ester + NaOC2H5

[joonkimdds]

CH3-(C=O)-CH2-(C=O)-O-C2H5 + 1)NaOC2H5 2) CH3CH2Br


what kind of reaction is this?

I thought ester + NaOC2H5 means we have to imagine 2 of the same ester,
remove alpha proton from one, remove OR from the other one and then connect
these two C together as if I am making aldol condensation product.[Claisen Reaction]

but the final product is
CH3-(C=O)-C(CH2CH3)-H-(C=O)-O-C2H5

This means CH2CH3 from CH3CH2Br or C2H5 from NaOC2H5 attacked the acidic carbon of the ester and that's different from Claisen rxn I thought it was.

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[jbrenn4]

Correct me if i'm wrong but i think you need heat to remove the OR group and heat is not given in the reaction that might explain the final product

 

[pbure9]

Correct me if i'm wrong but i think you need heat to remove the OR group and heat is not given in the reaction that might explain the final product

it looks like what you are starting with is already a product of a claisen reaction (a beta-keto-ester). In a claisen its an ester with a base like OC2H5-.

Here you have a double alpha, that just gets pulled off and then you're CH3CH2 joins in.

 

[ethanyoo726]

Ok so compound that they give you....Carbon #3 is between two carbonyl groups, making the Hydrogen attached to Carbon #3 REALLY acidic, making it really easy to pull off.

So for the first step of the reaction, -OC2H5, removes that Hydrogen, forming a carbocation on the Carbon 3. Then the CH3CH2Br comes in and reacts with the compound to give you your product.

 

[nze82]

Not sure, what the name of the reaction is...
But here is how it will proceed:

1-Notice that the the H atoms connected to the CH2 located between the two carbonyl groups are extremely acidic, due to the presence of the carbonyl groups. NaOC2H5 is a strong base, and it will abstract one of these protons, leaving a negative charge on CH.

2-The negatively charged CH is a strong nucleophile, and it will attack CH3CH2Br via SN2 mechanism.

The end product will look like:

CH3-(C=O)-CH-CH3CH2-(C=O)-OC2H5

Hope this helped!