12 -- Part 1: INFRARED SPECTROSCOPY AND MASS SPECTROMETRY
To propose structures for possible compounds, use the chemical shifts, splitting patterns, and integrations shown in a proton NMR spectrum.
The number of peaks and their chemical shifts in a 13C NMR spectrum can be used to determine the number of carbon atoms in a compound.
Predict the major features of the 13C NMR spectrum with a chemical structure.
The structures of unknown organic compounds can be determined by combining the information from NMR, IR, and mass spectrums.
The left occipital lobe of a human brain is shown to have a tumor in it.
The largebore, broadband NMR instrument produces a rapidly varying field to obtain spatial information that is used to form an image.
NMR can be used with a very small sample, and it does not harm it.
Many structures can be determined using only the NMR spectrum, which provides a lot of information about the structure of the compound.
More often than not, the structures of complicated organic molecules can be determined with the help of other forms of spectroscopy and chemical analysis.
A wide variety of nuclei, including 1H, 13C, 15N, 19F, and 31P, are studied using NMR.
Nuclear Magnetic Resonance (NREM) was first used to study protons, and the most common is the 1H NMR spectrometer.
Unless a different nucleus is specified, "Nuclear magnetic resonance" is assumed to mean "proton magnetic resonance".
We begin our study with 1H and end it with 13 C.
The nucleus of a protons has an odd atomic number of 1.
The spinning protons can be seen as a sphere of positive charge.
The charge is moving in a loop of wire.
When a small bar magnet is placed in the field of a larger magnet, it twists to align itself with the field of the larger magnet, which is a lower energy arrangement than an orientation against the field.
There have been dom orientations in the absence of an external magnetic field.
There are more spins when the a@spin state is lower.
The Earth's magnetic field is about half a gauss.
The energy difference between the two spin states of a protons is small.
For a strong magnetic field of 25,000 gauss, it is only about 10-5 kcal>mol.
The small energy difference can be detected by the NMR.
The spin of the protons can change from a to b or from a to a when they interact with a photon.
A nucleus aligned with the field can absorb the energy needed to flip.
When a nucleus is subjected to the right combination of magnetic field and electromagnetic radiation to flip its spin, it is said to be "in resonance", and its absorption of energy is detected by the NMR spectrometer.
The RF region of the spectrum contains the fields of currently available magnets.
The radio frequencies needed for resonance are calculated based on the field and the most powerful magnets are usually designed for the most practical price range.
The most common operating frequencies for student spectrometers have been 60 MHz, corresponding to a magnetic field of 14,092 gauss.
Higher-resolution instruments operate at frequencies of 200 to 600 MHz, corresponding to fields of 46,972 to 140,918 gauss.
We have considered the resonance of a naked protons in a magnetic field, but real protons in organic compounds are not naked.
They are protected from the external magnetic field by electrons.
A loop of wire is moved into a magnetic field.
The principle of the electric generator is that the electrons in the wire flow around the loop in the direction shown in Figure 13-4.
The electric current creates a magnetic field.
In a molecule, the electron cloud around each nucleus acts like a loop of wire.
This current is caused by the magnetic field opposing the external field.
If the external field is increased to 70,460 gauss, the effective magnetic field at the proton is increased to 70,459 gauss.
If all protons were protected by the same amount, they would all be in resonance.
protons are protected by different amounts in different environments
A magnetic field can induce a current in a wire.
This current causes its own magnetic wire loop to move in the opposite direction of the applied field.
The nucleus feels a slightly weaker field because of the magnetic field set up by the current.
A protons is protected by electrons.
The hydroxy protons absorb at a lower field than the methyl protons, but still at a higher field than a naked protons.
The shielding effects of electrons at different positions are different because of the diverse and complex structures of organic molecule.
Let's look at what happens in an NMR spectrometer.
The sample compound has particles placed in a magnetic field.
While still in the magnetic field, the protons are subjected to radiation of a Frequency they can absorb by changing the orientation of their magnetic moment relative to the field.
If protons were isolated, they would all absorb the same magnetic field.
The shielding of protons in a molecule depends on their environment.
The radiation at different magnetic field strengths is absorbed by the protons in the molecule.
The original idea was to vary the magnetic field and plot a graph of absorption energy as a function of the magnetic field strength.
