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The addition of $\ce{HBr}$ to $\ce{CH2=CHBr}$ produces $\ce{CH3-CHBr2}$. Why is this so?

According to me, halogens have a stronger $-I$ effect than a $+M$ effect (for example a halogen substituent on benzene, deactivates the ring), and hence the $\ce{CH3-CHBr+}$ carbocation should be more unstable as compared to $\ce{CH2+-CH2Br}$. Where am I going wrong?

Edit : I have seen this question but it doesn't answer my query.

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    $\begingroup$ What are the reaction conditions ? Addition of HBr proceeds via a radical mechanism if intitiated by photolysis. $\endgroup$ – J. LS Apr 16 '15 at 8:37
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    $\begingroup$ @J.LS I'm not sure of the conditions, but I'm talking about the ones that will cause the reaction to proceed through a carbocation mechanism. $\endgroup$ – Binary Geek Apr 16 '15 at 8:39
  • $\begingroup$ How does the linked question not answer your's? What are you still confused about? $\endgroup$ – bon Apr 16 '15 at 8:51
  • $\begingroup$ @bon As I have written in the question, my understanding is that halogens destabilise a carbocation due a stronger $-I$ effect. However this does not reflect in the product formed on addition of $\ce{HBr}$ to vinyl bromide. The link does not talk about why halogens are stabilising the carbocation instead of destabilising it. $\endgroup$ – Binary Geek Apr 16 '15 at 8:55
  • $\begingroup$ @BinaryGeek Ok sorry I misread your question slightly. $\endgroup$ – bon Apr 16 '15 at 8:59
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The reaction of $\ce{HBr}$ with olefins is a relatively fast reaction, in other words, the activation energy required is small. Further, formation of the carbocation intermediate is the rate determining (slow) step in the overall process. Considering both of these facts and applying the Hammond postulate we can say that the transition state leading from the reactants to the carbocation intermediate will be an early transition state, it will look more like the reactants than the carbocation.

This suggests that resonance and inductive effects need to be examined with regard to the reactants, rather than the transition state, in order to determine how the molecule will react with an electrophile (the proton).

enter image description here

As the above diagram illustrates, both resonance and inductive effects (in an early transition state that resembles the reactant) suggest that the beta carbon will be more nucleophilic than the alpha carbon. Therefore addition of the proton (electrophile) will occur at the beta carbon. This will be followed by rapid addition of $\ce{Br^{-}}$ at the carbon already bearing a bromine atom (the alpha carbon) to yield a geminal dibromide.

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According to me, halogens have a stronger −I effect than a +M effect (for example a halogen substituent on benzene, deactivates the ring

The -I effect of a bromo substituent does indeed reduce the rate in a further electrophilic aromatic substitution.

However, the +M effect of the same substituent is ortho/para-directing, as to be seen in the bromination of bromobenzene.

This effect, namely the stabilisation of a positive charge through interaction with the lone pair of the halogen ($\ce{H3C-CH+-Br <-> H3C-CH=Br+}$) controls the addition of $\ce{HBr}$ to vinyl bromide.

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    $\begingroup$ I'm still not able to understand. I follow that the $+M$ effect will stabilise the positive charge. However halogens exhibit a $-I$ effect which is stronger than the $+M$ effect. So why does the $+M$ effect control the addition of $\ce{HBr}$ to vinyl bromide when the $-I$ effect is stronger? $\endgroup$ – Binary Geek Apr 16 '15 at 10:35
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An additional point that I wanted to state from this source.

The same source has an image which shows the reaction mechanism:


2nd equivalent

Note the 2nd reaction in the above image.

I quote some relevant text:

Note that the carbocation in this case bears a chloride ion. Since carbocations are electron poor, and chlorine is quite an electronegative element, it’s interesting to point out that the electron releasing ability of the alkyl group [and the ability of chlorine to donate a lone pair to the carbocation] “win out” here over the electron-withdrawing character of chloride ion.

This is the genaralized case while in your case the R=H and it no longers qualifies as an alkyl group. But it is important to note that in the actual pathway, 3 H contribute in the hyperconjugation while in the other pathway , only 2. This effect, combined with the +M of the Cl overpower the -I effect.

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