Question

The major product formed in the following reaction is: \[ CH_3CH=CH_2 \xrightarrow{HBr,\ peroxide} \ ? \]

• A. \(CH_3CHBrCH_3\)
• B. \(CH_3CH_2CH_2Br\)
• C. \(CH_2BrCH=CH_2\)
• D. \(CH_3CHBrCH_2Br\)

Hint:

Peroxide effect is shown only by HBr and not by HCl or HI. \[ HBr + \text{Peroxide} \Rightarrow \text{Anti-Markovnikov addition} \]

Answer 1

The Correct Option is B

Concept: Addition of HBr in presence of peroxide follows: \[ \text{Anti-Markovnikov rule} \] This reaction is known as: \[ \text{Peroxide effect or Kharasch effect} \] Free radical mechanism causes bromine to attach to less substituted carbon atom.

Step 1: Identify the alkene. Given alkene: \[ CH_3CH=CH_2 \] This is propene.

Step 2: Understand peroxide effect. In presence of peroxide: \[ HBr \] adds through free radical mechanism. Bromine attaches to terminal carbon. Thus: \[ CH_3CH=CH_2 \rightarrow CH_3CH_2CH_2Br \]

Step 3: Identify the product. The product formed is: \[ 1\text{-Bromopropane} \] Structure: \[ CH_3CH_2CH_2Br \] Hence correct option is: \[ \boxed{(B)} \]