See image — Haloalkanes and Haloarenes Chemistry Question

💡 Solution & Explanation

Step 1 - Identify the transformation: The starting material is 2-bromopropane (a secondary alkyl bromide, CH3CHBrCH3) and the product is 2-propanol (CH3CH(OH)CH3). The OH ends up on the same carbon (C2) as the Br, so this is effectively a substitution of Br by OH at the secondary carbon. Step 2 - Analyze the two-step sequence: The reaction requires reagent A first, then reagent B. We need to identify a pathway that converts 2-bromopropane to 2-propanol. Step 3 - Consider option (d): A = NaNH2, B = Hg(OAc)2/H2O; NaBH4. - NaNH2 is a strong base that induces E2 elimination from 2-bromopropane to give propene (CH3CH=CH2). This is Zaitsev elimination, producing the more substituted alkene, but since the molecule is symmetric the only alkene possible is propene. - Step 2: Oxymercuration-demercuration of propene with Hg(OAc)2/H2O followed by NaBH4 gives Markovnikov addition of water, placing OH on the more substituted (secondary) carbon, yielding 2-propanol. - Net result: 2-bromopropane → propene → 2-propanol. This correctly gives the observed product. Step 4 - Evaluate other options: - Option (a): H3O+ does not react with an alkyl bromide to give an alcohol cleanly; H3O+ is not a suitable nucleophile/base for this transformation. BH3-THF/H2O2/NaOH on the starting alkyl bromide makes no sense as reagent B after H3O+. - Option (b): NaOH with 2-bromopropane (secondary substrate) would give either SN2 substitution (directly forming 2-propanol) or E2 elimination (giving propene). If NaOH directly gives 2-propanol via SN2, then no second step would be needed. However, with a secondary alkyl halide and NaOH, a mixture of substitution and elimination occurs, and B = BH3-THF; H2O2/NaOH (hydroboration-oxidation) applied after NaOH treatment on the alkene would give the anti-Markovnikov alcohol (1-propanol), NOT 2-propanol. So this does not give the correct product. - Option (c): HBr in ether adds HBr to an alkene, but the starting material is already an alkyl bromide, so this reagent would not facilitate conversion to an alcohol. Step 5 - Confirm option (d): NaNH2 causes E2 elimination giving propene, then oxymercuration (Markovnikov addition) gives 2-propanol. This is the correct two-step sequence. Therefore, the correct answer is D.