The H shielded hydroxy proton appears to be in the left field.
Information about the electronic environment of the protons in a molecule is provided by the NMR spectrum.
The relative strength of the magnetic field that causes the protons to absorb energy from the RF transmitter is what determines each environment.
There is a difference between the resonance field of the protons being observed and that of TMS.
CH3 is enough to distinguish individual protons because the signals often differ by a few thousandths of a gauss.
The methyl groups of TMS are electron-rich, and their protons are protected.
One strong absorption is given by the 12 protons in TMS.
A small amount of TMS is added to the sample, and the instrument measures the difference in magnetic field between where the protons in the sample are absorbed and where the TMS are absorbed.
The chemical shift of the protons is what determines the distance downfield of TMS.
The chemical shift in a sample is measured by a constant magnetic field by newer spectrometers.
The horizontal axis of the NMR spectrum is adjusted to show the chemical shift between the signals of a protons and TMS.
A chemical shift in parts per million can be calculated by dividing the shift inhertz by the frequencies measured in millions ofhertz.
In a 300-MHz spectrum, 1 part per million is 300 part per million.
The use of absorptionless chemical shifts standardizes the values of all NMR spectrometers.
The d scale is the most common scale for chemical shifts.
The d scale goes to the left of the spectrum when most protons are deshielded.
The spectrum is adjusted in two different ways.
Downfield toward 60 MHz more deshielded protons.
The chemical shift of a A 300-MHz spectrometer records a protons absorption at a 2130 Hz downfield, which is the same in any shielding from TMS.
Determine the chemical shift.
The fraction spectrometer is the chemical shift.
The chemical shift is not changed at 60 MHz.
There are two signals from methanol and a reference peak in the 300-MHz NMR spectrum.
The protons are absorbed by the methyl protons.
We say that the methyl protons absorb at d 3.4 because of the chemical shift.
It has a chemical shift of d 4.8.
The deshielding effects of the oxygen atom are shown by the hydroxy protons and the methyl protons.
The chemical shift in an alkane is about d 0.9.
The methyl protons are protected by an additional 2.5 parts per million.
Similar deshielding effects can be produced by other eronegative atoms.
The table compares the chemical shifts of the two compounds.
The chemical shift of the protons depends on the electronegativity of the substituent.
The chemical shift depends on the distance from the protons.
The chemical shift of the hydroxy protons is d 4.8, and it is separated from oxygen by one bond.
The chemical shift of the protons is 3.4 and they are separated from oxygen by two bonds.
The effect of an electron-withdrawing substituent decreases with increasing distance, and the effects are usually negligible on protons that are separated from the electronegative group by four or more bonds.
The decreasing effect can be seen by comparing the chemical shifts of the various photos.
The metal container at the back has a negative shielding effect.
In 1-bromobutane, protons on the magnet, cooled a carbon, and b protons are all protected by a liquid bath.
The b protons are protected by a negligible used to control the spectrometer and amount.
The third chlorine moves the chemical shift to d 7.2 for chloroform.
The peak is slightly less moved by each additional chlorine.
The chemical shift of the remaining methyl protons is changed by the addition of chlorine atoms.
The chemical shift of a protons is determined by its environment.
The reasons for some of the more interesting and unusual values can be found in the table of representative chemical shifts.
There is a table of chemical shifts in Appendix 1.
Predict the chemical shifts of the protons using Table 13-3
The carbonyl group in acetic acid is next to the methyl group.
COOH should absorb between d 10 and d 12.
The table shows how aromatic rings and double bonds affect the vinyl and aromatic protons.
The same type of circulation of electrons that shields the nucleus from the magnetic field is what causes these deshielding effects.
The induced field is at the center of the ring.
Most aromatic protons absorb in the range of d 7 to d 8.
The chemical shift for its protons is an average of all the possible orientations because benzene is tumbling in the solution.
If we were able to hold benzene problems.
All chemical shifts are affected by neighboring substituents.
The numbers assume that alkyl groups are the only other substituents present.
Appendix 1 has a more complete table of chemical shifts.
The one with the benzene ring edge-on to the magnetic field would absorb at a higher field.
The figure shows the spectrum of toluene.
The aromatics absorb protons d 7.2.
The methyl protons are absorbed by d 2.3.
