Chemistry - Mechanism of the reaction of an alkene with excess HI in presence of CCl4

Solution 1:

According to [1], 1,2-diiodopropane reacts with hydrogen iodide to give 2-iodopropane.

So the reaction path might be

$$\ce{CH2=CH\bond{-}CH2\bond{-}I + HI -> CH3\bond{-}CHI\bond{-}CH2\bond{-}I}$$

$$\ce{CH3\bond{-}CHI\bond{-}CH2\bond{-}I + HI -> CH3\bond{-}CHI\bond{-}CH3 + I2}$$

The mechanism of the second part is unclear to me. It may be elimination of $\ce{I2}$ followed by addition of $\ce{HI}$.

^{1 Th. M. Schmitt, Analysis of Surfactants, Second Edition, p. 67}

Solution 2:

According to Markovnikov's rule we would get a vicinal diiodide, which is unstable.

The iodine atoms would undergo elimination and once again form a double bonded structure that is 1-propene.

As to why (excess) has been mentioned, 1-propene reacts with this $\ce{HI}$ undergoing an addition reaction. As per Markovnikov's rule, this would in fact form 2-iodopropene.