A primary alkyl halide `C_(4)H_(9) Br` (A) reacted with alcoholic KOH to give compound B . B on reaction with HBr yields C which is an isomer of A . When A is reacted with sodium metal forms `D , C_(6) H_(8)` which is different from the compound formed when n-butylbromide is reacted with sodium . Give the structural formulae of A-D and write equations for all the reactions .

Explanation

Given `1^(@)` -alkyl halide molecular formula `C_(4) H_(9) Br`
Two isomers possible with molecular formula `C_(4) H_(9) Br`(i-alkyl halides )
`CH_(3) - CH_(2) -underset(I)(C)H_(2) - CH_(2) - Br ` and `CH_(3) -overset(CH_(3))overset(|)underset(II)(C)H - CH_(2) -Br`
`to` Given that compound A when reacted with Na does not forms the same product produced by n-Butyl bromide .
`therefore` isomer I cannot be 'A' .
`2CH_(3) underset(n-"Butyl bromide")(-CH_(2) - CH_(2)-)CH_(2) - Br + 2 Na overset("Ether")(to ) underset(D-"Octane")(C_(8)H_(18))`
(Given n- actane does not formed in the reaction)
So isomer II must be A