organic lab.

INFRARED SPECTROPHOTOMETRYI. HOW TO INTERPRET INFRARED SPECTRA
Infrared (IR) Spectral Area Absorptions
Divide each spectrum into 4 areas:
4000 – 2700 cm-1
2700 – 1900 cm-1
1900 – 1200 cm-1
1200 – 400 cm-1
Interpret IR spectra by noting absorptions absent or present in each region. Functional groups have
characteristic absorptions.
The most conclusive evidence is negative: the absence of an absorption characteristic of a certain
functional group is conclusive proof that that functional group is not present in the molecule.
Positive evidence, the presence of characteristic absorptions in a particular area, must be confirmed
by supporting evidence in other areas.
It’s important to know roughly where major absorbances occur, but DO NOT MEMORIZE THESE
NUMBERS!
Individual Absorptions
3800 – 3200 cm-1
Major absorptions in this region are due to X—-H stretching of:
O—H
N—H
C—H
alcohols
phenols
acids
amides
amines
terminal alkynes
alcohols
RO—H
3400 – 3200 (vs, br)
phenols
ArO—H
3400 – 3200 (vs, br)
1o amines
RNH2
2 bands
3500, 3400
1) symmetric stretching
R-N
H
H
2) asymmetric stretching
R-N
H
H
2o amines
R2N—H
1 band (m)
1o amides
O
R-C-NH2
2 bands
O
1) symmetric
RCN
H
H
O
2) asymmetric
RCN
H
H
2o amides
O
R-C-N—H
R
1 band (m)
acids
O
RCO—H (s, br)
terminal
alkyne
R-C  C–H (s, sharp)
The absence of any of these bands is conclusive evidence for the absence of these functional
groups in the molecule.
3200 – 2700 cm-1
Major absorptions in this region are due to C—H stretches.
above 3000 alkenyl (olefinic) C—H stretch
C=C
(w – m)
H
aromatic C—H stretch
C=C
(Ar—H)
(w – m)
below 3000 aliphatic (alkyl) C—H stretch
H
RC——H
H
(w – m)
H
R2C
R3C——H
H
aldehydic C—H stretch
O
RC—H
2 bands: 2800, 2700
2700 – 2000 cm-1
Major absorptions in this region are due to 3 functional groups:
mercaptans (thiols) RSH RS——H
2600 -2550
alkynes RC  CR
-CC-
2260 – 2100 (w – m)
RC  N
2250 – 2200
nitriles
RC  N
(w – m)
sometimes extremely
weak or missing
(w – m) depending on symmetry
2000 – 1630 cm-1
This region is the carbonyl region.
Important factors determining position are:
1) Type of carbonyl containing molecule: ester, aldehyde, acid, ketone, amide
O
2) Whether or not the carbonyl is conjugated, eg., -C=C-C- ,
O
-C  C-C- ,
nonconjugated
conjugated
ester
O
R-C-OR
1740 cm-1
1725 cm-1
aldehyde
O
R-C-H
1730 cm-1
1700 cm-1
acid
O
R-C-OH
1720 cm-1
1700 cm-1
ketone
O
R-C-R
1715 cm-1
1685 cm-1
amide
O
R-C-NR2
1665 cm-1
1650 cm-1
O
Ar-C-
Note: conjugation weakens the carbonyl bond and moves the stretch to a lower frequency.
1630 – 900 cm-1 (1200 – 900 cm–1 = “fingerprint region”)

