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Chemistry is a highly investigative science. When a compound is
synthesized and purified by an organic chemist, it is essential
that the molecular structure (and
3D shape) of the compound is
determined. These are qualities that affect the physical properties
and reactivity of the molecule. Luckily chemists today have a vast
array of spectroscopic "gadgets"- instruments which allow for a
remarkably in depth structural assessment. In this article, we will
cover some of the most useful of these techniques.
There are many possible methods by which structure may be
elucidated by placing a chemical under a set of experimental
circumstances and seeing the outcome, and is important to note
which are the most useful and widely used in present day science.
X-ray diffraction is the ultimate technique in terms of assigning
structure to a molecule as it can be used to find the exact
position of atoms in a molecule. However, this requires a single
crystal of the pure compound and not all compounds can be
crystallised in the required way, or structure in solution is
required. Of the most widely used techniques, nuclear magnetic
resonance (NMR) is by far the most important for a chemist as it
uses the difference in the electronic environments of atoms to
elucidate structure with good efficiency.
Mass spectrometry
This tool works on the concept that ions can be separated
according to their mass to charge ratio by application of electric
and magnetic fields. Molecules are effectively weighed by this
technique and they must be positively or negatively charged
depending on the technique used. The plot generated by a typical
mass spectrometer is of m/z ($\frac{\text{relative molecular
mass}}{\text{relative charge}}$) against relative intensity.
A single peak corresponding to the molecular ion, M$^+$, or
pseudomolecular ion (the original molecule with an ion attached) in
some cases, is not seen due to the fragmentation of the molecule
within the instrument itself. The peak with the highest m/z value
is usually the molecular ion (or pseudomolecular ion) but this is
not always the case.
A mass spectrometer is a highly variable instrument but it does
include 3 main components: an ion source, a mass analyser to
separate the ions, and a detector. A high vacuum is used to avoid
collisions between sample ions.
What would the
difference be between a conventional mass spectrum and one
conducted in an environment where there is no vacuum?
Ionization
The crudest
technique is electron
ionization (EI) where high energy electrons are fired and
the sample vapour to generate positively charged molecular ions.
The electrons removed are usually from lone pairs or multiple bonds
(i.e. those that are less "core-like" and in higher energy
orbitals. This method can cause large scale fragmentation. Chemical ionization (CI)
techniques involve the use of a reagent gas in the ionization
chamber (e.g. CH$_4$) at a partial pressure of 1000 times that of
the sample. The reagent gas is more likely to be ionized and
collisions between the ionized reagent and the sample can produce
ions usually by some kind of proton transfer giving [M+H]$^+$.
Fast atom bombardment (FAB)
is useful for compounds which are hard to volatilize and involves
bombardment of a matrix produced by mixing the sample with viscous
glycerol by high energy atoms or ions of krypton or xenon.
Ionization of large compounds such as proteins if facilitated by
matrix assisted laser
desorption (MALDI) which involves co-crystallization of the
sample with an organic compound to give a matrix. This is then
ionized by a laser set to a wavelength that is strongly absorbed by
the organic compound.
The technique of
choice for most labs in electrospray ionization. The
sample is made up in solution with an added ionic compound such as
a methoxide ion (MeO$^-$) and is sprayed to produce a fine aerosol
which evaporate in the vacuum giving charged pseudomolecular ions.
The advantages of this technique is that is is far more gentle and
reduces the amount of fragmentation.
Mass
separation
Early instruments
used a series of magnetic and electric fields to deflect ions based
on their mass and charge in an arc. More modern instruments use
time of flight or quadrupole mass analyzers. These
methods are quite simple to understand. The time of flight analyser
accelerates packets of ions from the source to the same kinetic
energy using an electric field.
What is the
formula for kinetic energy of a particle? What can you say about
the velocity of different fragments based on this relationship?
As ions drift into
a field free region the different times that ions reach the
detector give a measure of the m/z value of each fragment ion at
this point.
Quadrupole mass analysis inolves
altering the applied voltages to an arragement of 4 oppositely
charged, equally polarised rods and an applied radio frequency
(that causes the ion to accelerate in an alternating manner to and
from the rods) to allow only ions of a desired mass : charge ratio
to pass through the analyser. This works on the principle that
unstable trajectories cause collisions with charged rods that
neutralise the ions. Simultaneously altering the values of voltage
and radio frequency separates ions according to m/z.
