Chemical Forums
Chemistry Forums for Students => Organic Chemistry Forum => Organic Spectroscopy => Topic started by: bhuiyan85 on November 28, 2012, 06:18:26 PM
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MS : 158 (molecular weight)
IR peaks : 1755 , 1820, 3000
H NMR peaks : 1.0 (6H), 1.7 (4H), 2.4 (4 H)
C NMR peaks : 12.0 ppm, 18.0 ppm, 38.0 ppm, 78.5 ppm, 181 ppm
Please help me in finding the compound.
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Okay, What is the minimum number of carbons present? Maximum?
What function group(s) may be present (IR and 13C-NMR)?
How many hydrogens are present (easy)?
You could do this as follows, take the number of carbons and hydrogens you know are present and subtract that from the MW. What is missing? I don't know if the proton NMR peaks are singlets or not. If singlets, it would appear that symmetry is part of the problem. If you can come up with a MF, then you can calculate the number of rings and/or double bonds present.
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I got C10H6O2 but there needs to be 14H. How can that be possible?
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m/z =158 ; so i found that the initial possible formula is C12H14.
The peaks at IR are 1755 (C=O); 1820 ; 3000 (Ch3 Or OH)
The peaks of H NMR are 0.9 ( R- CH3); 1.7 (C=C ); 2.4 (benzene ring or CH-C=O)
The peaks of C NMR are 10 (R-CH3) ; 20 (R2- CH2); 180 (C=O)
According to H NMR there should be 14 H but if i add O then how can that be possible?
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If it were one oxygen, then it could be C11H10O. Are there any other possibilities?
Try this 158-14 = 144
144/12 = 12 (C12H14 = 158, but not it)
144 - n*12 -m*16 = 0
What are possible n and m?
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One additional hint. You notice C12H14 give MW 158 and 4*12 = 3*16. There are five signals in the 13C NMR, so you need to have at least C5 and max C12.