Aldol Condensation

[Caffine]

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For the Aldol Condensation reaction, I know you treat an Aldehyde or Ketone with a moderately strong base and water to form an enolate. This enolate can then react as a nucleophile with the electrophilic carbon of the parent ketone / aldehyde.

My question though is regarding the dehydration aspect of this reaction. In TPR, they explain dehydration occurs in the presence of strong acid (not base) and heat. This results in hydrolysis. This doesn't make sense to me since this is essentially an elimination reaction and elimination reactions require a weak / strong base...

And not only do they mention that, in some of their examples, they treat the beta-hydroxy aldehyde/ketone with NH4Cl (for the dehydration step). I was wondering if these were errors, and if not ...how exactly do they fit in the mechanism.

Thanks.

 

[justhanging]

For the Aldol Condensation reaction, I know you treat an Aldehyde or Ketone with a moderately strong base and water to form an enolate. This enolate can then react as a nucleophile with the electrophilic carbon of the parent ketone / aldehyde.

My question though is regarding the dehydration aspect of this reaction. In TPR, they explain dehydration occurs in the presence of strong acid and heat. This results in hydrolysis. This makes sense since this is essentially an elimination reaction... but then when I glance at the examples, for the dehydration step, they treat the beta-hydroxy aldehyde/ketone with NH4Cl. I was wondering if this was an error, and if not ...how exactly does this work in the mechanism.

Thanks.

If I remember correctly, you can dehyrate the aldol product in either basic or acidic media, though in basic media the leaving group is a hydroxide ion rather than water. I don't understand your problem though because NH4Cl is an acid you you said acids dehydrate.

 

[Caffine]

My question though is regarding the dehydration aspect of this reaction. In TPR, they explain dehydration occurs in the presence of strong acid (not base) and heat. This results in hydrolysis. This doesn't make sense to me since this is essentially an elimination reaction and elimination reactions require a weak / strong base...

Wait I realized the acid could convert the OH into a good leaving group (H2O), forming a carbocation. Then the conjugate base can deprotonate the alpha hydrogen forming a pi bond in place. An E1 elimination.

I can't seem to make sense of why NH4Cl was also used.

 

[Caffine]

If I remember correctly, you can dehyrate the aldol product in either basic or acidic media, though in basic media the leaving group is a hydroxide ion rather than water. I don't understand your problem though because NH4Cl is an acid you you said acids dehydrate.

Oh right. It's an ammonium salt which can act as a weak acid. Blondie moment!

 

[justhanging]

Wait I realized the acid could convert the OH into a good leaving group (H2O), forming a carbocation. Then the conjugate base can deprotonate the alpha hydrogen forming a pi bond in place. An E1 elimination.

I can't seem to make sense of why NH4Cl was also used.

I don't think the mechanism involves the formation of a carbocation. The enol kicks out the protanated water by moving the double bond. Check your mechanism again.

 

[Etorphine]

The second part works without a strong base is because the molecule REALLY wants to get a second conjugated double bond. This molecule will be very thermodynamically stable, and allows weaker bases (like water) to work.

If there was no conjugation aspect, water (the only base in a solution devoid of other bases) would not be able to perform that elimination so easily; it is only because the final product is so preferred that water pulls off the alpha hydrogen.

 

[MalibuPreMD]

I am pretty confident that the last step is an E1 elimination, where in the presense of strong base -OH will be a leaving group and a temporary carbocation is formed.

 

[justhanging]

I am pretty confident that the last step is an E1 elimination, where in the presense of strong base -OH will be a leaving group and a temporary carbocation is formed.

I don't think so. I remember getting a quesition about the mechanism of a base catalyzed condensation of an aldol in my orgoIII class last semester. Their was no E1.

 

[RogueUnicorn]

it's been a while, but as i understand it you can do the condensation via base or acid catalysis, and neither forms a carbocation.

 

[sibitrum]

It actually goes via an "e1cb" mechanism, where (e.g.) you deprotonate, forming the enolate. Then the enolate attacks the leaving group.