Free Radical Bromination of Alkanes
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 occurs only at tertiary carbons.
Notes: Light (hν) is the initiator for this reaction.
Examples:
Notes: Note that each of these reactions produces HBr as a byproduct. If this reaction occurs at a stereocenter, a mixture of alkyl bromides will be obtained.
Mechanism: When treated with light, Br2 fragments to bromine radicals (Step 1, arrow A). At any given time only a small amount of these radicals is present. There’s lots of leftover Br2! Remember this as it comes up in step 3. Bromine radical then abstracts a hydrogen from the tertiary carbon, leaving behind the tertiary radical (Step 2, arrows B and C). The tertiary radical then reacts with Br2, giving the tertiary bromide (Step 3, arrows D and E).
Notes: Avoid the common mistake of having the radical react with Br• in the third step – that’s not a propagation reaction, that’s termination!
Note that tertiary radicals are more stable than secondary or primary radicals and thus require the least energy to form (another way of saying this is that tertiary C-H bonds are weaker than secondary and primary C-H bonds). The higher selectivity of this reaction for tertiary C-H (when compared with chlorination) is a reflection of the fact that formation of the weaker H–Br bond provides less of a driving force for this reaction as compared to H–Cl (which is a stronger bond).
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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
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.