[Kate]

Hello.

How do I do these synthesis?

1-

So here I thought about using LDA to create the enolate from propanal and then use Cl-CO-CH₃. But I'm quite sure this is wrong, because aldehydes tend to condensate with themselves, right?

2-

In this one I have no idea how to start. I have to break the cyclohexene ring somehow, but I don't know how.

I'd appreciate any help. 

[Archer]

Hello.

How do I do these synthesis?

1-

So here I thought about using LDA to create the enolate from propanal and then use Cl-CO-CH₃. But I'm quite sure this is wrong, because aldehydes tend to condensate with themselves, right?

If you add LDA to propanal you quite rightly pointed out that you form the enolate.

How could an enolate then react with the aldehyde? There shouldn't be any aldehyde present if it has all reacted with the LDA.

[Altered State]

Btw, for these kind of reactions using aldehydes you can consider the posibility of forming an enamine first:

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

[kriggy]

If you add LDA to propanal you quite rightly pointed out that you form the enolate.

How could an enolate then react with the aldehyde? There shouldn't be any aldehyde present if it has all reacted with the LDA.

Could you control this by adding stechiometric amount of base to deprotonate propanal and then react it with acetaldehyde?

[Kate]

If you add LDA to propanal you quite rightly pointed out that you form the enolate.

How could an enolate then react with the aldehyde? There shouldn't be any aldehyde present if it has all reacted with the LDA.

Well, I thought that the first molecules of enolate that form could condensate with the remaining aldehyde. That was my worry. But since LDA is a strong and non-catalytic base, the enolate formation is probably too fast for that to happen.

Is the second part of the 1st reaction alright (using methyl chloroformate)?

Btw, for these kind of reactions using aldehydes you can consider the posibility of forming an enamine first:

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

Thanks. I thought I could only use a ketone when forming an enamine. 

[Archer]

Is the second part of the 1st reaction alright (using methyl chloroformate)?

No, you have drawn acetyl chloride

[Kate]

No, you have drawn acetyl chloride

Yes, sorry.

But I can use acetyl chloride to get the end product, right?

[Archer]

But I can use acetyl chloride to get the end product, right?

I don't see any reason why this wouldn't work.

[Kate]

Ideas for the 2nd synthesis?

I don't see any reason why this wouldn't work.

Ok, thanks.

[spirochete]

For the second one you can start with ozonolysis.

[Archer]

Ideas for the 2nd synthesis?

I was deliberately avoiding this one as I was at a bit of a loss because I had missed a vital difference between the starting material and the product and was barking up the wrong tree looking for some kind of rearrangement reaction, however...

For the second one you can start with ozonolysis.

..is quite clearly the way to go.

Start by sequentially numbering the carbon atoms in the starting material and then numbering the atoms in the product.

As you go through each step keep the numbering system of the starting material.

On the last step to get your product you will undoubtedly need to change your original numbering of the product, but I find this is a good technique to follow what it happening to each carbon throughout the synthesis.

[Kate]

Ok. So, first I use ozonolysis to get 6-oxo-heptanal. Then I use NaOH to form the thermodynamic enolate. This enolate then self-condensates at the aldehyde carbonyl group. Then with acid I get an aldol and with heat this aldol dehydrates and I get an alpha-beta insaturated carbonyl. Then I add CH3MgBr and after acidic work-up I get the final product.

Is this correct?

[Archer]

Yes, what type of ozonolysis work up would produce this, oxidative or reductive?

I had missed the fact that there was an extra carbon in the product. This is why numbering is helpful as it indicates that it is not simple rearrangement.

[Kate]

Reductive ozonolysis, with Zn in AcOH or with Me₂S.

I had never even heard of oxidative ozonolysis before, it's not even in my textbook so I googled it and it gives carboxylic acids instead of aldehydes.

Yeah, I always number carbons, it's so easy to get lost if I don't. 

[Archer]

I had never even heard of oxidative ozonolysis before, it's not even in my textbook so I googled it and it gives carboxylic acids instead of aldehydes.

Ozonolysis is an oxidative technique, it is the work up which is oxidative or reductive, but yes oxidative gives the acid rather than the aldehyde.