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Author: Subject: 2,5-Dibromo-1,4-benzoquinone Reactivity?
APO
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[*] posted on 22-11-2013 at 11:20


That has nothing on reactivity of non-nucleophilic bases with bromoquinones, so I don't really think that's relevant, but thanks, anyways.

Again Vulture, If you don't think aprotic solvents, DABCO, or TEA are suitable for any of this, do you have any ideas?




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[*] posted on 22-11-2013 at 11:42


No it does not have anything on the reactivity of non-nucleophilic bases per say, just simply supplying the reference as requested, that certain amines are known to add to the 2 and 5 positions of quinones coupled with a reduction of the quinone and then reoxidation by unsubstituted quinones. It's not so easy to find... I had no luck when googling it, for example. I promised before that I would give it a better go after my hectic week was over and so now I could :) I had to pull it from an old presentation of mine and then find some online source for it.

[Edited on 22-11-2013 by deltaH]




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[*] posted on 22-11-2013 at 20:33


Oh, I see.

Also I feel it is important to note that specifically in this case, from what I've read, ketal derivatives do not work, and neither do 2,2,5 bromine groupings.

From what I know, the only possible brominated derivative of Cyclopentanone that will work for this reaction is the ketone form with 2,3,4 bromine groupings, so using 2,2,5-Tribromocyclopentanone or it's Ethylene Ketal isn't possible.

So, even though there are other derivatives that seem like they would work, it apears that only one will do the job.

Additionally, on the reaction scheme I posted the last step is actually two, hv, and then NaOH, not at the same time.

Below is the revised, more comprehensive version, including steps for 2-Cyclopenten-1-one, as I will have to make it.



So, any suggestions on what will work for the DABCO step if this doesn't seem likely to work?

DeltaH thinks it has good probability, but Vulture, not so much.


[Edited on 23-11-2013 by APO]




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[*] posted on 23-11-2013 at 00:32


Are you following published steps up to the cyclopentenone or proposing? Do you know the conditional details and yields?

[Edited on 23-11-2013 by deltaH]




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[*] posted on 23-11-2013 at 03:33


In the second step, are you sure you will only selectively epoxidize one double bond?

I don't understand where the other isomer of the dialcohol comes from.

How do you convert the dialcohol to the ketone with p-TsOH?

I have no idea if DABCO will work or not.




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[*] posted on 23-11-2013 at 11:06



DeltaH, all steps up to 2,3,4-Tribromocyclopentanone have been successfully done before, as you know, the part with 2,5-Dibromo-1,4-benzoquinone, DABCO, and Diethyl Ether is the sketchy part. The total yield up to the mixture of Cyclopentendiols is 70%, and the yield of conversion to 2-Cyclopenten-1-one is 60%, so the total yield up 2-Cyclopenten-1-one is 42%. Step one is carried out at 270C, step two at 15C, step three at 5C, and step four at 60C while under vacuum. However, I don't feel the need to go into detail right now.

Vulture, the yield of 1,2-Epoxycyclopentene is nearly quantintative, so I think that it's favored for some reason. I'm not sure why the hydrolysis of the 1,2-Epoxycyclopentene gives two isomers though. I believe the conversion of the Cyclopentendiols to 2-Cyclopenten-1-one is an acid catalysed cycloaddition of some sort.



[Edited on 23-11-2013 by APO]




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[*] posted on 23-11-2013 at 12:05


...and that Vulture... is why you ask "Are you following published steps up to the cyclopentenone or proposing? Do you know the conditional details and yields?" first :P

Thanks APO, the reason I asked is because the same concerns occured to me as did to Vulture, except the 1,3 product made sense to me, but I was also worried about the selectivity of the partial oxidation and acid catalysed dehydration.




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[*] posted on 23-11-2013 at 12:37


You're welcome.

Speaking of conditional details, do you have an idea of what temperature the DABCO step would preferably be conducted at?

Also any idea of work up/purification?




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[*] posted on 23-11-2013 at 14:10


Answered by U2U.



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[*] posted on 23-11-2013 at 16:15


Thanks.



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