Question

Identify ‘A’ and ‘B’ in the following reaction and rewrite the complete reaction : \[ \text{CH}_3\text{-CH=CH}_2 \xrightarrow[\text{Peroxide}]{\text{HBr}} \text{A} \xrightarrow[\text{-KBr}]{\text{alcoholic KCN}} \text{B} \]

Hint:

The addition of HBr to an alkene can yield two different products. Without peroxide, it follows Markovnikov's rule (electrophilic addition). In the presence of peroxide, it follows a free-radical mechanism, leading to the anti-Markovnikov product. This effect is specific to HBr.

Answer 1

Step 1: First reaction (Propene to A)
This is the addition of HBr to propene in the presence of peroxide. This follows the anti-Markovnikov's rule (Kharasch effect), where the negative part of the addendum (Br\(^-\)) attaches to the carbon atom with more hydrogen atoms. \[ \underset{\text{Propene}}{\text{CH}_3\text{-CH=CH}_2} + \text{HBr} \xrightarrow{\text{Peroxide}} \underset{\text{ 1-bromopropane}}{\text{CH}_3\text{-CH}_2\text{-CH}_2\text{-Br}} \] So, A is 1-bromopropane.
Step 2: Second reaction (A to B) This is a nucleophilic substitution reaction where 1-bromopropane reacts with alcoholic potassium cyanide (KCN). The cyanide ion (CN\(^-\)) displaces the bromide ion (Br\(^-\)).
\[ \underset{\text{ 1-bromopropane}}{\text{CH}_3\text{-CH}_2\text{-CH}_2\text{-Br}} + \text{KCN (alc.)} \rightarrow \underset{\text{ Butanenitrile}}{\text{CH}_3\text{-CH}_2\text{-CH}_2\text{-CN}} + \text{KBr} \] So, B is Butanenitrile.
Complete Reaction: \[ \text{CH}_3\text{-CH=CH}_2 \xrightarrow[\text{Peroxide}]{\text{HBr}} \text{CH}_3\text{-CH}_2\text{-CH}_2\text{-Br} \xrightarrow[\text{-KBr}]{\text{alcoholic KCN}} \text{CH}_3\text{-CH}_2\text{-CH}_2\text{-CN} \]