Answer:
Bond dissociation enthalpy of \[{{I}_{2}}\] is less as compared to that of \[C{{l}_{2}}\] and \[B{{r}_{2}}\] which means that the homolysis of \[{{I}_{2}}\] is the easiest. However, \[HI\] formed in a reaction is a strong reducing agent and it makes the reaction reversible. Therefore, iodine does not reaction with ethane.
\[\underset{\text{Ethane}}{\mathop{C{{H}_{3}}-C{{H}_{3}}+{{I}_{2}}}}\,\rightleftharpoons \underset{\text{Iodoethane}}{\mathop{C{{H}_{3}}-C{{H}_{2}}-I+HI}}\,\]
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