[Svenm14]

I need help with this synthesis problem.  I can kind of see how the parts work, but im not sure how to put them all together.  The resitrcitions are in the picture below.  I see why the enol is prefered over the ketone, i see that part of it can be made by a claisen condesation of the two esters...i am just not sure how to put this all together.  any help would be apperciated.



Edit: edited title for better indexing. Mitch

[GCT]

I'll give you some hints.  The intermediate is an acyl chloride CH3COCl.  Why?  Well as you may see the reaction consist of three molecules.  The left end terminal carbonyl as is evident has been through a nucleophilic addition and then eliminatino of the hydroxyl to form the alkene by the claisen condensation through CH3COCl.  However, we see that the carbonyl group belonging to the latter molecule has kept its carbonyl character meaning that it has gone through an sequential addition-elimination reaction, where the nucleophile was is the molecule motif at the right terminal; Cl is the leaving group.

[Svenm14]

Can someone clarify what was just said.  It kind of confused me.

[Mitch]

Do you understand there is more carbons in the products than in the reactants?

[Svenm14]

yeah, it looks to me as if there are to equivs of the ester and one equic of the pyridine carboxylic acid.

[Mitch]

You would end up with too many carbons that way?

[GCT]

Alright...there are three reactants

The carboxlyic acid, a acyl chloride CH3COCl (the reagent that is not listed), and the ester.

The acyl chloride reacts with the acid through a claisen condensation.  The ester reacts acts as the nucleophile adding to the carbonyl carbon of the acyl chloride and the chloride subsequently leaves.  There you have it, the final compound.  Of course you will have to figure out any other side reagents that are needed; just ask if you need help with this also.

[Svenm14]

I really appreciate all of the help.  There is nothing in my book about how acid chlorides react with caboxylic acids in the claisen condensation.  Is there another way this can be done, perhaps by another intermediate?

[Svenm14]

This is something I came up with.  I have no idea if it is right.  If it is, I was wondering how can you enolize the ketone nearest the pyridine preferentially?  I guess foremost, can someone tell me if this synthesis is plausible.

Thanks.

[GCT]

You're making things more complicating.  The acyl chloride reacts through the claisen condensation with the acid, since it has the alpha carbonyl alkyl group, this can be deprotonated with a base and serve as the nucleophile.  Be sure to remember that this acyl chloride is the reagent that is not listed in your original problem but suffices for the requirement of a two carbon compound.  Be sure to draw it out correctly.

The acyl chloride with the ester does not react through a claisen condensation, it is not technically called the claisen condensation although it has similar mechanisms.  The difference is in the elimination aspect, the chloride is eliminated after the deprotonated alkly group attacks the carbonyl carbon of the acyl chloride, subsequent rearrangement eliminates the chlorine.  You should know that this is a addition elimination reaction of carbonyl compounds.  Review transacylations in your organic chemistry book.

Hope this helps.

[Svenm14]

Actually, I think both of us were making it more complicated than it really is.

it can be realized through a series of claisen condensation of esters  by the following:

[GCT]

I am pretty sure that the teacher/professor wanted a specific reaction in this case.  Your reaction takes too four as opposed to two that I proposed; mine is more practical and less expensive.  Also it is a wretched mess.  There are so many problems with it even without acknowledging the complete inefficiency.  I don't mean to sound condenscending however this is the way to do organic chemistry...one does not want to be wasteful.  

For your first step.  This reaction won't work at least not appreciably.  The carbonyl group of the acid is not very electrophilic and thus a strong base needs to be used.  Other problems occur if you were to deprotonate the alcohol with sodium (or any other method) since the resulting base will react with the hydrogen of the acid; think conjugate base and acid...which is the stronger pair products or reactant set?  Many other problems apply just with this step and I will point them out for you one by one if you wish.

Step 2...why?  Keep in mind that this would not be a claisen condensation.  It is a nucleophilic-elimination reaction.  The reactants are too bulky and elimination will be difficult to carry out since it involves a transition to a tetrahedral molecule.

Step 3.  alright, where can I begin.  I have already pointed out the fundamental mistake to this reaction in the above paragraph.  If you use a base, the methylene group is the most acidic (the hydrogen between the two carbonyl groups) this will probably act as a nucleophile.  Also decarboxylation can occur.  Also by this point you are using too many molecules.  Many combinations of reaction can occur.  

I will stop here.  What you should learn from this is that chemistry is difficult; one does not just jumble molecules together and call it a reaction.  There is so many room for error.

[Svenm14]

How about then, you show me how it works with a schematic.

[GCT]

I don't have a chemical drawing software and I tried downloading one (chemsketch) unsuccessfully.  What I said is basic chemistry; it isn't an ingenious idea on my part.  

-React the acyl chloride with the acid through a claisen.

-React this compound with the ester.

You should know which reagents to use.

[AWK]

I don't have a chemical drawing software and I tried downloading one (chemsketch) unsuccessfully.  What I said is basic chemistry; it isn't an ingenious idea on my part.  
Isis/Draw is also free. Start from free registering at: http://www.mdli.com/downloads/search_by_si/index.jsp