Conversion of primary alcohol to secondary bromide via dehydration and addition

Steps:

• Recognize butan-1-ol as a primary alcohol, which undergoes SN2 substitution to give primary halides directly.
• Note 2-bromobutane is secondary, requiring carbocation rearrangement or alkene intermediate.
• Consider dehydration of butan-1-ol with conc. H2SO4 to form but-1-ene or but-2-ene.
• Apply Markovnikov addition of HBr to the alkene, yielding 2-bromobutane.

Why D is correct:

• Concentrated H2SO4 dehydrates butan-1-ol to butene (E1 mechanism via carbocation), and HBr adds per Markovnikov's rule (H to less substituted carbon, Br to more substituted), forming the secondary bromide.

Why the others are wrong:

• A: Br2/UV light performs free radical substitution on alkanes, not suitable for alcohols without prior conversion to alkane.
• B: Conc. H2SO4 with KBr generates HBr in situ for SN2 on primary alcohol, yielding 1-bromobutane, not rearranged product.
• C: Conc. H2SO4 forms alkene, but Br2 addition gives vicinal dibromide (anti addition), not monobromide.

Final answer: D