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Free Radical Addition Reactions

The Peroxide Effect, Anti-Markovnikov Rule, and Thermodynamic feasibility.

By chemca Team • Updated Jan 2026

While alkenes typically undergo electrophilic addition, the presence of Organic Peroxides changes the mechanism to Free Radical Addition. This phenomenon is famously known as the Peroxide Effect or Kharasch Effect, leading to Anti-Markovnikov products.

1. The Reaction

Anti-Markovnikov Addition

When HBr adds to an unsymmetrical alkene in the presence of peroxide (e.g., Benzoyl Peroxide), the negative part ($Br^\bullet$) attaches to the carbon with more hydrogen atoms.

$$ R-CH=CH_2 + HBr \xrightarrow{\text{Peroxide}} \underset{\text{1-Bromoalkane (Major)}}{R-CH_2-CH_2-Br} $$

Important Constraint: This effect is observed ONLY with HBr. It is NOT observed with HF, HCl, or HI, even in the presence of peroxide.

2. Mechanism (Chain Reaction)

Step-by-Step Breakdown

Step 1: Initiation (Homolysis)
Peroxide undergoes homolytic fission to generate alkoxyl radicals, which abstract Hydrogen from HBr to form Bromine free radicals ($Br^\bullet$).

$$ R-O-O-R \xrightarrow{\Delta / h\nu} 2RO^\bullet $$ $$ RO^\bullet + H-Br \rightarrow R-OH + Br^\bullet $$

Step 2: Propagation (The Determining Step)
The $Br^\bullet$ attacks the alkene double bond. It attaches to the less substituted carbon to generate the More Stable Carbon Free Radical.

$$ R-CH=CH_2 + Br^\bullet \rightarrow \underset{2^\circ \text{ Radical (More Stable)}}{R-\dot{C}H-CH_2Br} $$

If $Br^\bullet$ attacked the middle carbon, a less stable $1^\circ$ radical would form.

Step 3: Propagation (Product Formation)
The carbon radical abstracts a Hydrogen atom from another HBr molecule.

$$ R-\dot{C}H-CH_2Br + H-Br \rightarrow R-CH_2-CH_2Br + Br^\bullet $$

3. Stability of Intermediates

Free Radical Stability Order

The direction of addition is governed by the stability of the intermediate carbon radical. The order is similar to carbocations.

$$ \text{Benzyl} > \text{Allyl} > 3^\circ > 2^\circ > 1^\circ > \text{Methyl} > \text{Vinyl} $$

In the reaction of Propene, the secondary radical ($CH_3-\dot{C}H-CH_2Br$) is formed preferentially over the primary radical ($CH_3-CH(Br)-\dot{C}H_2$).

4. Thermodynamics: Why only HBr?

Bond Energies

For a free radical chain reaction to sustain, both propagation steps must be Exothermic.

HX	Step 1 (Attack of $X^\bullet$)	Step 2 (H abstraction)	Conclusion
HF	Exothermic	Strongly Endothermic	H-F bond too strong ($566 kJ/mol$).
HCl	Exothermic	Endothermic	H-Cl bond too strong ($431 kJ/mol$).
HBr	Exothermic	Exothermic	Both steps favorable.
HI	Endothermic	Exothermic	C-I bond is weak; $I^\bullet$ combines to form $I_2$ instead.

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