Question:medium
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)
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)
(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)
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Peroxide-induced additions in organic reactions usually follow anti-Markovnikov's rule. Alcoholic KOH induces elimination reactions to form alkenes.
Updated On: Jan 13, 2026
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Solution and Explanation
(i) Propene (CH$_3$CH=CH$_2$) reacts with HBr and peroxide to undergo anti-Markovnikov addition, yielding 1-bromopropane (A). Subsequent reaction with KOH eliminates HBr, forming propene (B).
(ii) Ethyl chloride (CH$_3$CH$_2$CHCl) reacts with alcoholic KOH to eliminate HCl, producing propene (A). Hydration of this product primarily yields ethanol (B).
(ii) Ethyl chloride (CH$_3$CH$_2$CHCl) reacts with alcoholic KOH to eliminate HCl, producing propene (A). Hydration of this product primarily yields ethanol (B).
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