Combined Spectra
Reference: Smith, Chapters 13 and 14 and the Introduction to UV presented during lab period.
Pre-lab assignment: Study all materials on mass spectroscopy, IR, UV, and 1H NMR
spectroscopy.
1. This lab period consists of 10 spectral problems worth 2 points each.
2. Each problem contains a mass spectrum, IR spectrum, and a 1H NMR spectrum of an
unknown compound. In addition, UV data are given along with any other pertinent information.
3. The numbers above each signal (the integral) in a spectrum give the relative number of
hydrogen atoms. In most cases, the relative number of hydrogen atoms is the exact number of
hydrogens. When the sum of the integrals do not add up to the actual number of hydrogens
necessary from the other data, then the relative numbers must be multiplied by the whole number
that makes the sum of integrals equal the sum of hydrogens. For example, if you can only
account for half of the needed mass, then you must double everything in the NMR spectrum, etc.
4. Always compare the molecular weight of your proposed compound to the M+ you find in the
mass spectrum. If they do not match, your structure is wrong or you are comparing to the wrong
M+. Remember, certain atoms such as Br and Cl have isotopes that show up in the mass
spectrum.
Lab 13
1
Stu No___ Sec___ Last Name_________________________________. First ______________
Compound 1 UV:  > 1000 and  < 100
M+ = 132
Lab 13
2
1H (d)
5H (m)
1H (d)
1H (dd)
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3
Compound 2 UV: end absorption
M+ = 160
Lab 13
4
6H (t)
2H (s)
4H (q)
Lab 13
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Compound 3 UV:  > 1000
M+ = 166
Lab 13
6
3H (s)
2H (s)
5H (m)
Lab 13
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Compound 4 UV: log  < 2
M+ = 100
Lab 13
8
For the NMR:
Compound 4 exists in two forms called tautomers. Ignore the signals marked through by an X.
Analyze only the two signals labeled 2H and 6H.
x
6H (s)
x
2H (s)
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Compound 5 UV: log  > 3
MS: Don’t overlook the small peak at m/e172.
M+ + 2 = 172
M+ = 170
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2H (s)
5H (m)
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Compound 6 UV: none
M+ = 74
M+ + 2 = 76
Lab 13
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2H (s)
1H (s)
Although shown as singlets on the spectrum, the two signals in the 1H NMR spectrum of
Compound 6 are coupled via a long-range coupling that is outside the scope of our course.
Consider both signals to be singlets.
Lab 13
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Compound 7 UV:  > 1000
M+ = 107
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14
3H (s)
5H (m)
1H
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Compound 8 UV: log  > 3
M+ = 117
Lab 13
16
3H (s)
4H
(d) (d)
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Compound 9 UV:  > 1000
M+ = 122
Lab 13
18
5H (m)
2H (t)
1H (s)
2H (t)
Lab 13
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Compound 10 UV: log  > 3
M+ = 136
Lab 13
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6H (d)
4H
(d) (d)
1H (s)
1H (septet)
Note: One H signal is very faint.
Lab 13
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