1 H NMR Spectroscopy: Background

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1 1 NMR Spectroscopy: Background Molecules are too small to be observed with the naked eye. In fact, it was only recently that the technology was developed to directly observe molecules by a specialized microscope. The molecules that can be observed by such a technique are still quite limited and the instrumentation is prohibitively expensive. As a result, most characterization of organic molecules is done by Nuclear Magnetic Resonance Spectroscopy. Nuclei have a quantum property called spin. The 13 C nuclei and 1 nuclei (protons) have two separate spin states (-1 and 1) as their spin quantum number is 1. The 12 C nuclei have only one spin state (0) as their spin quantum number is 0. The spin quantum number is a measured property related to a nucleus's angular momentum and cannot be calculated based on number of neutrons and protons in a nucleus. Normally, the nuclei do not care what spin state they reside in as both states have equal energy. owever, when a magnetic field is applied the two spin states start to have different energies as the two spin states do not have the same inherent magnetic field. If the magnetic field is substantially large enough (~ 7-12 Tesla roughly 200 thousand times that of Earth's magnetic field) a noticeable energy difference between the two spin states occur. The slight energy difference is approximately equal to that of a radiowave. It should be noted that modern day spectrometers work differently than described below, but the principle is still the same. Since the energy difference is similar to that of a radiowave, a sample can be bombarded with a wide range of radio frequencies. When the energy of the gap and the radiowave match, there is an absorbance and the spectrometer registers a peak. Moreover, depending on symmetry, the nuclei in a molecule are in different environments

2 each with a small local magnetic field. As a result, a different peak is observed for each different nucleus allowing a unique spectrum for each molecule. Furthermore, this pattern can be interpreted to ascertain the basic structure of a molecule. Since the 12 C nucleus only has one spin state, no separation can be seen and thus no peaks can be seen, rendering 12 C nuclei invisible to NMR instruments. 1 NMR, more commonly known as proton NMR, can lead to far more insight into the structure of a molecule than 13 C NMR, especially when the molecules are small. There are three characteristics to a 1 NMR spectrum: chemical shift, integration, and coupling. Chemical Shift: Chemical shift works the same in proton NMR spectroscopy as it does in 13 C NMR. Chemical shift is relative and is based on the distance in ppm of a signal from the peak at 0 ppm (tetramethylsilane). For every signal you have a unique type of proton. The chemical shift depends on the localized environment of a proton. Chemical shift tables have been made utilizing empirical data collected by a wide range of functional groups. A more extensive chemical shift table can be found at the back of the lab manual. These tables just serve as guidelines. It is conceivable that protons associated with the various functional groups can be found outside the ranges due to structural features of the molecule itself. The number of signals that are present depends on the types of protons present. When comparing two protons, it is important to evaluate how they are related.

3 The protons can be homotopic, enantiotopic, diastereotopic, or heterotopic. omotopic and enantiotopic protons are in the same chemical environment and have the same chemical shifts. Protons that are heterotopic or diastereotopic are in different chemical environments and will have different chemical shifts. To determine the relationship between two protons, the substitution test is the most reliable method to use. Identify the two protons that are to be compared. Make two molecules by changing one and then the other proton into a deuterium. The two molecules are compared, if the molecules are the same, then the protons are homotopic. If the molecules are enantiomers, then the protons are enantiotopic. If the molecules are diastereomers, the protons are diastereotopic. If the molecules are constitutional isomers, then the protons are heterotopic. g vs f e vs f a vs b D D D C C C a b c d D g e C C C C f h D D As practice, let s examine methylcyclopropane. Methylcyclopropane has 8 protons, but how many signals should be expected in the 1 NMR spectrum? What is the relationship between each pair of protons? These two molecules are just conformations of the same molecule. Therefore, the f and g are homotopic and have the same chemical shift. h is also homotopic to f and g for the same reason. These isomers have different connectivity, so they are constitutional isomers. Therefore, e and f are heterotopic and will have different chemical shifts. e is very unique and is heterotopic with all of the other protons. These isomers have the same connectivity and the same interatom distances but are not superimposable. The first isomer is (R,R) while the second is (S,S). These isomers are enantiomers. Therefore, a and b are enantiotopic and will have the same chemical shift. As far as the NMR instrument can see, they are equivalent. other protons. c and d are also examples of a pair of enantiotopic protons.

