An alkyl halide `C_(5)H_(11)` (A) reacts with ethanolic KOH to give an alkene 'B' which reacts with `Br_(2)` to give a compound 'C' which on dehydromination gives an alkyne 'D' . On treatment with sodium metal in liquid ammonia one mole of 'D' give 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 the reactions involved.

Explanation

The outline of reaction scheme involved in the given problem is
`underset"Alkyl halide (A)"(C_5H_11Br) overset"Alc. KOH"to underset"Alkene (B)"(C_5H_10) overset(Br_2//CS_2)to underset"(C )"(C_5H_10Br_2) underset"-2 HBr"overset"Alc. KOH"to underset"Alkyne (D)"(C_5H_8) overset(Na- liq. NH_3)to underset"Sod. Alkynide"(C_5H_7Na+1//2H_2)`
(i)Since 1 mole of alkyne 'D' reacts with 1 mole of Na in liquid `NH_3` to form half a mole of `H_2`, therefore, (D) is a terminal alkyne. This means that triple bond is at the end of the carbon chain. The two structures for alkyne (D) are either (I) or (II)
`underset"1-Pentyne (I)"(CH_3CH_2CH_2-C-=CH) " " underset"3-Methylbut-1-yne (II)"(CH_3-oversetoverset (CH_3)|CH-C-=CH)`
Since alkyne 'D' on complete hydrogenation yields a straight chain alkane, therefore , the alkyne (D) is a straight chain alkyne, i.e., alkyne (D) is 1-pentyne (I).
(ii)Since alkene (B) on reaction with `Br_2` forms a compound 'C' which one dehydrohalogenation gives the alkyne, i.e., 1-pentyne (D), therefore, (C ) must be 1,2-dibromopentane and alkene (B) must be 1-pentene .
(iii)Further since alkene (B), i.e., 1-pentene is obtained by dehydrogenation of alkyl halide with M.F. `C_5H_11Br`, therefore, alkyl halide (A) must be 1-bromopentane.
All the reactions involved in this question may now be explained as follows :
`underset"1-Bromopentane (A)"(CH_3CH_2CH_2CH_2CH_2Br) underset"-HBr"overset(Alc. KOH, Delta)to underset"1-Pentene (B)"(CH_3CH_2CH_2CH=CH_2) overset(Br_2 "in" Cs_2)to underset"1,2-Dibromopentane (C )"(CH_3CH_2CH_2-overset2CHBr-overset1CH_2Br) underset"-2HBr " overset(Alc. KOH, Delta)to underset"1-Pentyne (D)" (CH_3CH_2CH_2C-=CH) overset("Na in liq." NH_3)to underset"Sodium 1-pentynide"(CH_2CH_2CH_2C-=CNa +1//2H_2)`
Please not that alkyl halide (A) cannot be 2-bromopentane because dehydrobromination of (A) would have given 2-pentane as the major product in accordance with Markovnikov's rule.