An alkyl halide C_{5}H_{11}Br (A) reacts with ethanolic KOH to give an alkene ‘B’, which reacts with Br_{2} to give a compound ‘C’, which on dehydrobromination gives an alkyne ‘D’. On treatment with sodium metal in liquid ammonia one mole of ‘D’ gives one mole of the sodium salt of ‘D’ and half a mole of hydrogen gas. Complete hydrogenation of ‘D’ yields a straight chain alkane. Identify A,B, C and D. Give the reactions involved.

Answers (1)

To identify A, B, C and D, the reactions involved are-

C_{5}H_{11}­Br(A) + alc. KOH \rightarrow C_{5}H_{10}(B)

C_{5}H_{10}(B) + Br/CS_{2} \rightarrow C_{5}H_{10}Br_{2}(C)

C_{5}H_{10}Br_{2}(C) + alc. KOH \rightarrow C_{5}H_{8}(D) Alkyne

2C_{5}H_{8 }+ 2Na \rightarrow 2C_{5}H_{7}Na + H_{2}

All the compounds above must be straight-chain as hydrogenation of alkyne (D) gives straight-chain alkane. It is clear that D is terminal alkyne as alkyne gives sodium alkenyde.

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