In an aldol condensation during the dehydration part, what role does NH4Cl play?
I'm a bit confused. Thanks!!

In an aldol condensation during the dehydration part, what role does NH4Cl play?
I'm a bit confused. Thanks!!
I'm not super fresh on the Aldol Condensation, but anytime you see an ammonium salt (NH4+Cl-) you can be assured that the purpose of it is to protonate something.

So, I'd assume that the NH4Cl dissociates in an aqueous solution, and donates a proton to the carbonyl oxygen.

I'll look it up tonight if I get time.

edit: Sorry, I looked and all I found was the mechanism involving NaOH.

In an aldol condensation during the dehydration part, what role does NH4Cl play?
I'm a bit confused. Thanks!!

A) It serves to protonate the OH and make it a better leaving group, so that the ß-hydroxy-ketone can undergo an E1 elimination.
B) It serves to protonate the OH and make it a better leaving group, so that the ß-hydroxy-ketone can undergo an E2 elimination.
C) It forms an imine intermediate from the ß-hydroxy-ketone, which makes OH a better leaving group.
D) It readily absorbs water, preventing hydrolysis of the conjugated ketone to form an alpha-hydroxyketone.

A) It serves to protonate the OH and make it a better leaving group, so that the ß-hydroxy-ketone can undergo an E1 elimination.
B) It serves to protonate the OH and make it a better leaving group, so that the ß-hydroxy-ketone can undergo an E2 elimination.
C) It forms an imine intermediate from the ß-hydroxy-ketone, which makes OH a better leaving group.
D) It readily absorbs water, preventing hydrolysis of the conjugated ketone to form an alpha-hydroxyketone.
I want to say A....but how can you determine whether its exclusively E1 vs E2?

I want to say A....but how can you determine whether its exclusively E1 vs E2?

To undergo an E2 reaction, you need a strong base to pull the proton off of the alpha carbon. NH4Cl is a weak acid, so the mechanism of choice is to protonate the hydroxyl group.

In the original question though, it wasn't specified if the original basic conditions (with hydroxide) were at high enough temp to get E2 elimination. In that case, the NH4Cl is there to simply work up the product mixture by neutralizing the excess base. Strong acid work up could potentially hydrolyze the pi-bond and regenerate the ß-hydroxyketone.

I was assuming the OP was talking about forming the B-hydroxyketone because he/she said dehydration...

oooo...one more question: Can an aldol condensation occur under acidic conditions?
Thank you BRTeach!