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Offline Heory

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Share an deduction problem
« on: December 04, 2009, 10:51:22 AM »
I kept deliberating on this deduction problem for several days and finally had to give up. However, when I saw the answer, I felt it was not so complicated as I thought. A very interesting problem.

From A to K, every letter presents an organic compound, while M presents an inorganic compound. G can react with NaOH. [H] means another reductant.

Offline Ulfsaar

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Re: Share an deduction problem
« Reply #1 on: December 05, 2009, 05:29:04 AM »
That is so difficult..  is it the chemistry contest problem?i have seen some disgusting  problem in the chemistry contest like this.

Offline Dan

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Re: Share an deduction problem
« Reply #2 on: December 05, 2009, 01:32:47 PM »
Haha, this is tough, and driving me slightly crazy. Can you tell me if M is a halogen (ie. if the first reaction is a Haloform reaction)?

Also, can I assume the number of carbon atoms is conserved throughout the reaction - ie. if I do a decarboxylation does the carbon dioxide count as another organic product - or are we defining organic as a molecule with a C-C bond?
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Offline Heory

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Re: Share an deduction problem
« Reply #3 on: December 05, 2009, 09:21:55 PM »
Sorry I can only answer one question now. Definitly CO2 is not organic. Definition of organic in every textbook is fit for this problem.
If nobody has a try in several days I will give another hint. :)

Offline tmartin

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Re: Share an deduction problem
« Reply #4 on: December 06, 2009, 11:33:46 AM »
So a few initial thoughts on my part here...  this has been driving me crazy the past few days as well, so I've chosen to avoid it  :P

I'm getting an IHD of 7 for the final product, thinking a lot of rings with maybe a carbonyl or alkene?   With I--> J, the first reaction is a Clemmensen reduction, correct?  So I'm assuming [H] is some sort of reduction of a C-C multiple bond?

Offline Heory

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Re: Share an deduction problem
« Reply #5 on: December 06, 2009, 11:43:39 AM »
Yes, M is a halogen. And yes, I--->J includes Clemmensen reduction. Go on! :)

Offline Dan

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Re: Share an deduction problem
« Reply #6 on: December 07, 2009, 06:14:23 AM »
Oh no! I just realised that B->D is achieved with nitrosyl chloride, I read it as sodium hypochlorite... Back to the drawing board...
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Offline Dan

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Re: Share an deduction problem
« Reply #7 on: December 07, 2009, 07:11:01 AM »
Ok, probably a bit silly but here's an idea....

Not sure my double bond would survive I2 though...
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Offline Heory

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Re: Share an deduction problem
« Reply #8 on: December 07, 2009, 08:22:58 AM »
Dan, you are always too modest. ;D
Tumbs up. That's the anwswer.
Could explain why you placed the double bond at the gamma position of the carbonyl group?

Offline Dan

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Re: Share an deduction problem
« Reply #9 on: December 07, 2009, 10:39:30 AM »
Wow, I thought the final structure looked too strange to be the right one!

The gamma decision was based on this thought process:

1. If the double bond was alpha, the cyclopropanone I will be cleaved to give a carboxylate in the presence of hydroxide - given the Clemmensen step, I wanted to keep the ketone intact.

2. If the double bond was beta, then it would probably isomerise in the presence of base
(first step) to the alpha alkene, which would be in conjugation with the carbonyl - more stable. This then brings up the problem in 1.

3. The only other option is gamma.
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Offline Heory

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Re: Share an deduction problem
« Reply #10 on: December 07, 2009, 09:46:16 PM »
Well done

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