May 7, 2009
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With regards to halogenation of 2-methylbutane, why doesn't a tertiary radical form after the free radical rips off a proton from one of the terminal methyl groups. I thought a tertiary radical would form (after a carbocation rearrangement) because tertiary radicals are the most stable....

Instead, a primary radical forms and we get 1-chloro-2-methylbutane instead of 2-chloro-2-methylbutane.

Thanks for all of your help.
 

wanderer

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With regards to halogenation of 2-methylbutane, why doesn't a tertiary radical form after the free radical rips off a proton from one of the terminal methyl groups. I thought a tertiary radical would form (after a carbocation rearrangement) because tertiary radicals are the most stable....

Instead, a primary radical forms and we get 1-chloro-2-methylbutane instead of 2-chloro-2-methylbutane.

Thanks for all of your help.
Free radicals aren't carbocations, and they don't typically undergo the sorts of shifts that carbocations do.
 

Doodl3s

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AH very good and IMPORTANT question for MCAT orgo. For radical halogenation

Reactivity: F>CL>BR>I ; So, the MORE reactive it is, the LESS selective it is, since it will grab whatever hydrogen it can gets its hands onto.

So since there are like 6 times as many end hydrogens as that one tertiary, the end hydrogens get ripped. On the other hand, Bromine WOULD have gone on that tertiary position. Iodine would too, Except Iodine cannot initiate a radical reaction its so weak, it can only propogate.
 

Caesar

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For Cl the rule is tert H x5, secondary H x 4, primary H x 1. Whichever has the highest number is the most likely.

I can't imagine the MCAT being this specific though. However, this is covered in any organic Chem class.

Also, as stated above, free radicals don't do methyl and hydride shifts as they already posses one electron.