The lab report below was submitted as part of the coursework for CM2102 Spectroscopic Applications. Please do not plagiarise from it as plagiarism might land you into trouble with your university. Do note that my report is well-circulated online and many of my juniors have received soft copies of it. Hence, please exercise prudence while referring to it and, if necessary, cite this webpage.
Abstract
This
experiment aims to: 1) match the given Infrared (IR) spectrum to the respective
functional groups, 2) identify 5 compounds through their IR spectrum and 3)
obtain the isosbestic point of 4-methoxy-2-nitrophenol through Ultraviolet-Visible
(UV) spectroscopy.
Reagents
Samples of
compounds A, B, C, D and E; Solutions of pH 7, 8 and 13; One 100mL and three
10mL volumetric flasks; One 50mL beaker; Two potassium bromide (KBr) plates; IR
and UV spectrophotometers
Introduction
Infrared(IR) spectroscopy involves absorption of IR radiation to
elucidate structures of molecules. Radiation of this energy range corresponds
to the stretching and bending of bonds. Stretching involves a displacement
along the bond axis with a change in the interatomic distance while bending
involves a change in bond angles between 2 bonds and an atom common to both. Only
bonds that have a dipole moment that changes with time are capable of absorbing
IR radiation.
Functional groups have characteristic IR absorption.
The stretching and bending frequencies of a bond depends on its multiplicity
and aromaticity. Stronger bonds and asymmetric stretches absorb energy of
higher wavenumbers than weaker bonds and symmetric stretches. The presence of
bands in regions gives direct information while their absences may be used to
deduce absence of functional groups.
In
ultraviolet-visible (UV) spectroscopy, transitions between electronic energy
levels result in the absorption of electromagnetic radiation. The most probable
transition is from the highest occupied molecular orbital (HOMO) to the lowest
unoccupied molecular orbital (LUMO). UV spectroscopy can detect the presence of
conjugated chromophores and determine concentration of solutions (according to
Beer-Lambert’s law).
Experimental
Infrared
Spectroscopy
A preliminary
background scan was obtained by scanning dry KBr plates. The KBr plates were
cleaned with chloroform, dried, before one drop of the unknown sample was added
between them. It was ensured that no air bubbles were present, prior to
conducting an IR scan from 400 nm to 4000 nm.
Ultraviolet Spectroscopy
To prepare
0.001M solution of 4-methoxy-2-nitrophenol, 0.0173 g was weighed in a clean,
dry 50 mL beaker. 2 mL of ethanol was then added. The solution was stirred for 7
– 10 minutes before transferring into a 100 mL volumetric flask and made up to
the mark with deionised water. 1.0 mL of the solution was pipetted into a 10 mL
volumetric flask and it was made up to the mark with pH 7 buffer solution.
Similarly, solutions of 4-methoxy-2-nitrophenol in pH 8 and pH 13 were prepared
in 2 separate 10 ml volumetric flasks. The
UV spectrums for all three solutions were obtained from 300 nm to 600 nm on the
same paper and their isosbestic point, determined.
Results
Exercise 1-Compounds containing oxygen
Band Region / cm-1
|
Vibration Mode
|
Functional Groups
|
i) 2700 – 3600
|
O-H Stretch
|
O-H (of alcohols, phenols
and carboxylic acids)
|
ii) 1850 – 1650
|
C=O Stretch
|
C=O (of aldehydes, ketones
and carboxylic acid deriatives)
|
iii) 1300 – 1000
|
C-O Stretch
|
C-O (of ethers , esters,
alcohols, carboxylic acids and anhydrides)
|
Spectrum [a]
|
Wavenumber/ cm-1
|
Peak Intensity
|
Vibrational Mode
|
About 3000
|
Strong, sharp
|
C-H stretching
|
|
About 1750
|
Strong, sharp
|
C=O stretching
|
|
About 1250
|
Strong, sharp
|
C-O stretching
|
Since both C-O
and C=O groups are present and O-H stretching is absent, the compound is an
ester, C5H11COOCH3.
Spectrum [b]
|
Wavenumber/ cm-1
|
Peak Intensity
|
Vibrational Mode
|
About 3000
|
Strong, sharp
|
C-H stretching
|
|
About 1100
|
Strong, sharp
|
C-O stretching
|
Since C-O
stretching is present while both C=O and O-H stretching are absent, this
compound is an ether, C4H9OC4H9.
