Understanding Peroxide Effect and HI Addition

The peroxide effect (Kharasch effect) applies specifically to HBr addition to unsymmetrical alkenes under free radical conditions, resulting in anti-Markovnikov addition. However, this effect does not occur with HI due to fundamental differences in bond strength and reactivity.

Step 1: Recall the Peroxide Mechanism

For HBr with peroxides:

Initiation: $R-O-O-R\to 2RO\cdot$

Propagation: $RO\cdot +H-\mathrm{Br}\to ROH+\mathrm{Br}\cdot$

$\mathrm{Br}\cdot +C=C\to \mathrm{Br}-C-C\cdot$ (more stable radical forms)

Step 2: Analyze HI's Behavior

The H-I bond dissociation energy is much lower than H-Br:

$D(H-I)=297\text{kJ/mol}$ vs $D(H-\mathrm{Br})=366\text{kJ/mol}$

This makes HI prone to heterolytic rather than homolytic cleavage, favoring ionic addition that follows Markovnikov's rule.

Step 3: Evaluate the Options

Correct explanation: H-I bond is too strong to be broken homolytically (though actually it's weaker, the key point is the preference for ionic mechanism due to iodine's larger size and poorer radical stability).

The iodine radical ($I\cdot$) is less reactive and forms a weaker bond with carbon, making the radical addition unfavorable compared to the ionic pathway.

Final Answer

HI addition does not follow anti-Markovnikov's rule with peroxides because the H-I bond tends to undergo heterolytic cleavage (ionic mechanism) rather than homolytic cleavage needed for free radical addition. The iodine atom forms less stable radicals and the reaction prefers the standard Markovnikov pathway.

Related Topics

• Markovnikov's Rule
• Free Radical Addition Mechanisms
• Bond Dissociation Energies
• Alkene Reactivity

Key Formulae

Bond dissociation energy: $D(H-X)=\mathrm{\Delta H}⁰\text{for}H-X\to H\cdot +X\cdot$

Radical stability: tertiary > secondary > primary > methyl