See image — Haloalkanes and Haloarenes Chemistry Question

💡 Solution & Explanation

Starting material: isobutyl alcohol = (CH3)2CHCH2OH Reagent key: A = Hg(OAc)2/H2O (oxymercuration) B = PBr3 & heat C = NaBH4 in alcohol D = LiAlH4 in THF E = NaCN in alcohol F = PCC in CH2Cl2 G = Jones' reagent H = HBr in CH2Cl2 I = H3PO4 & heat J = (CH3CO)2O + pyridine K = NaN3 in aqueous alcohol L = peracid M = NaH in ether & heat N = C2H5OH + acid catalyst & heat --- Part a: (CH3)2CHCH2OH → (CH3)2CHCH2Br (one step) --- PBr3 converts primary alcohol directly to primary alkyl bromide with inversion but retention of carbon skeleton. Reagent B (PBr3 & heat). --- Part b: (CH3)2CHCH2OH → (CH3)2C=CH2 (one step) --- Dehydration with H3PO4 and heat (acid-catalyzed elimination). The primary alcohol loses water; the carbocation rearranges (1,2-hydride shift) to give tertiary carbocation (CH3)3C+, which loses a proton to give isobutylene (CH3)2C=CH2. Reagent I. --- Part c: (CH3)2CHCH2OH → (CH3)2CHCH=O (one step) --- Oxidation of primary alcohol to aldehyde without over-oxidation requires PCC. Reagent F (PCC in CH2Cl2). --- Part d: (CH3)2CHCH2OH → (CH3)2CHCO2H (one step) --- Jones' reagent (CrO3/H3O+) oxidizes primary alcohol all the way to carboxylic acid. Reagent G. --- Part e: (CH3)2CHCH2OH → (CH3)3CBr (two steps) --- Step 1: Dehydration with H3PO4/heat gives (CH3)2C=CH2 via carbocation rearrangement (product b). Step 2: Addition of HBr to the alkene via Markovnikov addition gives (CH3)3CBr. So: I, then H. Answer: I, H. Alternatively, use previously prepared compound b (the alkene) then add HBr: written as 2H (using compound b as starting material for second step). Answer: I, H or 2H. --- Part f: (CH3)2CHCH2OH → (CH3)2CHCH2C≡N (two steps) --- Step 1: Convert alcohol to alkyl bromide using PBr3 (reagent B) → (CH3)2CHCH2Br (compound a). Step 2: SN2 with NaCN (reagent E) → (CH3)2CHCH2CN. Answer: B, E. Alternatively, use previously prepared compound a (the bromide) then NaCN: 1 (compound a), E. Answer: B, E or 1, E. --- Part g: (CH3)2CHCH2OH → (CH3)2CHCH2OCOCH3 (one step) --- Acetylation of alcohol with acetic anhydride in pyridine. Reagent J [(CH3CO)2O + pyridine]. --- Part h: (CH3)2CHCH2OH → (CH3)2CHCO2C2H5 (two steps) --- Step 1: Oxidize primary alcohol to carboxylic acid with Jones' reagent (G) → (CH3)2CHCO2H (compound d). Step 2: Fischer esterification with C2H5OH/acid catalyst (N) → (CH3)2CHCO2C2H5. Answer: G, N. Alternatively, use compound d then N: 4N. Answer: G, N or 4N. --- Part i: (CH3)2CHCH2OH → (CH3)2CHCH2OCH2CH3 (two steps) --- Etherification: The alcohol can be converted to an ether with ethanol under acid catalysis. However, since isobutyl alcohol is primary and ethanol is primary, intermolecular dehydration with acid catalyst (N) could work but selectivity is an issue. Alternatively: Use N (C2H5OH + acid catalyst & heat) in two steps. Actually, the simplest approach is acid-catalyzed etherification: isobutyl alcohol + ethanol under acid catalyst at moderate temperature gives the mixed ether. This is a two-step consideration but can also be viewed as: step 1 use reagent N directly (C2H5OH + acid catalyst & heat). The answer given is simply N, meaning both steps involve reagent N or the transformation requires two applications. Given answer: N. --- Part j: (CH3)2CHCH2OH → (CH3)3COH (three steps) --- Path 1: Step 1: Dehydration (I) → (CH3)2C=CH2. Step 2: Oxymercuration (A) with Hg(OAc)2/H2O gives Markovnikov addition of water → (CH3)3COH after reduction. Step 3: NaBH4 (C) for demercuration. Answer: I, A, C. Alternatively using compound b (the alkene from step b=2): 2, A, C = 2AC. Another path: I, then LiAlH4 doesn't apply directly. Also possible: I (dehydration to alkene b), then oxymercuration-reduction: A then C. Other paths noted: ILD or 2LD — this would mean: dehydration to alkene, then epoxidation with peracid (L), then LiAlH4 (D) reduction of epoxide to give tertiary alcohol. I, L, D or 2LD. Answer: I, A, C or 2AC or ILD or 2LD. --- Part k: (CH3)2CHCH2OH → (CH3)2CHCH2NH2 (three steps) --- Step 1: Convert to bromide with PBr3 (B) → (CH3)2CHCH2Br (compound a). Step 2: SN2 with NaN3 (K) → (CH3)2CHCH2N3. Step 3: Reduce azide with LiAlH4 (D) → (CH3)2CHCH2NH2. Answer: B, K, D. Alternatively starting from compound a (1): 1, K, D = 1KD. Answer: B, K, D or 1KD. --- Part l: (CH3)2CHCH2OH → (CH3)2CHCH2CH2NH2 (two steps) --- This product has one more carbon than starting material. Step 1: Convert to nitrile via bromide then NaCN: but that's two steps already. Since compound f = (CH3)2CHCH2C≡N is available, and is a nitrile with the extra carbon: Step 1: Use compound f (nitrile, step f). Step 2: Reduce nitrile with LiAlH4 (D) → (CH3)2CHCH2CH2NH2. Answer using compound f: 6, D = 6D. Alternatively, build from scratch: B (to make bromide a), then E (NaCN to make nitrile f), then D (LiAlH4 reduction): B, E, D. But that's 3 steps, yet problem says two steps. So likely: use previously prepared compound f (=6) then D. Or: 1 (compound a = bromide), E (NaCN → nitrile), D (reduction) — but that again is 3 reagents. Since the problem says two steps and allows reference to previously prepared compounds, using compound f directly: 6, D is two steps. Also B,E,D if counting differently. Answer: B, E, D or 1ED or 6D. Therefore, the correct answer is {"a": "B", "b": "I", "c": "F", "d": "G", "e": "I, H or 2H", "f": "B, E or 1, E", "g": "J", "h": "G, N or 4N", "i": "N", "j": "I, A, C or 2AC or ILD or 2LD", "k": "B, K, D or 1KD", "l": "B, E, D or 1ED or 6D"}.