Primary alkyl halide `C_4 H_9Br` :
(a) reacted with alcoholic KOH to give compound (b) is reacted with HBr to give : (c) which is an isomer of (a). When (a) is reacted with Na metal, it gives a compound (d), `C_8H_18` hich is different from the compound formed when n-butyl bromide is reacted with Na metal. Give the structural formula of (a) and write the equations for all the reactions.

Explanation

(i) There are two primary alkyl halides the molecular formula, `C_(4)H_(9)Br`.

(ii) Since compound (a) when reacted with Na metal gave a compound (d) with molecular formula `C_(8)H_(18)` which was different from die compound obtained when n-butyl bromide was reacted with Na metal, therefore, (a) must be isobutyl bromide and compound (d) must be 2,3-dimethylhexane.

(iii) If compound (a) is isobutyl bromide, than the compound (b) which it gives on treatment with alcoholic KOH must be 2-methyl-1-propane.

(iv) The compound (b) on treatment with HBr gives compound (c) in accordance with Markownikoff rule. Therefore, compound (c) is tert-butyl bromide which is an isomer of compound (a) ,i.e, isobutyl' bromide.

Thus
(a) is isobutyl bromide,
(b) is 2-methyl-1 -propane,
(c) is tert-butylbromide, and
(d) is 2,5-dimethylhexane.