# Why is vicinal diiodopropane unstable, leading to deiodination?

I came across this reaction where simple electrophilic addition mechanism will result in a vic-diiodide.

$$\ce{H2C=CH-CH2-I ->[HI, excess][CCl4]}$$

But the solution says that vicinal iodine groups are unstable and the following reaction occurs:

$$\ce{CH3-\underset{(unstable)}{\underset{\underset{\huge I}{|}}{C}H-\underset{\underset{\huge I}{|}}{C}H2} ->[][-I2] CH3-CH=CH2}$$

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?

• As for the question, I doubt one can really comment on the stability of the compound without knowing what else is present besides the compound. Apparently, the source also omitted the presence of any reagent that could have been used to cause the dehalogenation, only showing the $\ce{(-I2)}$. Mar 25 at 16:09

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.