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Topic: End of the acetone and cyclobutane scheme  (Read 5560 times)

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Offline Rutherford

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End of the acetone and cyclobutane scheme
« on: February 14, 2013, 02:01:30 PM »
Here is the last scheme from the problem.
Hints: C and D are isomers; J has two planes of symmetry.

I will attach where I've got in the next post.
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Offline Rutherford

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Re: End of the acetone and cyclobutane scheme
« Reply #1 on: February 14, 2013, 02:03:13 PM »
I think that C and D are optical isomers (pseudoassymetric centre at -OH group). I don't know to continue.
« Last Edit: February 14, 2013, 03:04:24 PM by Raderford »
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Offline discodermolide

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Re: End of the acetone and cyclobutane scheme
« Reply #2 on: February 14, 2013, 02:05:22 PM »
Cis and trans isomers, OH either up or down. Not optical isomers, there has been no chirality introduced in any of the schemes.
How do alcohols react with methanesulfonyl chloride? Think back to earlier in the thread.

p.s. It is Curtius!
« Last Edit: February 14, 2013, 02:24:01 PM by discodermolide »
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Offline Rutherford

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Re: End of the acetone and cyclobutane scheme
« Reply #3 on: February 14, 2013, 03:04:54 PM »
Quote from: discodermolide on February 14, 2013, 02:05:22 PM
p.s. It is Curtius!
Oops. So I end up with the attached compound. What is the usage of Et3N?
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Offline discodermolide

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Re: End of the acetone and cyclobutane scheme
« Reply #4 on: February 14, 2013, 03:08:12 PM »
Triethylamine removes the HCl formed.
Now what does NaN3 do? Think back to earlier.
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Offline Rutherford

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Re: End of the acetone and cyclobutane scheme
« Reply #5 on: February 14, 2013, 03:20:27 PM »
The -SO2CH3 is a good leaving group and it gets substituted by the azide?
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Offline discodermolide

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Re: End of the acetone and cyclobutane scheme
« Reply #6 on: February 14, 2013, 03:22:05 PM »
Yes, that is correct. Now what does the hydrogenation do?
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Offline Rutherford

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Re: End of the acetone and cyclobutane scheme
« Reply #7 on: February 14, 2013, 03:33:36 PM »
Azide to amine? Then TFA makes the amine group on the other side of the molecule?
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Offline discodermolide

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Re: End of the acetone and cyclobutane scheme
« Reply #8 on: February 14, 2013, 03:34:58 PM »
That's correct, so the final products are
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Offline Rutherford

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Re: End of the acetone and cyclobutane scheme
« Reply #9 on: February 14, 2013, 03:39:49 PM »
Is the right -NH2 group below or above the plane? J has two planes of symmetry so it need to be the cis isomer.
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Offline discodermolide

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Re: End of the acetone and cyclobutane scheme
« Reply #10 on: February 14, 2013, 03:42:48 PM »
In one compound the right hand NH2 is below and the other above.
Make a model.
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Offline Rutherford

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Re: End of the acetone and cyclobutane scheme
« Reply #11 on: February 14, 2013, 03:47:43 PM »
It is a normal line so it seems like it is in the same plane as the two rings, but I got it. J is the one where both amino groups are above or under the plane (cis). I needs to be trans then. Thanks for the help. This was a tough synthesis.
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Offline discodermolide

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Re: End of the acetone and cyclobutane scheme
« Reply #12 on: February 14, 2013, 03:49:32 PM »
No problem. Go and have a rest now and think about everything.
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Offline Rutherford

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Re: End of the acetone and cyclobutane scheme
« Reply #13 on: February 15, 2013, 04:09:31 AM »
Only, there should be 1 ring, not 2.
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Offline discodermolide

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Re: End of the acetone and cyclobutane scheme
« Reply #14 on: February 15, 2013, 04:23:44 AM »
WHAT????
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