[Cooper]

Hi,

In the reaction of isobutane with chlorine, my books says that isobutyl chloride forms more than tert-butyl chloride. I know that the reaction happens because a radical is left on a carbon of the isobutane after a chlorine removes a hydrogen. Then the radical on the carbon takes the other chlorine and bonds to it.

Radicals on tertiary carbons are more stable than radicals on primary carbons, which means that hydrogens attached to tertiary carbons are weaker, because when they get taken off, the radical that remains it the most stable. So if this is true, why doesn't tert-butyl chloride form more than isobutyl chloride?

Thanks,
~Cooper

*the pic. is from my textbook, Organic Chemistry, 10th edition, by Solomons and Fryhle

[discodermolide]

Perhaps this is a question about the reaction rates. Because the tertiary radical is more stable it reacts slower then the primary radical?
What do you think?

[Cooper]

That makes sense! Thank you I hadn't thought about how the rates might occur.

~Cooper

[fledarmus]

Actually, in this case I believe it is more a simple numbers problem. The chlorine radicals are very reactive, and faced with a choice of nine primary and only one tertiary hydrogen to pull, they end up snagging a lot of primary hydrogens.

As you go to less reactive radicals you can increase the selectivity of the reaction.

[curiouscat]

My reasoning was in terms of numbers: How many ways do you have to lose a H leading to tert vs iso?

1:9. So a probabilistic win favoring iso.

Maybe if you multiply the relative stabilities by relative probabilities you'll get  better approximation?

[Cooper]

That makes sense too, thanks

~Cooper

[orgopete]

Actually, in this case I believe it is more a simple numbers problem. The chlorine radicals are very reactive, and faced with a choice of nine primary and only one tertiary hydrogen to pull, they end up snagging a lot of primary hydrogens.

As you go to less reactive radicals you can increase the selectivity of the reaction.

This is the correct answer. A more stable radical, will be more selective. Fluorine radicals are very nonselective, bromine are quite selective, and iodine are to stable to react. Chlorine are between fluorine and bromine. Chlorine has sufficient reactivity, that even though a tertiary is favored, and occurs more rapidly, it has enough reactivity to abstract a hydrogen from a methyl group. Given they are in greater abundance, they form a significant amount of the product.