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Author: Subject: Diels-Alder Reaction THEN Intramolecular Reaction
Magelia
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[*] posted on 22-11-2011 at 16:24
Diels-Alder Reaction THEN Intramolecular Reaction


Hello Colleagues,

So, I am faced with a dilemma. I did a diels-alder reaction between maleic anhydride and trans,trans,2-4-hexadiene-1-ol and I was expected to form the normal diels-alder compound that is seen when those two compounds react (Diagram shown below). However, I knew the OH could do an intramolecular reaction therefore the ACTUAL product of my reaction was the cyclic ester... the anhydride moiety was gone.

So, to confirm that the intramolecular reaction did happen... I reacted my product of maleic anhydride and trans,trans,2-4-hexadiene-1-ol with potassium carbonate and ethyl iodide with 2-butanone as a solvent and I got the spectrum shown below.

The thing is, HOW does the reaction I did with ethyl iodide and potassium explain that the product I obtained from the reaction of maleic anhydride and trans,trans,2-4-hexadiene-1-ol is not the "expected" product but the product AFTER the intramolecular nucleophilic attack of the OH at the carbonyl forming a cyclic ester? How do I explain that my reaction conditions of K2CO3/EtI PROVES/DEMONSTRATES that I did indeed get the cyclic ester product!

Any help would be very very appreciated! :)

Actual_Product.png - 21kB

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DJF90
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[*] posted on 23-11-2011 at 00:27


Try the ethylation conditions with maleic or succinic anhydride - it shouldnt work, only for free acids. If it doesn't work, then you must have had a free acid in your product, pointing to intramolecular attack. You could also show your product is not the expected anhydride by MS - there should be a mass difference between the hydroxy-anhydride and the lactone-acid.
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ziqquratu
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[*] posted on 23-11-2011 at 01:19


The NMR is proof enough - for the expected product, with the free alcohol and intact anhydride, the CH2OH protons will be MUCH further upfield than the corresponding protons in the actual product (my estimate is 3.5 ppm for the CH2 in the alcohol, vs 4-ish ppm for the CH2 in the ester).

As far as the ethylation goes, the same argument applies. Ethylation of the free alcohol of the expected product would give an ether, with peaks for both methylenes at around 3.5 ppm; methylation of the ester product would result in the formation of a diester, with two sets of CH2 peaks around 4 ppm.
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fledarmus
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[*] posted on 23-11-2011 at 04:41


I think the point that you are trying to demonstrate is that there is no free alcohol in your product. If there was an alcohol, it would have made an ether under these conditions, as Zigguratu pointed out.
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Magelia
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[*] posted on 23-11-2011 at 06:02


Guys... YOU DA BEST!

Great explanation. Clear and concise and to the point! :)
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Nicodem
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[*] posted on 23-11-2011 at 09:04


Quote: Originally posted by Magelia  
So, I am faced with a dilemma. I did a diels-alder reaction between maleic anhydride and trans,trans,2-4-hexadiene-1-ol and I was expected to form the normal diels-alder compound that is seen when those two compounds react (Diagram shown below). However, I knew the OH could do an intramolecular reaction therefore the ACTUAL product of my reaction was the cyclic ester... the anhydride moiety was gone.

Why do you think the acylation is intramolecular and that it occurs after the Diels-Alder? You provided no experimental details, but I would guess you heated the reaction mixture somewhere in between 120 and 200 °C, which is enough for the maleic anhydride to acylate the dienyl alcohol even before the Diels-Alder reaction starts. It is thus just as possible that it is the Diels-Alder that is intramolecular, rather than the acylation. The rate of intramolecular Diels-Alders is much faster than the intermolecular ones (when geometry allows).

Quote:
So, to confirm that the intramolecular reaction did happen... I reacted my product of maleic anhydride and trans,trans,2-4-hexadiene-1-ol with potassium carbonate and ethyl iodide with 2-butanone as a solvent and I got the spectrum shown below.

You might want to elaborate your thought on how this ethylation is supposed to demonstrate anything. Even if you obtained the so called "expected product" it would rapidly rearrange to the "actual product" under K2CO3 treatment. The ethylation reaction products from both would therefore be the same!

Quote:
The thing is, HOW does the reaction I did with ethyl iodide and potassium explain that the product I obtained from the reaction of maleic anhydride and trans,trans,2-4-hexadiene-1-ol is not the "expected" product but the product AFTER the intramolecular nucleophilic attack of the OH at the carbonyl forming a cyclic ester? How do I explain that my reaction conditions of K2CO3/EtI PROVES/DEMONSTRATES that I did indeed get the cyclic ester product!