4 b vs d These isomers have the same connectivity but have different interatom distances. These isomers are diastereomers, so b and d are diastereotopic. ne has the proton on the same face of the molecule as the methyl group. The other has the proton on the opposite face. As a result, two different physical environments are present and two signals arise, one for b and one for d. a and c are also examples of a pair of diastereotopic protons. Methylcyclopropane has eight protons, but the 1 NMR is expected to have only four signals, one for each type of proton. ne signal for a and b (these are enantiotopic to one another). ne signal for c and d (these are enantiotopic to one another). ne signal for e. ne signal for f, g and h (these are homotopic to one another). Because protons that are homotopic and protons that are enantiotopic are in the same environment, sometimes it is said that they are the same type of proton. General notice: Protons on methyl groups are almost always homotopic as free rotation allows the interchange of them. The NMR is said to have a slow timescale, meaning most molecular rotations about sigma bonds fast. As a result, instead of seeing the protons on a methyl group as three different protons, it sees a blur and they appear as the same proton. D C acetaldehyde 4 protons 2 types D N C N,N-dimethylformamide 7 protons 3 types (room temperature) 2 types (high temperature) Rotation about the carbonyl nitrogen sigma bond of an amide is actually slower than the NMR timescale. The slower rotation is due to the amide resonance structure. As a result, you cannot rotate an amide bond when determining the number of types of protons. At higher temperatures, the rotation of the amide bond is fast, and as a result only two signals are seen. Unless otherwise stated, assume the spectrum is taken at room temperature. Protons on methylenes (C 2 s) are VERY rarely (almost never) homotopic. They are either enantiotopic or diastereotopic. In general, if a stereocenter is present, they are diastereotopic. If no stereocenter is present, they are enantiotopic.

5 There are some examples of diastereotopic methylene protons when a stereocenter is not present. The best way to determine whether they are enantiotopic or diastereotopic is the substitution test. Since chemical shift is governed by the local magnetic field that the nucleus experiences, it is important to know the factors that increase or decrease that magnetic field. Electron density has the largest impact on the localized magnetic field, as electrons themselves have a magnetic field associated with them. This field is situated to directly oppose the large external magnetic field, so it partially "shields" the nucleus from the magnetic field. The higher the electron density, the more the nucleus is shielded. Signals that are shielded are on the right hand side of the NMR spectrum and are described as "upfield." Inductive effects When electronegative elements are present such as oxygen or nitrogen, the signals are shifted to the left of the spectrum and are regarded as "downfield." The electronegative elements pull electrons away from the protons (hydrogen nuclei) through single bonds, thus deshielding the protons. This effect is called the inductive effect and is a very short range effect. It is felt typically through one or two single bonds. Diethyl ether 1-butanol 2 bonds away from 3.47 ppm 3 bonds away from 1.21 ppm 5 bonds 0.94 ppm 4 bonds 1.39 ppm 2 bonds 3.63 ppm 3 bonds 1.57 ppm 1 bond 2.24 ppm As can be seen in the examples of diethyl ether and butanol, the oxygen's effect greatly diminishes when it is further than two bonds away. It is interesting to note that in the case of 1-butanol that the proton directly attached to the oxygen is less affected that the proton two bonds away. Protons directly attached to heteroatoms (, N, S, non carbon atoms) often have broad and sometimes unpredictable chemical shift ranges. This is mainly due to hydrogen bonding which can have a significant impact on the environment a proton resides in. The amount of water present in the NMR sample can cause different NMR of these protons. Protons attached to carbons rarely partake in hydrogen bonding, and their chemical shifts are consequently more predictable.

