Chem 8361/4361: Interpretation of Organic Spectra
Problem Set 1
Due on Sept. 28, 2009
1) Work the following problems in Silverstein:
Chapter 1, 1.6, 1.8, 1.9
Mass
56.0264
73.0896
74.0363
89.0479
94.0535
96.0572
98.0736
100.0893
102.0678
113.0845
114.1043
116.0841
116.1206
122.0733
122.0733
126.1041
138.0687
150.0041
152.0476
156.9934
161.9637
169.9735
208.0094
1.6 (Spectrum)
H
G
F
I
N
W
R
E
L
M
A
T,B
T,B
K,V
K,V
S
J
D
O
P
U
C
Q
1.8 (Formula)
C3H4O
C4H11N
C3H6O2
C3H7NO2
C5H6N2
C6H8O
C6H10O
C6H12O
C5H10O2
C6H11NO
C7H14O
C6H12O2
C7H16O
C8H10O
C8H10O
C8H14O
C8H10O2
C5H11Br
C8H8O3
C6H4NO2Cl
C6H4OCl2
C7H7Br
C7H13O2Br
1.9 (HDI)
2
0
1
1
4
3
2
1
1
2
1
1
0
4
4
2
4
0
5
5
4
4
1
NOTE: For the formulas with halogens the exact masses will not have a close neighbor with those in
Appendix A. To figure out those formula you must first identify what halogen and how many are present.
Then subtract the mass of the lowest isotope (see Table 1.4, pg 15) from the exact mass. This will give you a
new mass for looking up in Appendix A.
Example: For 150.0041 the prominent M+2 peak in Spectrum D tells us there is a bromine present. If we
subtract 78.9183 from 150.0041 we get 71.0858. This new mass corresponds to C5H11.
Chapter 2, 2.2
a. benzonitrile
b. isobutylamine
c. diphenyl sulfone
d. dioxane
e. 1-nitropropane
f. formic acid
g. biphenyl
h. benzoic acid
i. benzamide
Chapter 2, 2.7
Formula = C2H3NS -> C3H4S -> C3H4 -> HDI = 2
1) No stretches above 3000 cm-1 indicates no NH or C=CH.
2) The large peak ~2100 cm-1 indicates the presence of a sp-atom. There are a number of functional groups
that contain this type of carbon, but we can get rid of a number of these because of the limited number of
atoms we have to work with [No N=C=N (not enough N), no C=C=C (no enough C), no C=N=N (not
enough N), no N=N=N (not enough N), no N=C=O (no O)]. This leaves us with a nitrile, isonitrile,
thiocyanate, or isothiocyanate as a possibility.
HS
N
HS
N
C
H3C
N
C
S
H3C
S
C
N
The first two can be eliminated because there is no S-H stretch at ~2600 cm cm-1.
While comparison with known spectra of the remaining two structures would be the safest way to determine
what we have, isothiocyantes (structure 3) have another stretch in the fingerprint region. There is a good
stretch at 647 cm-1 that points to structure 3 being the identity of this compound.
Chapter 2, 2.9
There are multiple possible answers depending on how close you want to look. I have only identified a few
key ones. The formula and HDI from the table above should also be consulted.
Spectrum A – C=O (possibly ketone), no C=C, no OH
Spectrum B – OH, no C=C, no C=O
Spectrum C – Possible C=CH, no C=O possible aromatic
Spectrum D – No C=C, no C=O
Spectrum E – C=O (possibly ketone or aldehyde), no C=C, no OH
Spectrum F – Probably CO2H, no C=C
Spectrum G – N-H, no C=C, no C=O
Spectrum H – OH, probable terminal alkyne, no C=C, no CO2H
Spectrum I – NO2 likely, No NH, no OH, no C=O, no C=C
Spectrum J – Probable OH, probably aromatics, no C=O
Spectrum K – Probable aromatics, no OH, no C=O
Spectrum L – probable CO2R, no C=C, no OH
Spectrum M – NH or OH, C=O (acid or amide), possible C=C, no aromatics (HDI too low)
Spectrum N – aromatics likely, no NH, no C=O
Spectrum O – possible OH, possible CO2H, aromatics likely
Spectrum P – aromatic likely, likely NO2, no NH, no OH, no aliphatic C-H, no C=O,
Spectrum Q – possible OH, C=O (possibly CO2H), no C=C, no aromatics
Spectrum R – possible OH, terminal alkyne likely, no C=O, no C=C
Spectrum S – C=O (ketone or aldehyde), C=C unlikely, no OH, no aromatics
Spectrum T – OH, C=O (possible CO2H), C=C unlikely, no aromatics
Spectrum U – possible OH, aromatics, likely few aliphatic C-H, possible C=O
Spectrum V – OH, possible aromatics, no C=O
Spectrum W – C=O, possible C=C, alkyne unlikely,
2) Calculate the molecular formula, HDI, and propose structures for the following compounds.
