Friedel Craft Acylation Question (KP)

[sxld_cu]

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This is a question from the Kaplan Quiz Bank
Can someone help me to clarify this mechanism? I am looking to this link for reference: http://www.name-reaction.com/friedel-crafts-acylation

I had thought that the AlCl3 would serve as Lewis acid to abstract the Cl from the aromatic ring compound and leave a positive charge. Can you assume that electrons from the conjugated pi system attack the electrophilic carbonyl C of the anhydride? If that is the case, what is the purpose of the AlCl3?

Also, I know that halogens are ortho, para directing but simultaneously deactivating. I've taken a look at the resonance structures but am still a bit confused.

Many thanks for any help you can offer!

 

[ayocaptain628]

um, so you are correct that AlCl3 will serve as a lewis acid.
it will act as a lewis acid with the acylating agent, acetic anhydride

the purpose of alcl3 is basically to break down something like
an acid chloride or an acetic anhydride in order to create the acyl group for acylation
this is b/c alcl3 is electropositive due to being surrounded by all those chlorines and is susceptible to attack at the aluminum center. so yeah in this case it is acting as a lewis acid

the oxygen in the middle connecting the 2 acyl groups that compose the acetic anhydride will attack the electropositive aluminum center.
this will generate the acyl group ch3-c=o with a positive carbocation which can then be attacked by pi electrons in the benzene ring.
then a hydrogen will be abstracted so that those electrons can re-establish the system.

the reason why halogens are ortho/para directing is because they are electron donating groups.
if you draw chlorobenzene, and then draw two of the lone pair electrons from chlorine donating to create a double bond, you will see that there is increased electron density at the ortho/para positions b/c as that double bond is formed, electron density must make one of those carbons negatively charged .:. EAS more likely to occur there.
you will see it when you draw it out that the ortho and para positions have an extra electron pair above them that makes them negative

they are deactivating b/c of high electronegativity and inductive effect.
just keep in mind that electron donating groups are ortho/para whereas electron withdrawing groups are meta
and then use logic on stability/instability w/electronegativity and stuff when you're thinking about "activating/deactivating" the ring

 

[The Brown Knight]

ayocaptain has explained it perfectly, but I wanted to add a little bit. halogens are o/p directing because they technically CAN donate electrons from their 3 lone pairs; however, they are ring deactivating because the inductive effect due to their relatively high electronegativity overpowers the resonance stabilization - in other words, they DON'T WANT to donate electrons.

 

[Teleologist]

this is b/c alcl3 is electropositive

You mean Al is electropositive.

 

[sxld_cu]

Hi! Thanks so much for the explanation - that makes a lot of sense now! For some reason, I didn't fully realize that the electropositive Al could be attacked by any nucleophile, including the O in the anhydride! Great clarification 🙂

 

[justadream]

I thought FC-acylation was not on the MCAT?

(at least I hope it's not lol)

 

[sxld_cu]

I thought FC-acylation was not on the MCAT?

(at least I hope it's not lol)

Oh, really?? I heard that the ortho/para/meta directing stuff for EAS were not on the MCAT but did not know anything about FC