# Question #84121

If alkane reacts with iodine to form alkyl iodide then $H I$ is also produced as given by the equation.
$R H + {I}_{2} r i g h t \le f t h a r p \infty n s R I + H I$
This reaction is reversible one since $H I$ possesses reducing property.As a result the alkane is reproduced again and the yield of alkyl iodide is negligible,
But presence of of an oxidizing agent eg $H I {O}_{3}$ in the reaction mixture oxidizes the $H I$ produced to ${I}_{2}$and prevents the reaction to occur in backward direction and thus may help the production of RI
$H I {O}_{3} + 5 H I \to 3 {I}_{2} + 3 {H}_{2} O$