[Altered State]

I'm not sure of what really happens in these 2 steps o two synthetic pathways.

In the first case, compound A reacts with MeOH/H+ (forgot to drawn in the img), to give compound B, and I'm not sure if I can explain it well:
In my opinion, MeOH's oxygen attacks carbonyl, and after some protonation equilibriums (acid media), the O of the carbonyl at start, leaves the molecule as H₂O allowing elimination and generates the alkene + eter.
The main problem I find is that I think is that in this case, almost 100% of the 1,3-dicarbonyl compound would be in enol form, so adition to carbonyl of MeOH couldn't happen.


In the second case I have that 1,3-dicarbonyl compound and is treated first with MeOH/H+ to give, enol( or it can happen the same as in the last reaction and give alkene-eter??), and then Me₂CuLi, as weak organometallic nucleophile, would form the compound I've drawn. I don't know if that's what would happen or it would do it another way, like in the first case. The result seems quite obvious too.

[spirochete]

Remember that 1,3 dicarbonyl compounds will be less likely to favor the half enol form if geometry prevents intramolecular hydrogen bonding. The enol form of the five membered ring is certainly stabilized by additional resonance compared to a simple carbonyl compound, but it can't really hydrogen bond within itself like a non-cyclic version could.

Regardless, you can draw a reasonable looking mechanism that uses the enol form. Simply protonate methanol and use the lone pair on either oxygen to do an Sn2 reaction.

I added a picture of both possible mechanisms. Sorry the lighting was not so great. I am not sure what the correct mechanism is given the legitimate concern you brought up about the equilibrium possibly favoring the enol. Keep in mind, however, that it is common for a reaction to proceed via an unstable but reactive intermediate, because of a kinetic principle known as the curtin hammet postulate.

[orgopete]

The main problem I find is that I think is that in this case, almost 100% of the 1,3-dicarbonyl compound would be in enol form, so adition to carbonyl of MeOH couldn't happen.

Re: first problem
If you had acetone and the enol of acetone and protonated each, can you explain why they give the same protonated product?

Second problem, it will only have one carbonyl group, the methoxy is lost.

[Altered State]

Remember that 1,3 dicarbonyl compounds will be less likely to favor the half enol form if geometry prevents intramolecular hydrogen bonding. The enol form of the five membered ring is certainly stabilized by additional resonance compared to a simple carbonyl compound, but it can't really hydrogen bond within itself like a non-cyclic version could.

Regardless, you can draw a reasonable looking mechanism that uses the enol form. Simply protonate methanol and use the lone pair on either oxygen to do an Sn2 reaction.

I added a picture of both possible mechanisms. Sorry the lighting was not so great. I am not sure what the correct mechanism is given the legitimate concern you brought up about the equilibrium possibly favoring the enol. Keep in mind, however, that it is common for a reaction to proceed via an unstable but reactive intermediate, because of a kinetic principle known as the curtin hammet postulate.

I see, I didn't thought about hydrogen bonds not forming in 1,3-cyclopentadiones, and forming in 1,3-cyclohexadiones.
I've had thought of those two mechanisms, and seem to explain the product formation.


But what about the second problem I posted? What do you think? Will it occur as I've drawn? I still think it is so obvious.

[Altered State]

The main problem I find is that I think is that in this case, almost 100% of the 1,3-dicarbonyl compound would be in enol form, so adition to carbonyl of MeOH couldn't happen.

Re: first problem
If you had acetone and the enol of acetone and protonated each, can you explain why they give the same protonated product?

They both would be the acetone enol, wouldnt they?

Second problem, it will only have one carbonyl group, the methoxy is lost.

I'm afraid I'm not following you here...
You mean that in case there is no conjugation after "methoxylation" such reaction could not happen?

[spirochete]

Remember that 1,3 dicarbonyl compounds will be less likely to favor the half enol form if geometry prevents intramolecular hydrogen bonding. The enol form of the five membered ring is certainly stabilized by additional resonance compared to a simple carbonyl compound, but it can't really hydrogen bond within itself like a non-cyclic version could.

Regardless, you can draw a reasonable looking mechanism that uses the enol form. Simply protonate methanol and use the lone pair on either oxygen to do an Sn2 reaction.

I added a picture of both possible mechanisms. Sorry the lighting was not so great. I am not sure what the correct mechanism is given the legitimate concern you brought up about the equilibrium possibly favoring the enol. Keep in mind, however, that it is common for a reaction to proceed via an unstable but reactive intermediate, because of a kinetic principle known as the curtin hammet postulate.

I see, I didn't thought about hydrogen bonds not forming in 1,3-cyclopentadiones, and forming in 1,3-cyclohexadiones.
I've had thought of those two mechanisms, and seem to explain the product formation.


But what about the second problem I posted? What do you think? Will it occur as I've drawn? I still think it is so obvious.

Actually I was not meaning to comment on five membered vs. six membered rings. Ultimately I think it's more E vs. Z around the pi bond in the enol. In rings smaller than roughly 8 carbons you can only have one of those two geometries.

I don't have time to comment on the second half at the moment so someone else can jump in before me.

[orgopete]

Re: first problem
If you had acetone and the enol of acetone and protonated each, can you explain why they give the same protonated product?

They both would be the acetone enol, wouldnt they?

If you protonate the enol of acetone, it occurs on the carbon. You get the same product as protonating acetone on oxygen. Methanol adds, deprotonation, protonation, minus water, and deprotonation to product. No SN2 reaction.

Second problem, it will only have one carbonyl group, the methoxy is lost.

I'm afraid I'm not following you here...
You mean that in case there is no conjugation after "methoxylation" such reaction could not happen?

I think the prior problem was a clue for this problem. It will also form an allyl ester. That reacts with the cuprate the same as cuprates react with conjugated enones. (I presume thus is what was being discussed in class.) Draw the mechanism to there and think about what that enolate could do.