[Mandeep Deka]

If i have 2-bromo prop-1-ene, and i am asked to add HBr, through carbocation as intermediate, in which position will the carbocation be more stable? and why?

What i am confused with is that, in general, halogens as substituents have electron withdrawing character, so from that point of view the positive charge must reside on the terminal carbon atom i.e C-1. But due to lone pair available on bromine atom in C-2, if the positive charge resides on C-2 it can be resonance stabilized. Which of the two is correct then?

Thanks

[Doc Oc]

The simple answer is to consider Markovnikov's rule.  That should guide you to the answer.

[orgopete]

I think sometimes it is useful to think backwards. If you knew the product of the reaction is 2,2-dibromopropane, then what would that tell you about the electron availability of the propene? It may seem counter intuitive, but draw a resonance structure of the starting material using the non-bonded electrons of be bromine. If you protonate that resonance structure, what would the intermediate look like?

I too struggle with understanding this type of scenario. Bromine is a strong inductive electron withdrawing and it can be a (weak?) resonance donating group. Which effect should take precedence? If you replace the bromine with a nitrogen or oxygen, the same scenario exists, but those atoms are very good resonance donors and that chemistry predominates.

As a consolation, when this effect re-emerges in electrophilic aromatic substitution, you can use this principle again. Your chemical intuition should be rewarded to find that a bromobenzene will be less reactive than benzene. That is, bromine will still be a resonance donor, but because of its reluctance to do so, it will actually react more slowly than if the bromine were replaced with a hydrogen.

[Mandeep Deka]

Thank you for the post
So am i right in saying that, though bromine has a strong inductive electron withdrawing effect, it is because of the presence of a positive charge on a the carbon attached to the bromine that it can somehow donate a pair to the carbon atom to produce a resonating structure (even though quite unstable), and as there is a good prospect for the positive charge to get distributed over the bromine atom too, the positive charge resides on the C-2 atom and hence 2-2 dibromo- pentane is formed.

[Mandeep Deka]

OK, i got your point, rather than looking at the possibility of stabilization of the carbocation, i shall rather look at the ease of formation of carbocation.

Due to the possible resonating structure of the initial reactant (between the pi bond an lone pair of bromine), i can say that in the resonance hybrid, the C-1 carbon of 2 bromo propene, has somewhat a partial negative charge, because of which, the proton finds it easier to attack C-1 hence forming positively charged carbon, C-2, and giving us the product, 2-2 dibromo propane.

Thanks a lot