Question

Complete the following reactions by writing the structural formulae of ‘A’ and ‘B’:
(i) CH$_3$CH=CH$_2$ + HBr $\xrightarrow{\text{Peroxide}}$ ‘A’ $\xrightarrow{KOH}$ ‘B’
(ii) CH$_3$CH$_2$CHCl + alc. KOH $\xrightarrow{\Delta}$ ‘A’ $\xrightarrow{\text{H}_2\text{O}}$ ‘B’ (Main product)

Hint:

Peroxide-induced additions in organic reactions usually follow anti-Markovnikov's rule. Alcoholic KOH induces elimination reactions to form alkenes.

Answer 1

(i) The reaction of propene (CH$_3$CH=CH$_2$) with HBr in the presence of peroxide leads to anti-Markovnikov addition of HBr, producing 1-bromopropane (A). The subsequent reaction with KOH leads to an elimination of HBr, producing propene (B).
(ii) The reaction of ethyl chloride (CH$_3$CH$_2$CHCl) with alcoholic KOH results in elimination of HCl, forming propene (A). The main product after treatment with water (hydration) is ethanol (B).