See image — Alcohols Phenols and Ethers Chemistry Question

💡 Solution & Explanation

Step 1 - Analyze each reaction in Column (I): (a) 1-methylcyclohexan-1-ol (starred/chiral center marked on OH carbon) + CH3OH / H2SO4 (conc.): This is an acid-catalyzed etherification (Fischer-type) of a tertiary alcohol with methanol. The tertiary carbocation forms, and methanol attacks it. The oxygen of the ether comes from methanol (unlabeled). The product is 1-methoxy-1-methylcyclohexane where the OCH3 is derived from methanol (no isotope label). The stereocenter is lost (tertiary carbocation intermediate), so no asterisk on the OCH3 carbon and no 14C label. This matches (s): cyclohexane with CH3 and OCH3 (no label, no asterisk) at C1. (b) 1-methylcyclohexan-1-ol (starred OH carbon) + (1) NaH, (2) CH3I: NaH deprotonates the alcohol to form the alkoxide. CH3I then alkylates the oxygen via SN2. The oxygen stays attached to the same carbon (the starred carbon), so the stereochemistry/label at the carbon is retained, and the OCH3 comes from unlabeled CH3I but the oxygen center retains the asterisk from the original chiral carbon. The product is 1-methyl-1-methoxycyclohexane with the asterisk on the OCH3 oxygen center (i.e., the carbon bearing OCH3 retains the star). This matches (r): cyclohexane ring with CH3 and OCH3* at C1. (c) 1-methylcyclohexan-1-ol (starred OH carbon) + (1) HBr, (2) Mg, (3) CH3I: Step 1: HBr converts tertiary alcohol to tertiary bromide (1-bromo-1-methylcyclohexane). Step 2: Mg converts the bromide to a Grignard reagent (1-methylcyclohexylmagnesium bromide). Step 3: CH3I reacts with the Grignard via SN2, replacing MgBr with CH3. The product is 1,1-dimethylcyclohexane (no oxygen remains). This matches (q): cyclohexane with two CH3 groups at C1. (d) 1-methylcyclohexan-1-ol (unlabeled, normal OH) + (1) Na, (2) 14C-labeled CH3I: Na deprotonates the alcohol to form the sodium alkoxide. 14CH3I (isotopically labeled methyl iodide with 14C) alkylates the oxygen via SN2. The product is 1-methyl-1-(14C-methoxy)cyclohexane where the OCH3 group contains 14C. The original carbon bearing oxygen is unlabeled, but the methyl of OCH3 is 14C-labeled. This matches (p): cyclohexane with CH3 and 14OCH3 at C1. Step 2 - Summary of matches: (a) -> (s): simple acid-catalyzed etherification, no isotope, no stereo label retained (b) -> (r): Williamson ether synthesis retaining the starred carbon, OCH3 asterisk shown (c) -> (q): Grignard sequence removes oxygen entirely, gives 1,1-dimethylcyclohexane (d) -> (p): Williamson with 14C-labeled CH3I gives 14C-labeled OCH3 product Therefore, the correct answer is {"a": ["S"], "b": ["R"], "c": ["Q"], "d": ["P"]}.