In fact, the intensity ratios are Each of these different spin arrangements of the methyl protons except the sets of degenerate ones, which are effectively identical gives the methylene protons in that molecule a different chemical shift value. Each spin arrangement causes the methylene protons in that molecule to have a chemical shift different from those in a molecule with another methyl spin arrangement except, of course, when the spin arrangements are indistinguishable, or degenerate.
Examination of this figure makes it easy to explain the origin of the triplet for the methyl group and the intensity ratios of Now one can see the origin of the ethyl pattern and the explanation of its intensity ratios. The occurrence of spin—spin splitting is very important for the organic chemist as it gives additional structural information about molecules. Namely, it reveals the number of nearest proton neighbors each type of proton has.
From the chemical shift one can determine what type of proton is being split, and from the integral the area under the peaks one can determine the relative numbers of the types of hydrogen.
This is a great amount of structural information, and it is invaluable to the chemist attempting to identify a particular compound.
Each entry in the triangle is the sum of the two entries above it and to its immediate left and right. Notice that the intensities of the outer peaks of a multiplet such as a septet are so small compared to the inner peaks that they are often obscured in the baseline of the spectrum.
The distance between the peaks in a simple multiplet is called the coupling constant J. The coupling constant is a measure of how strongly a nucleus is affected by the spin states of its neighbor.
The spacing between the multiplet peaks is measured on the same scale as the chemical shift, and the coupling constant is always expressed in Hertz Hz. In ethyl iodide, for instance, the coupling constant J is 7. To see how this value was determined, consult Figures 3. The spectrum in Figure 3. Notice the top of the spectrum. Now examine the multiplets. The spacing between the component peaks is approximately 1. When the protons interact, the magnitude in ethyl iodide is always of this same value, 7.
The amount of coupling is constant, and hence J can be called a coupling constant. The invariant nature of the coupling constant can be observed when the NMR spectrum of ethyl iodide is determined at both 60 MHz and MHz. A comparison of the two spectra indicates that the MHz spectrum is greatly expanded over the MHz spectrum. Despite the expansion of the spectrum determined at the higher spectrometer frequency, careful examination of the spectra indicates that the coupling constant between the CH3 and CH2 protons is 7.
The spacings of the lines of the triplet and the lines of the quartet do not expand when the spectrum of ethyl iodide is determined at MHz. The extent of coupling between these two sets of protons remains constant irrespective of the spectrometer frequency at which the spectrum was determined Fig. Compare, for example, 1,1,2-trichloroethane Fig. These coupling constants are typical for the interaction of two hydrogens on adjacent sp3-hybridized carbon atoms.
Two hydrogen atoms on adjacent carbon atoms can be described as a three-bond interaction and abbreviated as 3J. Typical values for this most commonly observed coupling is approximately 6 to 8 Hz. The bold lines in the diagram show how the hydrogen atoms are three bonds away from each other. It is a simple matter of subtracting these values to determine the coupling constants in Hertz. See, for example, the spectra shown in Figures 3. Section 5. A study of the magnitude of the coupling constant can give important structural information see Section 5.
A more extensive list of coupling constants appears in Chapter 5, Section 5. Before closing this section, we should take note of an axiom: the coupling constants of the groups of protons that split one another must be identical within experimental error.
This axiom is extremely useful in interpreting a spectrum that may have several multiplets, each with a different coupling constant.
Which triplet is associated with the quartet? The protons in each group interact to the same extent. It is also clear that triplets B and D are related to each other in the interaction scheme.
There is a tendency for the outermost lines of a multiplet to have nonequivalent heights. This second group of protons leans toward the first one in the same fashion. If arrows are drawn on both multiplets in the directions of their respective skewing, these arrows will point at each other.
See Figures 3. Even though the shifts in Hertz increase, the chemical shifts in ppm of a given proton at low field and high field are the same because we divide by a different operating frequency in each case to determine the chemical shift Eq. If we compare the spectra of a compound determined at both low field and high field, however, the gross appearances of the spectra will differ because, although the coupling constant has the same magnitude in Hertz regardless of operating frequency, the number of Hertz per ppm unit changes.