Spectrum [c]
|
Wavenumber/ cm-1
|
Peak Intensity
|
Vibrational Mode
|
About 3400
|
Strong, broad
|
O-H stretching
|
|
About 3000
|
Strong, sharp
|
C-H stretching
|
|
About 1050
|
Strong, sharp
|
C-O stretching
|
Since there is a
strong and broad peak at about 3400 cm-1 which indicates the
presence of the O-H group and C-O stretching is also present, it can be deduced
that the compound is an alcohol, C4H9CH(C2H5)CH2OH.
Exercise 2 – Aromatic compounds
Wavenumber/ cm-1
|
Functional Groups
|
Vibrational Mode
|
i) 3100-3000
|
Alkene
|
sp2
C-H stretching
|
ii) 2000-1700
|
Phenyl ring
substitution overtones
|
C-H stretching,
depends on positions of substitution
|
iii) 1650-1430
|
Aromatic
|
Aromatic C=C
stretching
|
iv) 1275-1000
|
Aromatic
|
=C-H in plane
bending
|
v) 900-690
|
Aromatic
|
=C-H
out-of-plane bending
|
Spectrum of Compound A
3-methylbenzonitrile
|
Wavenumber/ cm-1
|
Peak Intensity
|
Vibrational Mode
|
2062.96,3030.17
|
Medium, sharp
|
sp2 C-H stretching
|
|
2956.87,2924.09,2864.29
|
Medium, sharp
|
sp3 C-H stretching
|
|
2227.78
|
Strong, sharp
|
-CºN stretching
|
|
2000 - 1750
|
Weak
|
Aromatic overtones
|
|
1600.92,1583.56,1485.19,1456.26
|
Strong
|
Aromatic C=C stretching
|
|
918.12,881.47,788.89
|
Medium, sharp
|
=C-H out-of-plane bending
(meta-disubstituted)
|
A peak is observed at 2227.78 cm-1.
This peak indicates the presence of nitrile group. The pattern of the out of
plane C-H bends and aromatic overtones suggest a meta-substituted aromatic ring.
Therefore, sample A is 3-methylbenzonitrile.
Spectrum of Compound B
4-bromoanisole
|
Wavenumber/ cm-1
|
Peak Intensity
|
Vibrational Mode
|
3093.82, 3070.68,3043.67
|
Weak
|
sp2 C-H stretching
|
|
3003.17,2956.87,2937.59, 2902.87,2835.36
|
Strong, sharp
|
sp3 C-H stretching
|
|
1870.95, 1845.88
|
Weak
|
Aromatic overtones (para)
|
|
1577.77,1492.90,1483.26
|
Strong
|
Aromatic C=C stretching
|
|
1288.45, 1251.80,1240.23
|
Strong, broad
|
Aromatic C-O stretching
|
|
1072.42,1031.92
|
Strong, sharp
|
Aryl C-Br stretch
|
|
823.60,804.32,790.81
|
Strong
|
=C-H out-of-plane bending (para)
|
The aromatic C=C
stretch bands and the sp2 C-H stretch slightly above 3000 cm-1
indicate the presence of an aromatic ring. The out of plane C-H bends and
aromatic overtones indicates a disubstitution in para orientation. Aromatic C-O
stretches at 1288.45, 1251.80 and 1240.23 cm-1 suggests an aryl
ether while The sharp peaks at 1072.42 and 1031.92 cm-1 corresponds
to the aryl bromide stretch. Therefore,
sample B is 4-bromoanisole.
Spectrum of Compound C
2-ethylaniline
|
Wavenumber/ cm-1
|
Peak Intensity
|
Vibrational Mode
|
3458.37 – 3215.34
|
Strong, broad
|
N-H stretching (primary
amine)
|
|
3061.03,3018.60
|
Medium, sharp
|
sp2 C-H
stretching
|
|
2858.51
|
Strong, sharp
|
sp3 C-H
stretching
|
|
1950 – 1700
|
Weak
|
Aromatic overtones (ortho)
|
|
1616.35, 1583.56
|
Strong, broad
|
N-H bending
|
|
1494.83,1456.26
|
Strong, sharp
|
Aromatic C=C stretching
|
|
1315.45,1296.16,1274.95
|
Strong, broad
|
C-N stretching
|
|
796.60
|
Strong, sharp
|
N-H out-of-plane bending
|
|
748.38,705.95,644.22
|
Strong, broad
|
=C-H out-of-plane bending
|
The two N-H stretches at 3458.37 cm-1
and 3215.34 cm‑1 indicate a primary amine. The presence of sp2
C-H stretches at frequency above 3000 cm‑1 and the aromatic C=C
stretch peaks at 1494.83 and 1456.26 cm‑1 suggests an aromatic ring.
