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

`underset("(A)")(underset("Alkyl halide")(C_(5)H_(11)Br)) overset("alc. КОН")to underset("(B)")(underset("Alkene")(C_(5)H_(10))) overset(Br_(2))to underset("(C)")(C_(5)H_(10)Br_(2))underset("alc. KOH")overset("dehydrobromination")to underset("(D)")underset("Alkyne")(C_(5)H_(8))`
`underset("(D)")(C_(5)H_(8)) overset("Na, liq." NH_(3))to underset("Sodium alkylide")(C_(5)H_(7)Na)+ (1)/(2)H_(2)`
The reaction suggest that (D) is a terminal alkyne. The possible structures are :
`underset("(I)")(CH_(3)CH_(2)CH_(2)C -=CH) " " underset("(II)")(CH_(3)underset(CH_(3))underset(|)CHC -=CH)`
Since alkyne (D) gives straight chain on complete hydrogenation, therefore, only structure (I) is possible. Hence the reactions may be written as:
`CH_(3)CH_(2)CH_(2)CH_(2)CH_(2)Br overset("alc. KOH")to CH_(3)CH_(2)CH_(2)=CH_(2) overset(Br_(2))to CH_(3)CH_(2)CH_(2)underset(Br)underset(|)CHCH_(2)- Br overset("alc. KOH")to CH_(3)CH_(2)CH_(2)C -=CH overset("Na. liq." NH_(3))to CH_(3)CH_(2)CH_(2)C -=CNa^(+) `
and `CH_(3)CH_(2)CH_(2)C -= overset(2H_(2))to CH_(3)CH_(2)CH_(2)CH_(2)CH_(3)`
Hence the structures of A, B, C and D are :
(A) `CH_(3)CH_(2)CH_(2)CH_(2)CH_(2)Br`
(B) `CH_(3)CH_(2)CH_(2)CH= CH_(2)`
(C) `CH_(3)CH_(2)CH_(2)CH(Br)CH_(2)Br`
(D) `CH_(3)CH_(2)CH_(2)-=CH`