An alkyl halide C5H11(A) reacts with ethanolic KOH to give an alkene 'B', which reacts with Br2 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.


The reaction scheme involved in the problem is
C5H11Br(A)Alkyl halideAlc. KOHC5H10(B)AlkeneBr2/CS2C5H10Br2(C)Dilbromo alkane-2HBrAlc. KOHC5H8(D)AlkyneH2C5H8Liq. NH3Na.C5H7-Na++12H2(straight chain alkene)
Hydrogenation of alkyne (D) gives straight chain alkane hence all the compounds (A), (B), (C) and (D) must be straight chain compounds. Alkyne (D) form sodium salt which proves that it is terminal alkyne. Involved reactions are as follows
CH3CH2CH2CH2CH2Br1-bromopentane (A)-HBrAlc.KOH, CH3CH2CH2CH=CH21-pentene (B)Br2 in CS2
It is important point that alkyl halide (A) can not be 2-bromopentane because dehydrobromination of (A) would have given 2-pentene as the major product in accordance with Markownikoff's rule.