12 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.

Groups of neighboring protons can sometimes be identified by measuring their coupling constants.

Multiplets may arise from groups of protons that are 888-609- 888-609- 888-609- 888-609- 888-609-

The strength of the external magnetic field does not affect the magnetic effect that one proton has on another.

The field strength of the spectrometer does not affect the coupling constant.

A 60 MHz instrument records the same constants as a 300 MHz one.

Figure 13-28 shows typical values.

The 7-hertz splitting of protons on adjacent carbon atoms is the most common constant observed.

The value of 7 Hz in an alkyl group is averaged for rapid rotation.

Other splitting constants may be observed if rotation is hindered by a ring or bulky groups.

The 7-Hz grid in the insert box is a little larger than the 8 Hz constant.

The magnetically coupled protons Ha and Hb are ortho to each other.

The para isomer of nitrotoluene has four different types of aromatic protons, which are more complex than the ortho isomer.

The constants help to distinguish stereoisomers.

The compounds you expect to see are the ones with the protons on the b carbon.

There is a 7-Hz grid in the insert box that shows how wide the 9-Hz coupling is.

The compound C3H2NCl shows strong IR absorptions around 1650 cm-1 and 2200 cm-1.

At d 5.9 and d 7.1.
The structure should be consistent with the data.

Two spectrums are shown.

The structure should correspond to each spectrum.

The formula of complex splitting could be used to estimate the chemical spectrum of styrene.

The phenyl ring of styrene is adjacent to the vinyl proton Ha.

The chemical shift of Ha is protected by the aromatic ring and the vinyl group shift.

The splitting tree is similar to Figures 13-32 and 13-33.

The five hydrogens on the carbon atoms appear to split the Hb signal into two.

A signal can be split by two or more different types of protons.

The spectrum shows the Hb signal as a sextet with five equivalent protons.

The pattern is not a perfect sextet according to the trace in the insert box.

To see which peaks in the spectrum are given rise to by protons, assign them peaks.

To see the splitting of the vinyl protons, draw a tree.

Estimate the values of the constants.

The spectrum of cinnamaldehyde is found in the NMR.

Determine the chemical shifts of Ha, Hb, and Hc.

The absorption of one of these protons is hard to see.

To analyze the splitting of the protons, draw a tree.

Consider the spectrum of the following ketone.

To see the splitting predicted for the circled protons, draw a tree.

Stereochemical differences can result in different chemical shifts for protons.

The two protons on C1 of allyl bromide are not the same.
Hb is trans.
Hb absorbs at d 5.1.
The structure in the margin shows the different types of protons in allyl bromide.

If you substitute another atom for each of the protons in question, you can determine whether similar-appearing protons are equivalent.

The imaginary products are different when the replacement test is applied to C1 of allyl bromide.

The cis diastereomer and the trans diastereomer are given by replacing the cis hydrogen and the trans hydrogen.

At different chemical shifts, diastereotopic protons can split.

There are stereochemical relationships in a system.

It absorbs between d 3 and d 5 depending on the solvent and concentration.
Hb is absorbing between d 3 and d 4.
He and Hf are diastereotopic because He iscis to the hydroxy group and Hf is trans.

Notice that cyclobutanol has an internal mirror plane of symmetry.

The hydroxy group iscis to the trans group.
The two sets of protons absorb at different magnetic fields and are capable of splitting each other.

To compare two protons, use the are diastereotopic.

The compound should be replaced with an imaginary atom.

If the replacement of two protons with two imaginary products, then those products are pro two imaginary products.
The protons are not the same and the NMR cannot distinguish them.

If the two allylic protons are diastereomers, then the 3 of allyl bromide are enantiotopic.

There are two Hc in cyclobutanol.

Under most circumstances, enantiotopic protons diastereomerism can also occur in saturated, acyclic compounds.

The compound contains diastereo protonstopic.

The imaginary replacement of the CH2Cl group gives diastereomers.

The diastereotopic, diastereotopic protons on C1 exist in different chemical environments.

Different environments have different magnetic fields.

The single protons on C2 appear to be a complex multiplet at d 4.15.