Problem 12:
DETERMINATION OF FORMULA:
Rule of 13:
From MS: simple looking spectrum with M–127 peak & m/z = 127
MW = 156
From IR and NMR: does not seem to have much in the way of heteroatoms
n = 12
156 / 13 = 12
probably has I
C12H12 – C10H7 + I = C2H5I is molecular formula
Base formula is C12H12
This is consistent with NMR data (two carbons in 13C spectrum)
HDI:
C2H5I
C2H6
HDI = 0
There is really only one way to put this together
I
***Note the distinctive chemical shift of the CH2 that is attached to the iodine***
Problem 25:
DETERMINATION OF FORMULA:
Rule of 13:
From MS: odd mass points to odd number of nitrogens
m/z = 77 and m/z = 91 is indicative of a PhCH2 group
m/z = 91 = M–26
points to loss of C N
MW = 117
117 / 13 = 9
n=9
IR and 13C NMR confirms presence of nitrile:
IR: peak ~2251 cm-1
Base formula is C9H9
13C
NMR: peak with no attached protons at ~118 ppm
Does not seem to be much else in the way of heteroatoms
C9H9 – CH2 + N = C8H7N is molecular formula
HDI:
C8H7N
C9H8
SUBSTRUCTURES:
CH2
HDI = 6
m/z = 91, 1H integrations, 13C shifts
IR stretch would be
a little lower 2150–2110
NC
There are two ways to put these together
C
N
N
C
13
C shift would be higher
~160 ppm
Most consistent with data
Problem 52:
DETERMINATION OF FORMULA:
From MS: pattern of M+ ion is indicative of two bromine atoms being present
small M–79 peak looks to be present, M–158 could be loss of two romine atoms
Rule of 13:
MW = 310
310 / 13 = 23.84615
n = 23
NMR and IR do not indicate much else in the way of heteroatoms
must add 11 H's
23 x 13 = 299
Base formula is C23H34
C23H34 – 2(C6H7) + Br2 = C11H20Br2 is molecular formula
BUT...this would give a HDI of 1
No aliphatic carbons or protons are present. So we must adjust the formula by taking away H's
C11H20Br2 – H12 + C = C12H8Br2 is molecular formula
HDI:
C12H8Br2
C12H10
HDI = 8 Consistent with having lots of carbons and no aliphatic protons
SUBSTRUCTURES:
Br
or
Must be highly symmetric with respect to protons and carbons
Both aromatic rings have to be the same
Br
Possible Structures:
Br
IR peak at ~800 cm-1 is more consistent
with para substitution
Br
Br
Problem 67:
DETERMINATION OF FORMULA:
Rule of 13:
IR: broad peak ~3420 cm-1 and stretch at 1742 cm-1
OH group and C=O; probably not CO2H, most likely CO2R
MW = 104
104 / 13 = 8
n=8
Base formula is C8H8
1H
NMR: exchangeable peak ~3.6 ppm; not CO2H
13C
NMR: peak with no attached protons at ~173 ppm
C8H8 – 3(CH4) + O3 = C5H-4O3 must add H's to get positive
C5H-4O3 – C + H12 = C4H8O3 is molecular formula
This is consistent with NMR data (8 H's and four carbons)
HDI:
C4H8O3
C4H8
HDI = 1
Consistent with CO2R and no other unsaturations or rings
SUBSTRUCTURES:
O
From 1H NMR this is a singlet
HO
CH2
CH3
CH2
O
From 1H NMR these are coupled to each other
O
There are two ways to put these together:
O
chemical shift of CH2 is most
consistent with this structure
O
OH
HO
O
hemi-acetal is probably
not too stable
Problem 74:
DETERMINATION OF FORMULA:
Rule of 13:
From MS: odd mass points to odd number of nitrogens
MW = 101
IR: broad peak ~3305 cm-1 N–H
101 / 13 = 7.76923
1H
7 x 13 = 91
n=7
NMR: two broad exchangeable protons at ~1.2 ppm
13C
must add 10 H's
NMR: not peaks above 60 ppm
Probably only one nitrogen atom and no other heteroatoms
Base formula is C7H17
C7H17 – CH2 + N = C6H15N is molecular formula
This is consistent with the six carbon atoms in the 13C spectrum
HDI:
C6H15N
C7H16
HDI = 0
Consistent with all of the carbon atoms being below 40 ppm
SUBSTRUCTURES:
CH3
H2N
CH2
5x
From DEPT spectrum
There are several possible amine that could be drawn with six saturated carbon atoms. However, only
one isomer will give us an NH2 and one methyl group.
NH2
Problem 78:
DETERMINATION OF FORMULA:
From IR and NMR: appears to be two CO2H
Rule of 13:
IR: very broad peak ~3000 cm-1 and stretch at 1729 cm-1
MW = 160
160 / 13 = 12.3077
12 x 13 = 156
1H
n = 12
NMR: taken in D2O so we cannot determine how many acidic H's there are
13C
must add 4 H's
NMR: two peaks at 170 and 175 ppm, consistent with CO2H or CO2R
no peaks in C–O bond area
Base formula is C12H16
C12H16 – 4(CH4) + O4 = C8H0O4 must add H's!!
C8H0O4 – C + H12 = C7H12O4 is molecular formula
This is consistent with NMR data (ten proton's are seen and two CO2H are inferred)
HDI:
C7H12O4
C7H12
HDI = 2
Consistent with two CO2H and no other unsaturations
SUBSTRUCTURES:
O
H3C
C
HO
2x
2x
1.0 (s, 6H)
CH2
CH2
From 1H NMR these are coupled to each other
13C
43 ppm
There is really only one way to put these together
HO2C
CO2H