The coupling constant does not change, but it becomes a smaller fraction of a ppm unit! When we plot the two spectra on paper to the same parts-per-million scale same spacing in length for each ppm , the splittings in the high-field spectrum appear compressed, as in Figure 3. The coupling has not changed in size; it has simply become a smaller fraction of a ppm unit.
At higher field, it becomes necessary to use an expanded parts-per-million scale more space per ppm to observe the splittings. The MHz multiplets are identical to those observed at 60 MHz. This can be seen in Figure 3. With MHz spectra, therefore, it is frequently necessary to show expansions if one wishes to see the details of the multiplets.
In some of the examples in this chapter, we have used MHz spectra—not because we are old-fashioned, but because these spectra show the multiplets more clearly without the need for expansions. In most cases, the expanded multiplets from a high-field instrument are identical to those observed with a low-field instrument. This simplification occurs because the multiplets are moved farther apart, and a type of interaction called a second-order interaction is reduced or even completely removed.
Chapter 5 will discuss second-order interactions. These guidelines can be consulted whenever you are trying to establish the class of an unknown compound. Coupling behaviors commonly observed in these compounds are also included in the tables. This coupling information was not covered in this chapter, but it is discussed in Chapters 5 and 6. It is included here so that it will be useful if you wish to use this survey later. Alkanes Alkanes can have three different types of hydrogens methyl, methylene, and methine , each of which appears in its own region of the NMR spectrum.
In long chains, all of the methylene CH2 absorptions may be overlapped in an unresolvable group. Note that methine hydrogens CH have a larger chemical shift than those in methylene or methyl groups. Hydrogens in methyl groups are the most highly shielded type of proton and are found at chemical shift values 0.
In long hydrocarbon chains, or in larger rings, all of the CH and CH2 absorptions may overlap in an unresolvable group. Methyl group peaks are usually separated from other types of hydrogens, being found at lower chemical shifts higher field.
However, even when methyl hydrogens are located within an unresolved cluster of peaks, the methyl peaks can often be recognized as tall singlets, doublets, or triplets clearly emerging from the absorptions of the other types of protons.
Methine protons are usually separated from the other protons, being shifted further downfield. Note that the integral can be used to estimate the total number of hydrogens the ratio of CH3 to CH2-type carbons since all of the CH2 hydrogens are in one group and the CH3 hydrogens are in the other. The NMR spectrum shows the lowest whole-number ratios. You need to multiply by 2 to give the actual number of protons. Each type has a characteristic chemical shift region.
Hydrogens attached to a carbon adjacent to a double bond allyic hydrogens are also deshielded by the anisotropy of the double bond, but because the double bond is more distant, the effect is smaller. Both types of hydrogens are deshielded due to the anisotropic field of the p electrons in the double bond. The effect is smaller for the allylic hydrogens because they are more distant from the double bond.
A spectrum of 2-methylpentene is shown in Figure 3. Note the vinyl hydrogens at 4. The splitting patterns of both vinyl and allylic hydrogens can be quite complex due to the fact that the hydrogens attached to a double bond are rarely equivalent and to the additional complication that allylic hydrogens can couple to all of the hydrogens on a double bond, causing additional splittings.
These situations are discussed in Chapter 5, Sections 5. Aromatic Compounds Aromatic compounds have two characteristic types of hydrogens: aromatic ring hydrogens benzene ring hydrogens and benzylic hydrogens those attached to an adjacent carbon atom.
Benzylic hydrogens are also deshielded by the anisotropic field of the ring, but they are more distant from the ring, and the effect is smaller. It is often possible to determine the positions of the substituents on the ring from these splitting patterns and the magnitudes of the coupling constants.
