My professor noted that bromination with light and methylene chloride gives you exclusively the monobromination product.

However, bromination without light will give you the multi-brominated product.

I have no found any reference to bromination in my textbook nor online. Is the second statement an accurate statement?

  • 2
    $\begingroup$ bromination of what? $\endgroup$ – ron Aug 5 '14 at 20:54
  • $\begingroup$ Bromination of a hydrocarbon $\endgroup$ – Dissenter Aug 5 '14 at 21:58

Bromination of alkanes with Br2 will not occur in a dark environment. In the presence of light, multiple halogenations can occur given enough time.

For more information, including the full mechanism, check out: https://en.wikipedia.org/wiki/Free-radical_halogenation

  • $\begingroup$ So he's wrong about monobromination? $\endgroup$ – Dissenter Aug 5 '14 at 18:43
  • $\begingroup$ Let's just say my first guess is you misheard him. $\endgroup$ – Lighthart Aug 5 '14 at 23:36
  • $\begingroup$ My first guess is bromination using $\ce{Br2}$ in methylene chloride as a solvent. $\endgroup$ – Ben Norris Aug 6 '14 at 1:13

The rate determining step (rds) in the free radical bromination of an alkane involves hydrogen abstraction to yield the corresponding alkyl radical. Whether the chain initiating bromine radical is generated in the dark (thermally or by using N-bromosuccinamide for example) or light (where the Br-Br bond is homolytically cleaved using light energy), does not particularly matter; the same mechanism and (roughly the same) product distribution will result.

Selectivity (which hydrogen will be abstracted) will be determined by the relative strengths of the various C-H bonds in the molecule. A tertiary C-H bond (tBu-H) has a bond dissociation energy (BDE) of around 93 kcal/mol, a secondary C-H bond (iPr-H) has a BDE around 95 kcal/mol, while a primary C-H bond (Et-H) has a BDE around 98 kcal/mol and methane has a BDE of 103 kcal/mol. I wouldn't expect a bromine atom attached to carbon to increase the BDE for a hydrogen also attached to this carbon, so multibromination at that carbon should also occur. Further, even in a simple molecule like ethane, once bromoethane is formed, a second bromination at the non-brominated carbon should occur as readily as bromination at the initially brominated carbon. So both 1,1 and 1,2-dibromoethane should result as the secondary products.

In my opinion, both reactions will proceed through the same free radical - hydrogen abstraction mechanism, and, if 1) there is not a large excess of alkane to bromine and 2) the reaction is not terminated shortly after initiation, then both reactions will produce multi-brominated products


My first guess is that in light alkanes forms alkyl halides and in dark alkenes form dihalo-alkanes with bromine in non polar solvent as $CCl_4$; anyways:

Bromine is less reactive than other halogens toward alkanes in general but bromine is more selective in the site of attack.

Bromine shows a much greater ability to discriminate among the different types of hydrogen atoms. The reaction of 2-methylpropane and bromine, for example, gives almost exclusive replacement of the tertiary hydrogen atom.


Abstraction of hydrogen by a bromine atom is endothermic in both cases. The transition states are more product-like and possess more radical character; therefore, the difference in radical stability is more strongly expressed, and $\Delta E_{act}$is larger. The larger $\Delta E_{act}$ is associated with greater product selectivity, since the tertiary bromide is obtained from the tertiary free radical.

In the laboratory it is more convenient to use light, either visible or ultraviolet, as the source of energy to initiate the reaction. Reactions that occur when light energy is absorbed by a molecule are called photochemical reactions. Photochemical techniques permit the reaction of alkanes to be performed at room temperature.


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