Question

Addition of HBr to propene yields 2-bromopropane, while in the presence of benzoyl peroxide, the same reaction yields 1-bromopropane. Explain and give mechanism.

Answer 1

(A) Addition of HBr to propene (in absence of peroxide)

In the absence of peroxide, addition of HBr to propene follows Markovnikov’s rule.

According to Markovnikov’s rule, hydrogen atom adds to the carbon atom already carrying more hydrogen atoms, and bromine adds to the carbon atom carrying fewer hydrogen atoms.

Thus, propene gives 2-bromopropane as the major product.

Mechanism (Electrophilic addition):

Step 1: Formation of carbocation

CH₃–CH=CH₂ + H⁺ → CH₃–C⁺H–CH₃

(A more stable secondary carbocation is formed.)

Step 2: Attack by bromide ion

CH₃–C⁺H–CH₃ + Br⁻ → CH₃–CH(Br)–CH₃

Thus, 2-bromopropane is formed.

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(B) Addition of HBr to propene (in presence of benzoyl peroxide)

In the presence of benzoyl peroxide, the reaction follows a free radical mechanism and gives anti-Markovnikov addition.

This is known as the peroxide effect or Kharasch effect.

Thus, propene gives 1-bromopropane as the major product.

Mechanism (Free radical addition):

Step 1: Initiation

Benzoyl peroxide → 2 C₆H₅COO· → 2 C₆H₅· + 2 CO₂

C₆H₅· + HBr → C₆H₆ + Br·

Step 2: Propagation

Br· + CH₃–CH=CH₂ → CH₃–ĊH–CH₂Br

(The more stable secondary radical is formed.)

CH₃–ĊH–CH₂Br + HBr → CH₃–CH₂–CH₂Br + Br·

Step 3: Termination

Combination of free radicals stops the reaction.

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Conclusion:

In the absence of peroxide, HBr adds to propene according to Markovnikov’s rule giving 2-bromopropane. 
In the presence of benzoyl peroxide, anti-Markovnikov addition occurs due to free radical mechanism, giving 1-bromopropane.