The H molecule has an asymmetric carbon atom.

Depending on the differences in their environments, they may or may not be resolved in the NMR.

Predict the number of different signals produced by each compound.

These hydrogen atoms are not equivalent.

Evidence has already been seen that the picture of a molecule is not instantaneous.

A terminal alkyne doesn't give a spectrum where the molecule oriented along the field absorb at a high field and those oriented to the field absorb at a lower field.
The signal we see has its position averaged over all the orientations of the molecule.
Any movement or change that takes place in less than a hundredth of a second will produce an average spectrum.

The principle is illustrated by the spectrum.

There are two types of protons in the chair.
The hydrogens become equatorial by chair-chair interconversions.
The interconversions are very fast at room temperature.
There is only one peak at room temperature in the NMR spectrum of cyclohexane.

The chair-chair interconversion of cyclohexane is retarded by low temperatures.

The spin-spin splitting between the axis and the equator can be used to determine the conformations of tons on the same carbon atom and on adjacent carbons.

Chemical processes that are used as drugs occur faster than the NMR technique can observe them.

There is no acid, base, or water in this sample.

There is a trace of an acidic (or basic) impurity in the spectrum of ethanol.

A single, unsplit absorption is produced by this rapidly exchanging protons.

There is no splitting between the two particles.

During the measurement, the hydroxy protons are attached to a large number of different alcohols and experience spin arrangements of the methylene group.
What we see is a single, unsplit hydroxy absorption, corresponding to the averaged field, which the protons experience from bonding to many different alcohols.

Most alcohols and carboxylic acids and many amine and amides have protons exchange.

One sharp averaged signal is what we see.
We see splitting if the exchange is very slow.
If the exchange is slow, we may see a broadened peak that is not cleanly split or averaged.

Because of moderate rates of exchange and the magnetic properties of the nitrogen nucleus, trons on nitrogen often show broadened signals in the NMR.

H may give absorptions that are sharp and cleanly split, sharp and unsplit, or broad and shapeless.
The shapeless peak of NH2 is very broad.

A complex splitting pattern can often shake the sample with an excess of deuterium oxide, D2O.

Any exchangeable hydrogens can be simplified by replacing a hydrogen with a deuterium atom.

The signals from any exchangeable protons are either absent or less intense when a second NMR spectrum is recorded.

Draw the spectrum expected from the shake of the alcohol.

Propose chemical structures that are in line with the formulas.

Explain why the peaks around d 1.65 and d 3.75 are not clean multiplets.
Explain why some of the protons are likely to be missed.

There are hints in this section that can help make the analysis simpler.

The elements of unsaturation suggest rings, double bonds, or triple bonds.

The numbers of protons are represented by the individual peaks by matching the integrated peak areas with the number of protons in the formula.

The OH group is likely.

A signal around d 3 to d 4 suggests protons on a carbon bearing element.

Less strongly deshielded pheros are those that are more distant from the atom.

There are signals around d 7 to d 8.

An electron-withdrawing substituent may be attached if some of the aromatic absorptions are farther downfield.

There are signals around d 5 to d 6. cis and trans isomers can be differentiated by splitting constants.

To determine the numbers of adjacent protons, use the splitting patterns to assemble pieces of the molecule.

You can learn to recognize ethyl groups and isopropyl groups by their splitting patterns.

There are signals around d 2.1 to d 2.5 that suggest protons next to an aromatic ring.

A singlet at d 2.1 often results from a carbonyl group.

An aldehyde is suggested by signals in the range d 9 to d 10.

A terminal alkyne is suggested by a singlet around d 2.5.

These hints are not complete or exact.

They are easy ways to make educated guesses about the major features of a com pound.
The hints can be used to draw partial structures to look at how the hints might be combined to make a molecule.
Appendix 1 has a more complete table of chemical shifts.

One element of unsaturation is implied by the formula known as C4H8O2.

There are three types of protons in this spectrum.
The signals at d 4.1 and d 1.3 are similar to an ethyl group.

An ethoxy group is suggested because the formula contains oxygen.