6 The Effect of Aromatic Rings An important chemical shift region to remember is the aromatic region. It is located from 6.5 ppm to 8.0 ppm. Protons directly attached to an aromatic ring (benzene) can be found in this region. The aromatic region is a strongly shifted region due to the ring current that is present in aromatic molecules. Think of the electrons has circulating about the ring. They can do so clockwise or counter clockwise. Both situations have equal energy. owever, when a magnetic field is applied (such as in an NMR instrument), they go around in one direction generating a minute electrical current. Such a current creates a small magnetic field. The magnetic field that is generated goes against the external magnetic field in the center of an aromatic ring. Now, magnetic fields are circular. So, if it goes against the magnetic field inside the ring, it goes with the magnetic field outside the ring. The aromatic protons are outside the ring and as a result experience a greater magnetic field (net effect: the protons are strongly deshielded).

7 Resonance effects Inductive effects are weak effects with a limited range and are transmitted through the sigma bond network. When an oxygen or nitrogen is attached to a sp 2 or sp hybridized carbon, a much more powerful effect takes precedence due to resonance. This resonance effect, sometimes referred to as a mesomeric effect, can drastically increase or decrease the electron density about a carbon and subsequently protons attached to that carbon. Consider the cases of p-xylene and p-dimethoxybenzene. Even though there are ten protons present, only 2 signals are observed for each molecule due to symmetry. p-xylene p-dimethoxybenzene 2.30 ppm A 7.05 ppm B 7.05 ppm B 2.30 ppm A 3.75 ppm C 6.83 ppm D 6.83 ppm D 3.75 ppm C B B D D As expected, the presence of the oxygen has a deshielding effect on C. owever, the same oxygen has a shielding effect on D. This is due to resonance. D D D D C C C C D D D D D D D D C C C C D D D D

8 The oxygen is a strong electron donating group. By donating electrons into the aromatic ring, it effectively adds negative charge on groups ortho and para to itself. The increase in negative charge means an increase in electron density, and thus a strong shielding effect for the protons attached to the carbons with the negative charge. C is directly attached to an sp 3 carbon that has no p orbitals and consequently does not feel the effect of the resonance. Electron withdrawing groups have the opposite effect: p-xylene p-acetylbenzene 7.05 ppm 7.05 ppm 8.03 ppm 8.03 ppm 2.30 ppm B B 2.30 ppm 2.64 ppm D D 2.64 ppm A A C C B B D D C feels a deshielding effect due to the carbonyl inductively drawing electrons away. The carbonyl has a much weaker inductive effect than the oxygen in p- dimethoxybenzene as oxygen has a higher inherent electronegativity than the carbonyl group. The carbonyl does have a substantial resonance effect on D as it is a strong electron withdrawing group when it comes to resonance. D D D D C C C C D D D D D D D D C C C C D D D D Instead of placing negative charge into the ring like the oxygen, the carbonyl group pulls electrons out of the ring creating positive charge within the ring. As

9 a result, the carbons that bear that partial positive have a reduced electron density, and therefore the protons attached to those carbons are deshielded (shifted to the left of the spectrum). acetophenone 7.9 ppm 7.3 ppm Deshielded C B 2.59 ppm A anisole 6.88 ppm 7.26 ppm Shielded G F 3.75 ppm E D 7.6 ppm Deshielded 6.92 ppm Shielded C B G F The resonance structures using electron withdrawing and electron donating groups also remove or donate electrons to and from the ortho and para positions. Both positions are affected, but in proton NMR these groups have more effect in the ortho position then they do in the para position. This is seen in the examples of acetophenone and anisole above. Common Electron Donating Groups Common Electron Withdrawing Groups N alkoxy nitro carbonyl N amino nitrile N

