Question

What will be formed when \(CH_{3}-CH=CH_{2}+HBr\) reacts in the presence of peroxide?

• A. \(CH_{3}-CH_{2}-CH_{2}-Br\) (n-propyl bromide)
• B. \(CH_{3}-CHBr-CH_{3}\) (isopropyl bromide)
• C. \(BrCH_{2}-CH=CH_{2}\) (allyl bromide)
• D. No reaction will occur.

Hint:

Remember: Without peroxide, \[ \boxed{\text{HBr follows Markovnikov's rule}.} \] With peroxide, \[ \boxed{\text{HBr follows Anti-Markovnikov's rule}.} \] The peroxide (Kharasch) effect is shown only by HBr and not by HCl or HI.

Answer 1

The Correct Option is A

Concept: Normally, the addition of HBr to an unsymmetrical alkene follows Markovnikov's rule. However, in the presence of organic peroxides, the reaction proceeds through a free-radical mechanism and follows the anti-Markovnikov rule. This phenomenon is known as the peroxide effect or Kharasch effect. The peroxide effect is observed only with HBr and not with HCl or HI. \[ \boxed{\text{Peroxide + HBr } \Rightarrow \text{Anti-Markovnikov addition}} \]

Step 1: Identify the alkene.
The given alkene is propene, \[ CH_{3}-CH=CH_{2}. \] It is an unsymmetrical alkene.

Step 2: Apply the peroxide effect.
In the presence of peroxide, \[ HBr \] adds according to the anti-Markovnikov rule. Therefore,

• Bromine atom attaches to the carbon having more hydrogen atoms.

• Hydrogen attaches to the other carbon atom.
Thus, the product formed is \[ CH_{3}-CH_{2}-CH_{2}Br. \]

Step 3: Identify the product.
The product \[ CH_{3}-CH_{2}-CH_{2}Br \] is \[ \boxed{\text{n-propyl bromide (1-bromopropane).}} \] Hence, \[ \boxed{\textbf{Option (A)}} \] is the correct answer.

Reaction: \[ CH_{3}CH=CH_{2} \xrightarrow[\text{Peroxide}]{HBr} CH_{3}CH_{2}CH_{2}Br \]