Unfortunately, due to the mentioned reason, it can demonstrate absolutely nothing.
Like ziqquratu said, the 1H NMR spectra of the product would be enough, but you did not provide that spectra, so it is not up to us to draw conclusions.
Quote: Originally posted by DJF90  
Try the ethylation conditions with maleic or succinic anhydride - it shouldnt work, only for free acids. If it doesn't work, then you must have had a free acid in your product, pointing to intramolecular attack. You could also show your product is not the expected anhydride by MS - there should be a mass difference between the hydroxy-anhydride and the lactone-acid.

The anhydride and the lactone would give the same HRMS result!
Also, maleic anhydride can not rearrange to a lactone under the same ethylation conditions (it has no alcohol group), so the absence of a reaction would prove nothing.
Quote: Originally posted by fledarmus  
I think the point that you are trying to demonstrate is that there is no free alcohol in your product. If there was an alcohol, it would have made an ether under these conditions, as Zigguratu pointed out.

If there was a free alcohol as in the "expected product", then this compound would rearrange into the lactone (see the cis configuration dictating a rearrangement), the carboxylate of which would then be ethylated to the same ester.
Also, with K2CO3 you can have a chemoselective O-alkylation of the carboxylate and no O-alkylation of the alcohol function (the absence of the ethyl ether therefore demonstrates nothing). It depends on the reaction solvent, time and temperature, but it is not easy to O-alkylate alcohols using K2CO3 as a base.
Quote: Originally posted by Magelia  
Guys... YOU DA BEST!

Great explanation. Clear and concise and to the point! :)

No, we all lousy, the explanation was no good, totally missed the point, and you would gain a lot by learning some critical thinking. Otherwise, welcome to the forum.




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Magelia
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[*] posted on 23-11-2011 at 11:05


Well Nicoderm, I forgot to include some important experiment details.

I first did my Diels-Alder reaction between maleic anhydride and trans,trans,2-4-hexadiene-1-ol and took its IR and H-NMR Spectrum. I then went home, watched some soccer on TV, and went to bed.

The next day I went back to the lab and did the ethylation on my Diels-Alder product and got its H-NMR Spectrum that I had posted.

Therefore, the earlier mentioned points by the other authors did indeed make a lot of sense and were correct.

Thank you though for your answers, they were well said.

Of course, thank you for welcoming me to the forum :).
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Nicodem
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[*] posted on 24-11-2011 at 10:21


Well, how about sharing the information? You get us interested and then you don't tell us the conclusion you got from the NMR. In the context of my cultural background, that is quite an impolite behavior.
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[*] posted on 24-11-2011 at 13:27


Interesting: The original poster is unable to look up answers to questions like https://www.sciencemadness.org/whisper/viewthread.php?tid=17... in a book, but has access to weird dienes and time on the NMR spectrometer?

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Magelia
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[*] posted on 24-11-2011 at 13:54


Quote: Originally posted by Nicodem  
Well, how about sharing the information? You get us interested and then you don't tell us the conclusion you got from the NMR. In the context of my cultural background, that is quite an impolite behavior.


Sorry Nicodem, I am still learning. I'll try to do my best next time I post/say something.

Quote: Originally posted by turd  
Interesting: The original poster is unable to look up answers to questions like https://www.sciencemadness.org/whisper/viewthread.php?tid=17... in a book, but has access to weird dienes and time on the NMR spectrometer?



Thanks for the investigative work, and trying to put me down. Appreciate it.
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Nicodem
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[*] posted on 25-11-2011 at 12:29


Quote: Originally posted by Magelia  
Sorry Nicodem, I am still learning. I'll try to do my best next time I post/say something.

I just wanted to know whether the NMR confirmed the anhydride-type or the lactone-type of the product? I find it hard to deduce anything in this regard from what you wrote up to now. I guess you found out it was the lactone?
If you refuse to give an answer because you are still unsure, then I suggest you to post the spectra and we'll determine what it is together.
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Magelia
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[*] posted on 25-11-2011 at 13:11


Quote: Originally posted by Nicodem  
Quote: Originally posted by Magelia  
Sorry Nicodem, I am still learning. I'll try to do my best next time I post/say something.

I just wanted to know whether the NMR confirmed the anhydride-type or the lactone-type of the product? I find it hard to deduce anything in this regard from what you wrote up to now. I guess you found out it was the lactone?
If you refuse to give an answer because you are still unsure, then I suggest you to post the spectra and we'll determine what it is together.


Well, again I forgot to mention I got the IR spectrum of my product as well :( So based on that it was pretty easy to see what type of carbonyl I had :)
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