[hpl912]

Here are the reactions:


From what i understand, BuLi in these 3 reactions acts as strong bases and deprotonates the acidic alpha proton.

For the first reaction, does the product A results in an enolate? or the negative charge will stay at the alpha position?
NaH is a strong base as well, so i assume it will deprotonate the second alpha hydrogen? but it's weird to have a -2 charge at the alpha position...

For the second reaction, does it reacts the same forming an enolate as product E? If so, i assume LDA as a bulky base will deprotonate now the H from the OH terminal which results now in two negative charges, one on each oxygen? but then how the BuBr will possibly attach to the negative oxygen?

For the third reaction, does it takes both the alpha protons since it's not a bulky base and the alpha position is more acidic than the OH hydrogen? but then again, it will result in -2 charge at the alpha position which seems to be weird.

Thanks for any help/suggestion

[Dan]

The first base treatment will remove the most acidic proton, the second equivalent of base will remove the second most acidic proton. You need to identify these protons in each of your starting structures.

For the first reaction, does the product A results in an enolate? or the negative charge will stay at the alpha position?

The negative charge will be shared between the carbon and the oxygen - it is delocalised - the two scenarios you refer to are resonance structures, the "real" structure is a superposition of the two.

NaH is a strong base as well, so i assume it will deprotonate the second alpha hydrogen?

After the first deprotonation, the alpha position will become much less acidic since there is now considerable electron density at the alpha carbon. Can you think of any other sites from which LDA may remove a proton?

the alpha position is more acidic than the OH hydrogen

Are you sure about that? The pKa of the alpha position is probably about 25 (cf. ethyl acetate) whereas the OH of the carboxylic acid is about 5 (cf. acetic acid)... The acidities of these positions are wildly diferent, by a factor of around 100,000,000,000,000,000,000 (or 10²⁰).

[hpl912]

1) The first base treatment will remove the most acidic proton, the second equivalent of base will remove the second most acidic proton. You need to identify these protons in each of your starting structures.

2) The negative charge will be shared between the carbon and the oxygen - it is delocalised - the two scenarios you refer to are resonance structures, the "real" structure is a superposition of the two.

3) After the first deprotonation, the alpha position will become much less acidic since there is now considerable electron density at the alpha carbon. Can you think of any other sites from which LDA may remove a proton?

4) Are you sure about that? The pKa of the alpha position is probably about 25 (cf. ethyl acetate) whereas the OH of the carboxylic acid is about 5 (cf. acetic acid)... The acidities of these positions are wildly diferent, by a factor of around 100,000,000,000,000,000,000 (or 10²⁰).

1) do you mean that in the first reaction, the second deprotonation by NaH wouldn't  be the second alpha hydrogen? Is it because after the first deprotonation by BuLi, leaving a negative charge at the alpha position makes the remaining alpha hydrogen not acidic? but isn't the negative charge be delocalized right away to the carbonyl oxygen?

2) if it's shared between the carbon and oxygen, then where would the electrophile EtI in the first reaction attach to? the carbon or the oxygen? i assume it's the oxygen because the oxygen is  better at stabilizing the negative charge?

3)I think that LDA will deprotonate the alcohol hydrogen on the second reaction since it's too bulky and woulnd't be able to take the alpha position even if it could. but for the first reaction, the second deprotonation by NaH isn't as bulky as LDA, so why doesn't it take the second alpha hydrogen?

4)for the second reaction, then do you mean the first deprotonation will take the OH hydrogen? but then the second deprotonation by LDA it won't have any more protons to take since it's too bulky to take any non terminal protons?

[Dan]

1) do you mean that in the first reaction, the second deprotonation by NaH wouldn't  be the second alpha hydrogen? Is it because after the first deprotonation by BuLi, leaving a negative charge at the alpha position makes the remaining alpha hydrogen not acidic?

Yes, exactly.

but isn't the negative charge be delocalized right away to the carbonyl oxygen?

The first one is, but the second is not. After the first deprotonation, you form an anion which will occupy a p orbital to allow it to overlap with the pi system of the carbonyl - this is a requirement for resonance stabilisation. Removal of a second proton from the alpha position would leave a second anion occupying an sp² orbital - this would be orthogonal to the pi system (ie no overlap at all) and have no resonance stability. Alternatively, the bis-anion could rehybriside to sp³ in which case both might overlap weakly. The main point though is that deprotonating at a negatively charged position is extremely difficult because it is already negatively charged. 

2) if it's shared between the carbon and oxygen, then where would the electrophile EtI in the first reaction attach to? the carbon or the oxygen? i assume it's the oxygen because the oxygen is  better at stabilizing the negative charge?

That's a good point to raise. Enolates can alkylate at O or C, but it can be controlled and usually enolates are used to form new C-C bonds. The alpha carbon of the enolate is the softer site of the two, and softer electrophiles (alkyl iodides being the classic example) will tend to react at C rather than O. Solvents that H-bond to O will also favour C-alkylation (eg, alcohols), as do harder counterions/chelating agents (such as Li⁺).

3)I think that LDA will deprotonate the alcohol hydrogen on the second reaction since it's too bulky and woulnd't be able to take the alpha position even if it could.

First off, it's not an alcohol it's a carboxylic acid. It is about one hundred million million times more acidic than the alpha position - this is why it is deprotonated first (BuLi). By the time it sees LDA there won't be any OH bonds left.

but for the first reaction, the second deprotonation by NaH isn't as bulky as LDA, so why doesn't it take the second alpha hydrogen?

After the first deprotonation, the second alpha H is not the most acidic in the molecule

4)for the second reaction, then do you mean the first deprotonation will take the OH hydrogen?

Yes

but then the second deprotonation by LDA it won't have any more protons to take since it's too bulky to take any non terminal protons?

I don't think it is too bulky, LDA is a pretty common reagent for generating enolates - even non-terminal ones. I could be wrong on that one though. LDA can deprotonate Ph₃CH, and that proton is more hindered than the alpha proton in this system.

[hpl912]

so to summarize from what i understood, for the first reaction, the product A have a negative charge on the oxygen but wouldn't the negative charge come back making an oxygen double bond and displacing OEt since it's a good leaving group?
if not, then the product B would be the deprotonation of the previously alpha position's last hydrogen (now with a negative charge on the oxygen and the alpha position).
then the product C will be the attachment of ethyl iodide's ethyl group to the negatively charged carbon (since softer reacts with softer sites). now leaving only a negative charge still on the oxygen.
finally with the workup to attach a hydrogen to the negatively charged oxygen.
  
below the attached resulting structure D... is that correct?

[hpl912]

anyone?