Solution:
(a) Anti-Markovnikov addition (the peroxide effect) does not apply to \text{HCl} and \text{HI} because one of the propagation steps in their free-radical chain mechanisms is energetically unfavorable (endothermic), whereas both propagation steps for \text{HBr} are exothermic. [1, 2]
The Free-Radical Mechanism
The anti-Markovnikov addition of hydrogen halides (\text{HX}) to alkenes relies on a two-step propagation cycle involving free radicals: [1, 2]
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Step 1: The halogen radical (\text{X}^{\bullet }) adds to the alkene double bond to form a carbon radical intermediate.
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Step 2: This carbon radical abstracts a hydrogen atom from \text{HX} to yield the final addition product and regenerate a new halogen radical. [1, 2]
Why \text{HCl} and \text{HI} Fail
For the overall radical chain reaction to proceed, both propagation steps must be exothermic (release energy). \text{HBr} satisfies this condition, but \text{HCl} and \text{HI} fail due to bond energy mismatches: [1]
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For \text{HCl}: The H–Cl bond is very strong (high bond dissociation energy). As a result, the second propagation step—where the carbon radical abstracts hydrogen from \text{HCl}—is endothermic and unfavorable. [1]
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For \text{HI}: The C–I bond formed in the first step is relatively weak, making the first propagation step—the addition of an iodine radical (\text{I}^{\bullet }) to the alkene—endothermic and reversible. Furthermore, iodine radicals (\text{I}^{\bullet }) prefer to combine with each other to form stable \text{I}_{2} molecules rather than react with the alkene. [1]
Because these energetic bottlenecks prevent an effective radical chain, \text{HCl} and \text{HI} bypass the radical pathway and default to normal electrophilic ionic Markovnikov addition, even in the presence of peroxides. [1, 2, 3]
(b) The most common peroxides used in the lab are the tert- butyl peroxide and benzoyl peroxide. Sometimes, you can see that some instructors and even some textbooks use the hydrogen peroxide, which is a bit of a white lie, as H_2O_2 doesn’t break into radicals as easily as they make it seem.
Next, our newly formed peroxide is going to scavenge a hydrogen from HBr. This step gives us the Br-radical, which will actually kick-start our propagation cycle.
So, the first step of the propagation is going to be the interaction between the Br-radical and the alkene.
In this reaction, one electron will come from bromine. Next, another electron comes from the pi-bond to make a bond between bromine and carbon. And finally, the second electron from the pi-bond stays with another carbon of what used to be a double bond. This gives us the product in which we get the new bond between carbon and bromine, and we’ll also see a tertiary radical.
These reactions will always give you the most stable radical. And when it comes to the radical stability, they follow the same trend as carbocations: resonance stabilization beats the 3° radical, which is more stable than the 2° radical, which in turn, is more stable than the 1° radical.
Next, our organic radical will grab a hydrogen from another molecule of HBr.
This regenerates the Br-radical that can go back into another round of the propagation cycle, and the final product.
And, of course, like in any radical reaction, we’ll also need a termination step that terminates the propagation cycle.
Here only shows one of the many possible termination steps.
