Home >
Community >
Why is vicinal diiodopropane unstable, leading to deiodination?
Upvote
VOTE
Downvote
+ Halides
+ Chemistry
Posted by
Adam Wu
Why is vicinal diiodopropane unstable, leading to deiodination?
That 1,2-diodopropane is unstable does not indicate whether it is as the neat halide or in solution. Breaking (BDE) two $\ce{C-I}$ bonds costs$\pu{+114 kcal/mol}$ while forming the $\pi$-bond of propene is worth approximately $\pu{-62 kcal/mol}$ and the formation of iodine, $\pu{-36 kcal/mol}$. The net reaction is endothermic by $\pu{+16 kcal/mol}$. The reaction is entropically favored and the products are volatile. Therefore, the unfavorable equilibrium can be shifted to the right.
That 1,2-diodopropane is unstable does not indicate whether it is as the neat halide or in solution. Breaking (BDE) two $\ce{C-I}$ bonds costs$\pu{+114 kcal/mol}$ while forming the $\pi$-bond of propene is worth approximately $\pu{-62 kcal/mol}$ and the formation of iodine, $\pu{-36 kcal/mol}$. The net reaction is endothermic by $\pu{+16 kcal/mol}$. The reaction is entropically favored and the products are volatile. Therefore, the unfavorable equilibrium can be shifted to the right.
Hey i found an interesting observation, C-C single bond length is 134 pm. While van der wall radius of Iodine is 198 pm and that of chlorine is 175 pm. That means even if we count opposite orientation in Iodine, van der wall repulsions are too significant unlike in case of Chlorine.More
That 1,2-diodopropane is unstable does not indicate whether it is as the neat halide or in solution. Breaking (BDE) two $\ce{C-I}$ bonds costs $\pu{+114 kcal/mol}$ while forming the $\pi$-bond of propene is worth approximately $\pu{-62 kcal/mol}$ and the formation of iodine, $\pu{-36 kcal/mol}$. The net reaction is endothermic by $\pu{+16 kcal/mol}$. The reaction is entropically favored and the products are volatile. Therefore, the unfavorable equilibrium can be shifted to the right.
That 1,2-diodopropane is unstable does not indicate whether it is as the neat halide or in solution. Breaking (BDE) two $\ce{C-I}$ bonds costs $\pu{+114 kcal/mol}$ while forming the $\pi$-bond of propene is worth approximately $\pu{-62 kcal/mol}$ and the formation of iodine, $\pu{-36 kcal/mol}$. The net reaction is endothermic by $\pu{+16 kcal/mol}$. The reaction is entropically favored and the products are volatile. Therefore, the unfavorable equilibrium can be shifted to the right.
More
VOTE
VOTE
VOTE
VOTE
VOTE
VOTE