Question about halogenation of benzene

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Hey,

I didn't understand too much ( when reading my text book) why benzene undergoes polyhalogenation when it contains a strong activator ( i.e: NH2) and reacted with bromine and FeBr3 as a catalyst.

I was thinking because of the fact that there is a carbocation in every resonance structure at the ortho/para location, however, that would only give me 3 different product in a reaction, no? because the carbocation only occurs one place at the time ( I mean , the reaction either gives one resonance structure or the other, no?)

Thanks for the help!
 
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Was there a particular context in which the textbook is discussing polyhalogenation? Perhaps, reactions of aniline? Or, as an interference with some other reaction? All the possible polyhalogenated benzene and aniline compounds are listed in 60 year old CRC Handbook, and nothing noteworthy to any of them, nor do any of my immediately available texts remark on any heroic efforts necessary to synthesize them.
 
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Hey,

I didn't understand too much ( when reading my text book) why benzene undergoes polyhalogenation when it contains a strong activator ( i.e: NH2) and reacted with bromine and FeBr3 as a catalyst.

I was thinking because of the fact that there is a carbocation in every resonance structure at the ortho/para location, however, that would only give me 3 different product in a reaction, no? because the carbocation only occurs one place at the time ( I mean , the reaction either gives one resonance structure or the other, no?)

Thanks for the help!
Hi MarcL.
If I understand your question, you want to know why aniline is poly-halogenated when treated with Br2+FeBr3, rather than mono-halogenated in each possible position.
First, as you say NH2 is indeed an ortho/para directing group, and this implies that the (main) products of mono-halogenation are 2, not 3.
Second, it's quite difficult to have two positive charges on the same molecule, so I suppose that yes, poly-halogenation occurs via a sequence of carbocationic (Wheland) intermediates, each bearing one positive charge. The first will be mono-brominated aniline, the second di-brominated etc.
This leads automatically to the answer to your main question.
You would ~only get the two isomeric mono-brominated anilines (2-Br and 4-Br) if aniline reacted with Br2+FeBr3 several orders of magnitude faster than mono-bromo-anilines or 2,4-dibromoaniline react with it. Experimentally, it is found that this not the case: due to the strong activating effect of NH2, even a mono-bromo aniline is still sufficiently reactive (compared to aniline) to start to give 2,4-dibromo-aniline before all the aniline is consumed.
So it's a matter of kinetics. Even if you use only one equivalent of brominating agent, you can't decide which reactions must run and which must not. They will each proceed according to their own rate constants and the varying concentration of each intermediate in the mixture.

This happens classically in the alkylation of amines. If you react 1 Eq of NH3 and 1Eq of an alkyl halide RX, most of the time you will get a mixture of NH3, RNH2, R2NH and R3N (and even quaternary ammonium if RX is very reactive).

Many ways to navigate around the issue exist, but this is another chapter.
 

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