Q.In the presence of peroxide addition of HBr to propene takes place according to anti Markovnikov's rule but peroxide effect is not seen in the case of HCl and HI. Explain.

Concept: Anti-Markovnikov Addition via Free-Radical Mechanism

In the presence of peroxides, HBr adds to alkenes through a free-radical mechanism rather than the usual electrophilic addition, leading to anti-Markovnikov products. This peroxide effect depends critically on the bond dissociation energy and stability of the halogen radical formed.

Why HBr shows the peroxide effect:

1. The peroxide initiates a radical chain by homolytically cleaving H–Br to form $\ce{Br·}$. The $\ce{Br·}$ radical then adds to the alkene at the less substituted carbon (forming the more stable secondary radical on propene), which abstracts hydrogen from another HBr molecule—yielding the anti-Markovnikov product (1-bromopropane).

2. The H–Br bond dissociation energy (~366 kJ/mol) is moderate: low enough that the radical chain propagates efficiently, yet the $\ce{Br·}$ radical is reactive enough to add to the double bond.

Why HCl and HI do not show the peroxide effect: …

When an alkene reacts with a hydrogen halide, the usual outcome follows Markovnikov's rule: the hydrogen attaches to the carbon with more hydrogens, and the halogen goes to the more substituted carbon. This happens through a carbocation intermediate, where the more stable (more substituted) carbocation forms preferentially.

But in the presence of peroxides, HBr behaves differently. The reaction switches to a free-radical mechanism, and the product flips to anti-Markovnikov orientation. The question is: why does this peroxide effect work for HBr but not for HCl or HI?

The answer lies in the energetics of each step in the radical chain mechanism.

The Free-Radical Mechanism

When peroxides are present, they decompose to generate radicals that initiate a chain reaction:

1. Initiation: Peroxide breaks homolytically to form radicals, which then abstract a hydrogen from HBr to produce a bromine radical $\text{Br}^\cdot$.

2. Propagation Step 1: The bromine radical adds to the alkene. Radicals, unlike carbocations, are more stable when they have more alkyl groups around them. So $\text{Br}^\cdot$ adds to the less substituted carbon of propene, forming the more stable secondary radical on the middle carbon:

3. Propagation Step 2: This carbon radical abstracts a hydrogen from another HBr molecule, regenerating $\text{Br}^\cdot$ and giving the anti-Markovnikov product:

For this chain to be efficient, both propagation steps must be exothermic (or at least not too endothermic). If either step requires too much energy, the chain breaks down and the radical mechanism cannot compete with the ionic pathway.

Why HCl Fails

The H–Cl bond is very strong (bond dissociation energy $\approx 431 \, \text{kJ/mol}$).

In the second propagation step, the carbon radical would need to abstract hydrogen from HCl:

This step is highly endothermic because breaking the H–Cl bond costs far more energy than is released by forming the new C–H bond. The reaction simply doesn't proceed at a useful rate, so the radical chain never gets going.

Why HI Fails

The H–I bond is weak (bond dissociation energy $\approx 297 \, \text{kJ/mol}$), so hydrogen abstraction from HI is easy and exothermic. The problem is in the first propagation step. …