They are found in a region of their own 6. Occasionally, a highly deshielded vinyl hydrogen will have its absorption in this range, but this is not frequent. The hydrogens on an aromatic ring are more highly deshielded than those attached to double bonds due to the large anisotropic field that is generated by the circulation of the p electrons in the ring ring current.
See Section 3. The largest chemical shifts are found for ring hydrogens when electron-withdrawing groups such as INO2 are attached to the ring. These groups deshield the attached hydrogens by withdrawing electron density from the ring through resonance interaction. Conversely, electron-donating groups like methoxy IOCH3 increase the shielding of these hydrogens, causing them to move upfield. Nonequivalent hydrogens attached to a benzene ring will interact with one another to produce spin—spin splitting patterns.
The amount of interaction between hydrogens on the ring is dependent on the number of intervening bonds or the distance between them. It is frequently possible to determine the substitution pattern of the ring by the observed splitting patterns of the ring hydrogens Section 5.
One pattern that is easily recognized is that of a para-disubstituted benzene ring Fig. The spectrum of a-chloro-p-xylene is shown in Figure 3. The highly deshielded ring hydrogens appear at 7. The chemical shift of the methyl protons at 2. The large shift of the methylene hydrogens is due to the electronegativity of the attached chlorine.
The acetylenic hydrogen will be absent if the triple bond is in the middle of a chain. Protons on a carbon next to the triple bond are also affected by the p system. It is shifted upfield because of the shielding provided by the p electrons Fig. A spectrum of 1-pentyne is shown in Figure 3.
The peaks in the expansions have been labeled with Hertz Hz values so that coupling constants can be calculated. Note that the acetylenic proton c at 1. This coupling constant is calculated by subtraction: Values less than 7.
Sections 5. The type of pattern is referred to as a triplet of doublets. The 3J coupling constant is calculated by subtraction, for example, counting from left to right, peak 6 from peak 4 The 4J coupling constant can also be calculated from the triplet of doublets, for example, peak 6 from peak 5 The sextet for the CH2 group b at 1.
Finally, the triplet for the CH3 group a at 1. Alkyl Halides w. This deshielding effect is due to the electronegativity of the attached halogen atom. The extent of the shift is increased as the electronegativity of the attached atom increases, with the largest shift found in compounds containing fluorine. The other halogens I, Cl, Br do not show any coupling. Other halogens do not cause spin—spin splitting of hydrogen peaks. The spectrum of 1-chlorobutane is shown in Figure 3.
Note the large downfield shift deshielding of the hydrogens on carbon 1 due to the attached chlorine. Alcohols w In alcohols, both the hydroxyl proton and the a hydrogens those on the same carbon as the hydroxyl group have characteristic chemical shifts.
The peak may be broadened at its base by the same set of factors. Protons on the a carbon are deshielded by the electronegative oxygen atom and are shifted downfield in the spectrum. This peak can be found anywhere in the range of 0.
The variability of this absorption is dependent on the rates of IOH proton exchange and the amount of hydrogen bonding in the solution Section 6. Exchange is promoted by increased temperature, small amounts of acid impurities, and the presence of water in the solution. A freshly purified and distilled sample, or a previously unopened commercial bottle, may show this coupling.
On occasion, one may use the rapid exchange of an alcohol as a method for identifying the IOH absorption. After shaking the sample and sitting for a few minutes, the IOH hydrogen is replaced by deuterium, causing it to disappear from the spectrum or to have its intensity reduced.
If exchange of the OH is taking place, this hydrogen will not show any coupling with the IOH hydrogen, but will show coupling to any hydrogens on the adjacent carbon located further along the carbon chain. A spectrum of 2-methylpropanol is shown in Figure 3. Note the large downfield shift 3. The hydroxyl group appears at 2. The methine proton at 1. There are nine 1.
Ethers In ethers, the a hydrogens those attached to the a carbon, which is the carbon atom attached to the oxygen are highly deshielded. RIOICHI he m 4a In ethers, the hydrogens on the carbon next to oxygen are deshielded due to the electronegativity of the attached oxygen, and they appear in the range 3.