10 Integration Unlike the rare 13 C nucleus, the 1 nucleus represents 99% of all hydrogen nuclei. As a result of this abundance, proton NMR is not as sensitive allowing for faster accumulation times and more importantly integration. In proton NMR the area underneath is directly proportional to the number of protons that signal represents. This can make assignments of signals to protons a trivial process. In the case of trimethylsilylacetate, the assignments could be made based on chemical shift (the numbers to the left of the signal) or they can be assigned by using integration (the number to the right of the signal). In this case, the signal at δ 3.76 ppm has an integration of 2 and corresponds to the C 2. The signal at δ 2.04 ppm has an integration of 3 and corresponds to the single methyl group. The 3 methyl groups attached to the silicon all rapidly interconvert with each other sure to rotation about C-Si and C-C sigma bonds causing them to all overlap and appear as a single peak with an integration of 9 (0.08 ppm). n the NMR spectrum, we would simply draw the molecule out and write a, b, and c next to the protons and match them to the corresponding signals.

11 NTE T TSE TAKING CEM 344: The NMR spectra that you will interpret in the chemistry labs will be spectra of the compounds that you make in lab. As a result, they may not be as clean as some of the spectra that you see here. The format will be slightly different as well. The chemical shift value will not be above the peak. Instead, you will have to write it out based on where the signal falls on the x axis. Also, the integration will not be written out on the sheet to the right of the signal. It will be located under the signal itself. It is also very likely the integration will not be whole numbers. When dealing with integration, the relative ratio is important. The integration in the example above (2:3:9) could have been 4:6:18 or even 14.29:21.43:64.29 as the NMR instrument frequently calculates integration as a ratio totaling 100. Integration is a very useful tool not only for identifying compounds and assigning peaks, but also can be used to calculate the purity of a sample should the identity of the impurities be known. The proton NMR spectrum of a mixture of p-xylene and methanol is shown on the next page. This spectrum was taken using the instrument that will process all of your laboratory samples, and the integral values are located beneath the peaks. Notice that they add to 100. We can assign the signals at 7.1 ppm and 2.3 ppm to p-xylene and the signals at 3.4 ppm and 3.0 ppm to methanol using a chemical shift table. The integration also confirms this. For p-xylene, the peak ratio should be 1:1.5 (4:6) and is 1:1.64 on the spectrum, which is fairly close. For methanol, the ratio should be 1:3 and is actually 1:2.8. The numbers are not exact, as there are some errors having to do with the how the instrumentation acquires and how the software processes the data.

12 With the peaks correctly assigned, the ratio of methanol to p-xylene can be calculated by comparing the integration values. First you must choose which values to compare. For the best results, you should compare peaks that are close to one another, but are still far enough away for other peaks so that the integration is as accurate as possible. In this case, let's use the signals for the two sets of methyl groups (3.4 ppm for methanol and 2.3 ppm for p-xylene). Before you can compare them, you must first normalize the peaks, because the p-xylene methyl signal consists of 6 protons and the methanol methyl signal consists of only 3 protons. That is just the molar ratio. To get a mass ratio or volume ratio, you would need to incorporate the corresponding molecular weights and densities.

13 Coupling Coupling (sometimes referred to as splitting) is the probably the most useful characteristic of an NMR spectrum and is at the same time the hardest to understand. When protons that do not have the same chemical shift (ie: inequivalent protons) are within a certain range (2-4 bonds), they can influence the shape of the signals. The biggest influence occurs when they are three bonds away. For example, isopropyl ether has two types of protons: the C (methyne) and the C 3 (methyl). The four methyl groups are related by a symmetrical mirror plane and the two methynes are related by a mirror plane as well. isopropyl ether Methyne 3.64 ppm C 3 Methyl C ppm 3 C 2 3 The methyl protons are located three bonds away from the methyne proton. As a result coupling occurs and the signals representing the methynes and the methyl groups no longer consist of single peaks but multiple peaks. From the integration it is easy to assign which peak is which. The methyl groups have integration of 12 and the methynes have integration of two. The number of the peaks within each signal is important and is due to coupling. The methyl signal consists of two peaks, while the methyne signal consists upwards of five peaks. Both patterns are the result of the same effect.