The aromatic overtones and out of plane C-H bend at 748.38cm-1 shows
that the aromatic ring is ortho substituted.
The band from 1315.45 – 1274.95 cm-1 corresponds to C-N stretch.
Therefore, sample C is 2-ethylaniline.
Exercise 3 – Compounds containing methyl groups
Spectrum of Compound D
isobutylamine
|
Wavenumber/ cm-1
|
Peak Intensity
|
Vibrational Mode
|
3373.50,3300.20
|
Strong, broad
|
N-H stretching (primary
amine)
|
|
2954.05 – 2870.08
|
Strong, sharp
|
sp3 C-H
stretching
|
|
1600.92
|
Medium, broad
|
N-H bending (primary
amine)
|
|
1469.76
|
Strong, sharp
|
Asymmetric deformation of
the H-C-H angles of a CH3 group
|
|
1386.82, 1365.60
|
Medium, sharp
|
Isopropyl C-H bending
|
|
1064.71
|
Medium, sharp
|
C-N stretching
|
|
786.96
|
Strong, broad
|
N-H out-of-plane bending
|
The presence of 2 the 2 N-H stretch
peaks at 3373.50 and 3300.20 cm-1 and N-H bending at 1600.92 cm‑1
indicate that a primary amine group is present. The antisymmetric deformation
of H-C-H angles of a CH3 group gives rise to very strong absorption
at 1469.76 cm‑1. A doublet at 1386.82 cm-1 and 1365.60 cm-1
of equal intensity is due to the C-H bending in an isopropyl group. Hence,
compound D is isobutylamine.
Spectrum of Compound E
tert-butylamine
|
Wavenumber/ cm-1
|
Peak Intensity
|
Vibrational Mode
|
3400 – 3200
|
Medium, broad
|
N-H stretching (primary)
|
|
2960.73,2872.01
|
Strong, broad
|
sp3 C-H
stretching
|
|
1600.92
|
Medium, broad
|
N-H bending
|
|
1469.76
|
Strong, sharp
|
Asymmetric deformation of
the H-C-H angles of a CH3 group
|
|
1388.75,1363.67
|
Strong, sharp
|
tert-butyl C-H bending
|
|
1244.09,1220.94
|
Strong, sharp
|
C-N stretching
|
The presence of broad N-H stretch
peaks from 3400 – 3200 cm‑1 and N-H bending at 1600.92 cm‑1
suggests a primary amine. Two peaks of unequal intensity can be observed at 1388.75
and 1363.67 cm-1, with the lower energy peak being more intense,
shows that a tert-butyl group is present. Hence, compound E is tert-butylamine.
Exercise 4 – Aromatic compounds
a)
Aniline has an amino substituent –NH2 that carry
nonbonding (n) electrons that can delocalize into the benzoic ring (as seen
below) resulting in greater conjugation. This causes a decrease in energy gap
for the transition of electrons and an increase in the probability of such
transitions. Less energy is needed for the 1Lb transition;
therefore, absorption shifts to a higher wavelength (red shift, also known as
bathochromic shift). When the n electrons are more available, the shifting of
the bands is greater.
Under low pH (high
H+ concentration), aniline is protonated and exists mainly as the
anilinium ion. The nitrogen atom in the anilinium ion has no unshared pairs of
electrons for conjugation with the benzene ring. As a result, the observed
bands should closely resemble that of unsubstituted benzene. An increase in pH, on the other hand, will
result in an increase in availability of n electrons and conjugation; this
correspondingly leads to greater bathochromic and hyperchromic effect for the 1Lb
transition.
b) The isosbestic point is defined as the wavelength at which two or more species have the same molar
absorptivity during a reaction. At this point, the absorbance value is
independent of the ratio of concentration of each individual absorbing
species. The presence of isosbestic
point implies that only 2 dominant species, 4-methoxy-2-nitrophenol and its
conjugate base 4-methoxy-2-nitrophenolate are in the solution.
