Mechanism for the addition of hydrogen iodide to 3,3‐dimethylbut‐1‐yne

Question

The following reaction mechanism was given as a solution to a solved problem in my textbook¹ for the addition of hydrogen iodide to 3,3‐dimethylbut‐1‐yne:

It can be seen that that 2,2-diiodo-3,3-dimethylbutane (a geminal dihalide) is the product according to this mechanism. I arrived at a different product 2,3-diiodo-2,3-dimethylbutane (a vicinal dihalide) by using the following mechanism:

The difference is due to the methanide shift I did on the second step, converting 3,3-dimethylbut-1-en-2-ylium cation to 2,3‐dimethylbut‐3‐en‐2‐ylium cation. I consider this to be a reasonable carbocation rearrangement because after the methanide shift, the 2,3‐dimethylbut‐3‐en���2‐ylium cation is tertiary as well as in conjugation with the double bond. Moreover, the initial carbocation has a positive charge on a more electronegative $\mathrm{sp^2}$ hybridized carbon atom. Due to this, I believe, the methanide shift produces a more stable carbocation. But, why did the author proceed without making this rearrangement?

When I discussed this in chat, there was a consensus that the product of this reaction must be a geminal dihalide as obtained by the author. The only way, I could think of, by which I can obtain the geminal dihalide even after the rearrangement discussed earlier is to do another rearrangement during the addition of second molar equivalent of hydrogen iodide as given below:

The problem with this mechanism is the carbon bearing the postitive charge after the methanide shift also has an iodine atom attached to it. Earlier, I learnt that chlorine atom is the only halogen for which the positive mesomeric effect is stronger than the negative inductive effect thereby stabilizing the positive charge. But here, due to the presence of iodine, I think the carbocation is not stable after rearrangement.

Even after neglecting this fact, there seems to be a major difference betweeen the author's mechanism and the modified mechanism, which I've emphasized using a carbon-12 labelled reactant as given below:

It can be seen that even though we obtain geminal dihalides through either of the mechanism, the products formed aren't exactly the same. One has the iodine attoms attached to the normal carbon whereas in the other they are attached to the carbon-12 atom.

In short, what happens when hydrogen iodide is added to 3,3‐dimethylbut‐1‐yne?

Reference

1. Solomons, et al. Organic Chemistry for JEE (Main & Advanced). Edited by MS Chouhan, Third Edition; Wiley India Private Limited. ISBN 978-81-265-6065-3

Answer 1

There seems to be an issue with both mechanisms on the front that an S_N1 reaction would not take place since the carbocation formed is a vinylic carbocation which is highly unstable.

The actual reaction follows a termolecular mechanism where the rate of reaction is given to be:

$$\text{Rate}=[\ce{HX}]^2[\text{alkyne}]$$

Now, according to Advanced Organic Chemistry by Francis A. Carey, the reaction mechanism would be as follows:$^1$

Step $1$:

A concerted termolecular reaction...

This involves an acid/base reaction, protonation of the alkyne developing positive charge on the more substituted carbon. The π electrons act pairs as a Lewis base.

The other part is attack of the nucleophilic bromide ion on the more electrophilic carbocation creates the alkenyl bromide.

Step $2$:

In the presence of excess reagent, a second protonation occurs to generate the more stable carbocation.$^2$

Step $3$:

Attack of the nucleophilic bromide ion on the electrophilic carbocation creates the geminal dibromide.

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[$1$]: The reaction mechanism stated above uses $\ce{HBr}$ and not $\ce{HI}$ instead. However this reaction takes place for $\ce{HX}$.

[$2$]: The same reaction using $\ce{HI}$ would have a comparatively lower yield of the geminal product since geminal diiodides are unstable due to the steric hinderance posed by the large size of the iodine atoms.