The aromatic ring of electrons protects the aromatic protons from the pi electrons of an alkene.
The effect in the alkene is smaller than in benzene because there is not a large ring of electrons.
The applied field at the middle of the double bond is opposed by the magnetic field generated by the motion of the pi electrons.
The vinyl protons are on the edge of the field, where the field bends around and reinforces it.
Most vinyl protons absorb in the range of d 5 to d 6.
The applied magnetic field along the axis of the induced field ring is opposed by the magnetic field of the circulating aromatic electrons.
External field lines curve around and reinforce the applied field.
The aromatic protons absorb at chemical shifts near d 7.2.
There is a pi bond.
The magnetic field was caused by H.
The triple bond has two pi bonds.
Acetylenic hydrogens absorb around d 2.5, compared with d 5 to d 6 for vinyl protons.
As the molecule tumbles in solution, a cylinder of electrons can circulate to produce a magnetic field.
The result is a resonance around d 2.5 when this shielded orientation is averaged.
The axis of this field has the acetylenic protons along it.
Between d 9 and d 10, CHO absorb at even lower fields than vinyl protons and aromatic protons.
The carbonyl group gives a resonance between d 9 and d 10.
The concentration of H in amines is dependent on the concentration of protons.
H when alcohol or amine is mixed with a non-hydrogen-bonding solvent such as CCl4.
The signals are observed around d 2.
protons exchange from one molecule to another during the resonance During this exchange, the protons absorb over a wide range of frequencies and field strengths.
The positive character of carboxylic acid protons is due to their bonding to an oxygen next to a carbonyl group.
O absorbed at chemical shifts greater than d 10.
The C R exchange broadens the absorption of acid protons.
The carbonyl group protects the protons of acetic acid.
An offset trace shows the acid protons at d 11.8.
The chemical shift is not scanned in the usual range of the NMR spectrum.
The sum of d 9.8 read from the trace and the d 2.0 offset make up the acid protons.
The number of NMR signals is related to the number of different kinds of protons present in the molecule.
There are two types of protons in -butyl ether.
The three methoxy protons give rise to a single absorption at d 3.2.
The butyl protons are different from the methoxy protons.
There are two types of protons in butyl ether.
The same shielding has the same chemical shift.
The butyl protons are similar.
Butyl acetoacetate has three different types of protons.
There are different types of protons in the molecule and there may be fewer signals in the NMR spectrum.
There are only two distinct signals in the spectrum.
The amount of shielding felt by any of the substituents on the ring is not influenced by the amount of aromatic protons.
The aromatic protons produce two signals, but they happen to be nearly the same chemical shift.
There are only two absorptions of xylene in the spectrum.
The peak of the aromatic Hb and Hc is accidentally equivalent.
Determine the number of different kinds of protons.
The aromatic region around d 7.2 has more than one sharp absorption.
You don't know how many protons there are.
When it goes over a peak, the second trace rises.
The area of the peak affects the amount of the trace rising.
Some of these integrals can be measured using a millimeter ruler.
The area of each peak is not represented by the other integrals.
The numbers correspond to the heights of hydrogens and the rises in the trace.
The peaks at d 1.2 and d 3.2 represent the number of hydrogens that would look for a compound.
We have to use 6 : 8 : 12 or 3 : 1 ratio to understand the structure.
The blue trace rises by an amount that is proportional to the area under the peak.
The integrated spectrum of a compound is shown in Figure 13-20.
The 12 protons in the molecule have been integrated vertically by the integrator.
The p@donors of electron density represent the protons.
They protect the protons on the carbon atom.
The signal at d 3.8 has an integral of 3.0mm.
The integrator moves close to two protons at d 2.6.
There are signals in the spectrum when attached to aromatic rings.
The system of the ring is 6 p.
The ratios of the peak areas are determined.
To pair up the compounds with their spectrum, use this information, shielding of ortho and para together with the chemical shifts.
Nuclear Magnetic Resonance Spectroscopy peaks in each spectrum to the protons they represent.
The shielding electrons are subjected to the external magnetic field while the protons are subjected to the internal magnetic field.
The absorption frequencies of the protons we are observing are affected if there are other protons nearby.
The small magnetic fields affect nearby protons in predictable ways.
Consider the spectrum of 1,1,2-tribromoethane.
