Free Radical Bromination of Alkanes

by James

Description: When treated with bromine (Br2) and light (hν) alkanes are converted into alkyl bromides. In the absence of any double bonds, with Br2 this is selective for tertiary carbons.
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{ 9 comments… read them below or add one }

vu quoc November 19, 2011 at 3:28 pm

Dear Sir,
I see that exciting topic. But you give me some example for “heat may also suffice”. What difference hv and heat are inside this reaction ?
“Notes: Although light (hν) is commonly employed as the initiator, heat may also suffice.”

thanks

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james November 19, 2011 at 4:29 pm

Thanks for asking for clarification – what I should have said is that one can also use heat and peroxides. A common example of a peroxide for these types of reactions is tert-butyl peroxide, (CH3)3C–O–O–C(CH3)3 . When this is heated, it fragments to give the free radical (CH3)3CO• , which can fragment Br–Br homolytically to give Br• and start the chain reaction.

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MzF December 6, 2013 at 10:41 pm

this may seem silly, but i just cant remember–are rearrangements possible in radical bromination?

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James Ashenhurst December 8, 2013 at 10:08 am

If you’re asking if H and alkyl shifts are possible, for our purposes the answer is no.

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stefani johnson April 11, 2014 at 11:18 am

what about chain termination?

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James April 15, 2014 at 10:51 am

Chain termination can be drawn by showing ANY of the radicals depicted here combining with the carbon radical. Any time you have two radicals combining to form a new bond, that is chain termination.

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Petr Menzel May 22, 2014 at 1:58 am

Hi, there is table with comparison of selectivity. For Br-: 0,007; 1; 220; 19,4. You write that Br- “likes” tertiary carbon. Why there is not the highest number? Thx, P.

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James May 22, 2014 at 2:49 pm

Did you mean to say “19,400 ” ?

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Petr Menzel May 22, 2014 at 3:07 pm

Sorry, I can see 19 and there is “comma”, not “dot” —> 19 400. It is clear for me now.

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