The singlet at d 2.1 might be a carbonyl group.

The element of unsaturation would be accounted for by a carbonyl group.

There are eight hydrogen atoms in the spectrum.

We arrive at a structure after putting together all the clues.

At this point, the structure should be reexamined to make sure it is in line with the formula.

The Hc protons give a singlet of area 3 at d 2.1.

Explain how the spectrum of ethyl acetate differs from the expected spectrum of methyl propionate.

The structure of the compound of C4H10O should be proposed.

There are no elements of unsaturation in the formula C4H10O.

There are four types of hydrogens in this spectrum.
The singlet at d 2.4 might be a hydroxy group, and the signal at d 3.4 might be a protons on a carbon atom.
The carbon atom is implied to have one hydrogen by the d 3.4 signal.

The pattern for an isopropyl group is similar to the signals at d 1.8 and d 0.9.

The isopropyl group must be bonded to a carbon atom because it absorbs methine at a high field.

Two of the carbon atoms appear in both partial structures, which adds to the total of six carbon atoms.

The structure needs to be checked to make sure that it has the correct formula and that it accounts for all the structural evidence provided by the spectrum.

The protons are in isobutyl alcohol.

There are five protons and five NMRs given here.
The structure should be consistent with the spectrum.

Both of these groups are not visible.

The signals between d 2.1 and d 2.5 are suggestive, but we can't see the carbonyl group.
An internal alkyne is more difficult because there are no distinctive absorptions in the IR or the protons.

Carbon NMR (13C NMR or CMR) was made possible by the development of Fourier transform NMR, and high-field superconducting spectrometers allowed it to be as convenient as protons.

The magnetic environments of the carbon atoms are determined by carbon NMR.
Carbonyl carbon atoms, alkyne carbon atoms, and aromatic carbon atoms all have chemical shifts in the 13C NMR spectrum.

The electronics in the early instruments couldn't detect the weak carbon signal, so it took longer to become a routine technique.

The isotope has an even number of protons and neutrons, so it can't give off a magnetic spin.
13C has a magnetic spin of one-half and has an odd number of neutrons.
The sensitivity of 13C NMR is decreased by a factor of 100 because only 1% of the carbon atoms are magnetic.
The gyromagnetic ratio of 13C is only one-fourth that of the protons, so the 13C resonance frequency is only one-fourth of that for 1H NMR.
The smaller gyromagnetic ratio leads to a decrease in sensitivity.

Special techniques are needed to obtain a spectrum because 13C is less sensitive than 1H.

The random noise cancels when the desired signals are reinforced.
The spectrum can be averaged and plotted by the computer if several are taken.
The 13C NMR technique is much less sensitive than the 1H NMR technique, so hundreds of spectra are averaged to produce a usable result.
The procedure of averaging the CW spectrum is long and tedious.
There is a better way to go.

When magnetic nuclei are placed in a uniform magnetic field and irradiated with a pulse of radio frequencies close to their resonance frequencies, the nuclei absorb some of the energy and precess like little tops.

This precession of many nuclei at slightly different frequencies produces a complex signal that decays as the nuclei lose the energy they gained from the pulse.
Many FIDs can be averaged in a few minutes and can be recorded by a radio receiver and a computer.
The averaged Transients are converted into a spectrum by a computer.

A Fourier transform spectrometer needs sophisticated electronics that can accurately receive the complicated Transients.

A good 13C instrument can do 1H NMR as well.
The small amount of sample produced by the Fourier transform technique is what makes it good for protons.

The radio-frequency is close to the resonance of the pulse.

Each nucleus precesses at its own resonance frequencies.
Transient FIDs can be accumulated and averaged in a short period of time.
The spectrum recorded on the printer is produced by a computer using a Fourier transform.

Figure 13-41 gives a typical range of chemical shifts for carbon atoms.

Appendix 1C has a more detailed table of carbon chemical shifts.

Carbon chemical shifts are larger than compa solution.