Methoxy groups are especially easy to identify as they appear as a tall singlet in this area. Ethoxy groups are also easy to identify, having both an upfield triplet and a distinct quartet in the region of 3. An exception is found in epoxides, in which, due to ring strain, the deshielding is not as great, and the hydrogens on the ring appear in the range 2. The spectrum of butyl methyl ether is shown in Figure 3. The absorption of the methyl and methylene hydrogens next to the oxygen are both seen at about 3.
The methylene hydrogens are split into a triplet by the hydrogens on the adjacent carbon of the chain. Amines vn Two characteristic types of hydrogens are found in amines: those attached to nitrogen the hydrogens of the amino group and those attached to the a carbon the same carbon to which the amino group is attached.
More commonly, this coupling is obscured by quadrupole broadening by nitrogen or by proton exchange. See Sections 6. This coupling is usually not observed. Due to chemical exchange, this coupling is usually not observed. The a hydrogen is slightly deshielded due to the electronegativity of the attached nitrogen. This hydrogen is deshielded due to the anisotropy of the ring and the resonance that removes electron density from nitrogen and changes its hybridization.
These peaks are extremely variable, appearing over a wide range of 0. The position of the resonance is affected by temperature, acidity, amount of hydrogen bonding, and solvent. In addition to this variability in position, the INH peaks are often very broad and weak without any distinct coupling to hydrogens on an adjacent carbon atom. This condition can be caused by chemical exchange of the INH proton or by a property of nitrogen atoms called quadrupole broadening see Section 6.
The amino hydrogens will exchange with D2O, as already described for alcohols, causing the peak to disappear. Reliable prediction is difficult. A spectrum of propylamine is shown in Figure 3. Notice the weak, broad NH absorptions at 1. Nitriles w In nitriles, only the a hydrogens those attached to the same carbon as the cyano group have a characteristic chemical shift. Hydrogens on the adjacent carbon of a nitrile are slightly deshielded by the anisotropic field of the p-bonded electrons appearing in the range 2.
A spectrum of valeronitrile is shown in Figure 3. The hydrogens next to the cyano group appear near 2. Hydrogens on the carbon adjacent to the CJ O group are also deshielded due to the carbonyl group, but they are more distant, and the effect is smaller. Protons appearing in this region are very indicative of an aldehyde group since no other protons appear in this region.
The aldehyde proton at 9. NMR is far more reliable than infrared spectroscopy for confirming the presence of an aldehyde group. The other regions have also been expanded and shown as insets on the spectrum and are summarized as follows: vn Proton a 1. The CH group b adjacent to the carbonyl group appears in the range of 2.
In the present case, the pattern at 2. In a ketone, the hydrogens on the carbon next to the carbonyl group appear in the range 2. If these hydrogens are part of a longer chain, they will be split by any hydrogens on the adjacent carbon, which is further along the chain.
Methyl ketones are quite easy to distinguish since they show a sharp three-proton singlet near 2. Be aware that all hydrogens on a carbon next to a carbonyl group give absorptions within the range of 2. Therefore, ketones, aldehydes, esters, amides, and carboxylic acids would all give rise to NMR absorptions in this same area.
Infrared spectroscopy would also be of great assistance in differentiating these types of compounds. Absence of the aldehyde, hydroxyl, amino, or ether stretching absorptions would help to confirm the compound as a ketone. A spectrum of 5-methylhexanone is shown in Figure 3. Notice the tall singlet at 2. This is quite characteristic of a methyl ketone.
Since there are no adjacent protons, one observes a singlet integrating for 3 H. Typically, carbon atoms with more attached protons are more shielded. Thus, the methyl group appears further upfield than the methylene group e , which has fewer attached protons. The quartet for the methylene group b is clearly visible at about 1. The doublet for the two methyl groups at about 0. Remember that the doublet results from the two equivalent methyl groups seeing one adjacent proton 3J.
