See image — Haloalkanes and Haloarenes Chemistry Question

💡 Solution & Explanation

Step 1 – Identify the transformation: The starting material is a branched alkane (2,2-dimethylbutane: (CH3)3CCH2CH3). The product is the primary alcohol 2,2-dimethyl-1-butanol: (CH3)2C(CH2CH2OH) — meaning the OH ends up at the terminus of the longer chain. Comparing structures, the conversion requires: introducing unsaturation (an alkene) at the correct position, then hydrating it with anti-Markovnikov selectivity. Step 2 – Evaluate option (d): (1) Br2, hv: Free-radical bromination at the most reactive (tertiary or allylic) C–H, placing Br at the tertiary carbon (C3 of 2,2-dimethylbutane, or the carbon bearing two methyls and adjacent to CH2CH3), giving a tertiary alkyl bromide. (2) t-butoxide (a bulky, strong base): E2 elimination of HBr. The bulky base favors the less-substituted (Hofmann) alkene, producing the terminal alkene: (CH3)2C=CHCH3 or more precisely the less hindered alkene placing the double bond away from the gem-dimethyl group, yielding a terminal or less substituted alkene. (3) BH3, THF: Hydroboration adds boron to the less hindered (terminal) carbon of the double bond (anti-Markovnikov). (4) H2O2, OH-: Oxidation of the C–B bond gives the primary alcohol at the terminal carbon. This sequence correctly converts the alkane to the primary alcohol shown. Step 3 – Why other options fail: (a) Uses NBS (allylic/benzylic bromination reagent) on an alkane with no allylic position, then a long sequence involving NH2- and disiamyl borane — overly complicated and the first step is inapplicable to a simple alkane. (b) Uses Cl2/hv (chlorination) giving a mixture of products; the subsequent steps with HgSO4/H2SO4 (Markovnikov hydration) would not give the primary alcohol. (c) NBS, ROOR followed by OH-/DMSO is only two steps (SN2 to give alcohol) and would not lengthen or correctly functionalize the chain to a primary alcohol at the right position. Therefore, the correct answer is D.