Techniques using rable protons.
The carbon atom is 13C and 1H and gives it closer to a shielding group than its attached hydrogen.
A family of lowest-energy structures (a), an aldehyde proton (bonded to the oxygen through the carbon) and the carbonyl carbon atom all absorb around d 9.4 from which a computer calculates a in the 1H NMR spectrum.

The first estimate of a carbon atom's chemical shift can be made using this approximation.

The position is at d 180.

The carbon signal for deuterated chloroform is split into three equal-sized peaks.

The CDCl3 solvent signal can be used as an internal reference instead of the TMS if desired.

An electron-withdrawing group has a large effect on the chemical shift of a carbon atom.

The 1H and 13C NMR spectrum of 1,2, 2-trichloropropane is shown in Figure 13-43.
In accordance with our prediction, the chemical shift of this carbon is 15 times that of its attached protons.

It is 15 times more powerful than the protons.

The group absorbs around d 5.8.
The absorption of carbon is 15 times greater than the shift.

The characteristics of 13C NMR are similar to those of the 1H technique.

There are differences.

The table shows the approximate chemical shift values.

The carbon spectrum has a width of 200 parts per million, while the proton spectrum has a sweep width of 10 parts per million.

The gyromagnetic ratio for 13C is less than that of the protons.

A 70,459-gauss magnet needs a 300-MHz transmitter for protons and a 75.6-MHz transmitter for 13C.
A spectrometer with a 14,092-gauss magnet needs a 60-MHz transmitter for protons and a 15.1-MHz transmitter for 13C.

Carbon atoms with two or three protons attached Appendices 1A, 1B, and 1C usually give the strongest absorptions, and carbons with no protons tend to give weak absorptions.

Decoupling tech interpretation and prediction of niques can be used to equalize the absorptions of different carbon atoms.
The mode makes peak carbon and protons.

13C NMR splitting patterns are very different from 1H NMR.

There is only a small chance that a 13C nucleus is adjacent to another 13C nucleus because only 1% of the carbon atoms are magnetic.
Carbon-carbon splitting can be ignored.
It is common.
Carbon atoms can beBonded directly to hydrogen atoms or close to hydrogen atoms for carbon-hydrogen spin-spincoupling to be observed.
Splitting patterns can be difficult to interpret.

The protons quickly flip their spins when they are in resonance.

Any carbon-hydrogen splitting has been eliminated so each signal appears as a single peak.

Draw the 13C NMR spectrum of the compounds.

The chemical shifts can be estimated using Figure 13-41.

Some valuable information is lost in the process of pliton spin decelerating.

The 13C nuclei are split by the protons.
A carbon atom with a methine appears as a doublet, a carbon with two attached protons gives a triplet, and a carbon with a methylene splits into a quartet.
The off-resonance-decoupled spectrum is easily recognized by the appearance of a quartet of 0 parts per million.

The number of nonequivalent types of carbon atoms and their chemical shifts are shown by the singlets in the broadband-decoupled spectrum.

The signals in the off-resonance-decoupled spectrum show the number of hydrogen atoms in each carbon atom.
There are two traces, one broadband and the other off-resonance.
It is usually broadband decoupled if only one trace is given.

The spectrum is compared with Figure 13-43.

You should draw the spectrum for butan-2-one.

Modern, computer-controlled Fourier transform spectrometers make it easier to run DEPT.

The peaks remain decoupled and it avoids overlap.

The 13C nucleus is coupled with the protons.

The transfer of polarization from the protons to the carbon nucleus is possible under the right circumstances.
The 13C nucleus has a number of protons.

There are carbons that are not protons.

This information allows us to distinguish between carbons with 0, 1, 2, or 3 hydrogen atoms and carbons with no H's.

Normal positive peaks are given by methine carbons.

There are no peaks in the normal spectrum, no peaks in the DEPT-90 spectrum, and negative peaks in the DEPT-135 spectrum.

There are no peaks in the normal spectrum, no peaks in the DEPT-90 spectrum, and no peaks in the DEPT-135 spectrum.

The normal decoupled 13C NMR spectrum is shown in Figure 13-46.

The carbonyl carbon is only visible in the regular spectrum.
Cc has 1 protons and appears in all the spectrum.

