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.
$$\small\text{2-methylpropane}\overbrace{\longrightarrow}^{\text{Br}_2,h\nu,127^oC}\text{tertiarybutylbromide}+(99\%)+\text{sec-butylbromide(trace)}$$
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.