The isosbestic
point for 4-methoxy-2-nitrophenol occurs at 407.0 nm with absorbance = 0.257.
Discussion
Infrared spectroscopy
In exercise 1, IR absorption ranges
were used to identify compounds with different types of oxygen bonding. We have
to look for not only the presence, but also the absence of absorption bands of certain
functional groups in order to deduce the unknown molecules. For example, ester
has bands due to both C=O and C-O-C groups but none due to the OH group unlike
in alcohol which contains O-H and C-O groups.
For exercise 2, the
positions of substituents in aromatic compounds can be determined from their
C-H out-of-plane bending and aromatic overtones. The out of plane C-H bending
vibrations (690-900cm‑1) are intense because of strong coupling between
adjacent H atoms. The relative number and patterns of aromatic overtones, from
1650 – 2000cm-1, enables the extent of substitution and the position
of substituents around the aromatic ring to be known. In addition, stretching
vibrations from the substituents were also utilised to identify the unknown
compounds.
Exercise 3 uses
the IR spectrum to identify isopropyl and tert-butyl group based on the C-H
bend in the region 1390 -1360 cm-1. Since they are structural
isomers, they will have the same functional group vibration. By analysing the
splitting of two peaks at around 1375 cm-1 corresponding to the -CH3 asymmetric
and symmetric bending, the germinal dimethyl group may be differentiated from
tert-butyl group. If isopropyl group is present, the
band is split into 2 peaks of nearly equal intensity; however, a wider split of
peaks with differing intensity will occur if ter-butyl group is present.
Before
performing scans of samples, a preliminary background scan was conducted. This allows
for the subsequent removal of ambient absorptions by any infrared active
atmospheric gases. The KBr cells was thoroughly washed with chloroform first as the
presence of impurities may cause inaccuracies in the spectrum. In addition, the plates should
only be held by its sides. KBr disks were chosen as it does not react with
sample and it does not absorb in the typical IR region of 4000 – 400 cm-1.
Ultraviolet (UV) spectroscopy
UV-Vis
spectroscopy is used to study electronic transitions with compounds. The
electronic transitions are basically of the HOMO-LUMO π→π*
type,
with an intense band at around 185 nm. The longest wavelength transition is a
low-intensity system centered near 225 nm; it is known as the benezoid band.
4-methoxy-2-nitrophenol
is highly conjugated and absorb strongly in the violet part of UV-visible
region; hence, it appears in the complementary colour of yellow.
Both -OH and -OMe
are electron donating groups that increase both the wavelength and the
intensity of the benezoid band while the electron withdrawing -NO2
group have no effect on the position of the B band. When pH is raised, the
compound is deprotonated to form 4-methoxy-2-nitrophenolate; it has an
additional negative charge on the oxygen atom that is more available for
interaction with the benzene ring, As a result, the π→π* transition
shifts the spectrum to longer wavelength and increases its intensity, resulting
in bathochromatic and hyperchormatic shifts in absorption. As pH increases,
there is an increasing red shift and greater peak intensities.
The solutions used must be diluted to
reduce the scattering of light during spectroscopy. After
filling the curvettes with sample compounds, the curvettes were wiped dry and
clean as any fingerprints will affect the absorbance. In addition, before a
next set of solutions were to be scanned, the curvettes were rinsed several
times with the new solutions to reduce contamination. All conditions such as
temperature, slit, path differences must be held constant for quantitative
analysis using UV spectroscopy.
Conclusion
Spectrum [a]
shows an ester, C5H11COOCH3, spectrum [b]
shows an ether, C4H9OC4H9
and spectrum [c] shows an alcohol, C4H9CH(C2H5)CH2OH.
The unknown compounds
A, B, C, D and E were identified by analyzing their IR spectra. Compound A is
3-methylbenzonitrile, compound B is
4-bromoanisole while compound C is
ethylaniline. Compound D is
isobutylamine and compound E is tert-butylamine.
The
isosbestic point of 4-methoxy-2-nitrophenol was found to be at λ = 407.0
nm with an absorbance of 0.257.
References
Spectroscopy,
4th edition. Lampman, Pavia Kriz, Vyvyan.
Inorganic
Chemistry 4th Edition, Shriver & Atkins, Oxford University
Press.
Comments
Post a Comment