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analyze
problem set .pdf (page 11 of 20) 3421 IR Spectrum (KBr disc) 1684 4000 3000 2000 1600 1200 800 V (cm) 100 107 Mass Spectrum
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Answer #1

IR: 3421: -NH stretching
-OH is not present as this peak is not broad.
1684: amide C=O stretching
Mass: M+ = 149
odd mass, hence one of odd numbers of nitrogens present.
Rule of 13: divide 149 by 13: it gives 11 and remainder is 6 so the CH formula is: C11H11+6 = C11H17
Now add one nitrogen by replacing CH2: C10H15N
Add one oxygen by replacing CH4: C9H11NO
Degree of unsaturation = ((2C+2)+N-H-X)/2 = ((20)+1-11-0)/2 = 5
Out of 5 sites of unsaturation 1 accounts for C=O and 4 for benzene ring.
As there is a peak at 106/107 means it is losing C=O(CH3) (acyl) group.
Thus the compound is N-(4-methylphenyl)acetamide. There are other possibilities as well, the methyl group could be on ortho or meta position.

NH NH, CH CH3 Học Hąc N-(4-methylphenyl acetamide m/z=106/107 m/z=43

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