Another very
interesting application of mass analysis is its use in tandem mass spectrometry, in
which the first analyser allows the selection of an ion with a
particular m/z value, the fragmentation of which can then be
followed.
Detection
Electron
multipliers are commonly used as detectors as they give a
recordable electrical output. Array detectors may be used which
detect ions with a range of masses simultaneously.
A well set up mass
spectrometer is easily capable of measuring the mass of an ion to
within 5 ppm or better. The
value of a particular m/z peak is hugely improved due to this
precision as using the atomic masses of specific isotopes, it makes it possible
to determine the molecular formulae of ions through their exact
masses. For example, consider a peak recorded at m/z = 30.0469.
Potential candidates include:
$^{12}$C$_2$H$_6$
=
$^{15}$N$_2$ =
$^{14}$N$^{16}$O
=
[$^{12}$C = 12
gmol$^{-1}$, $^{15}$N = 15.00011 gmol$^{-1}$, $^{14}$N = 14.00307
gmol$^{-1}$, $^{16}$O = 15.99491 gmol$^{-1}$, $^{1}$H = 1.00782
gmol$^{-1}$]
Calculate the relative molecular masses
for each of these species using the accurate masses for each of the
isotopes given. Which species corresponds to the peak?
Fragmentation
We spoke briefly
about the usefulness of tandem mass spectroscopy earlier and the
value of following the fragmentation of a specific ionic species.
The fragmentation of a molecule is very useful as it gives
information about the structure of the molecule (though there are
better techniques available for structure determination).
The complexity of
the fragmentation pattern means that it is highly specific to the
compound. Mass spectroscopy is thus a useful tool in identifying
compounds whose mass spectra have already been recorded. Often a
molecule may show systematic fragmentation - for example peaks
spaced at m/z 14 in long chain alkanes, corresponding to systematic
losses of CH$_2$ groups. This is certainly useful in assigning each
of the peaks to a specific ionic species. A tandem mass spectrum
can be processed in such a way to show ions which fragmented to
certain products which is useful in mixture analysis. The major
advantage of mass spectroscopy is that the sample size for
meaningful data is very small. However, the pattern of peaks is
often more complex than expected and difficult to interpret.
Spectroscopy
The basis of all
spectroscopy techniques is the quantisation of electron energy
levels and the vibrational and rotational energy levels a molecule
possesses. Promotion to another energy level can occur if a
wavelength of light, corresponding to the difference in energy
between two energy levels, is absorbed by a molecule ($\Delta$E =
h$\upsilon$). The exact frequency of light absorbed is highly
dependent on the particular molecule. DIfferent frequencies of
light also cause different types of transition (again
dependent on the difference in energy between two energy
levels).
Nuclear magnetic
resonance (NMR)
Nuclei can possess
a spin and thus nuclei can
have an associated weak magnetic field. In an applied magnetic
field, an interaction with this spin gives rise to a set of nuclear
spin energy levels. Absorption of radiowaves of the correct
frequency facilitates a transition between energy levels that
produces an NMR signal.
The nuclear spin
quantum number, I, of a nucleus gives rise to (2I + 1) energy levels in an
applied field. The value of I for a particular nucleus depends on
the atomic number and the number of neutrons it contains. It may be
taken as a guide that:
- nuclei with odd mass number have
$\frac{1}{2}$ an integer spin value
- nuclei with odd atomic number and an odd number of neutrons have a
spin value that is an integer
- nuclei with even atomic number and an even number of neutrons have 0
spin
What type of spin do the following
species have?
$^{12}$C
$^{13}$C
$^{1}$H
$^{2}$H
$^{19}$F
$^{10}$B
$^{11}$B
Research what particular nuclear
spin quantum number each has. How many different energy levels does
each species give rise to in an applied magnetic field?
Environments
The exact
difference in energy between various spin states depends upon the
local magnetic field around a nucleus. This local field is
determined mainly by the electron density surrounding a nucleus.
