# Peroxide effect in reagents apart from HBr and CHCl₃

Is the peroxide effect observed only with addition of hydrogen bromide and chloroform? If so, why not with other reagents in electrophilic addition on alkenes?

• What is the "Peroxide effect"? – Jori Apr 26 '15 at 8:34
• @Jori anti markovnikov rule – Freddy Apr 26 '15 at 13:53

Is the peroxide effect observed only with addition of hydrogen bromide and chloroform?

No, the "peroxide effect" or "anti-Markovnikov" addition of a free radical to an unsaturated bond is observed with other compounds too, $\ce{CCl4}$ is another example.

Also, just to be clear, these reactions involve the addition of radicals ($\ce{X^{.}}$), not electrophiles, to alkenes (or alkynes).

Let's examine the radical addition of $\ce{HX}$ to an olefin. I'll use a general method that you can later use to examine any radical addition to an unsaturated bond.

There are two propagation steps that are key to our analysis: $$\ce{X* + H2C=CH2 -> X-H2C-CH2. ~~(1)}$$ $$\ce{HX + X-H2C-CH2. -> X-H2C-CH2-H ~~(2)}$$

If two halogen radicals combine, or if a halogen radical and a halo-alkyl radical (formed in the propagation steps) combine, the reaction is terminated. The propagation steps are in competition with the termination steps. If the propagation steps are endothermic, then they compete less effectively with the termination steps and the radical addition will not proceed in a useful manner.

In propagation step (1) we are breaking a pi bond and making a $\ce{C-X}$ bond, let's look at the thermodynamics of this step.

\begin{array}{|c|c|c|c|} \hline \ce{C-X} & \text{Bond Strength} & \text{Pi Bond Strength} & \text{Overall} \\ & \text{(kcal/mol)} & \text{(kcal/mol)} & \text{(kcal/mol)}\\ \hline \ce{C-F} & 116 & 66 & -50\\ \hline \ce{C-Cl} & 81 & 66 & -15\\ \hline \ce{C-Br} & 68 & 66 & -2\\ \hline \ce{C-I} & 51 & 66 & 15\\ \hline \end{array}

We see that this step is exothermic for the first 3 hydrogen halides, but is endothermic for the $\ce{HI}$ case. Already we see that hydrogen iodide will not add to olefins via a radical pathway at room temperature.

In propagation step (2) we are breaking an $\ce{HX}$ bond and making a $\ce{C-H}$ bond. Let's perform a similar thermodynamic analysis of this step.

\begin{array}{|c|c|c|c|} \hline \ce{H-X} & \text{Bond Strength} & \ce{C-H}~ \text{Bond Strength} & \text{Overall} \\ & \text{(kcal/mol)} & \text{(kcal/mol)} & \text{(kcal/mol)}\\ \hline \ce{H-F} & 135 & 99 & 36\\ \hline \ce{H-Cl} & 103 & 99 & 4\\ \hline \ce{H-Br} & 88 & 99 & -11\\ \hline \ce{H-I} & 71 & 99 & -28\\ \hline \end{array}

Both propagation steps (1) and (2) must occur for the reaction to proceed. Only in the case of the addition of $\ce{HBr}$ is the reaction exothermic for both steps. For the hydrogen halide series, only $\ce{HBr}$ will add to an olefin by a free radical mechanism.

• Your answer is wrong because Markonikov addition $\ne$ Peroxide Effect. And I would be happy if you could provide some source for CCl4's addition – RE60K Apr 27 '15 at 5:19
• Plus CCl4 is not an electrophile. – RE60K Apr 27 '15 at 11:13
• Ok Now I can agree, but still I learnt that Peroxide -anti markonikov, please check it again. – RE60K Apr 27 '15 at 13:14
• Sorry but you gave wrong lectures, I am really sorry but please see where it is written "Mechanisms that avoid the carbocation intermediate may react through other mechanisms that are regioselective, against what Markovnikov's rule predicts, such as free radical addition. Such reactions are said to be anti-Markovnikov" I hope you don't argue before checking, sorry. – RE60K Apr 27 '15 at 13:20
• @ADG You are correct, it shoulkd be anti-Markovnikv. I have corrected the answer. Free radicals are generated and a free radical addition to a double bond will follow when peroxides are used to initiate a reaction. Such reactions result in an anti-Markovnikov addition pattern to the double bond. Therefore terms like "peroxide" or "anti-Markovnikov" are often used to describe them. None of these reactions involve electrophiles, they involve free radicals. Here is a link to an example of a free radical addition of $\ce{CCl4}$ to a double bond. Look about half-way down the page. – ron Apr 27 '15 at 13:27