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?

  • 2
    $\begingroup$ What is the "Peroxide effect"? $\endgroup$
    – Jori
    Apr 26, 2015 at 8:34
  • 5
    $\begingroup$ @Jori anti markovnikov rule $\endgroup$
    – Freddy
    Apr 26, 2015 at 13:53

2 Answers 2


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: $$\ce{X. + H2C=CH2 -> X-H2C-CH2. ~~(1)}$$ $$\ce{HX + X-H2C-CH2. -> X-H2C-CH2-H + X. ~~(2)}$$

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}

(source for bond strengths)

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.

  • $\begingroup$ 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 $\endgroup$
    – RE60K
    Apr 27, 2015 at 5:19
  • $\begingroup$ Plus CCl4 is not an electrophile. $\endgroup$
    – RE60K
    Apr 27, 2015 at 11:13
  • $\begingroup$ Ok Now I can agree, but still I learnt that Peroxide -anti markonikov, please check it again. $\endgroup$
    – RE60K
    Apr 27, 2015 at 13:14
  • $\begingroup$ 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. $\endgroup$
    – RE60K
    Apr 27, 2015 at 13:20
  • $\begingroup$ @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. $\endgroup$
    – ron
    Apr 27, 2015 at 13:27

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

No, there are in fact a few more of them that I know. For example, $\ce{CHBr3}$, $\ce{CCl4}$ and $\ce{CBrCl3}$.

Reagent positive part negative part
$\ce{CHCl3}$ $\ce{H+}$ $\ce{CCl3^-}$
$\ce{CHBr3}$ $\ce{H+}$ $\ce{CCl3^-}$
$\ce{CCl4}$ $\ce{Cl+}$ $\ce{CCl3^-}$
$\ce{CBrCl3}$ $\ce{Br+}$ $\ce{CCl3^-}$

Some useful examples: $$\ce{R-CH=CH2 ->[\ce{H-CCl3}][R-O-O-R]R-CH2-CH2-CCl3}$$

$$\ce{R-CH=CH2 ->[\ce{Br-CCl3}][R-O-O-R]R-CHBr-CH2-CCl3}$$


  1. Apart from using H₂O₂, peroxide effect (or Kharasch addition) is also observed when other peroxides like R₂O₂ or (RCO)₂O₂ are used.
  2. Wikipedia's article on Morris S. Kharasch is worth reading to understand peroxide effect.
  • 1
    $\begingroup$ Welcome to CH SE site! Note that using photos/screenshots of text instead of typing text itself is highly discouraged. The image text content cannot be indexed nor searched for, nor can be reused in answers. Specifically handwritten scripts can be difficult to decipher. Consider copy/pasting or rewriting of at least essential parts. Suitable formatting can be done according to formatting math/chem expressions/equations. $\endgroup$
    – Poutnik
    Apr 7, 2022 at 8:40
  • $\begingroup$ Thanks @poutnik, but i think I have jotted down all of the things given in the screenshot, so it must contribute to a better search result... $\endgroup$ Apr 7, 2022 at 10:42
  • $\begingroup$ You could for the table easily use the ---|---|--- SE Markdown syntax. $\endgroup$
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  • $\begingroup$ In addition to @Poutnik 's comment, I don't understand what meaning five (six counting the background, which is for some reason not transparent/white) different colors are supposed to convey. In scientific writing every mean of formatting signifies something. Random colors only distract. Also, please refrain from abbreviations such as "+ve"/"-ve", they are illiterate, confusing and the latter doesn't even make semantic sense since you've used a hyphen, not a minus (cf. -,). $\endgroup$
    – andselisk
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  • $\begingroup$ @andselisk well, it is just a screenshot from my class notes... And my organic chemistry teacher likes to make colorful eye-catching notes... Btw, i am removing the image. $\endgroup$ Apr 7, 2022 at 12:49

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