Name: ____________
Chem 434
Instrumental Analysis
Vocabulary -Everybody must do this problem.
1. What is the difference between the following terms:
Hypsochromic shift (blue shift) and Bathochromic shift (red shift)
Hypsochromic or blue shift. Shift to shorter wavelengths associated with polar solvents.
Observed in n6B* transitions. Thought to occur because n electrons like to form hydrogen bonds
with solvent, and this H bond lowers the E of the n state but does not affect the E of the B state,
so the overall E of the transition increases and the wavelength gets shorter. This effect is fairly
large usually around 30 nm.
Bathochromic or red shift, shift to longer wavelengths is observed in B6B* transitions as you go
more polar solvents. Usually smaller than blue shift, only about 5 nm. Polar solvents tend to
lower the energy of the B orbitals, both B and B*, but the B* more, so the overall transition E is
lowered make it a longer 8
Singlet State, Doublet State and Triplet State
Singlet State - A ground singlet state is where all electrons paired and in their lowest energy
levels - An excited singlet state is when there are 2 electrons are different orbitals but they still
have opposite spins from each other. Doublet state - A molecule contains an unpaired electron
Triplet state - one in which there are 2 electrons in different orbitals but they have the same
spin.
External conversion and Internal conversion
External conversion - E of an electronically excited molecule is transferred to a solvent
molecule.
Internal conversion - E of an electronically excited molecule is passed to a lower electronic
energy state of the same molecule without emission of radiation.
predissociation and dissociation
Predissociation - An internal conversion in which E of molecule is transferred to a vibrationally
excited, lower electronic state, but vibration E is so high that a bond in the molecule ruptures.
Dissociation - E of light directly ruptures a bond in the molecule, so there is no absorption and
internal conversion before the bond breaks.
Harmonic oscillator and Anharmonic oscillator
Harmonic oscillator - An oscillator in which the potential energy field with respect to
displacement is symmetric.
Anharmonic oscillator - An oscillator in which the potential energy field is not symmetric with
respect to displacement.
Near IR and Far IR
Near IR - 78-2.5:m or 12800-4000 cm-1 or 3.8x1014 Hz- 1.2x1014Hz
Far IR - 500-1000:m or 200-10 cm-1 or 6.0x1012 Hz- 2x1013Hz
Chemical Shift and Spin-Spin Coupling
Chemical shift- Shift of a nucleus’ resonance frequency due to electrons around the nucleus
‘shielding it from the external magnetic field. It is proportional to external magnetic field.
Spin-Spin coupling - Shift of a nucleus’ resonance frequency due to magnetic fields of nearby
nuclei . These effects are relayed by electrons bonding the nuclei to each other. This effect is
independent of the external magnetic field.
In Class Problems
You may skip any one problem from 2-8
2.) One of the types of instrumental problems you looked at in the chapter on Molecular
Spectroscopy was scattered light. What exactly is scattered light, and when does it become an
important source of error?
Scattered light - Light the hits the detector that has not gone through the sample and the
monochrometer. It typically becomes a problem when the sample absorbance is very high, hence
very little light from the sample is hitting the detector, so the scattered light makes the detector
act as if more light was hitting it. This yields an artificially low absorbance reading. In the
lambda 2 this is why the absorbance reading in the <220 region are incorrectly low.
3.) In the chapter on Visible on UV spectroscopy you found that four kinds of transitions could
occur in UV/visible spectroscopy: F6F*, n6F*, n6B* and B6B*. How do these transitions
compare in approximate wavelengths, and expected molar absorptivities. Some of these
transitions also undergo characteristic shifts in polar solvents; which one shift, and how do they
shift.
F6F* - The highest E molecular transition, hence the farthest into the UV region (typically <150
nm) This is so far into the UV that it is not ordinarily studied.
n6F* - Next highest E, typically in the 150-250 nm range
Intermediate absortivities (100-3000)
Has a hypsochromic (Blue shift) in polar solvents
n6B*
In the 200-700nm region.
Low absorptivities (10-100)
Typically has a blue shift in polar solvents
B6B* - tied with B6B* for lowest E.
In the 200-700 nm region.
High absorptivities (1000-10,000)
Typically undergoes a bathochromic (red shift) with polar solvents.
4.) Not all molecules fluoresce. What special structural features does a molecule need to exhibit
fluorescence.
Fluorescence is associated with low energy B6B* transitions. Low energy B6B*, transitions, in
turn, are associated with aromatic groups or highly conjugated double bonds. Fluorescence is
observed in most unsubstituted aromatic compounds. Fluorescence is usually not observed in
heterocyclic aromatic systems. Fluorescence is also usually not observed in polycyclic systems,
unless the rings are merged into a single large rigid system.
5.) Describe how an FTIR works. Why does our FTIR contain a red laser?
A diagram like figure 7-42 really helps. IN this figure there is a beam splitting mirror that
sends
½ of the light to a fixed mirror, and ½ of the light goes to moving mirror. The light from these
two paths are then recombined, and focused on the detector.
When the moving mirror is exactly same distance as the fixed mirror, the two paths have
exactly the same length, and all the light recombines in phase so the detector ‘sees’ a maximum
in light intensity.
As the moving mirror is moved away, the light along this path now travels a longer
distance. As a result when it recombines, the light of certain 8’s are now out of phase so
destructive interference occurs when the light is recombined, and the detector ‘sees’ a smaller
signal. Each different frequency goes though oscillations from in phase and hitting the detector
and out of phase and not hitting detector, and all of these signals are piled on top of each other in
the total detector response. We use the Fourier Transform to sort out the different frequencies in
the signal, the intensity of the light with and without a sample in the light beam, to calculate our
absorbance spectrum.
The Red laser is NOT an IR light source. It is used to keep track of the position of the
moving mirror.
6.) I have a linear molecule with 4 atoms. How many normal mode vibrations should I expect in
an IR spectrum ?
Linear molecule: use the equation: modes = 3N-5
3(4)-5 = 12-5 = 7
If I had asked for a nonlinear molecule, then you would have used the equation 3N-6
12-6 =6
7.) Convert 488 nm into cm-1
1 cm = .01 m so
.01m/488e-9 m = 20491.8 waves /cm or simply cm-1
8.) Explain how spin-spin coupling arises in an NMR spectrum.
Coupling. The magnetic field that a nucleus ‘sees’ is either slightly augmented, or slightly
reduced, by the magnetic field that is produced by a nearby nucleus. This is not a direct
interaction; the magnetic field of the neighboring nucleus interacts with bonding electrons around
that nucleus, those electrons, in turn, interact with bonding electrons around neighboring atoms,
to relay the coupling effect through one or more chemical bonds (usually not more than 3 bonds
unless it is a rigid system)
Take home Problems
Here are the empirical formulas, IR, 1H and 13C spectra of three compounds.
What are the characteristic group frequencies you observe in each of the IR spectra?
Write a proposed structure for each molecule
Based on the structure, assign as many proton and carbon resonances in the NMR spectra
as possible.
Hint: I will put the ‘Aldrich Library of Infrared Spectra’ on reserve in the library. You
can use this to identify to help identify the IR spectrum of the compound.
Compound 1. C4H9Cl 1-chloro-2-methyl propane (135)
Compound 2
C4H8O Tetrahydrofuran (661)
Compound 3 C8H9BrO p bromophenetole (689)