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Question

Acetylene can be converted to 1-Bromo pentane via coupling of 1-Bromopropane. The set of reagents required are :

The correct answer is

NaNH2, CH3CH2CH2Br; H2/Lindlar cut; HBr, H2O2

Step 1 (chain extension): NaNH2 deprotonates acetylene to the acetylide ion, which displaces bromide from 1-bromopropane by SN2: HC≡CH → HC≡C–Na+ → HC≡C–CH2CH2CH3 (1-pentyne, five carbons).

Step 2 (partial reduction): H2 over Lindlar catalyst (Pd/CaCO3 poisoned with lead acetate and quinoline) stops at the alkene: 1-pentyne → CH2=CH–CH2CH2CH3 (1-pentene).

Step 3 (anti-Markovnikov addition): HBr in the presence of a peroxide (H2O2) adds by a free-radical chain; Br· adds to the terminal CH2 to give the more stable secondary radical, so bromine ends up on C-1: CH2=CHCH2CH2CH3 → BrCH2CH2CH2CH2CH3 (1-bromopentane).

Using H2, Pd/C reduces the alkyne all the way to pentane, which does not react with HBr; LiAlH4 does not reduce a simple terminal alkyne to the alkene; and HBr without peroxide adds by Markovnikov's rule to give 2-bromopentane.

Therefore the sequence is NaNH2/1-bromopropane, then H2/Lindlar catalyst, then HBr with H2O2.

Hence, NaNH2, CH3CH2CH2Br; H2/Lindlar cut; HBr, H2O2.

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