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Problem

3‐iodoprop‐1‐ene + HI (excess) + CCl4

Answer

2-iodopropane
2-iodopropane

Question

I thought of Markovnikov's rule and suggested 1,2‐diiodopropane as the the final product:

1,2‐diiodopropane

Apparently, this is wrong. I can not figure out the mechanism of this reaction (will $\ce{HI}$ dissociate in the presence of $\ce{CCl4}?)$ And what is the significance of excess $\ce{HI}$ here?

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    $\begingroup$ Probably rather 2-iodopropene. $\endgroup$ – aventurin Apr 7 '17 at 17:47
  • $\begingroup$ @aventurin although the answer is an alkane but it's entirely possible that it may be wrong,I don't see how this reaction is any different from an ordinary addition reaction. $\endgroup$ – Daenys Targaryen Apr 8 '17 at 5:10
  • $\begingroup$ Theoretically, it could also be a rearrangement reaction. $\endgroup$ – aventurin Apr 8 '17 at 11:42
  • $\begingroup$ what kind of rearrangement might take place here? Please give a hint, I'm stuck with this one for 4 days $\endgroup$ – Daenys Targaryen Apr 11 '17 at 16:09
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I am at a loss to supply the mechanism, but based on the problems I've worked, and based on the theory that my instructor has taught, a general thumb-rule is that iodine on adjacent carbon atoms (vicinal di-iodo, if you will) are unstable, and they are eliminated as $\ce{RCHICH2I -> RCH2=CH2 }$

An idea for this can be picturised (though this may possibly be the incomplete or even incorrect reasoning) by considering that iodine atoms are bulky, and steric factors cause strain in the main chain. Moreover, there are possibilities of van der Waals' forces between vicinal iodine atoms, and also the good leaving tendency of iodine, that may favour the formation of $\ce{I2}$ by elimination.

A brilliant example that lucidly explains this is the reaction of glycerol with excess of HI; after substitution of all the hydroxyl groups, there is an elimination of $\ce{I2}$ as discussed above, leading to the alkene of your original question. Subsequently, there is an addition of HI, indeed following Markovnikov sensibilities, that leads to the formation of 1-propene. It subsequently adds HI to yield your final product. Thus, 1 mole glycerol consumes 5 moles of HI; 3 via substitution, 2 via addition.

Now, the last example of glycerol wasn't strictly necessary, but I personally find the sequence quite beautiful; a single pathway leading to 3 rounds of substitution, two of addition and all interspersed with a unique type of elimination, and all this when we barely had a handful of reactants to begin with.

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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

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  • $\begingroup$ even if I2 is eliminated, I will be left with a carbocation, how will HI add to it? $\endgroup$ – Daenys Targaryen Apr 23 '17 at 15:04
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    $\begingroup$ $\ce{I2}$ is eliminated from a neutral molecule, 1,2-diiodopropane (see step 2 above). So there's no carbocation left but propene to which $\ce{HI}$ may be added. $\endgroup$ – aventurin Apr 23 '17 at 19:38
  • $\begingroup$ 2 moles of $\ce{HI}$ are used up in total and in the second step vicinal $\ce{I2}$ are unstable so they get eliminated.After this one more mole has to be used. $\endgroup$ – shreya Jul 30 '20 at 18:01
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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.

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