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The peroxide effect is seen with hydrogen bromide and not with hydrogen chloride or hydrogen iodide…

Try it on paper first. A hint, the key idea and the worked solution are here when you need them.

The peroxide effect is seen with hydrogen bromide and not with hydrogen chloride or hydrogen iodide. One propagation step of the radical chain decides this for hydrogen chloride. An alkyl radical $\text{R}^{\bullet}$ takes a hydrogen atom from $\text{HX}$ :

\displaystyle \text{R}^{\bullet}\ +\ \text{H}\text{-}\text{X}\ \longrightarrow \ \text{R}\text{-}\text{H}\ +\ \text{X}^{\bullet}

The energy change of this step is the energy to break the H-X bond, less the energy given out when the new C-H bond forms. Take the average C-H bond as $413\ \text{kJ/mol}$ . The bond enthalpies of H-Cl, H-Br and H-I are $431$ , $366$ and $299\ \text{kJ/mol}$ . A radical chain can run only if every propagation step passes the radical on at a useful rate. Usually this means that no step may take in a large amount of energy.

Find the energy change of the step for each of the three acids.

For which acid is the step endothermic, and what happens to the chain?

Hydrogen iodide gives a very favourable step here. Why is it still without a peroxide effect?

Need a reminder? The key ideaShow

An alkene adds HX by Markovnikov's rule: the more stable carbocation forms, so the halide goes to the carbon with more alkyl groups. A carbocation can rearrange to a more stable one by a hydride or methyl shift. With a peroxide, HBr adds by a radical chain to give the anti-Markovnikov product, because the more stable radical forms.

The worked solution

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Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

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