[Heli0x]

Hi,

I'm having trouble figuring out the next step in my reaction mechanism, aiming towards creating a 1,9-Diphenyl-nona-1,3,6,8-tetraen-5-on. (Dicinnamalacetone)

The notes I have from class suggest that the next step would be a reaction with water, but I'm unsure as to the possibility of applying that on this particular reaction.

My reaction so far:



Cheers

[MissPhosgene]

Hello, the mechanism written is incorrect. Instead, try deprotonating acetone and drawing arrows for reaction of the resultant enolate with cinnamaldehyde followed by elimination.

[cupid.callin]

MissPhosgene is right.

Deprotonate acetone from both sides and then they will attack 2 benzene substituted compounds ... AND YOU HAVE THE REQUIRED PRODUCT !!
 

@ MissPhosgene

I know by seeing the product that acetone is attacked ... but i have no reason that why it is attacked ... can you explain why is it attacked?

[Heli0x]

Does that mean that the alfa-carbon(s) is in fact on the Acetone, and not on the Cinnamaldehyde?

[MissPhosgene]

MissPhosgene is right.

Deprotonate acetone from both sides and then they will attack 2 benzene substituted compounds ... AND YOU HAVE THE REQUIRED PRODUCT !!
 

@ MissPhosgene

I know by seeing the product that acetone is attacked ... but i have no reason that why it is attacked ... can you explain why is it attacked?

Does that mean that the alfa-carbon(s) is in fact on the Acetone, and not on the Cinnamaldehyde?

Looking at the product, you can see that the carbonyl corresponding to acetone is extant. If cinnamaldehyde were deprotonated at the a-position, the product would look completely different.

First, cinnamaldehyde is drawn incorrectly in the scheme. It is supposed to have a double bond between the a- and b-positions to the aldehyde. In this case, you can generate the enolate of acetone in the presence of the aldehyde, because the a-position to the aldehyde is sp2. You won't generate the dianion of acetone.

Secondly, if there were sp3 centers a-to the aldehyde, you could generate the enolate first and then add it to a solution of the aldehyde. Isolate the elimination product, and then generate the enolate and add it to another solution of the aldehyde.

 

[cupid.callin]

@ MissPhosgene

you didn't answer my question. Or maybe you did but i didn't understood!!!

Like in this rxn:

CH3-CHO + CH3-CO-CH3 ---> CH3-CH(OH)-CH2-CO-CH3

Both the carbons have sp3 a-H but the one with Acetone is deprotonated.

Please Answer!!!

[MissPhosgene]

Hello, No, I didn't understand your question. I can't find pKa values for acetone and acetaldehyde in the same solvent so this is based off of 1HNMR chemical shifts for the methyl group protons (SDBS, acetone = 2.162, acetaldehyde = 2.206).

Both would get deprotonated if they were thrown in a flask together with some base. The ratio would depend on the base being used... LDA would fully deprotonate both, but something  like pyridine would result in the enolized species being enriched in the enolate of acetaldehyde because, as can be seen from the NMR data, the methyl protons are very slightly more deshielded and therefore a bit more acidic.

This can be understood by making use of the inductive effect. Methyl has a larger I+ effect than H.

[MissPhosgene]

Another correction to the drawing: the hydrogen labeled as alpha is a to the aldehyde carbon. However, when discussing enolization, the appropriate atom to label as aplha is a carbon atom adjacent to the carbonyl carbon atom.

[Doc Oc]

There are some egregious errors in your mechanism, I strongly advise you to go back and study the chapter on aldol condensations (Ms. Phosgene has noted most of the problems).

Specifically, draw the correct structure of cinnamaldehyde and then read the section on Michael reactions.  You won't get the product you're trying to get with cinnamaldehyde and acetone.

[MissPhosgene]

I think you do get this with cinnamaldehyde. I did a lab like this when I had ochem lab class.

[Heli0x]

Ok, thank you very much!

The product I got was in fact the one within the box in the first post, Dicinnamalacetone. I did this in a lab.

[Heli0x]

After browsing and reading through my notes and handouts for this particular lab, it came to me that perhaps this could be a Cross-Coupled (Claisen-Schmidt) type of reaction.

I do have an aromatic aldehyde, and a keton (acetone).

My reworked reaction:



Now, the thing is I'm having some trouble drawing the actual product from this reaction.
The general idea seems to be a repetative reaction here, if I have understood correctly that is?
The first product from the above reaction should be somewhat near half of the desired molecule, then the reaction repeats whereupon another Cinnamaldehyde-molecule bonds to the 2nd (deprotonated alfa-H) position.

What happens with the two aldehyde-oxygens? Is the loss of this considered the actual "condensation", or does that refer to the first formation of H2O in step one?

[Doc Oc]

That's MUCH better, good work.  Loss of H₂O is the condensation step, otherwise it'd be an aldol addition.

I'm surprised that this doesn't yield Michael addition products, but I haven't done an aldol on this substrate either so I'll have to trust that it doesn't.

[Heli0x]

Oh ok, I see.

Thank you for your help and encouraging words.