# Carbocation formation in electrophilic addition of hydrogen halides to alkenes

Question:

My attempt:

I made the carbocation on the carbon labelled $1$ because I think that the positive charge would be slightly more stable on $1$ compared to $2$ due to the inductive effect of $\ce{Cl}$.

But my teacher made the carbocation on $2$ giving the explanation that after rearrangement, the positive charge will be stabilised by resonance with $\ce{Cl}$.
His approach:

Now, my question is, comparing the stabilities of the carbocation intermediates without rearrangement, the positive charge will be stable on $1$ compared to $2$. After rearrangement too, I think that the tertiary carbocation will be more stable than compared to the one stabilized by resonance with $\ce{Cl}$.

So, what is wrong in my reasoning?

Or as suggested by Mithoron, will there be intramolecular attack by $\ce{Cl}$?

• en.wikipedia.org/wiki/Neighbouring_group_participation Nov 3 '17 at 15:24
• @Mithoron I don't understand how will be NGP here? If it was that way, the carbocation will be formed on $3$ without $\ce{Cl}$ to stabilise it. It would be very unstable. It would also lead to a different product, $\ce{Cl}$ bonded to $2$. Nov 3 '17 at 15:34
• Protonation on "1" should be preferred because it leads to cyclic chloronium cation and indeed a product is different then both of your suggestions. Nov 3 '17 at 18:55
• @Mithoron I don't think that due to the following reasons. 1) The cyclic chloronium transition state will be 4 membered. It would be highly unstable. It is not even aromatic. 2) The resulting cation after rearrangement of $1$ carbocation will be a tertiary carbocation. So, the intermediate will be much stable compared to the one on $3$ which will be there if it did NGP. This carbocation is primary with nothing to stabilise it. Nov 4 '17 at 3:20
• 3) I think it 1,2 - Hydride shifts would be faster than the $\ce{Cl}$ attacking from a distance of 2 carbons in between. $\ce{Cl}$ attacking will require more work compared to $\ce{H}$ as $\ce{Cl}$ is much heavier than $\ce{H}$. Nov 4 '17 at 3:21

## 1 Answer

The rate determining step is carbocation formation. After which, depending on conditions, the trapping of the cation may or may not be slow enough for [1,2]-shifts.

The [1,2]-shifts are probably under thermodynamic control if there is time for them to take place. This means that if there's time for the shifts, we're hunting for the most stable cation. Otherwise, you trap the most readily formed cation in a kinetic sense (which is not necessarily the most stable).

Assuming shifts are fast, I think your solution is better because the tertiary cation is more stable. Dichloromethane doesn't undergo substitution as easily as t-butyl chloride, so I would have to argue that the tertiary carbocation is more stable.

• I think in this case, both kinetic and thermodynamic factors help the formation of carbonation at $1$. Nov 3 '17 at 16:50
• You've already assumed the kinetic factor doesn't matter because it would mean that you would trap the cation where it's first formed without any migration.
– Zhe
Nov 3 '17 at 18:19
• As suggested by Mithoron, is there any chance of intramolecular attack by $\ce{Cl}$? Nov 4 '17 at 15:31
• That's actually a really good suggestion, and probably better than either of the proposals you've presented to pick from. The chloronium ion is the accepted intermediate for halogenation of alkenes. This is an interesting way to access the intermediate.
– Zhe
Nov 4 '17 at 19:47
• @ApoorvPotnis Experience? I mixed dichloromethane and alchoholic solvents but there's been minimal reaction. On the other hand, t-butyl chloride reacts fairly readily with water/alcoholic solvent mix.
– Zhe
Nov 8 '17 at 14:11