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  #1  
Unread 09-06-2005, 07:25 AM
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Default Answered: Can a chemist please help with NMR?

Please outline for me the chemical shift patterns of all the protons in NMR spectrum of SALICYLIC ACID. Thanks!
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Best Answer - Posted by romanwahoo
You should have 4 protons in the aromatic region between 6.5 and 8, which include two triples and two doublets. If you have the 1h nmr and you are just assigning the peaks, then simply figure out which carbons are more positively charged and which are more negatively charged. Since the two groups' effects are additive and not interfering, just go with the carboxylic acid, which is an electron withdrawing group. The carbon to which the -COOH is attached therefore is more negative, the next more positive, then more neg, and so forth. The protons attached to carbons that are more positive are more deshielded, so they have a higher shift. Using this method, you should be able to assign all four aromatic protons.The -COOH and the -OH are generally much harder to see, and they are concentration dependent. A broad peak around 6 though is usually indicative of the -OH peak. Because the oxygens in the -COOH can resonate charges, the carboxylic acid proton is usually shifted waaay downfield, around 14 or so, I believe, and which is usually not shown in a normal proton spectra.

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Default Can a chemist please help with NMR?

Below is the spectrum for acetasalicylic acid. There will only be a small difference between the chemicals shifts in the aromatic hydrogens between the two compounds. Also, you will obviously not have the singlet from the 3 methyl hydrogens.The second link shows the spectrum in methanol, but its kinda hard to see.
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Unread 09-06-2005, 11:13 AM
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Default Can a chemist please help with NMR?

You should have 4 protons in the aromatic region between 6.5 and 8, which include two triples and two doublets. If you have the 1h nmr and you are just assigning the peaks, then simply figure out which carbons are more positively charged and which are more negatively charged. Since the two groups' effects are additive and not interfering, just go with the carboxylic acid, which is an electron withdrawing group. The carbon to which the -COOH is attached therefore is more negative, the next more positive, then more neg, and so forth. The protons attached to carbons that are more positive are more deshielded, so they have a higher shift. Using this method, you should be able to assign all four aromatic protons.The -COOH and the -OH are generally much harder to see, and they are concentration dependent. A broad peak around 6 though is usually indicative of the -OH peak. Because the oxygens in the -COOH can resonate charges, the carboxylic acid proton is usually shifted waaay downfield, around 14 or so, I believe, and which is usually not shown in a normal proton spectra.
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Default Can a chemist please help with NMR?

Assuming you are in chloroform dthere are 4 aromatics. Starting ortho to the acid group and moving clockwise....1 H doublet 2 H doublet of doublets2 H doublet of doublets1 H doubletAll of these will be in the 6.8-7.6 range. There will probably be overlap unless you are at 300 MHz or higher.The phenol proton will be shifted much higher than phenols normally do. This is because it H-bonds to the carboyxlic acid carbonyl, causing it to go out to about 10-11 ppm.The acid peak will be broad and in the 12-14 range.Keep in mind exchangeable protons (phenol/acid/alcohols etc)can vary greatly depending on solvent, concentration, etc.

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