How does the product change with varying $\ce{AlCl3}$ concentration? enter image description here

Answer: In excess, meta bromination is observed and if not, then alpha hydrogen is substituted by ketone halogenation mechanism

I am trying to understand why the answer is so using mechanisms. I know that the first step of brominating involves the attack of di bromine bromine onto aluminium chloride, and then, the attack of the benzene's pi electrons onto the one of the bromines, the reaction ends with removal of a hydrogen by the remaining bromine- aluminium chloride molecule.

In the ketone halogenation mechanism, it begins with an enol with negative charge formation into attack of bromine molecule. And finally the product.

In the first case, we have that the Bromine is made more electrophilic and in the second we have that the compound is made more nucleophilic. So, how does concentration control which pathway the reaction takes?

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    $\begingroup$ If you have less than 1 eq of AlCl3 then there will still be molecular bromine in the reaction mixture for the enol to react with giving alpha bromination $\endgroup$
    – Waylander
    Aug 3 at 9:23
  • $\begingroup$ Oh so is the 'excess' relative to bromine? $\endgroup$
    – 666User666
    Aug 3 at 13:15
  • $\begingroup$ Yes. It is reasonable to assume that the bromine reacts preferentially with any AlCl3 present 1:1. When all tthe free AlCl3 is consumed the residual bromine will react with the enol. $\endgroup$
    – Waylander
    Aug 3 at 13:49
  • $\begingroup$ Perhaps a noob question, but in the wikipedia for the ketone halogenation reaction, the first step involves acetic acid/ OH attack, I see neither here, so how does the reaction begin with the enolate? @Waylander $\endgroup$
    – 666User666
    Aug 4 at 19:49
  • $\begingroup$ This is not any ketone, there is a greater enol content because of the adjacent benzene ring $\endgroup$
    – Waylander
    Aug 4 at 19:53

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