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Alicia's Masters Thesis
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Ugi NMR Analysis
To synthesize a
by an acid catalyzed transamidation reaction. (
1998 Hulme Tet. Lett.
To a solution of
ugi product 108C
(47.7mg, 0.1 mmol) in CDCl3 (500uL), trifluoroacetic acid (500uL) was added in an NMR tube. The reaction was monitored by NMR. After a complete conversion the solution was poured into a centrifuge tube and neutralized with 10% NaOH solution, extracted from with dichloromethane, dried with anhy. MgSO4.
(20uL/ 500uL in 500uL CDCl3) [clear solution]
t = 02min
t = 06min
t = 08min
t = 11min
t = 20min
t = 35min
t = 50min
t = 1h 15min
t = 2h 11min
t = 2days
t = 5days
The first step in this reaction is the loss of the BOC group within a few minutes. This can be monitored by the appearance of the t-butyltrifluoroacetate (t-Bu) peak at 1.6 ppm, and is consistent with the behavior of similar compounds (e.g. EXP101).
After removal of the BOC group, two possible pathways are expected
According to the pathway
, the molecule would cyclize via a transamidation reaction. This would produce a benzylammonium salt, which is expected to appear as a quartet at 4.3 ppm, based on the spectrum obtained from a solution of benzylamine in an excess of 50% TFA in CDCl3. The quartet at 4.64 ppm which forms after a few minutes is consistent with the CH2 next to the NH3+ formed after Boc cleavage.
According to the pathway
, loss of the furan ring may take place and is expected to be detectable by the appearance of a doublet at ~8.4 ppm. as was previously observed in
After 5 days, it is apparent that neither pathways are operating and the reaction likely stops at the ammonium salt stage after boc deprotection of the amine (structure II-see below)
The possible products of the above reaction include I) diketopiperazine formed as a result of a transamidation reaction. The side product of the transamidation reaction would be benzylammonium chloride. Product II) deprotected Ugi product- a result of just a plain de boc reaction. Side products include t-butyltriluoroacetate, this product should be seen in all the above three case, so not a good peak to infer any thing important, except in the where all signals remain as is from the starting Ugi product (108C). The final possible product III) would be the furfuryl cleavage product with a distinct loss of furan from the system. This should be the most easily decipherable product based on a) the loss of furan protons b) appearance of doublets at 8.45 ppm and 5.66 ppm.
Ruling out product III) the furfuryl cleavage product.
From the HNMR of t=5days, it is evident that a fufuryl cleavage does happen based on the doublets related to the red arrow, the chiral proton at 5.66 ppm and the adjacent amide proton at 8.45ppm. However the intensity of these peaks very clearly show that this process is extremely slow. If at all the (less than 5% based on 5.66ppm and the benzylic CH2 at 4.01 ppm formed after 5 days )
Ruling out product 1) - Why is the benzylic system still connected to the Ugi product (deprotected)
a) From the HNMR of benzyl amine in TFA /CDCl3- a quartet at 4.32ppm is expected, this is not observed in the 5th day HNMR. Instead all three CH2s, with similar integration pattern appear in the region between 4.1 and 4.5.
It would definitely help if the reaction mixture was subjected to aqueous extraction procedure.
Very small amount of furfuryl cleavage observed the product likely stays stable at the deprotected ammoinium salt stage.
14:10 Weighed 47.7 mg of the ugi product 108C-1 from
in an NMR tube and dissolved it in CDCl3 (500 uL)
HNMR of 108C-1
14:40 Added TFA (500uL) to the NMR tube, making solution 110A
14:42 Obtained H NMR of 110A (t = 02 min)
14:46 Obtained H NMR of 110A (t = 06 min)
14:48 Obtained H NMR of 110A (t = 08 min)
14:51 Obtained H NMR of 110A (t = 11 min)
15:00 Obtained H NMR of 110A (t = 20 min)
15:15 Obtained H NMR of 110A (t = 35 min)
15:30 Obtained H NMR of 110A (t = 50 min)
15:55 Obtained H NMR of 110A (t = 1 h 15 min)
16:51 Obtained H NMR of 110A (t = 2h 11min)
16:17 Obtained H NMR of 110A (t = 2days)
15:40 Obtained H NMR of 110A (t = 5days)
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