14 Let's take a look at the methyl group first. The methyl protons are three bonds away from the single methyne proton. The methyne proton can be spinning one way or another (with or against the external field). ne spin state shields, the other spin state desshields. Since it is a near equal chance either way, two different energy transitions appear equal in magnitude and each offset from the original frequency by the same amount. Two different energy transitions mean two different peaks. The peaks are separated by the coupling constant abbreviated J. J is measured in hertz (z) and is typically 7 z in the case of linear alkyl groups. A way to show this is called a coupling tree also known as a splitting diagram. All of the original integration of the methyl signal is divided almost equally into two peaks. Each peak offset from the original signal by 3.5 z (half of the coupling constant). Therefore, if the integration of the original signal is 12, then each of the two peaks would have integration of 6. The methyne pattern though is a bit more complicated (> 5 peaks), but arises through the same mechanism. Except instead of having one neighboring proton three bonds away it has six. Each of them has a spin state that shields or deshields the original signal by 3.5 z. The result is actually a seven line pattern called a septet with each peak 7 z from the next.

15 Unlike the case of the doublet, these peaks do not have equal intensity. Instead, the peaks at the center have the highest intensity, because the center peaks consist of a series of overlapping peaks. If the peak heights were carefully measured, the ratio would be approximately 1:6:15:20:15:6:1. When dealing with coupling constants of similar magnitude, an important rule to remember is the n+1 rule, where n equals the number of protons three bonds away (or protons on neighboring atoms for short). In the case of the methyl group, it has one neighboring proton so 1+1=2 a doublet. In the case of the methyne, it has six neighboring protons so 6+1=7 a septet. The intensities of these multiplets can even be predicted by the mathematical construct called Pascal's triangle.

16 It should be noted that the septet, octet, and nonets can be tricky to identify. Most of these have low integration values to begin with as they are invariably connected to two or three carbons each bearing multiple protons. Most of that integration is in the central peaks and very little of it is on the outer edges. ne outer peak of the septet has 1/64 of the total integration. Such a peak is very small and may blend into the baseline. Therefore, when you count five peaks, there may actually be seven or even nine present. When you count six peaks, there may be eight present. You can tell the even number patterns from the odd number patterns as the odd number pattern have a single tall central peak. The even number pattern has two central peaks of approximately the same intensity. Let's take a look at nitroethane. nitroethane N The methyl group is attached to a methylene (C 2 ) so its signal should be a triplet (2+1=3). The methylene is attached to a methyl group and a nitro group which does not have a proton, so it will be a quartet (3+1=4). The triplet will have peaks with intensity ratio of 1:2:1 and the quartet will have a peak intensity ratio of 1:3:3:1. Since the methylene is directly attached to the nitro group it is the most deshielded so it will be on the left. The quartet will be on the right. This is indeed what the spectrum looks like. Another way to look at these multiplets is in term of the spin of the neighbors. The methyl group has two neighboring protons each with the possibility of an up spin or a down spin. Let's say the down spin causes shielding and the up spin causes deshielding. So the possibilities are:

17 one deshielded combination two "neutral" combinations 1:2:1 triplet one shielded combination ne the other hand the methylene has three neighboring protons. This gives the following situation: one strongly deshielded combination three deshielded combinations three shielded combinations one strongly shielded combination 1:3:3:1 quartet The n+1 rule does have a couple of caveats. It works when the coupling constants are the same. Typically, when the carbon atoms are sp 3 hybridized and are linear (not locked in a ring) the coupling constant is 7 z. Protons that have the same chemical shift (homotopic or enantiotopic) do not visibly split each other. Protons can be equivalent based on symmetry or rapid rotation about a single bond. The C 2 groups in 1,2-dimethoxyethane do not couple with each other even though they are on adjacent carbons. Protons on heteroatoms may or may not couple depending on the conditions in the NMR sample. Generally, they do not couple but instead appear as a broad peak. When peaks broaden, they shorten. The width of carboxylic acid signals can span over an entire ppm, and as a result appear as baseline.