Electrons shield the
nucleus by opposing the applied magnetic field and so the apparent
magnetic field felt by a nucleus is reduced. The energy separation
of spin states depends on the magnetic field. The greater the strength of the magnetic
field, the greater the energy separation.
Thus species which are highly electronegative and remove
electron density from around a nucleus causes an effective
deshielding of that nucleus. The applied magnetic field experienced
by the nucleus increase, as does the separation in the energies of
the spin states. Thus the frequency of radio wave required for
resonance is shifted to a higher frequency.
The number of peaks in
an NMR spectrum corresponds to the number of different environments. Thus with
NMR spectra, it is important to acknowledge and look for symmetry
in the molecules that you encounter. For example, the $^{13}$C
spectrum of 5-nonanone contains 5 peaks whereas that for 2-nonanone
contains 9 peaks.
5-nonanone:
2-nonanone:
Can you draw the structures of
5-nonanone and 2-nonanone?
Given the above information, what
would the $^13$C NMR spectrum of the following molecules look like?
Draw in any lines of symmetry you see on the molecular
structures.
The NMR shift scale itself is established relative to the resonant
frequency of a reference compound, meaning that any differences in
the external magnetic field strength will affect all frequencies by
the same factor. Thus the shift scale remain unaffected by any
differences in effective magnetic field experienced by nuclei in a
sample, whether this is due to the actual applied magnetic field or
external variations, allowing for experimental comparison.
TMS or tetramethylsilane is the
reference compound of choice, being chemically inert, producing one
signal in either $^13$C or $^1$H NMR that is far to the right of an
observed spectrum.
$^{13}$C and $^{1}$H
NMR
By far the most commonly used types of NMR are $^{13}$C and $^{1}$H
NMR. Both of these nuclei have nuclear spin quantum number, I =
$\frac{1}{2}$ and consequently possess 2 energy levels. Note that
the resonant frequency of each nucleus in a certain magnetic field
differs on the MHz level (c.f. environmental differences on the
range of hundreds of Hz). For example in a 4.7 tesla magnetic field
the resonant frequency of an $^{1}$H nucleus is 200 MHz whereas
that of $^{13}$C is 50 MHz. Note that other types of high-field NMR
spectroscopy are also possible including $^{19}$F and $^{31}$P
NMR.
The advantage of $^{13}$C is its simplicity. There are usually
fewer carbon environments in a molecule compared with hydrogen
environments. The low natural abundance of $^{13}$C (1.1%) means
that no C-C
coupling is
observed in a $^{13}$C NMR spectrum. $^{1}$H NMR is something you
may already have encountered. $^{1}$H NMR gets very complicated due
to the degree of coupling to many H atoms. This does have its
advantages in elucidating more information about the structure of
the molecule.
Coupling
Coupling is a very important phenomenon in NMR but what exactly is
it and how does it come about?
We know that the frequency at which a nucleus resonates depends
upon the local magnetic field it experiences. Coupling refers to a
situation where a signal which would usually be a single peak is
split about the original
chemical shift position. The degree of separation gives the value
of a
coupling constant J,
measured in Hz. The coupling constant is written
$^x$J$_{\text{Y-Z}}$ where $x$ represents the number of bonds
through which Z is coupling to a peak in a Y NMR spectrum.
In general, when a nucleus couples to $n$ equivalent nuclei with a
certain spin value, $I$, its resonance signal is split into
($2nI + 1$) peaks. Coupling
occurs due to the various spins of the coupling nucleus. Take for
example a $^19$F nucleus of $I = \frac{1}{2}$. We know that the F
nucleus thus has (2I + 1) = 2 energy levels. The separation between
the 2 energy levels is so small that it is almost equally likely
for a given $^19$F nucleus to be either spin up or spin down.
In the spin up state, the fluorine atom reinforces the magnetic
field experience by the nucleus it is coupling to and shifts the
resonant frequency to one slightly higher than the frequency the
nucleus would normally resonate at.
What happens if the $^19$F
nucleus is in the spin down state?
Coupling to 2 $^19$F nuclei gives a, ($2(2)(\frac{1}{2}) + 1$) = 3,
triplet of peaks. This can be seen through a consideration of the
possible spin states of fluorine.
