Chemical Forums
Chemistry Forums for Students => Organic Chemistry Forum => Topic started by: nj_bartel on March 05, 2009, 04:55:08 PM
-
I'm trying to sythesize this
http://img15.imageshack.us/my.php?image=randomp.png
using anisole as my only organic carbon source.
So far, I cleaved anisole to give phenol, nitrated that to give para-nitrophenol, then reduced to give para-aminophenol. I then set this product aside for later.
Then, I cleaved another anisole to give phenol, treated with PBr3 to give bromobenzene, treated with Mg to form a grignard, reacted with CO2 followed by acid to give benzoic acid, then treated with SOCl2 to give the acyl chloride of benzoic acid. I then reacted this with the product I had set aside.
Unfortunately, that gives the product pictured, except without the methoxy substituent in the bottom left-hand corner, and I'm drawing blanks on where I can work it in.
Thanks for any help.
-
I'm not sure where this would react on your molecule but could you try trichloroacetyl chloride in a first reaction then in a second time react what you have with sodium methoxide.
-
Unless you have a way of synthesizing the trichloroacetyl chloride, I wouldn't be allowed to use that due to it being an organic molecule that's adding carbons. Thanks though.
-
Got it, nvm :)
-
Share share! :D
-
Went about the second part in the wrong way.
It's still anisole -> p-aminophenol as stated above
Then anisole is brominated, and the para product is isolated to give p-bromoanisole, which is used to form a grignard, which reacts with CO2 followed by acid to give p-methoxybenzoic acid, which is treated with thionyl chloride to give the acyl chloride, followed by p-aminophenol to product
-
Can we use protecting groups? Even then I think the last step is doubtful, giving us a mixture of amides...
-
What do you mean by a mixture, I think the Beckmann is fairly stereospecific (I think that's the right word) - what I mean is that if you follow the mechanism as at
(https://www.chemicalforums.com/proxy.php?request=http%3A%2F%2Fupload.wikimedia.org%2Fwikipedia%2Fcommons%2F0%2F0d%2FBeckmann_rearrangement_mechanism.PNG&hash=e846e02f72819b4c170c730a4d5a7725e3630de8)
if R = Et, and R' = Me (i.e. the oxime from butanone), then you get the major product as N-ethylacetamide, rather than N-methylpropionamide. Forming the required oxime isomer selectively may be a different problem, though ;)
S
-
Forming the required oxime isomer selectively may be a different problem, though ;)
That was exactly my concern...the second to last step will give us a pair of diastereomers (in my scheme..) and hence we will get a mixture in the final step.
-
That was exactly my concern...the second to last step will give us a pair of diastereomers (in my scheme..) and hence we will get a mixture in the final step.
OK just checking, I read your initial post as suggesting that you get the (E)-oxime exclusively, then the scrambling occurs with the Beckmann. I suppose that if you're allowing separation of protected 4-hydroxy-4'-methoxybenzophenone and protected 4-hydroxy-2'-methoxybenzophenone then this may be OK.
Presumably we have a workable route from anisole to benzenesulfonyl chloride?
-
then this may be OK.
But won't we have to separate the 2 amides after the last step? I used PhSO2Cl as it did not add carbons, as it is only performing the work of an assistant.