18 When the molecule is in a ring or there is a pi bond present, the rotation about some of the bonds is restricted. As a result, different coupling constants can occur. First case to look at is 4-isopropylaniline. The N 2 and the isopropyl group is straightforward to assign as the N 2 is a broad peak (3.5 ppm), the C is a multiplet (2.8 ppm), and the two methyl groups form the doublet at 1.2 ppm. The isopropyl coupling constant is about 7

19 z as the carbons are sp 3 hybridized. The ring protons are a different situation. The para substituted benzene ring gives a very distinct pattern of two pairs of doublets in the aromatic region. The protons attached to the ring are rigidly held in one geometry and do not freely rotate. As a result, the coupling constant is different from the freely rotating 7 z coupling constant. Instead, it is the much larger ortho coupling constant. J ortho can range from 6-12 z, but typically lies around 10 z. If you examine the patterns closely, you will see that the peaks are not really doublets. Instead each peak consists of six lines. These lines represent part of a very complex pattern (referred to as a second order pattern). For the purposes of this course, these peaks can be thought of as pseudodoublets. To actually assign the peaks to the appropriate protons, resonance structures can be drawn. The protons ortho to the N 2 group bear a partial negative charge and are thus more shielded. It becomes more complicated when a meta substitution pattern is present. When two protons are meta to each other, they can split each other even though they are four bonds distant. This coupling constant (J meta ) is smaller than the ortho and ranges from one to three hertz. This can lead to some interesting patterns. Consider the aromatic region of 3-nitroanisole:

20 The aromatic region has 4 signals. Two of which appear to be triplets, though one is wider than the other. The other two patterns have eight lines apiece. To assign this spectra, it will be necessary to make a table of the coupling constants. A B C D J ortho J meta With the table in hand, we can generate a series of splitting diagrams to predict the shapes of the signals. There are two possibilities for A. If the two meta coupling constants are not equal then A is a doublet of doublets (4 lines). If they are equal, then A is a triplet (3 lines). (A triplet is just a doublet of doublet where the magnitude of the two coupling constants are equal). The signal for C will have similar possibilities as A, except C will be wider. After all, the width of the signal is just the sum of the coupling constants, and the J ortho is greater than J meta.

21 A A J AD J AD J AB J AB four line pattern (dd) doublet of doublets 1:1:1:1 ratio When you compare the predicted patterns to the observed patterns, it becomes clear that the triplet at 7.71 ppm is A and the triplet at 7.48 ppm is C. B and D give rise to a more complicated pattern as both have one large J ortho and 2 small J meta. If all three coupling constants are different, an eight line pattern called a doublet of doublets of doublets (ddd) is formed. Indeed, this is the case, and the multiplets at 7.79 ppm and 7.23 ppm are best described as ddd. three line pattern (t) triplet 1:2:1 ratio B Since both give identical patterns, the only way to assign the two structures is by looking closely at the structure of the molecule itself and relying on chemical shift. B is ortho to an electron withdrawing group, therefore it will be strongly deshielded. Conversely, D on the other hand is ortho to an electron donating group, thus it will be strongly shielded. With this information, the ddd at 7.79 ppm can be assigned to B and the ddd at 7.23 ppm belongs to D. J BC eight line pattern (ddd) doublet of doublets of doublets 1:1:1:1:1:1:1:1 ratio J BD J AB A majority of the 1 NMR spectra for this guide came from the Sigma-Aldrich company ( There is a lot of help available on the internet. ne good place is: This site offers a lot of NMR practice problems. (Skip compound 2 in the beginning section. It really should be in the intermediate section).