Consider the $^{13}$C spectrum of
trifluoromethane.What possible combination of spin statescould the
3 fluorine atoms have? How many peaks is the $^{13}$C signal split
into? Can you determine the relative intensity of these
peaks?
You may be able to see the pattern with respect to $I =
\frac{1}{2}$ nuclei gives a pattern of peak corresponding to the
nth row of the
Pascal's
triangle. This can be reasoned in terms of combinatorics.
Take for example the central peak of the spectrum representing
coupling to 2 $^19$F nuclei. We have 2 nuclei of which one needs to
be spin up and the other spin down. So $^2\textbf{C}_1$ = 2.
Of course coupling can be between different nuclei, in which case
the
relative values of the
coupling constant $^x$J$_{\text{Y-Z}}$ requires consideration when
sketching the shape of the NMR.
In terms of coupling it is important to note that:
- No C-H coupling is usually seen in $^{13}$C NMR as should be
expected. This is because of decoupling. Broadband proton
decoupling involves the irradiation of the sample over a large
frequency range (to target all protons) in such a way that the
nuclear spins of the $^1$H atoms oscillate so rapidly between spin
"up" and "down" that their coupling contribution averages to
0.
- Splittings due to couplings between equivalent nuclei are not
seen, though equivalent spins do interact with each other.
The NMR scale
Taking the $^{13}$C NMR spectrum as an example, the scale can be
conveniently broken down. Note that electron withdrawing groups
cause a deshielding of the nucleus in question, increasing the
effective magnetic field it experiences and thus increases the
energy separation and frequency at which the nucleus
resonates.
Natural abundances
If there is coupling occurring due to a spin active species not
present in large proportions in the sample, then satellite peaks
are often seen. This is because for a proportion of the sample,
coupling of the original signal occurs and splitting is seen, but
for a significant proportion of the sample, no coupling is seen.
Thus the original peak position contains a signal on the spectrum,
superimposed with the splitting pattern due to the spin active
proportion of the sample.
The relative intensities of these peaks are dictated by the
terrestrial abundances of the coupling nucleus. Take for example
the $^1$H NMR spectrum of chloroform as an example of this.
Consider the $^15$N NMR spectrum
a molecule which contains a Pt-N bond and no other species bonded
to the nitrogen that would make an impact on the splitting. The
following isotopes of platinum exist in the following
abundances:
$^{190}$Pt = 0.014%
$^{192}$Pt = 0.782%
$^{194}$Pt = 32.967%
$^{195}$Pt = 33.832%
$^{196}$Pt = 25.242%
$^{198}$Pt = 7.163%
Research the spin values of the
following isotopes. What would the $^15$N NMR look like? What are
the intensities of the observed peaks?
Infrared spectroscopy
IR spectroscopy works on the investigation of transition between
vibrational energy levels in molecules. A typical IR plot describes
the the percentage transmission across a range of wavelengths of
applied infrared radiation. Absorption peaks thus correspond to
downward pointing features. Frequency is given in wavenumbers
where:
wavenumber, $\tilde\nu$ = $\frac{1}{\lambda}$
$\nu$ = $\frac{c}{\lambda}$
$\therefore\nu = c \times \tilde\nu$
Thus, $\nu$ $\alpha$ $\tilde\nu$
and as $\Delta E = h\nu$
hence $\Delta E$ $\alpha$ $\tilde\nu$
Large wavenumber values correspond to vibrational transitions which
have a large associated energy change.
The
harmonic oscillator is
a useful model in this scenario and, mechanically, consists of a
weight hanging from a spring whose displacement from an equilibrium
position causes a
restoring
force to be generated. The weight therefore oscillates about
an equilibrium position.
Hooke's law, $\nu$ $\alpha$ $\sqrt{\frac{k_f}{m}}$ where $\nu$ is
the frequency of oscillation, ${k_f}$ is the force constant
(difficulty of stretching in Nm$^{-1}$) and m is the mass, shows
what the frequency of oscillation depends upon.
This model can be applied to a diatomic, modelled as two masses
connected by a spring. It turns out there is a fair correlation
between bond strength and the force constant. As there are now 2
masses involved we can take both into account by calculating the
reduced mass of the system,
$\mu$.