Esters vn Two distinct types of hydrogen are found in esters: those on the carbon atom attached to the oxygen atom in the alcohol part of the ester and those on the a carbon in the acid part of the ester that is, those attached to the carbon next to the CJ O group. Hydrogens on the carbon attached to the single-bonded oxygen are deshielded due to the electronegativity of oxygen.
The anisotropic field of the carbonyl group deshields these hydrogens. The peak in the 3. The large chemical shift of these hydrogens is due to the deshielding effect of the electronegative oxygen atom, which is attached to the same carbon.
Either of the two types of hydrogens mentioned may be split into multiplets if they are part of a longer chain. A spectrum for isobutyl acetate is shown in Figure 3. Note that the tall singlet c at 2.
If the methyl group had been attached to the singly bonded oxygen atom, it would have appeared near 3. The ICH2I group d attached to the oxygen atom is shifted downfield to about 3. That group integrates for 2 H and appears as a doublet because of the one neighboring proton b on the methine carbon atom. That single proton on the methine carbon appears as a multiplet that is split by the neighboring two methyl groups a and the methylene group d into a nonet nine peaks, at 1.
Finally, the two methyl groups appear as a doublet at 0. Carboxylic Acids ll. Carboxylic acids have the acid proton the one attached to the ICOOH group and the a hydrogens those attached to the same carbon as the carboxyl group.
This usually broad signal is a very characteristic peak for carboxylic acids. Hydrogens adjacent to the carbonyl group are slightly deshielded. With the exception of the special case of a hydrogen in an enolic OH group that has strong internal hydrogen bonding, no other common type of hydrogen appears in this region.
A peak in this region is a strong indication of a carboxylic acid. Since the carboxyl hydrogen has no neighbors, it is usually unsplit; however, hydrogen bonding and exchange many cause the peak to become broadened become very wide at the base of the peak and show very low intensity.
Sometimes the acid peak is so broad that it disappears into the baseline. In that case, the acidic proton may not be observed.
Infrared spectroscopy is very reliable for determining the presence of a carboxylic acid. As with alcohols, this hydrogen will exchange with water and D2O. However, when this is done the —COOH absorption will disappear from the spectrum. Notice that this peak is very broad due to hydrogen bonding and exchange. Also notice that proton c is shifted downfield to 3. The normal range for a proton next to just one carbonyl group would be expected to appear in the range 2.
Amides he N. The a hydrogens in amides absorb in the same range as other acyl next to CJ O hydrogens. They are slightly deshielded by the carbonyl group. Hydrogens on the carbon next to the nitrogen of an amide are slightly deshielded by the electronegativity of the attached nitrogen.
In most cases, either the quadrupole moment of the nitrogen atom or chemical exchange decouples this interaction. Usually not seen for the same reasons stated above. Exchange of the amide NH is slower than in amines, and splitting of the adjacent CH is observed even if the NH is broadened. The INH absorptions of an amide group are highly variable, depending not only on their environment in the molecule, but also on temperature and the solvent used.
Because of resonance between the unshared pairs on nitrogen and the carbonyl group, rotation is restricted in most amides. Without rotational freedom, the two hydrogens attached to the nitrogen in an unsubstituted amide are not equivalent, and two different absorption peaks will be observed, one for each hydrogen Section 6.
Nitrogen atoms also have a quadrupole moment Section 6. If the nitrogen atom has a large quadrupole moment, the attached hydrogens will show peak broadening a widening of the peak at its base and an overall reduction of its intensity. Hydrogens adjacent to a carbonyl group regardless of type all absorb in the same region of the NMR spectrum: 2.
The spectrum of butyramide is shown in Figure 3. Notice the separate absorptions for the two INH hydrogens 6. This occurs due to restricted rotation in this compound. The hydrogens next to the CJ O group appear characteristically at 2. Nitroalkanes w O. In nitroalkanes, a hydrogens, those hydrogen atoms that are attached to the same carbon atom to which the nitro group is attached, have a characteristically large chemical shift. The electronegativity of the attached nitrogen and the positive formal charge assigned to that nitrogen clearly indicate the deshielding nature of this group.
