Master Organic Chemistry Reaction Guide

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 is selective for tertiary carbons.
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Comment section

15 thoughts on “Free Radical Bromination of Alkanes

  1. 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.”


    1. 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.

    1. 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.

  2. 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.

  3. Is there a mistake in Step 3? You listed that in arrow D, a Br-Br bond is formed, and in arrow E, a C2-Br bond is broken. Shouldn’t it be the reverse?

  4. So is there an explanation why Bromine prefers tertiary carbons while Chlorine rather takes primary or secondary ones?

  5. Sorry i just saw the explanation but I still don’t understand why the relative weakness of H-Br-Bonds is causal for the tertiary selectivity.

    1. Using Hess’ law you can calculate delta H for the reaction. Breaking bonds has an energy cost, formation of bonds has an energy gain. The “gain” in energy from forming H-Br is small relative to H-Cl and H-F. Therefore delta H will be less negative for the reaction. Therefore only weak C-H bonds like tertiary and allylic will participate due to these unfavorable thermodynamic properties.

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