$\mu = \frac{m_1m_2}{m_1 + m_2}$ which has units of kg
molecule$^{-1}$
Consider the situation where m_1
is far greater than m_2. What does the reduced mass simplify
to?
The frequency of vibration for a diatomic is given by:
$\tilde\nu$ = $\frac{1}{2\pi c}\sqrt{\frac{k_f}{\mu}}$
So, the smaller $\mu$ becomes, the higher the frequency of
oscillation. The higher the value of $k_f$ the higher the frequency
of oscillation. Here is brief outline as to the division of a
typical infrared spectrum.
Of course the actual vibration of a molecule is far more
complicated, but can be broken down to
normal modes composed of cis and
trans bends and various symmetric and asymmetric stretches. So a
single absorption can correspond to many bond vibrations, but it is
often a case that a given vibration
largely corresponds to a certain
part of a molecule vibrating. This allows for functional group
identification which is very useful.
The larger the
dipole of a
vibrating bond the stronger the absorption so the larger the peak
on the spectrum. This is because light interacts with the vibrating
dipole because it is an oscillating electromagnetic field.
Can you classify the strength of
absorption when the following bonds vibrate?
Symmetrical vibrations with no change in the dipole moment may be
determined using a technique called
Raman spectroscopy which analyses
the scattering of light by vibrating bonds. Determining whether a
given vibrational mode is IR active, Raman active or both is
something beyond the scope of this article but we can say
that:
- Homonuclear diatomics and heteronuclear diatomics are both
Raman active.
- Homonuclear diatomics will clearly only be Raman active as they
have no change in dipole moment on vibration.
Preparation
For gaseous samples, a glass gas cell is filled. The windows of
the cell through which IR passes are composed of cut and polished
single crystal inorganic salts, not glass as this is an absorber of
IR radiation. Salts have characteristic absorbances outside of the
investigative range.
Liquid samples are applied as a thin film between sodium chloride
plates. Thus liquid samples are quick and easy to analyse.
Solid samples can be prepared by grinding them up in a pestle and
mortal witha viscuous hydrocarbon oil, nujol to make a mull that is
sandwiched as a thin layer between sodium chloride plates. The
solid may also be ground with dry KBr and the mix put into a mould
and compressed in a hydraulic press. KBr exhibits cold flow under
these conditions and the sample becomes incorporated into a
transparent disk. The advantage of the latter method is it avoids
the introduction of a hydrocarbon which typically absorbs over the
investigative range.
Characteristic
absorptions
Analysing an IR spectrum involves assigning absorption peaks to
certain characteristic bond stretches. Bending vibrational modes
are low frequency and found mainly in the fingerprint region- a
useful area of the spectrum to give identity to samples that are
one of a possible known range. Tables of typical absorbances are
readily available and thus assignment of absorptions is an
elementary task. Such analysis can often identify functional groups
which means along with NMR, IR can prove a powerful complementary
tool in structure determination.
We shall consider some of the more interesting examples in terms of
absorption frequencies.
O-H absorption
O-H absorption peaks are typically broad due to the presence of
hydrogen bonding within the
sample. The strengths of O-H bonds vary slightly across the range
of absorption frequencies. The peak is found at ~3200- 3600
cm$^{-1}$.
Consider the case of the above
molecule. What would the the O-H absorption peak look like in this
case? Give your reasoning, no matter how speculative it may
be!
NO$_2$ group
absorptions
The NO$_2$ group has 2 stretches corresponding to a symmetric and
antisymmetric bond stretch as seen above.
How easily is each stretch
generated? Consider what this means about force constants and
bonding considering our previous modelling.
Carbonyl groups
Patterns of absorption in the IR specturm for the C=O group are
highly interesting as the exact frequency of the stretch can point
to the sort of functional group present and can even give a guide
to its chemical reactivity.
The ketone carbonyl gives a reference of the standard frequency of
the C=O group as ~1715 cm$^{-1}$. As we saw previously an increase
in bond strength corresponds broadly to an increase in $k_f$,
meaning that a molecular environment can cause carbonyl absorption
to
rise to a higher
frequency or
lower
frequency by respectively
strengthening or
weakening the carbonyl
bond.
