I came across a radical substitution which Methane reacts with $\ce{IBr}$ to form iodomethane and hydrogen bromide and many more products. I understand the whole process but I don't quite see how the $\ce{IBr}$ molecule will break via homolytic fission instead of heterolytic. Since Bromine is more electronegative than Iodine, wouldn't $\ce{Br}$ takes both electrons and form $\ce{Br-}$ ions?
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The presumed stability of $\ce{Br-}$ is overcompensated by the instability of $\ce{I+}$. While some polyatomic iodine cations like $\ce{I2+}$ exist, $\ce{I+}$ is found only to be stable in the gas phase and in the absence of a counterion. Due to the low difference of electronegativities (0,5) of Br and I, the I-Br bond is predominantly covalent, which also makes heterolysis unlikely. Homolytic cleavage of the I-Br bond is therefore energetically more favorable.
answered May 1, 2014 at 21:19
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$\begingroup$ Thanks! Considering the instability of I+ the concept makes much more sense now $\endgroup$ Commented May 1, 2014 at 21:40