A spectrum of 1-nitrobutane is shown in Figure 3. Note the large chemical shift 4. What are the allowed nuclear spin states for the following atoms? Calculate the chemical shift in parts per million d for a proton that has resonance Hz downfield from TMS on a spectrometer that operates at 60 MHz. A proton has resonance 90 Hz downfield from TMS when the field strength is 1. Acetonitrile CH3CN has resonance at 1. The larger dipole moment for the cyano group suggests that the electronegativity of this group is greater than that of the chlorine atom.
Explain why the methyl hydrogens on acetonitrile are actually more shielded than those in methyl chloride, in contrast with the results expected on the basis of electronegativity. Hint: What kind of spatial pattern would you expect for the magnetic anisotropy of the cyano group, CN? The position of the OH resonance of phenol varies with concentration in solution, as the following table shows. On the other hand, the hydroxyl proton of ortho-hydroxyacetophenone appears at Phenol The chemical shifts of the methyl groups of three related molecules, pinane, a-pinene, and b-pinene, follow.
In benzaldehyde, two of the ring protons have resonance at 7. Explain the patterns and intensities of the isopropyl group in isopropyl iodide. What spectrum would you expect for the following molecule? What arrangement of protons would give two triplets of equal area?
Predict the appearance of the NMR spectrum of propyl bromide. The following compound, with the formula C4H8O2, is an ester. Give its structure and assign the chemical shift values. The following compound is a monosubstituted aromatic hydrocarbon with the formula C9H The following compound is a carboxylic acid that contains a bromine atom: C4H7O2Br.
The peak at What is the structure of the compound? The following compounds are isomeric esters derived from acetic acid, each with formula C5H10O2. Each of the spectra has been expanded so that you will be able see the splitting patterns.
With the first spectrum 17a as an example, you can use the integral curve traced on the spectrum to calculate the number of hydrogen atoms represented in each multiplet pp. Alternatively, you can avoid the laborious task of counting squares or using a ruler to measure the height of each integral! It is far easier to determine the integral values by using the integral numbers listed just below the peaks. These numbers are the integrated values of the area under the peaks.
They are proportional to the actual number of protons, within experimental error. The process: Divide each of the integral values by the smallest integral value to get the values shown in the second column 1.
The values shown in the third column are obtained by multiplying by 2 and rounding off the resulting values. If everything works out, you should find that the total number of protons should equal the number of protons in the formula, in this case 10 protons.
What are the structures of the two esters? Draw the structure. Provide a structure for each. The downfield protons appearing in the NMR spectra at about Draw the structures of the isomers.
Make no attempt to interpret the aromatic proton area between 7. The infrared spectrum shows a doublet at about cm—1. Draw the structure of the compound. Their infrared spectra show strong bands near cm—1. Draw the structure of the compounds. Draw the structures of the compounds.
Along with the following NMR spectrum, this compound, with formula C5H10O2, shows bands at cm—1 broad and cm—1 strong in the infrared spectrum. Draw its structure. The infrared spectrum shows medium-intensity bands at and cm—1. The infrared spectrum has strong absorption at cm—1 and has several strong bands in the range to cm—1. Draw the structure of this compound. Highway, Suite , Plymouth, NH , www. Schatz, P. This database includes infrared, mass spectra, and NMR data proton and carbon for a large number of compounds.
They provide links to other sites with problems for students to solve. Berger, S. Crews, P. Rodriguez, and M. Friebolin, H. Gunther, H.
Jackman, L. Lambert, J. Shurvell, D. Lightner, and R. Macomber, R. Sanders, J. Silverstein, R. Webster and D. Kiemle, Spectrometric Identification of Organic Compounds, 7th ed. Williams, D. Fleming, Spectroscopic Methods in Organic Chemistry, 4th ed. Yoder, C. Compilations of Spectra Ault, A. Pouchert, C. Pretsch, E. Buhlmann, and C. Affotter, Structure Determination of Organic Compounds.
