Question

Primary alkyl halides C₄H₉Br 

(a) reacted with alcoholic KOH to give compound 

(b) Com-pound is reacted with HBr to give 

(c) which is an isomer of (a) When (a) is reacted with Na metal it gives compound 

(d) C₈H₁₈ which is different from the compound formed when n- butyl bromide is reacted with sodium. Give the structural formula of (a) and write the equations for all the reactions.

Answer 1

(a) The two primary alkyl bromides are possible from the molecular formula (a) C₄H₉Br. 

These are:

Thus compound (a) is either n-butyl bromide or isobutyl bromide.

(b) Since compound (a) when reacted with Na gives a compound (d), C₈H₁₈ which is different from the compound formed when n-butyl bromide is reacted with Na, therefore compound (a) must be isobutyl bromide and compound (d) must be 2,5-dimethyl hexane.

(c) If compound (a) is isobutyl bromide then compound (d) which it gives on treatment with alcoholic KOH must be 2-methyl-1-propene.

(d) The compound (b) on treatment with HBr gives compound (c) according to Markownikov’s rule. Therefore, compound (c) is tert-butyl bromide which is an isomer of compound (a).

Thus, compound (a) is isobutyl bromide, (b) is 2-methylprop-1-ene, (c) is tert-butyl bromide and (d) is 2, 5-dimethyl hexane.