By Anonymous on Wednesday, June 30, 2004 - 07:39 am:
hi, i know this is a LC forum. But i do have a question aoubt HNMR of HDI(O=C=O=N-CH2-CH2-CH2-CH2-CH2-CH2-N=C=O).The areas of the three different kinds of H are different,about 20% difference. Could anybody tell me why? thank you!
By Constantin Sychov on Wednesday, June 30, 2004 - 08:39 am:
Its allright. Mmmm.. Overhauser effect, as far as I remember.
By Alexander on Thursday, July 1, 2004 - 06:17 pm:
N=C=O is an electon acceptor, so the closer your CH2 groups to the substituent, the higher the shift to the low field. The central CH2CH2 must be in the strongest field.
By Anonymous on Sunday, July 4, 2004 - 08:56 am:
hi,alexander, what i meant by that was not the shift,but the area. say, the areas of the 3 different type of Hs are different,the lowest 4 and the bigget 5.3.The structure is O=C=N-CH2-CH2-CH2-CH2-CH2-CH2-N=C=O,not O=C=O=N-CH2-CH2-CH2-CH2-CH2-CH2-N=C=OŁ¬sorry about that!
By Anonymous on Sunday, July 4, 2004 - 02:03 pm:
what's your solvent and what are the shifts and integrals of all the methylene signals? shame you can't scan the spectrum and attach it..
By Anonymous on Monday, July 5, 2004 - 03:40 am:
I think you have a lot of overlapping in your spectrum, so differences accumulate. Is the spectrum well phased/integrated? Otherwise the compound is not pure. Check it with a C13 spectrum.
By Alexander on Wednesday, July 7, 2004 - 06:33 pm:
20% difference in integration maybe normal, depends on the instrument and concentration. Try higher concentration/more scans. This will give you a better integration number. You may also have some overlapping from impurities of partially hydrolyzed compound from the reagent or from moisture during the sample prep.