[Hazel]

I have no idea of this question.
Can someone give me some hints?I want to try it by myself anywhere.
Thank you.

[Dan]

(a) diisobutylaluminium hydride (aka DIBAL, DIBAL-H, DIBAH) is an important reducing agent to be aware of, see:

http://en.wikipedia.org/wiki/Dibal

for an introduction.

(b) The Claisen condensation is a classic exam paper reaction that is well worth familiarising yourself with, see:

http://en.wikipedia.org/wiki/Claisen_condensation

for an introduction.

[Hazel]

Thanks Dan,
I have studied the links and get the following solutions,but there are some points that I still doubt with.

(1) PhCH2COOC2H5  +  H-  --->  PhCh2COH  +  -OC2H5  ---(+H)--> PhCh2COH  +  C2H5OH

The hydride(Strong base) will attack the alpha H and become H2.Electron will attack the carbonyl carbon(Since there is no other electrophilic molecule) and cause the elimination of -OC2H5.


(2) CH3COOC2H5  +  CH3CH2OH   ---(C2H5O-)-->  CH3CH2CH2COOC2H5  +  -OH   --(+H)--> H20

The C2H5O-(act as base) attack the alpha hydrogen of CH3COOC2H5.Then the CH2(-)COOC2H5 will attack hydroxyl C of CH3CH2OH and give out -OH.

CH2(-)COOC2H5 will resonance and has a conjugation of CH2CO(-)OC2H5.Why CH2(-)COOC2H5 will be major in the reaction following.

Please point out if I have any mistake.Thanks!

[lutesium]

Yes you're totally mistaken at the 2nd reaction.

Adding NaOEt into an active carbonyl compound will just result in a Na salt of the Acetic Ester (now he's mostly called Ethyl Acetate) which can be condensed with another molecule of Acetic Ester to give Aceto Acetic Ester and its not the -OH that seperates at the condensations of such. Its called deprotonation so its the H that seperates so you're totally mistaken by telling that its the -OH that seperates. That H goes to the OC2H5- of the other molecule of Acetic Ester (whose carbonyl group has been attached to the first Acetic Ester's alpha carbon thus giving Aceto Acetic Ester) creating a moecule of EtOH.


Hope this helps


Lutesium...

[Dan]

The hydride(Strong base) will attack the alpha H and become H2.Electron will attack the carbonyl carbon(Since there is no other electrophilic molecule) and cause the elimination of -OC2H5.

Your products are correct, but your reasoning is questionable.
You do not have free hydride here, DIBAL is not a source of basic hydride (whereas NaH is, for example). DIBAL itsself is a lewis acidic reducing agent.
Collapse of the tetrahedral intermediate, eliminating ethanol, occurs during workup, not directly following reduction - this is the root of the reason why DIBAL will normally reduce esters only to aldehydes and not all the way to the alcohol.
You ought to look up the mechanism of this reduction, the wikipedia page is very brief and only meant as an introduction, any university level organic chemistry text will do, eg. Clayden

Your answer to 2 is wrong,

The C2H5O-(act as base) attack the alpha hydrogen of CH3COOC2H5.Then the CH2(-)COOC2H5 will attack hydroxyl C of CH3CH2OH and give out -OH.

You are correct in your generation of the enolate, but this enolate will not attack ethanol to displace hydroxide as you have described. Hydroxide is a very poor leaving group. Are there any more electrophilic compounds in your solution..?

If you read the wiki page on the Claisen condensation more carefully you will see that it contains the answer for this exact question, and the mechanism.

I strongly advise you to sit down with a good textbook and get a clearer notion of what's going on. If you're still unsure you will certainly get more help here.

[lutesium]

You're going to get Aceto Acetic Ester as the product of your 2nd reaction and a molecule of alcohol.


Lutesium...

[Hazel]

I know why I will confuse already!

May I know why we need to add ethanol in the reaction,doesn't sodium ethoxide can provide the weak base?

CH3COOC2H5  +  C2H5O-  ----->  CH2(-)COOC2H5  +  CH3COOC2H5  -----> CH3CH2COOC2H5CO(-)OC2H5  ---->  CH3COCH2COOC2H5  +  -OC2H5 ----->  CH3COCH(-)COOC2H5  +  C2H5OH  ---(H+)--> CH3COCH2COOC2H5

Therefore, the products I obtained is acetoacetic ester and ethanol.Between,why the acetoacetic ester is so named?

I am still working on the question 1. 

[lutesium]

May I know why we need to add ethanol in the reaction,doesn't sodium ethoxide can provide the weak base?

Not the weak base its the strong base. You don't need to add extra Ethanol. Its used to create the NaOEt.

Aceto Acetic Ester is so named because its the Acetylated Acetic Ester (Acetyl Ethyl Acetate).

You're right at the first reaction but its not the H- (hydride ion - Which is a very strong base and liberated from highly basic compounds such as NaH) that's liberated from DIBAL. Consider it as 2H+ s liberated from DIBAL deesterifying your molecule and creating an Aldehyde...


Lutesium...