- Electron withdrawing carbonyl substituents strengthen the
carbonyl as it increases the degree of polarity of the C=O bond
increasing the electrostatic contribution to bond strength.
- Electron donating substituents weaken the carbonyl as donation
occurs into the $\pi$* orbital, an orbital which has a large
coefficient on the carbon atom causing a reduction in the polarity
of the bond and also weakening the $\pi$ bond due to its
antibonding nature.
Using this information try to
explain the absorptions corresponding to the following C=O
stretches:
Of the above molecules, the acid chloride is generally the most
reactive with the amide being the least reactive. If we consider
that the mode of reaction at a carbonyl is generally by
nucleophilic attack, it is easy to see why this is the case. The
acid chloride with its highly polarised C=O bond makes the molecule
a good electrophile. The $\delta ^+$ contribution on the carbon is
increased by the
inductive
electron withdrawing effect of the Cl substituent, increasing its
reactivity to nucleophiles. The lone pair on the N atom within the
amide acts
conjugatively,
making the carbon less electrophilic by donation. This reduces the
reactivity of the molecule and its ability to act as a good
electrophile.
Conjugation and ring
strain
The carbonyl bond is weakened by a phenomenon known as conjugation.
Conjugation is said to be taking place if two double bonds are
separated by just one single bond. For example:
Conjugation lowers the frequency of the normaal stretch by ~30
cm$^{-1}$.
If the carbonyl forms part of a ring, then the
smaller the size of the ring the
higher the stretching
frequency.This is due to the fact that on vibration, the carbon of
the carbonyl group moves along with the oxygen but feels some
resistance due to the C-C bonds it has with other carbons in the
ring. The smaller the angle in the ring (and thus the smaller the
ring itself), the more these C-C bonds are
compressed during vibration. This
compression requires more energy meaning a higher absorption
frequency.
Using your newly gained knowledge
on conjugation and ring strain, can you predict the carbonyl
stretching frequencies of the following molecules?
Raman spectroscopy
This technique is so closely related that it can be considered
as complementary to IR spectroscopy and considers photon
scattering. By complementary we mean that normal modes of vibration
that are not IR active are often Raman active. A laser irradiates a
sample with monochromatic light in the visible red or infrared
regions of the spectrum. The wavelength is chosen such that
absorption does not occur but on a small scale, scattering occurs
due to some interaction.
In most cases, the frequency of the scattered photons is
identical to the initial light. This is termed Raleigh
scattering. However in some cases, a fraction of the
scattered photons are in fact higher or lower in energy than the
incident photons. The energy shift corresponds to losses and gains
in energy on interaction with the molecule. The amount of energy
lost or gained is equivalent to the energy difference between the
two molecular energy levels.
$E = h\nu$
If the molecule undergoes a transition to a higher energy
level, then the energy of a scattered photon interacting with such
a molecule under transition will be reduced in energy in a Raman
scattering known as a Stokes
scattering.
$h\nu_{Stokes}$ = $h\nu_{L} - (E_{upper} - E_{lower})$
The anti-Stokes scattering is the energy change accompanying a fall
in molecular energy level.
$h\nu_{anti-Stokes}$ = $h\nu_{L} + (E_{upper} - E_{lower})$
Measuring this energy shift enable us to measure the separation
between molecular energy levels.
Vibrational energy level $\nu$ = 0 is the only state significantly
populated, so the most intense scattering comes from the transition
between $\nu$ = 0 and $\nu$ = 1.
Stokes photons with energy $h\nu_{Stokes}$ = $h\nu_{L} -
\hbar\omega$, where $\hbar\omega$ is the energy separation between
these two layers. The frequency of these photons is
therefore:
$h\nu_{Stokes}$ = $h\nu_{L} - \hbar\omega$
$\nu_{Stokes}$ = $\nu_{L} - \frac{\omega}{2\pi}$
Raman spectroscopy is useful as there does not need to be a change
in dipole moment during vibration, only the polarizability of the
molecule has to change. The scattering is weak and so the detection
equipment is quite complex.