[Nanolysis]

Hi guys,

I have a couple of questions regarding conjugate addition. I hope you guys can help me out.

So with the first reaction: after attack of n-BuLi on the nitro compound. Does the nitro group leave? It is very electron withdrawing. I'd draw an arrow from the carbon-nitrogen bond to the nitrogen, and then O-N=O would leave? I don't understand how the nitro group will behave in this reaction...

The next question is really hard for me.

Do I let the ethoxide ion attack the epoxide first? Or the carbonyl carbon (and then I need to decide which one if it attacks the carbonyl carbon)? What happens after the epoxide is attacked by the nucleophile EtO^-? Is EtOH only a solvent?  How is the ether group restored than? It seems like epoxide, NaOEt and EtOH even don't particiapte since the product has the same amount of carbon as the starting product?

[discodermolide]

What do you know about the reaction of nitro compounds and beta-ketoesters with bases?
Hint, think about the acidity of the protons alpha to the nitro group and the acidity of the protons in the beta-keto ester.

[quantumnumber]

As 'discodermolide' said, the key is the acidity of the protons alpha to the nitro an keto groups respectively.
In the first case you are taking BuLi as a nucleophile... but it is also a strong base, and here it acts as one.
In the second case is more or less the same with EtONa... Is it acting as a nucleophile or as a base? By the way, EtOH is the solvent.

[AromaticAcrobatic]

Both of the above comments are good, and I would like to add to them by suggesting the poster think about what he/she knows about Iliads.

[Dan]

Note: conjugate addition is not involved in these reactions.

[Nanolysis]

Hey guys, thank you for taking your time to reply.

So in the first reaction (with the nitro compound) one alpha hydrogen is removed? The nitro group is electron withdrawing so it makes the alpha hydrogens (slightly) acidic, n-BuLi deprotonates the alpha carbon..what will happen with the carbocation we get after deprotonation? Does n-BuLi only snatch away a proton and leave a carbocation, or will the n-Butyl group be attached to the alpha carbon (where does the acidic proton go then?)?

For the second reaction, I'd say the ethoxide-ion is acting as a base removing one proton from the carbon between the two C=O bonds...what happens next?

[discodermolide]

Well you don't get a carbocation, you get an anion. The "H" disappears as butane in the case of BuLi. so it's effectively a H to Li exchange. The carbanion is also stabilised to an extent by the solvent, usually THF.
So with the carbanion present, how will it react with the electrophile?

Try opening the epoxide ring (the electrophile) with the anion you generate with NaOEt and see what you get.

[kriggy]

1) You got it almost right. How there could be carbocation when H⁺ leaves? Then, look at the benzyliodide, how it can react with the deprotonated compound?

2) You might want to read about chemical properties of oxirane
http://en.wikipedia.org/wiki/Ethylene_oxide#Chemical_properties
I think it explains the 1st step well, Im not sure what is happening next, I have some Ideas but cant verify them.

btw did you draw those products yourself? Because Im pretty sure the butyrolactone is correct but it is formed in different way than you think

edit: yea disco wrote it already

[Nanolysis]

Sorry for the late reply guys, statistics and other courses are killing me too 

So the carbanion that's formed will attack the C-I carbon and iodine leaves? You'll get the benzene compound with the R-NO2 group attached?

As for the last question, I didn't write the products myself, our professor gives us problem sets which we have to make.
The OEt anion attacks the CH₂ proton, which will give an enolate anion, the formed pi-bond will attack the symmetrical epoxide. But when it has attacked the epoxide, what will happen then? The R-O^- in the resulting molecule will attack where? Looking at the product it will attack the carbonyl carbon, but I don't have an explanation why it will attack the carbonyl carbon at the right side (looking at the product this will happen).

[discodermolide]

Your answer to the first question seems correct, can you just draw the structure of the product.

The second one, draw the structure of the product of the reaction of the enolate with ethylene oxide and see if you can get somewhere from there.