoalcohol. - (a) cannot be (q) or (s) because no chlorine is involved. - (b) cannot be (r) because hydroboration gives only one OH. - (c) cannot be (s) because Cl2/H2O gives chlorohydrin (Cl + OH), not dichloro. - (d) cannot be (q) because CCl4 conditions give dichloro, not chlorohydrin. Summary of matches: (a) → (r): syn dihydroxylation → cis-diol (b) → (p): hydroboration-oxidation → trans-alcohol (anti-Markovnikov, syn addition, overall trans to bulky group) (c) → (q): chlorohydrin formation → trans-Cl/OH (d) → (s): halogenation → trans-dichloride Therefore, the correct answer is {"a": "R", "b": "P", "c": "Q", "d": "S"}.