Tables of Spectral Data, 3rd ed. Translated from the German by K. Carbon spectra can be used to determine the number of nonequivalent carbons and to identify the types of carbon atoms methyl, methylene, aromatic, carbonyl, and so on that may be present in a compound. Thus, carbon NMR provides direct information about the carbon skeleton of a molecule. Typically, both techniques are used together to determine the structure of an unknown compound. Unfortunately, the resonances of 13C nuclei are more difficult to observe than those of protons 1H.
They are about times weaker than proton resonances, for two major reasons. First, the natural abundance of carbon is very low; only 1. If the total number of carbons in a molecule is low, it is very likely that a majority of the molecules in a sample will have no 13C nuclei at all. In molecules containing a 13C isotope, it is unlikely that a second atom in the same molecule will be a 13C atom.
Therefore, when we observe a 13C spectrum, we are observing a spectrum built up from a collection of molecules, in which each molecule supplies no more than a single 13C resonance. No single molecule supplies a complete spectrum. Second, since the magnetogyric ratio of a 13C nucleus is smaller than that of hydrogen Table 3. Recall that at lower frequencies, the excess spin population of nuclei is reduced Table 3. For a given magnetic field strength, the resonance frequency of a 13C nucleus is about one-fourth the frequency required to observe proton resonances see Table 3.
For example, in a 7. With modern instrumentation, it is a simple matter to switch the transmitter frequency from the value required to observe proton resonances to the value required for 13C resonances.
To compensate for the low natural abundance of carbon, a greater number of individual scans must be accumulated than is common for a proton spectrum. An important parameter derived from carbon spectra is the chemical shift. The correlation chart in Figure 4. Approximate 13C chemical shift ranges for selected types of carbon are also given in Table 4.
Notice that the chemical shifts appear over a range 0 to ppm much larger than that observed for protons 0 to 12 ppm. Because of the very large range of values, nearly every nonequivalent carbon atom in an organic molecule gives rise to a peak with a different chemical shift. Lampman, George S. Kriz and James R. This fourth edition of Introduction to Spectroscopy contains some important changes.
The discussion of coupling constant analysis in Chapter 5 has been significantly expanded. Long-range couplings are covered in more detail, and multiple strategies for measuring coupling constants are presented. Most notably, the systematic analysis of line spacings allows students with a little practice to extract all of the coupling constants from even the most challenging of first-order multiplets.
Chapter 5 also includes an expanded treatment of group equivalence and diastereotopic systems. Discussion of solvent effects in NMR spectroscopy is discussed more explicitly in Chapter 6. A new section on determining the relative and absolute stereochemical configuration with NMR has also been added to this chapter. All of the common ionization methods are covered, including chemical ionization CI , fast-atom bombardment FAB , matrix-assisted laser desorption ionization MALDI , and electrospray techniques.
Different types of mass analyzers are described as well. Fragmentation in mass spectrometry is discussed in greater detail, and several additional fragmentation mechanisms for common functional groups are illustrated.
Numerous new mass spectra examples are also included. Pavia Gary M. Lampman George S. Kriz James R. This is the fourth edition of a textbook in spectroscopy intended for students of organic chemistry. This book is also a useful tool for students engaged in research. Our aim is not only to teach students to interpret spectra, but also to present basic theoretical concepts. As with the previous editions, we have tried to focus on the important aspects of each spectroscopic technique without dwelling excessively on theory or complex mathematical analyses.
This book is a continuing evolution of materials that we use in our own courses, both as a supplement to our organic chemistry lecture course series and also as the principal textbook in our upper division and graduate courses in spectroscopic methods and advanced NMR techniques.
Explanations and examples that we have found to be effective in our courses have been incorporated into this edition. File Size: Pages: Please read Disclaimer. Free download Introduction to Spectroscopy by Donald L.
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