only useful for providing corroborative evidence

mostly C-O, C-C, C-N stretching; N-H, C-H bending
amines:
H
1
o
N
bending
1630 – 1560 (w – m)
H
2 o, 3 o
C
N
stretching
1360 – 1020 (w – m)
C
N
stretching
1360 – 1020 (w – m)
stretching
2 bands
1500, 1600 (w – m)
stretching
1 band
1667 – 1640 (w – m)
(if conjugated, 1600)
aromatics:
-C
C-
alkenes:
-C=C-
phenols:
Ar
OH
1260 – 1180 (m – s)
R
OH
1150 – 1050 (m – s)
Ar
O
R
1275 – 1200 (s, br)
1077 – 1020 (s, br)
R
O
R
1150 – 1070 (s, br)
alcohols:
ethers:
esters:
acids:
O
R-C
OR
1300 – 1060 -bands may
overlap
O
R-CO
R
1300 – 1160 -bands may
overlap
O
R-C
OH
O
R-C
alkanes:
stretching
1450 – 1400
bending
1350 – 1250
bending
twisting,
wagging
1465 – 1375
1350 – 1150
OH
H
-C
CH3
C
CH3
geminal
1385 -1380
disubstituted
bending
1370 – 1365
900 – 650 cm-1 Aromatic substitution patterns
monosubstituted
770 – 730
710 – 690
2 bands
ortho disubstituted
770 – 735
1 band
para disubstituted
860 – 800
1 band
meta disubstituted
910 – 870
810 – 760
710 – 690
3 bands
II. DETERMINING THE DEGREE OF UNSATURATION IN A COMPOUND
When asked to identify a compound from its IR spectrum and molecular formula, it is
useful to know how many rings or pi bonds are reasonable for the molecular formula given. To
do this you must calculate the degree of unsaturation in the molecule. Once you know this, you
will know whether it is reasonable to postulate the presence of certain functional groups, which
will greatly assist you in the interpretation of IR spectra. One unsaturation in a molecule
indicates the presence of a ring or a pi bond. A cycloalkyl group has 1 unsaturation, a double
bond has 1 unsaturation, a triple bond has 2 unsaturations. An alkane has no unsaturations
because it contains neither rings nor pi bonds. A benzene ring has 4 unsaturations because it
contains a ring and 3 double bonds. A carbonyl group(C=O) has 1 unsaturation. An imine
group (C=N) has 1 unsaturation. A nitrile (R-C  N) has two unsaturations. And so on.
The degree of unsaturation is determined by calculating the # of H atoms that the molecule
would have if it were saturated, subtracting the # of H atoms it actually has, and dividing the
difference by 2.
For hydrocarbons,
for C6H14,
for C5H8,
# of unsaturations
=
# unsaturations =
2(6) + 2 –14
 = 0;
2
# unsaturations =
2(5) + 2 – 8
 = 2;
2
For compounds containing O, # of unsaturations
(the O atoms are ignored)
2( # of C’s) + 2 – # of H’s

2
therefore, there are no
rings or pi bonds in the molecule.
therefore, there are 2 double bonds,
or 1 ring and a double bond or a
triple bond in a compound of this
formula.
=
For compounds containing N, # of unsaturations =
(1 H atom is added for every N)
# unsaturations =
2(5) + 2 – 5 + 1
 = 4
2
for CH4N2O, # unsaturations =
2(1) + 2 – 4 + 2
 = 1
2
for C5H5N,
2( # of C’s) + 2 – # of H’s

2
2( # of C’s) + 2 – # H’s + # N’s

2
2( # of C’s) + 2 – # of H’s – #X’s
For compounds containing halogen, #unsaturations = 
(1 H atom is subtracted for every X)
2
for C6H5Br, # unsaturations =
2(6) + 2 – 5 – 1
 = 4
2
for C2H2Br2, # unsaturations =
2(2) +2 – 2 – 2
 = 1
2
III.
Assignment
A. The following are copies of spectra of various compounds. For each spectrum label
(directly on the spectrum) the absorption associated with the functional groups noted on
the structure.
Label the acid OH absorption, the benzene ring C-H absorption, and the carbonyl absorption.
Label the benzene C-H absorption and the nitrile absorption.
Label the OH absorption and the alkyl C-H stretch.
B. For the following spectra draw a structure that is consistent with the molecular formula
given. If more than one structure is possible, draw them all. Indicate the degrees of
unsaturation for each molecule.
CH O
Degrees of unsaturation = _______
CH O
Degrees of unsaturation = _______
CHO
Degrees of unsaturation = _______
C H Cl
Degrees of unsaturation = _______
CHO
Degrees of unsaturation = _______
CH N
Degrees of unsaturation = _______