Chemical Forums
Chemistry Forums for Students => Organic Chemistry Forum => Topic started by: xshadow on May 15, 2021, 05:14:35 AM
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Have to write all the products of the following steps:
(https://i.imgur.com/zawWY5h.jpg)
I 've done the first three steps but I'm stuck on the last two!!
Is the piperidine here used to deprotonate the 1,3 decarboxylic compound and generates the enolate? Or will form an imine reacting with the keto group?
And then where willthe enolate attack?
I have an esteric and a carboxylic group on my substrate, if the previous steps are correct
Thanks...
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In the second scheme, the lactone will not open, the ringclosed compound is thermodynamicaly more stable then the open methyl ester.
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In the top scheme you use allylbromide but get a only a vinyl-substituent, you miss one carbon.
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In the second scheme, the lactone will not open, the ringclosed compound is thermodynamicaly more stable then the open methyl ester.
So what will the NaOMe can do in that step?
If it can't react with che C=O group it could act as a base and deprotonates the alpha hydrogen getting an enolate??
But anyway that enolate will attack the lactone and the only possible pathway should be again the ring openong?!?!
Thx
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In the top scheme you use allylbromide but get a only a vinyl-substituent, you miss one carbon.
Right!! :o
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Nothing? ???
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I am not sure what can hapen if you treat the allyl-lactone with NaOMe, it should not open, that would be thermodynamically unfavourable. The double bond can migrate so you get a methyl vinyl but generally this is accomplished with a stronger base like potssiumtert-butoxide or LDA. One possibility is that you get a little ester enolate and this attacks another molecule of lactone but then things get very complicated.
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i think I've done a mistake in the IBX step because IBX is a mild oxidizing agent so I should get an aldehyde.
Than in the last step the piperidine is a base and can deprotonate che other reactant ( the1,3 dicarboxylic compound, very acid , I think piperidine is good enough)
So I get an enolate that should react chemoselectively with the aldehydic group of the substrate (aldehyde much more reactive than ester)
Can it make sense??
thanks:)
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I am not sure what can hapen if you treat the allyl-lactone with NaOMe, it should not open, that would be thermodynamically unfavourable. The double bond can migrate so you get a methyl vinyl but generally this is accomplished with a stronger base like potssiumtert-butoxide or LDA. One possibility is that you get a little ester enolate and this attacks another molecule of lactone but then things get very complicated.
Perhaps the IBX is added at the "same time" so when the lactone is opened , the alcoholic group is oxidized very quickly to an aldehyde ( and so we can't have the ring closure anymore)
Than in the last step I suppose what could be happen in the post above this.
According to you???
THANKS) :)
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The lactone would not open at all and IBX would react with methoxide. Its a bit of a mystery, is something not correct in the structure of the startingmaterial? Or does the creater of the exercise thinking that the lactone would open incorrectly?
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The lactone would not open at all and IBX would react with methoxide. Its a bit of a mystery, is something not correct in the structure of the startingmaterial? Or does the creater of the exercise thinking that the lactone would open incorrectly?
I would guess that the creator of the exercise didn't want them to think too hard about it, and we are supposed to assume that the transesterification to open the ring would actually work. I wonder if a large excess methanol might allow the equilibrium to favor ring opening? Or if saponification first might work much better, followed by a different way of making the ester.
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If so its a fatal error, I have worked with lactones and hydroxy esters, its very easy to close this type of ring with both acid and base. It is a equilibrium but very little ringopened hydroxy ester would form, not practically amounts. Its just ignorant.
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I wondered about this issue of opening a lactone. I did find a couple of papers where it seems to work fine to do acid catalyzed transesterification to open a stable lactone like 5 or 6 membered ring:
"A Convenient and Mild Procedure for the Preparation of Hydroxyesters from Lactones and Hydroxyacids"
"Transesterification via Baeyer–Villiger oxidation utilizing potassium peroxydisulfate (K2S2O8) in acidic media"
The latter does a Bayer-Villiger first to make the lactone, but I don't think would effect the equilibrium for opening the lactone.
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If so its a fatal error, I have worked with lactones and hydroxy esters, its very easy to close this type of ring with both acid and base. It is a equilibrium but very little ringopened hydroxy ester would form, not practically amounts. Its just ignorant.
Introductory organic chemistry often doesn't care a lot about yield when it comes to theoretical exercises. The student would get bonus points for pointing something like that out, though.
