Predict the major product when $ ext{HBr}$ is added to propene in the presence of organic peroxides, and explain the mechanism that leads to it.

Question

Predict the major product when $	ext{HBr}$ is added to propene in the presence of organic peroxides, and explain the mechanism that leads to it.

Accepted Answer

Concept understanding — Markovnikov vs Anti Markovnikov Addition

Markovnikov vs Anti-Markovnikov Addition

Imagine you're adding something across a double bond — say, H–Br. The double bond is like a crowded room: one carbon has more friends (alkyl groups) attached, the other has fewer. Where does the hydrogen go? That's the core question.

The Intuition: Stability Wins

A carbocation (a carbon with a positive charge) forms as an intermediate in many addition reactions. The more alkyl groups attached to that positively charged carbon, the more stable it is — alkyl groups push electrons toward the positive centre, stabilising it.

So when H⁺ from H–Br attacks the double bond, it will choose the path that leads to the more stable carbocation. That means the hydrogen goes to the carbon that already has more hydrogens (the less substituted carbon), leaving the positive charge on the more substituted carbon (which has more alkyl groups to stabilise it).

Note

This is the Markovnikov rule in its original form: "The rich get richer." The carbon that already has more hydrogens gets the new hydrogen.

The Precise Statement

Markovnikov Addition: When H–X adds to an unsymmetrical alkene, the hydrogen attaches to the carbon with more hydrogen atoms already attached, and the halogen (or other group) attaches to the carbon with fewer hydrogen atoms.

Anti-Markovnikov Addition: The opposite — hydrogen attaches to the carbon with fewer hydrogen atoms, and the other group attaches to the carbon with more hydrogen atoms.

Why Does Anti-Markovnikov Happen?

Normal H–Br addition follows Markovnikov. But if you add peroxides (ROOR) to the reaction, the mechanism changes from ionic to free radical. In the radical mechanism, the bromine radical attacks first, and it prefers to form the more stable radical intermediate — which is the more substituted carbon radical. That forces the hydrogen to go to the other carbon, giving anti-Markovnikov product.

Watch out

Anti-Markovnikov addition only works reliably with H–Br. H–Cl and H–I do not give clean anti-Markovnikov products under radical conditions — H–Cl's bond is too strong, and H–I's radical is too unstable.

Quick Comparison Table

| Feature | Markovnikov | Anti-Markovnikov |

|---------|-------------|------------------| …

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Peroxides decompose to generate radicals that abstract a hydrogen from $\text{HBr}$ , producing a bromine radical, $\text{Br}^{\bullet}$ , which initiates a different, free-radical addition pathway instead of the normal ionic one. $\text{Br}^{\bullet}$ adds to the double bond at whichever carbon leaves the MORE STABLE carbon radical: adding to the terminal $\text{=CH}_2$ carbon leaves the radical on the internal carbon, giving the more stable secondary radical $\text{CH}_3\overset{\bullet}{\text{C}}\text{HCH}_2\text{Br}$ (stabilised by two alkyl groups, exactly as a carbocation would be), rather than the less stable primary radical that would result from the opposite regiochemistry. This secondary radical then abstracts a hydrogen from another molecule of $\text{HBr}$ , giving the product and regenerating a bromine radical to continue the chain. Because the bromine ends up on the TERMINAL carbon this time (opposite to the ionic mechanism's result), the product is 1-bromopropane, $\text{CH}_3\text{CH}_2\text{CH}_2\text{Br}$ -- the anti-Mar …

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Q.Predict the major product when HBr\text{HBr}HBr is added to propene in the presence of organic peroxides, and explain the mechanism that leads to it.

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Q.Predict the major product when $\text{HBr}$ is added to propene in the presence of organic peroxides, and explain the mechanism that leads to it.

Unlock everything free for 14 days