In halogenation of benzene, e.g. bromination, $\ce{AlBr3}$ and $\ce{Br2}$ is added in order to make $\ce{Br+}$ by forming delivery agent $\ce{AlBr5}$ (step 2 of picture).

Why is the terminal (outermost) Br atom serve as an electrophile instead of bromine with positive formal charge? I don't get it because the terminal Br atom has an octet and should be stable and not as electron-deficient, why is it bears a partial positive charge?

another question related is that can use the combination of $\ce{AlCl3}$, $\ce{Br2}$ to achieve bromination? Is the mechanism the same as adding $\ce{AlBr3, Br2}$ if it is possible? Will the electrophilicity of terminal halogen be altered and prevent the reaction? enter image description here

  • 1
    $\begingroup$ I don't get it what a+1 formal charge means, and yes, I think AlCl3 can act as catalyst, as it can still act as a Lewis acid for Br2. Only the rate of reaction my be altered. $\endgroup$ – ba-13 Jun 11 '20 at 12:08
  • $\begingroup$ Consider how large the FeBr4 complex is $\endgroup$ – Waylander Jun 11 '20 at 12:12
  • $\begingroup$ @Waylander Sorry I don't understand why size is related. I am thinking is it possible that the Br with positive formal charge is partially negative due to high EN difference with Fe, hence terminal Br is induced to become partially positive, is this a possible answer or induction effect falls quickly as size (distance) increase? Thank you $\endgroup$ – 234ff Jun 11 '20 at 13:30