Selectivity of free radicals in halogenation.

[hyddro]

Hi, can anyone please explain what selectivity of free radicals is? For example, there is an MCAT question that I recently came across and that can't seem to make sense of the answer.

Which of the following halogens will give the greatest percent yield of tertiary alkyl halide when reacted with isobutane in the presence of heat or light?
a. F₂
b. Cl₂
c. Br₂
d. Isobutane won't produce a tertiary alkyl halide

The answer is C. Br₂, but i don't understand why... I applying common sense. Here is what I thought it would work. F is the least selective and the most reactive when is in the free radical form. My common sense tells me that since is very reactive and less selective, it won't care what carbon it binds to, so the tertiary hydrogen in isobutane (which is surrounded by 3 alkyl groups, or 9 hydrogens), will also react with F. Br and Cl are both less reactive and more selective. Hence they will most likely bind to the other primary carbons cause they are readily there. But this seems to not have been the right approach, can you please help me out? Starting by what selectivity is...

Thank you.

 

[AGNuke]

Fluorine is well, fluorine. You simply don't mess with it. The only products you will see is CF₄ and HF.

As for Chlorine, the hierarchy between primary, secondary and tertiary is not wide. But since the probability of formation of primary halide is 0.9 (you can realize this) which is very high, you get primary halide with Chlorine as major product.

But in case of Bromine, the hierarchy is very very wide. Bromine prefers reacting with tertiary Carbon around 1800 times more than primary carbon, so even if the probability of formation of tertiary halide is 0.1, the tendency of formation increases its productivity nonetheless.

 

[hyddro]

thank you, but still don't understand about selectivity... :(

 

[AGNuke]

Selectivity is experimentally determined. We can only give suitable explanations. This depends upon the kinetics of the reaction. And you may know, the rate of the reaction can only be determined experimentally.

This is some data that I was taught. Hopefully this will help you. You can estimate the major product by using this term "Productivity", a product of relative reactivity and collision probability leading to formation of that compound.

For chlorine, relative reactivity is : Primary is (1), Secondary is (3.9) and Tertiary is (5).
For bromine, relative reactivity is : Primary is (1), Secondary is (80) and Tertiary is (1800).

Try finding the productivity of Primary and Tertiary Chloride and Bromide and see which halogen will tend to form which product as major one.

*I can't guarantee the authenticity of this method. But it worked for me.

 

[DeeNos]

Selectivity of free radicals in halogenation refers to the tendency of free radicals (highly reactive species with unpaired electrons) to preferentially react with certain types of carbon atoms in a molecule. In the context of halogenation, this refers to the preference of free radicals to react with tertiary carbon atoms over primary or secondary carbon atoms.

In the MCAT question provided, the answer is C. Br2 because bromine free radicals have a higher selectivity for tertiary carbon atoms compared to chlorine free radicals. This is because bromine free radicals are larger and less reactive than chlorine free radicals, allowing them to more easily attack the bulky tertiary carbon atom in isobutane.

Your approach of using common sense is a good start, but it's important to also consider the chemical properties of the halogen free radicals and how they affect their selectivity in halogenation reactions. I hope this helps clarify the concept of selectivity in free radical reactions.