Ca, the carbon with three protons, appears to be a normal peak in the DEPT-135 spectrum.

The 13C NMR spectrum of phenyl propanoate is shown here.

The same principles are used to interpret 13C NMR spectrum.

Carbon spectrum is often easier to interpret.

In the 13C NMR spectrum of d@valerolactone, the CH2 groups in the upper (off-resonance-decoupled) spectrum are split into triplets, but they appear as singlets in the lower.

The off-resonance-decoupled spectrum shows a singlet at 173 parts per million.

The next absorption has a chemical shift of 70 parts per million.

This is more than 20 times the chemical shift of a protons on a carbon bond.
The formula suggests that the element must be oxygen.

The signal at 30 ppm is related to a carbon atom.

The triplet in the off-resonance-decoupled spectrum shows the carbon atom as a methylene group.

The two signals at 19 and 22 ppm are from carbon atoms that are not directly bonding to any deshielding group, although the carbon at 22 ppm is probably closer to one of the oxygen atoms.

They correspond to methylene groups in the off-resonance-decoupled spectrum.

The complete structure is given when the partial structures are combined into a ring.

Only the broadband-decoupled spectrum is provided in the following problems.

The number of protons is given for each carbon atom, either zero (C), one (CH), two (CH2), or three (CH3).

A bottle of allyl bromide was found to have an impurity.

The molecule C3H O was separated by a careful distillation.

The structure for this impurity should be proposed.

A graduate student was making something.

The yield of a different compound is shown here.

The structure for this product should be proposed.

A student in a laboratory was converting cyclohexanol to cyclohexyl bromide by using one equivalent of sodium bromide in a large amount of sulfuric acid.

The structure for this product should be proposed.

Suggest changes to the reaction to get a better yield.

The most uniform magnetic field can be found within one part per billion.

They place small tubes of solutions in the magnetic field and spin them to get an idea of the magnetic field's variations.
They want the sample to behave as if it were all at a single point in the magnetic field, with every molecule subjected to the same external magnetic field.

A sample of a living human body is placed in the magnetic field of a large-bore superconducting magnet.

The magnetic field is nonuniform with a gradient that allows the protons in one plane of the sample to be in resonance at any one time.
The instrument can look at one point within the sample, a line within the sample, or a plane within the sample using a combination of field and Fourier transform techniques.
An image of a two-dimensional slice is generated by the computer.
The computer can give a three-dimensional plot of the proton resonances within the bulk of the sample.

There is no radioactive material in the instrument.

Magnetic resonance is the least dangerous method for analyzing the inside of the body.

The technique field is slightly more aligned with the lower-energy state than has been used to study the effects against it.

The number of spins that are oriented against the magnetic field can be inverted by a radio-frequency pulse of just the right duration.
Over a period of a few seconds, the spins gradually relax metabolism ofphosphate esters such as their normal state.
By following the free-induction.

This image shows how the spine has been pinched in a patient.

The intervertebral disc between L4 and L5 is no longer supporting the spine in their proper positions.

This is an image of a torn foot.

Over stress in athletics can lead to the most common tendon break injury, the tear of the Achilles tendon.

Valuable information about the tissues involved is given by the color or intensity of the image.

Cancerous tissues tend to have longer relaxation times than normal tissues, so tumors are readily apparent in the NMR image.
The second image is a slice through a patient's knee, which will require surgery.

We have learned how to use IR and mass spectrometry to determine the structures of unknown organic compounds.

These techniques usually provide a unique structure.
An effective strategy is more important than simply looking at the spectrum, hoping that something will show up.
The strengths and weaknesses of each technique should be taken into account.
The information provided by each technique is summarized in the table.

The process of identifying an unknown compound depends on what you already know about the chemistry of the compound and what you learn from each spectrum.

To keep track of mass numbers, formulas, possible functional groups, and carbon skeletons, always use scratch paper and a pencil during the process.

Determine a tentative molecular weight by looking for a molecule.

Most compounds don't contain nitrogen and will show odd-numbered fragments.
Some compounds may fail to give a visible ion.
If the weight is odd, consider the nitrogen atom.
If an HRMS is available, use the tables to find a formula with a mass close to the experimental value.

