Addition of HBr to propene yields 2-bromopropane, while in the presence of benzoyl peroxide, the same reaction

Exercises · 9.16

Q. Addition of HBr to propene yields 2-bromopropane, while in the presence of benzoyl peroxide, the same reaction yields 1-bromopropane. Explain and give mechanism.

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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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The regioselectivity of HBr addition to propene switches from Markovnikov (2-bromopropane) to anti-Markovnikov (1-bromopropane) in the presence of benzoyl peroxide because the peroxide initiates a free-radical chain mechanism that proceeds through the more stable primary radical intermediate, overriding the ionic pathway.

The key to understanding this problem lies in recognising that the reaction conditions determine the mechanism — and the mechanism determines the product. Without peroxide, HBr adds via an ionic pathway following Markovnikov’s rule. With benzoyl peroxide (a radical initiator), the reaction switches to a free-radical chain mechanism that gives the opposite regiochemistry.

Let’s break this down.

1. The ionic (Markovnikov) addition — no peroxide

In the absence of peroxides, HBr adds to propene through a polar, two-step mechanism.

Step 1: The $\pi$ bond of propene acts as a nucleophile and attacks the electrophilic proton of HBr. This forms a carbocation intermediate. The proton adds to the less substituted carbon (the terminal carbon) because that gives the more stable carbocation — a secondary carbocation (stabilised by two alkyl groups) rather than a primary one.

\text{CH}_3\text{–CH=CH}_2 + \text{H}^+ \rightarrow \text{CH}_3\text{–}\overset{+}{\text{C}}\text{H–CH}_3

Step 2: The bromide ion ( $\text{Br}^-$ ) attacks the positively charged carbon, giving 2-bromopropane.

\text{CH}_3\text{–}\overset{+}{\text{C}}\text{H–CH}_3 + \text{Br}^- \rightarrow \text{CH}_3\text{–CHBr–CH}_3

Markovnikov’s rule: In the ionic addition of HX to an unsymmetrical alkene, the hydrogen adds to the carbon with more hydrogens (less substituted), and the halogen adds to the carbon with fewer hydrogens (more substituted).

The product is 2-bromopropane (isopropyl bromide).

2. The free-radical (anti-Markovnikov) addition — with benzoyl peroxide

Benzoyl peroxide ( $\text{C}_6\text{H}_5\text{CO–O–O–COC}_6\text{H}_5$ ) is a radical initiator. It decomposes upon heating to generate phenyl radicals, which then abstract a hydrogen atom from HBr, producing bromine radicals.

Tip

The peroxide does not directly add to the alkene. Its role is to generate $\text{Br}^\bullet$ radicals from HBr. This is why only HBr shows the peroxide effect — HCl and HI do not (HCl’s bond is too strong, and HI’s bond is too weak, making radical abstraction inefficient).

The mechanism proceeds in three phases: initiation, propagation, and termination.

Initiation:

Benzoyl peroxide decomposes homolytically:

(\text{C}_6\text{H}_5\text{COO})_2 \xrightarrow{\Delta} 2\,\text{C}_6\text{H}_5\text{COO}^\bullet \rightarrow 2\,\text{C}_6\text{H}_5^\bullet + 2\,\text{CO}_2

The phenyl radical abstracts a hydrogen from HBr:

\text{C}_6\text{H}_5^\bullet + \text{H–Br} \rightarrow \text{C}_6\text{H}_6 + \text{Br}^\bullet

Now we have bromine radicals ready to attack the alkene.

Propagation — step 1:

The bromine radical adds to the double bond. Here’s the critical regiochemical choice: the radical can add to either carbon of the $\pi$ bond.

If $\text{Br}^\bullet$ adds to the terminal carbon (C1), we get a secondary radical (more stable, because the unpaired electron is on a carbon bearing two alkyl groups).
If $\text{Br}^\bullet$ adds to the internal carbon (C2), we get a primary radical (less stable).

The radical intermediate’s stability determines the outcome. The secondary radical is about 10–15 kJ/mol more stable than a primary radical, so the addition is reversible — but the more stable radical forms faster and accumulates.

Thus, the bromine radical adds to the less substituted carbon (the terminal carbon), giving the more stable secondary radical:

\text{CH}_3\text{–CH=CH}_2 + \text{Br}^\bullet \rightarrow \text{CH}_3\text{–}\overset{\bullet}{\text{C}}\text{H–CH}_2\text{Br}

Watch out

A common mistake is to think the radical adds to the more substituted carbon because “Markovnikov” is the default. In radical addition, the radical stability (not carbocation stability) governs the regiochemistry. The radical centre is stabilised by alkyl groups, so the more substituted radical is favoured — which here means the bromine ends up on the terminal carbon.

Propagation — step 2:

The secondary radical now abstracts a hydrogen atom from another molecule of HBr, regenerating a bromine radical and forming the final product: …

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Q.Addition of HBr to propene yields 2-bromopropane, while in the presence of benzoyl peroxide, the same reaction yields 1-bromopropane. Explain and give mechanism.

1. The ionic (Markovnikov) addition — no peroxide

2. The free-radical (anti-Markovnikov) addition — with benzoyl peroxide

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