Ochem question

[sweetpea2]

Advertisement - Members don't see this ad

I just took an OAT exam. It can be found here:

https://www.ada.org/oat/index.html

Question 85:

The question is which will undego an free radical bromination most rapidly.

The answer said it was CH4 but i thought it would be (CH3)3CH. Cause it forms a tertiary (most stable) carbocation. Can someone explain why free radical bromination is fastest with methane or if this is a mistake because it basically goes against everything iver learned in ochem, thanks!

 

[klutzy1987]

First off there is no carbocation by a free radical reaction. Second, I believe that a radical reaction is SN2 therefore it would work best on the primary.

 

[thebozz1975]

Sounds like a mistake to me. Chlorination might prefer the methyl, but not bromination.

 

[thebozz1975]

First off there is no carbocation by a free radical reaction. Second, I believe that a radical reaction is SN2 therefore it would work best on the primary.

I think you need to study some more...

 

[klutzy1987]

Sounds like a mistake to me. Chlorination might prefer the methyl, but not bromination.

All free radicals prefer the primary/methyl. They are all sn2.
http://en.wikipedia.org/wiki/Free_radical_reaction

 

[thebozz1975]

All free radicals prefer the primary/methyl. They are all sn2.
http://en.wikipedia.org/wiki/Free_radical_reaction

read up on free radical HALOGENATION

 

[futuredent22]

free radical reactions are not all sn2 and do not favor primary/methyl. In fact, reacting with Br radical favors tertiary 1600 times more than primary at 125 C.

A carbocation is not formed, but a carbocation-like intermediate is formed in a reaction resembling Sn1. In the first step, when radical Br removes a hydrogen, it will remove it from the carbon that will result in the most stable intermediate, one that is tertiary.

So CH(CH3)3 is MUCH more likely to react with a radical Br than CH4 is.

 

[klutzy1987]

free radical reactions are not all sn2 and do not favor primary/methyl. In fact, reacting with Br radical favors tertiary 1600 times more than primary at 125 C.

A carbocation is not formed, but a carbocation-like intermediate is formed in a reaction resembling Sn1. In the first step, when radical Br removes a hydrogen, it will remove it from the carbon that will result in the most stable intermediate, one that is tertiary.

So CH(CH3)3 is MUCH more likely to react with a radical Br than CH4 is.

Could be i am mistaken.

 

[thebozz1975]

Could be i am mistaken.

Uh, yes, as I said before.

 

[klutzy1987]

Uh, yes, as I said before.

Dude you need to get life everyone makes mistakes. Now take a chill pill and get back on your meds.

 

[thebozz1975]

Dude you need to get life everyone makes mistakes. Now take a chill pill and get back on your meds.

Pay close attention, and I'll teach you how to swing from the vines.

 

[sweetpea2]

so im guessing this is a mistake in the answer key and that the answer is in fact the CH(CH3)3 ... I know that chlorination would be the opposite..wikipedia decided to use chlorination as the example for free radical halogenation but bromination prefers the CH(CH3)3. I hope this is correct, my test is next week and i can see this coming up haha, thanks guys

 

[futuredent22]

chlorination isnt the opposite. Its just that chlorination does not prefer the tertiary over the primary nearly as much as bromination. As I said earlier, bromination is 1600 times more likely in tertiary than primary, while chlorination is only 5 times more.

This difference is due to enthalpy. Chlorination is exothermic, so the carbocation-like intermediates resemble the reactants and don't have much energy difference. Thus, there is not too much preference for tertiary over primary. On the other hand, bromination is endothermic. The carbocation-like intermediates resemble the products, and the tertiary is much more stable than the primary.

This is the whole reactivity vs. selectivity idea. Bromination is much more selective in its reacting.

 

[harrygt]

Don't worry it's just a type, but let me clarify something. There are no carbocations involved when you talk about radicals! SN2 and SN1 are nucleophilic substitution reactions. SN2 is a concerted reation which does not even produce any carbocations, however, SN1 does produce carbocations as intermediates.
You should have said the correct answer is A, because it makes tertiary RADICALS [not carbocations]

Harry