It could be S, Cl, or Br.

This is more time consuming than going on to other spectrums and looking at theMS fragmentation patterns to help determine the structure.

Once you have a structure, you can verify the patterns more easily.

The functional groups should be determined by the exact position of the peak and other characteristics.

The combination of IR and an odd ion in the mass spectrum should confirm amines, amides, and nitriles.

The OH absorption in the IR suggests that the mass spectrum could be low due to the loss of water.

Consider the number of signals and chemical shifts.

Look for carboxylic acids, aldehydes, and aromatic protons.
Moderately deshielded peaks might be vinyl protons or protons on a carbon bond to a negative atom.
A peak around d 2.1 to d 2.5 might be an acetylenic protons on a carbon next to a carbonyl group, a benzene ring, or a vinyl group.

The possibilities should be checked to make sure they are in line with the IR spectrum.

The relative numbers of protons responsible for the signals should be revealed by the peak integrations.
The spin-spin splitting patterns should be analyzed to see what the structures of the alkyl groups are.

If the 13C NMR spectrum is available, use the number of signals and their chemical shifts to provide information on how many types of carbon atoms are present, and their possible chemical environments, consistent with the functional groups suggested by the IR spectrum.

There should be at least one or two tentative structures once you have considered the whole spectrum.

The structure should be checked to see if it accounts for the major characteristics.

You can be confident that the structure is correct if it accounts for all of the features.

The two compounds are given a set of results.

The solution for compound 1 is given after the problem, but you should go as far as you can before looking at it.

There is an odd molecule weight at 121 and a large fragment at 106.

There is a possibility that these features indicate the presence of a nitrogen atom.

An aromatic ring is indicated by H absorptions above 3000 cm-1 and C stretch around 1600 cm-1.

The total integral of 5 suggests the ring is monosubstituted, as the NMR shows complex splitting in the aromatic region.

Part of the aromatic absorption is shifted upfield of d 7.2, suggesting that the substituent on the benzene ring is a pi electron-donating group such as an amine or an ether.
An ethyl group is seen at d 1.2 and d 3.1, which is appropriate for protons on a carbon atom.
We propose a nitrogen atom bonding to a hydrogen atom, a benzene ring, and an ethyl group.
The total weight for this structure is 121, which is in agreement with the mass spectrum.

The aromatic signals moved above d 7.2 in the NMR.

The ethyl signals in the NMR are explained by the ethyl group bonding to nitrogen.

If they absorb at the same chemical shift, they can't split.

There are atoms that can't be distinguished.

There is a difference between the resonance field of the carbon nucleus and that of the TMS.

The chemical shifts are usually given on the d scale.

Splitting by two or more different types of protons.

A method of running several 13C experiments with different sequence so that the carbon atoms appear different depending on whether they are one, two, or three protons.

A group withdraws part of the electron density from the nucleus.

Larger chemical shifts are caused by the movement of the de shield tions downfield.
There are diastereomeric positions that the nucleus occupies.
There is a replacement test for diastereotopic atoms.

The farther downfield the nucleus absorbs, the more deshielded it is.

When many nuclei are irradiated by a pulse of energy and precess at their resonance frequencies, the signal results.

The gyromagnetic ratio of a protons is 28,753.
The 13C nucleus has a gyromagnetic ratio of 6.

A magnetic field is created by the motion of electrons in a molecule or wire.

The area under a peak is measured by the number of protons giving rise to that peak.

Spin-spin splitting occurs when the magnetic fields of the nucleus influence each other.

A group of peaks arise from the spin-spin splitting of a single type of nucleus.

A magnetic field is used to measure the absorption of radio-frequency energy by nuclei.

Nuclear spin transitions are caused by the absorbed energy.

The spin-spin splitting is caused by the protons directly bonding to a carbon atom.

The nuclear spins return to their normal state after an RF pulse.

The electrons are surrounded by a magnetic field.

The applied magnetic field is larger than the effective magnetic field at the shielded nucleus.

