# 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 – Mithoron 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 $2$. – Apoorv Potnis 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. – Mithoron 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. – Apoorv Potnis 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 $\ce{H}$. – Apoorv Potnis Nov 4 '17 at 3:21

• I think in this case, both kinetic and thermodynamic factors help the formation of carbonation at $1$. – Apoorv Potnis Nov 3 '17 at 16:50
• As suggested by Mithoron, is there any chance of intramolecular attack by $\ce{Cl}$? – Apoorv Potnis Nov 4 '17 at 15:31