[Cooper]

Hi, I am having trouble with these three enolate chemistry problems, any help? None of the answers posted below in the picture are correct (according to my homework program); they are my attempts.

For the second one I get that they want me to apply a Michael Addition, but in reality, wouldn't the base deprotonate the primary carbon on the diene, since primary carbanions are more stable?

Just to be clear, it's three different problems.

Thanks

[Dan]

1. There is another based induced reaction of aldehydes that will compete with the aldol in this case due to the steric bulk of the enolates. Hint: The reaction I'm thinking of would also happen with benzaldehyde.

2. An enolate is a highly stabilised "carbanion" - it is probably more realistic to think of it as an oxyanion (enolate). Enolate stability is similar to alkene stability. Under thermodynamic control, you will see reaction of the more thromodynamically stable enolate (which will be the major enolate species present) with the vinyl ketone. Q: What is the most themodynamically stable enolate you can form from those reagents?

3. This is a strange question. If you take a dicarboxylic acid with NaOH, all you will get is the disodium carboxylate salt and no further reaction.

[Cooper]

1. There is another based induced reaction of aldehydes that will compete with the aldol in this case due to the steric bulk of the enolates. Hint: The reaction I'm thinking of would also happen with benzaldehyde.

2. An enolate is a highly stabilised "carbanion" - it is probably more realistic to think of it as an oxyanion (enolate). Enolate stability is similar to alkene stability. Under thermodynamic control, you will see reaction of the more thromodynamically stable enolate (which will be the major enolate species present) with the vinyl ketone. Q: What is the most themodynamically stable enolate you can form from those reagents?

3. This is a strange question. If you take a dicarboxylic acid with NaOH, all you will get is the disodium carboxylate salt and no further reaction.

2. Hm, so can it be said that weaker, smaller bases like hydroxide make the thermodynamic enolate, and larger, stronger bases like LDA make the kinetic enolate, because they react more quick?

3. I was thinking that a chain would form in an intramolecular reaction. So should I add ethoxide in ethanol first? Then I will be able to create a ring if I add a base, right? Because it's nonprotic?

I am still a little confused about 1., trying to figure out what else it can do.

Thanks for your help

[Dan]

2. Hm, so can it be said that weaker, smaller bases like hydroxide make the thermodynamic enolate, and larger, stronger bases like LDA make the kinetic enolate, because they react more quick?

That's not the point here. You get thermodynamic control if the reaction is reversible. Deprotonation of a ketone by hydroxide is readily reversible (because the enolate can deptotonate water, regenerating the ketone). If you used a very strong base in an aprotic solvent (LDA/THF, for example), the reaction is essentially irreversible because the enolate is not strong enough a base to deprotonate diisopropylamine.

3. I was thinking that a chain would form in an intramolecular reaction. So should I add ethoxide in ethanol first? Then I will be able to create a ring, right? Because it's nonprotic?

If you add ethoxide to a carboxylic acid, you still get the carboxylate salt. Ethanol is protic. The problem here is not the choice of base or solvent - whatever base you add you will get deprotonation of the carboxylic acid at O, not C, because the O-H is many orders of magnitude more acidic than the C-H. If you want to do an intramolecular Claisen, you will have to convert the carboxylic acid groups into something else...

I am still a little confused about 1., trying to figure out what else it can do.

Classic named reaction. Italian. Does not involve an enolate.

[Cooper]

2. Hm, so can it be said that weaker, smaller bases like hydroxide make the thermodynamic enolate, and larger, stronger bases like LDA make the kinetic enolate, because they react more quick?

That's not the point here. You get thermodynamic control if the reaction is reversible. Deprotonation of a ketone by hydroxide is readily reversible (because the enolate can deptotonate water, regenerating the ketone). If you used a very strong base in an aprotic solvent (LDA/THF, for example), the reaction is essentially irreversible because the enolate is not strong enough a base to deprotonate diisopropylamine.

3. I was thinking that a chain would form in an intramolecular reaction. So should I add ethoxide in ethanol first? Then I will be able to create a ring, right? Because it's nonprotic?

If you add ethoxide to a carboxylic acid, you still get the carboxylate salt. Ethanol is protic. The problem here is not the choice of base or solvent - whatever base you add you will get deprotonation of the carboxylic acid at O, not C, because the O-H is many orders of magnitude more acidic than the C-H. If you want to do an intramolecular Claisen, you will have to convert the carboxylic acid groups into something else...

I am still a little confused about 1., trying to figure out what else it can do.

Classic named reaction. Italian. Does not involve an enolate.

2. Oh okay I think I see, thank you. My textbook explained it but I didn't understand it the way it was written. So for weaker bases, their conjugate acids are going to be stronger, leading to the protonation-deprotonation reaction occuring multiple times, because the enolate is able to deprotonate the conjugate acid, and producing the thermodynamic enolate? Whereas when you use a strong base the deprotonation only occurs once, at the kinetic spot?

3. Oh right of course the ethoxide would still deprotonate the carboxylic acid. So I should use PCC first to turn the alcohol into an aldehyde. I think I've got it now, how does the picture below look? You need the acid step to quench the base and get rid of the salt, right?

Thank you again

[Cooper]

Actually, for the synthesis I think it would be better just to use LDA, a base, rather than the NaOH in water, right? To prevent addition?

[Dan]

I think it would be better just to use LDA, a base, rather than the NaOH in water, right? To prevent addition?

Prevent addition of what?

Your aldehyde intramolecular aldol route is conceptually fine, though you might have difficulty stopping dehydration of the beta-hydroxy aldehyde.

There is a more attractive route very close to what you already posted with the acid. Q: What are the usual substrates of the Claisen condensation and how are they prepared from acids?