# Friedel-Crafts synthesis - How is HCl produced?

In the mechanism for the acetylation of bromobenzene, how is $$\ce{HCl}$$ gas produced?

The reactants are acetic anhydride, bromobenzene, $$\ce{AlCl3}$$ and dichloromethane.

Initially, 4-bromoacetophenone is made, accompanied with (as far as I know) acetic acid and the regeneration of $$\ce{AlCl3}$$.

How then is $$\ce{HCl}$$ produced?

To remove the $$\ce{AlCl3}$$ at the end, 3 $$\ce{H2O}$$ are added and $$\ce{HCl}$$ gas is produced then.

I'm trying to think of ways that $$\ce{HCl}$$ gas is produced prior to that, however. Could a chlorine of the aluminate ion deprotonate the arenium ion?

Could dichloromethane (solvent) donate a $$\ce{Cl-}$$ to $$\ce{AlCl3}$$?

• I suspect your suggestion is correct. $\ce{[AlCl3(OOCCH3)]-}$ loses a chlorine instead of the acetate when deprotonating arenium ion to form $\ce{HCl}$ as a byproduct – bon Feb 2 '15 at 18:40

You could perform the electrophilic aromatic substitution with acetyl chloride as well instead of acetic anhydride - in the first case it is evident where $$\ce{HCl}$$ comes from. However, the mechanism is similar if you use acetic anhydride. The first step consists in the electrophile activation by $$\ce{AlCl3}$$, which generates in situ the "acylium" ion ($$\ce{H3C-C#O+}$$) - the actual acylating agent. The second step is the electrophilic aromatic substitution, in which the electrophile is preferentially "attacked" by the aromatic ring at the ortho- or para-position (ortho- and para-orienting effect of halides). The driving force for the reaction is the last step, which leads to restore the aromaticity and it is assisted by $$\ce{(AcO)(Al)Cl2}$$ with liberation of $$\ce{HCl}$$.
• I have a question..Are you sure $AlCl_3$ doesn't attack on the middle oxygen? – user35508 Dec 6 '18 at 11:17