There are two signals with areas in the ratio.
The bromine atoms shield the smaller signal at d 5.7.
There is a larger signal at d 4.1.
The signals are triplet and doublet and do not appear as single peaks.
The signal for the Hb protons of 1,1,2-tribromoethane at d 4.1 can be seen in Figure 13-22.
There is a small magnetic field of the adjacent protons.
Every molecule in the sample has a different orientation of Ha.
Ha is aligned against the field in some molecules and against the field in others.
The Hb protons feel a slightly stronger total field when Ha is aligned with the field.
The Hb protons absorb at a higher field when the magnetic moment of the Ha is aligned against the external field.
There are two absorptions of the doublet.
The two absorptions of the doublet are nearly equal in area because half of the molecule has Ha aligned with the field.
The Ha absorption occurs at a lower field when the Hb spins.
Two permutations, where the Hb spins of Hb protons allow Ha to absorb at its normal position.
The peak area ratios are 1 and 2.
There is a property called spin-spin splitting.
The second proton must split the first one.
The middle signal is twice as large as the others because it is related to two possible spin permutations.
The two Hb protons absorb the same chemical shift and do not split.
Because of their resonance at the same combination of frequencies and field strength, strontiums that absorb at the same chemical shift can't split.
More complicated systems can be analyzed after the splitting of 1,1,2-tribromoethane.
Consider splitting the signals for the ethyl group.
When you see splitting, look by three protons, as a quartet of areas 1 : 3 : 3.
This pattern is typical of the ethyl group.
ethyl groups are carbon atoms.
The alkyl substituent has a small effect on the chemical shifts of the aromatic protons.
In this high-resolution spectrum, the aromatic protons split in a complicated manner.
The aromatic protons would not be resolved at a lower field, and they would appear as a single peak.
There is no spin-spin splitting between the aromatic protons and the ethyl group in ethylbenzene.
The protons are too far away to be magnetically coupled.
The aromatic protons are a multiplet.
Spin-spin splitting involves the separation of protons that are separated by three bonds and bonding them to carbon atoms.
Most spin-spin splitting takes place between protons and carbon atoms.
Most of the time, protons on the same carbon atom are 888-609- 888-609- 888-609- 888-609- In most cases, the protons on the same carbon atom can't split.
If nonequivalent, spin-spin splitting is usually observed.
Normally, spin-spin splitting is observed.
The most common case is C 2 C.
Spin-spin splitting can't be seen when particles are separated by more than three bonds.
These cases are unusual and occasionally occur.
We only consider nonequivalent protons on adjacent carbon atoms to be magnetically coupled.
There are two multiplets in the upfield part of the ethyl benzene spectrum.
The quartet at lower field leans toward the triplet at a higher field, and often leans toward the protons that are causing the splitting.
There is a splitting pattern in the NMR spectrum.
A strong doublet at a higher field and a weak multiplet at a lower field are characteristic of the isopropyl group.
The group appears as a singlet.
There are no protons on the adjacent carbon atom, so the singlets around d 2.1 are characteristically given by methyl ketones and acetate esters.
The horizontal common is also shown in the insert box.
For clarity, learn to recognize them scale expanded.
The peaks fit in the box when the scale is adjusted.
The methine proton Hc is a multiplet of relative area 1.
If the spectrum is amplified, some small peaks of this septet can't be seen.
From this pattern, it's easy to recognize isopropyl groups.
The spin-spin splitting is caused by the chemical shift of protons.
You can predict the characteristics of an NMR spectrum by analyzing the structure of a molecule with these principles in mind.
Learning to recognize the features of the actual spectrum will help you.
If a systematic approach is used, the process is not difficult groups.
A stepwise method is illustrated by drawing the NMR with each deshielding spectrum of the following compound.
Determine how many types of protons are present.
The area ratios for the CH should be 6.
The chemical shifts of the protons can be estimated.
The splitting patterns need to be determined.
Use the information from your summary to draw the spectrum.
Work through the problem to become comfortable with it.
You would expect the following compounds to have the same result.
Additional structural information can be provided by the distances between peaks of multiplets.
The distances are in the middle of the spectrum.
The distance between the peaks of the Hc multiplet must be the same as the distance between the peaks of the Hb doublet.