In organic chemistry, the addition of hydrogen halides (HX) to unsymmetrical alkenes generally follows Markovnikov’s rule. However, an interesting exception occurs when hydrogen bromide (HBr) reacts with an alkene in the presence of organic peroxides. In this case, the addition takes place according to the Anti-Markovnikov rule.
This reaction is known as the Peroxide Effect or Kharasch Effect.
But an important question arises:
Why does the peroxide effect occur only with HBr and not with HCl or HI?
The answer lies in the energetics of the free-radical chain mechanism. For the peroxide effect to occur, both propagation steps must be energetically favourable (exothermic).
Let’s understand the mechanism step by step.
1. What Is the Peroxide Effect?
The peroxide effect is the anti-Markovnikov addition of HBr to an unsymmetrical alkene in the presence of organic peroxides.
For example, when propene reacts with HBr in the presence of peroxide, the major product is 1-bromopropane rather than 2-bromopropane.
Normal addition (Markovnikov):
CH₃–CH=CH₂ + HBr → CH₃–CHBr–CH₃
Peroxide effect (Anti-Markovnikov):
CH₃–CH=CH₂ + HBr → CH₃–CH₂–CH₂Br
The reaction follows a free-radical chain mechanism.
2. Mechanism of the Peroxide Effect
The free-radical mechanism consists of three stages:
A. Initiation
Organic peroxide undergoes homolytic cleavage in the presence of heat or light, producing free radicals.
RO–OR → 2RO•
These radicals react with HBr to generate bromine radicals.
RO• + HBr → ROH + Br•
B. Propagation
This is the most important stage of the reaction.
Step 1: Addition of bromine radical
The bromine radical attacks the double bond of the alkene, producing a carbon radical.
CH₃–CH=CH₂ + Br• → CH₃–ĊH–CH₂Br
The bromine atom attaches to the terminal carbon, producing the more stable secondary carbon radical.
Step 2: Hydrogen abstraction
The carbon radical abstracts a hydrogen atom from another HBr molecule.
CH₃–ĊH–CH₂Br + HBr → CH₃–CH₂–CH₂Br + Br•
A new bromine radical is generated, allowing the chain reaction to continue.
C. Termination
Two radicals combine to form a stable molecule, ending the chain reaction.
Br• + Br• → Br₂
3. Why Does HCl Not Show the Peroxide Effect?
HCl does not undergo anti-Markovnikov addition through the radical chain mechanism because its second propagation step is endothermic.
Let’s examine both steps.
Step 1: Addition of chlorine radical
Cl• + Alkene → Carbon radical
This step is generally exothermic.
Step 2: Hydrogen abstraction
Carbon radical + HCl → Product + Cl•
This step requires breaking the very strong H–Cl bond.
The bond dissociation energy of H–Cl is approximately 431 kJ/mol. The energy released by forming the new C–H bond is insufficient to compensate for the energy required to break the H–Cl bond.
Therefore, the second propagation step is endothermic.
As a result, the radical chain reaction cannot sustain itself efficiently, and the peroxide effect is not observed.
4. Why Does HI Not Show the Peroxide Effect?
HI fails for the opposite reason.
Its first propagation step is energetically unfavourable.
Step 1: Addition of iodine radical
I• + Alkene → Carbon radical
The addition of an iodine radical to the double bond is endothermic. The formation of the carbon–iodine bond does not release enough energy to compensate for the loss of the alkene π bond and the change in bonding.
Therefore, this step is energetically unfavourable.
Step 2: Hydrogen abstraction
Carbon radical + HI → Product + I•
This step is exothermic because the H–I bond is relatively weak (approximately 297 kJ/mol).
However, the reaction cannot proceed efficiently because the first propagation step is endothermic.
Thus, HI does not show the peroxide effect.
5. Why Does HBr Show the Peroxide Effect?
HBr has the right balance of bond strengths for both propagation steps to be exothermic.
Step 1: Addition of Br• to the alkene is exothermic.
Step 2: The carbon radical abstracts hydrogen from HBr. Breaking the H–Br bond (approximately 366 kJ/mol) and forming a C–H bond makes this step exothermic.
Since both propagation steps are energetically favourable, the radical chain reaction can continue.
This is why HBr is the only hydrogen halide that exhibits the peroxide effect.
6. Comparison of HCl, HBr and HI
Hydrogen halide First propagation step Second propagation step Peroxide effect HCl Exothermic Endothermic No HBr Exothermic Exothermic Yes HI Endothermic Exothermic No
7. Important Points for JEE and NEET
Remember these important facts:
- The peroxide effect is also called the Kharasch effect.
- It involves a free-radical chain mechanism.
- The peroxide effect is observed only with HBr.
- HCl fails because its second propagation step is endothermic.
- HI fails because its first propagation step is endothermic.
- Both propagation steps must be energetically favourable for an efficient radical chain reaction.
Conclusion
The peroxide effect is a good example of how bond dissociation energies and reaction energetics determine whether a free-radical chain reaction can occur.
HCl has a very strong H–Cl bond, making hydrogen abstraction unfavourable. HI has a weak H–I bond, but iodine radical addition to an alkene is unfavourable. HBr has the right balance, allowing both propagation steps to proceed exothermically.
Therefore, only HBr shows the peroxide effect, whereas HCl and HI do not.
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