Chemical Forums
Chemistry Forums for Students => Organic Chemistry Forum => Topic started by: Rutherford on February 16, 2013, 08:03:17 AM
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This scheme has to be done in reversed order, that's why it gives my a headache.
(http://img571.imageshack.us/img571/2660/scheme1x.gif)
Hint: A is a gaseous hydrocarbon with the density lower than that of air.
A is probably either C2H2 or C2H4. In the B reaction scheme, it gives a positive result to the iodoform reaction, so I is probably a methyl secondary alcohol, produced by acidification of B which is an alcoxide.
What reaction of A and B happens at the specified conditions?
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So F is an ester. N is a tosylate, H a bromide, M an alkene, L an acetylene, start with that.
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I understand all that, but I need the first reaction. Don't know how to determine anything without it.
Till now, I know that:
-F is an ester (know the structure), N is a tosylate, H a bromide, M an alkene, L an alkyne.
-K is an ester, J an acid salt, I a methyl secondary alcohol, B an alkoxide.
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I am not sure if the molecular formula for N is correct. The chrysanthemic acid has 10 C atoms, while N has 12, so F will have at least 12, too ???.
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Once again you are not presenting the full information!
Have you read the PDF you posted in the Chem Education forum? If so you would have read that the first reaction was discovered by Favorskii in 1905 http://en.wikipedia.org/wiki/Favorskii_reaction (http://en.wikipedia.org/wiki/Favorskii_reaction)
This is probably your first step.
I also note they present 2 routes to this compound. See if you can work it out now.
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Sorry for that. I thought that it is more informative than needed to know and I thought first to do one scheme and then the other one. I will try to solve it now when knowing the reaction.
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Compound E is probably this one.
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Yes, I finished the first scheme, now I am back to the second one. Why is the acid added to B (which I got that is acetone)?
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What did you get for A+B :rarrow:L ?
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Attached. Now I think HBr won't be added according to Markovnikof's rule. As N contains only 2 O atoms, what happened to the -OH group? Was it eliminated with HBr?
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I think this is H
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And then the tosilate is introduced. To finish the scheme I need K. I assumed that L is made from acid catalyzed aldol addition of acetone. I finished up with K being the attached compound. Could that be K? If yes, what happens then in reaction with the tosilate and NaOMe?
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I think this is N:
It's not a tosylate but a sulfone.
Please show me your scheme B :rarrow: I :rarrow: J :rarrow: K
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Here it is. Is it ok? If yes, what happens with K and N afterwards?
Didn't write that A is acetylene.
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OK, now what happens to give F?
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That's what bothers me. I can't think of the reaction :-\.
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Yes, I didn't think you would get it. OK, the sulfone makes the CH2 group next to it quite acidic, so you can generate a carbanion which attacks what? In a named reaction. I must say that I think the yield of this may be quite low. Also note you have the ethyl ester and are using methoxide as a base so it will transesterify to give product F as the methyl ester. If you can't figure it out I'll post my scheme of what I think goes on.
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I don't see a way to combine those three compounds ???.
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Two compounds. As I said the base abstracts a proton from the CH2 of the sulfone, this adds in michael reaction to the ester K, the intermediate enolate attacks the carbon carrying the sulfone forming the three membered ring and eliminating the sulfone. Have a look at my scheme.
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Shouldn't an ethyl ester be obtained?
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Using NaOMe, I would expect transesterification, if they use methanol as a solvent, if not then the ethyl ester remains.
It doesn't matter as it is hydrolysed anyway.
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Yes, but structure of F is important, too. As it appears in the previous scheme, where I got that it is an ethyl ester, it has to be the same here. Didn't think of this earlier ::).
Thanks for the help.
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No problem.
Then they must have used NaOMe in a solvent like THF or toluene. Then the ethyl ester remains an ethyl ester.
p.s. I have actually made some of these compounds, although not by this methodology. But I can't disclose their structures.
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Got one issue now. In the attached scheme, G is a natural alcohol. H is the same compound that was present in the previous scheme. What is G then? I have a formula, but I don't recognize this alcohol.
H is:
I got that G is:
Is this a natural alcohol?
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Yes, it is
http://en.wikipedia.org/wiki/Prenol (http://en.wikipedia.org/wiki/Prenol)
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Huh, thanks. It would be a mess if it wasn't.
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How to draw the (1R,3R)-configuration of chrysanthemic acid? I don't know how to apply the Cahn-Ingold system here. How to do it? I think that the stereocenters are in the ring.
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How to draw the (1R,3R)-configuration of chrysanthemic acid? I don't know how to apply the Cahn-Ingold system here. How to do it? I think that the stereocenters are in the ring.
There's no difference to if it were not in a ring. Which two atoms are the stereocentres?
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I think that the marked atoms are the stereocenters. Which one is 1R and which one is 3R? How to apply the Cahn-Ingold system?
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like this?
(1R,3R)-2,2-dimethyl-3-(2-methylprop-1-en-1-yl)cyclopropanecarboxylic acid
This should be easy for you, we've done enough of them in the past.
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I should've known that 1 is the one closer to the -COOH group.
My try is attached. I marked the priorities with numbers. On the top left, I assigned the configuration for the 3 chiral atom, and on the right, I assigned the configuration of the 1 chiral atom. The final structure is on the bottom. Did I do it correct?
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Well that compares with the one I posted, so it looks OK :o