Why is vicinal diiodopropane unstable, leading to deiodination? I came across this reaction where simple electrophilic addition mechanism will result in a vic-diiodide.
$$\ce{H2C=CH-CH2-I -&gt;[HI, excess][CCl4]}$$
But the solution says that vicinal iodine groups are unstable and the following reaction occurs:
$$\ce{CH3-\underset{(unstable)}{\underset{\underset{\huge I}{|}}{C}H-\underset{\underset{\huge I}{|}}{C}H2} -&gt;[][$-$I2] CH3-CH=CH2}$$
I am unable to figure out how the dihaloalkane is unstable. It might be due to steric reasons, but still the two iodine can arrange in a way that reduces the repulsions.
But even if it is unstable, how did $\ce{I2}$ leave the haloalkane to yield an alkene? This dehalogenation does happen when a metal is used though (Zn dust, Mg in acetone, I guess) but what would be the mechanism here?
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.