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I looked up the paper, it seems its possible with acid to get the hydroxy ester. I was wrong, sorry. Remains to see if I am wrong regarding basic catalysis. That would go against my own experience in the lab. If we assume its possible, what would the rest of the steps lead to? You need to draw the allyl-substituted compound and proceed.
I have worked with a butyrolakton, to open this we used ammonia to get the 4-hydroxy amide. We protected the hydroxy group and treated the amide with nitrogen dioxide to get the nitroso amine. This was hydrolyzed with lithium peroxide to get the avid. So this is my point of reference, it was very difficult to get the open-chain compound in this case without going via the amide. This was 25 years ago, maybe my memory is failing in some way.
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To Xshadow: The IBX should be making an aldehyde, not a carboxylic acid probably. Then, knoevenagle condensation followed by hydrolysis of both esters followed by decarboxylation of the beta-acid group. Edit: Also, you lost a carbon following the Sn2 adding the allyl group.
To Rolnor: in Clayden, et al (an awesome intermediate organic chem textbook, pages 617-618), a delta-lactone with two alpha hydrogens in shown doing a self-aldol process with a generic base. Not claisen, but aldol! If there is one alpha hydrogen, like in this example, I would imagine opening the lactone via transesterification should be fine because aldol condensation is not possible. But anybody in a basic introductory organic class should NEVER propose something like this with a lactone. There is a chance it will confuse and anger the professor. The proposal of ring opening via transesterification seems very common.
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A bit tricky!!
Thanks guys!! :)
-So IBX oxidize the alcohol to an aldehyde.
-In the second last step the base deprotonates the 1,3 dicarboxylic compound and I have an aldolic condensation on the aldehydic group of my substrate (much more reactive than the esteric group).
But in the final step what happens?
I have HCl + heating
I suppose two things can happen!!!
1) HCl + H2O: acid hydrolysis of the esteric group to give a carboxylic acid
2) Cl- 1,4 nucleophile addition to the michael acceptor generated before with the aldolic condensation
Can 1) and 2) both happen?
Thanks :)))
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Definitely don't be adding Cl- via conjugate addition. This type of product is very unstable and would only form with a full equivalent of HCl or more, and when there is no other reasonable thing that could be happening. If it does happen, the heat will cause the HCl to fall back off. HCl is very common as an acid catalyst. The chloride shouldn't be adding to anything.
H3O+ and heat will hydrolyze all esters to carboxylic acids, and then I think heat will cause one of the carboxylic acids to decarboxylate. If you are unfamiliar with decarboxylation, read about acetoacetic ester synthesis, or decarboxylation of carboxylic acids in general.
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Definitely don't be adding Cl- via conjugate addition. This type of product is very unstable and would only form with a full equivalent of HCl or more, and when there is no other reasonable thing that could be happening. If it does happen, the heat will cause the HCl to fall back off. HCl is very common as an acid catalyst. The chloride shouldn't be adding to anything.
H3O+ and heat will hydrolyze all esters to carboxylic acids, and then I think heat will cause one of the carboxylic acids to decarboxylate. If you are unfamiliar with decarboxylation, read about acetoacetic ester synthesis, or decarboxylation of carboxylic acids in general.
Oh yes!!
There is also decarboxylation
But after decarboxylation and ester hydrolysis I get an alpha-beta unsaturated aldehyde.
Can't (after those reactuonsl Cl- react with the C=C-CHO giving 1,4 product?
Because in my textbook I have an example of HCl (Cl-) that gives a 1,4 product (Cl- attacks the C=C double bond)with a α-β unsaturated ketone.
Can't have also that reaction , after all the other one?
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It's a ketone, not an aldehyde. But either way, a very good leaving group like a halogen at the beta position is very unstable. In heat, it should easily just do E1CB to eliminate HCl. Textbooks sometimes show examples of reactions that give fairly unstable products that will only work in a limited number of simple cases. I think addition of H-X to a conjugated enone falls into this category. In this example it is especially unreasonable because it is hydrolysis conditions, where there is a lot of water. Why not make a beta hydroxy? Beta hydroxy is also wrong though. With heat especially, the enone is favored.
I suppose if you suggest this answer your professor might accept it, based on your argument, but I don't think it's correct. Even more so because the decarboxylation requires heat.