It is possible to eliminate spin-spin splitting by irradiating one type of nuclei.

The standard's absorption is defined as d 0.00.

The farther upfield it absorbs, the more shielded a nucleus is.

Each skill is followed by problem numbers.

Explain which protons are equivalent and which are not.

The likely types of protons are suggested by the chemical shifts of absorptions.

The relative numbers of different types of protons can be determined using the integrations.

The general features of the 13C NMR spectrum of a compound can be drawn.

Predict the number of signals and chemical shifts of carbon atoms in a compound.

The likely types of carbons are suggested by the chemical shifts of 13C absorptions.

The number of hydrogen atoms in a given carbon atom can be determined using either the off-resonance-decoupled spectrum or the DEPT 13C spectrum.

The structures of organic compounds can be determined by combining the chemical shifts, integrals, and spin-spin splitting patterns in the NMR spectrum with the information from the mass and IR spectrums.

An unknown compound has a formula.

The structure for this compound should be proposed.

There is a compound of C3H O.

The structure for this compound should be proposed.

To see which signals are in the spectrum, assign peaks.

A compound (C10H O 12 2) was isolated from a mixture containing 2-phenylethanol and acetic acid.

The structure for this compound should be proposed.

To see which signals are in the spectrum, assign peaks.

You can sketch out your predictions for the following compounds.

Tell me how you would use the NMR to distinguish between the compounds.

Show which carbon atoms give rise to which peaks in the spectrum by proposing a structure for this compound.

A pilot plant was making 2,3-dibromobutane by adding bromine to the double bond of but-2-ene.

The controller malfunction allowed the temperature to rise.
The product had several impurities formed, including the one that appears below.
The peaks should be assigned to the protons in your structure.

A new chemist is working in an industrial lab on oxygenated gasoline Additives.

She found an old bottle with a clear, pleasant-smelling liquid that wasn't on the label.
She was able to determine the identity of the compound using the quick NMR spectrum.
Give a structure and assign peaks.

The National Science Foundation designated a major university as a national nuclear magnetic resonance center.

Several large superconducting instruments were being installed when a government safety inspector appeared and demanded to know what provisions were being made to handle the nuclear waste produced by these instruments.

If you are the manager of the center, you should give an explanation that a nonscientist could understand.

A compound was isolated as a minor component.

The spectrum is shown here.

List the structural characteristics you can determine from each spectrum.

A tentative structure can be proposed by looking at the set of spectrum as a group.

The major features of each spectrum should be accounted for in your proposed structure.

A compound is an important starting material for organic synthesis.

To determine the structure, first you have to consider each spectrum individually and then you have to consider all the spectrum together.
To show that your proposed structure accounts for the major features of each spectrum, assign peaks.
The information on the carbon NMR is blue.

The three isomers are difficult to identify using 13C NMR.

Tell us how carbon NMR distinguishes these three isomers.

Explain why they are difficult to distinguish.

The formula C3H6Br2 has five isomers.

For each structure, show how many different types of H are present.

For each structure, show how many different types of C there are.

One monobrominated product and four dibrominated products will be given.

These products are easily separated by GC-MS, but the dibrominated products are difficult to distinguish.
Explain how 13C NMR would be able to distinguish between the four dibrominated products and the monobrominated product.

Each structure has a formula called C4H8O2.

Match the structure with its signals.

In each case, show which carbon atoms are the same.

Secondary alcohol is shown with the addition of phenyl Grignard reagent.

The product alcohol, a racemic mixture, shows two different 3H doublets, one at d 0.75 and one at d 1.0, despite the fact that the two methyl groups are equivalent.

The product is projected along the C1-C2 axis.

Determine the structures of ethers and explain their absorption characteristics.

There are mechanisms showing the formation and reactions of ethers.

R and R may be alkyl groups or aryl groups.

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Both hydrogens are replaced by alkyl groups.

We will discuss how ethers are formed and how they react.

The unreactive nature of ethers makes them not frequently used as synthetic intermediates.
ethers are used for organic reactions because of their stable nature.
The properties of ethers are considered in this chapter.