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Problem

3‐iodoprop‐1‐ene + HI + CCl4

Answer > 1,2‐diiodopropane

I thought "I" can stabilizes carbocation formed intially by resonance so there can be a shift of (+) to the carbon close to "I" and hence forming the product 1,1-diiodopropane but given is 1,2 product , whats wrong in this ? Or is that H+ and +I combined effect of CH3 is slightly greater than compared to resonance of "I"?

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  • $\begingroup$ yes! the inductive effect of halogens is a special case. That it overweighs their +R $\endgroup$
    – Prajwal Tiwari
    Apr 12 at 4:32
  • $\begingroup$ No exception? In any case? $\endgroup$
    – WizardMath
    Apr 12 at 4:33
  • $\begingroup$ not that I know of, but I believe there must be some exception as in almost all concepts, we get to see at least one exception in chemistry. Someone who has more idea about this might be able to help $\endgroup$
    – Prajwal Tiwari
    Apr 12 at 4:42
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    $\begingroup$ Do you know two -OH groups on a single carbon are unstable? And here we have I atoms! In fact, how would you explain the formation of 2-iodopropane when excess HI is taken? $\endgroup$
    – Vilakshan Gupta
    Apr 12 at 4:42
  • $\begingroup$ What the OP is asking is that could the carbocation shift to the carbon to which the I is initially attached, then we could get a 1,1 di iodo product $\endgroup$
    – Prajwal Tiwari
    Apr 12 at 4:57
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There are three reasons as far as I can see as to why the 1,2 product is favored. There may be more.

  1. Steric hindrance

    The iodine atom is large in size compared to the carbon atom. Having two iodine atoms on successive carbon atoms in a hydrocarbon is quite unstable. In fact, should two successive carbon atoms possess iodine substituents, they usually undergo elimination reactions. And here the 1,1 product would have two iodine groups on the same carbon atom!

  2. Inductive effect of halogens

    While it is true that halogen atoms have a positive mesomeric effect, this effect is almost always dominated by their negative inductive effect. If you see a halogen atom, you can usually neglect it's +M effects in favor of it's −I effects.

  3. Carbocation stability

    The carbocation formed in the process of formation of the 1,2 product is more stable (due to hyperconjugation) than the one formed in the process of formation of the 1,1 product (again, due to the −I effect of iodine).

Either way, 1,2-diiodopropane is more stable than 1,1-diiodopropane.

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  • $\begingroup$ Ty i got your pt one more thing wanted to ask any exceptions uk when that -I of halogens fails over +M ? $\endgroup$
    – WizardMath
    Apr 12 at 8:20
  • $\begingroup$ @WayBig I have not heard on any exception. $\endgroup$
    – m-Xylene
    Apr 12 at 9:31
  • $\begingroup$ I see , Can u check this link out too its regarding same reaction type ? chemistry.stackexchange.com/q/149835/106350 $\endgroup$
    – WizardMath
    Apr 12 at 9:49

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