Structure Determination: Nuclear Magnetic Resonance Spectroscopy Chapter 13 Structure Determination: Nuclear Magnetic Resonance Spectroscopy Suggested Problems – 1-23,29,34,36-8,44-7,51

Nuclear Magnetic Resonance Spectroscopy Nuclei are positively charged, have spin, and interact with an external magnetic field, B0 Magnetic rotation of nuclei is random in the absence of a magnetic field In the presence of a strong magnet, nuclei adopt specific orientations In the absence of an external magnetic field, the spins of magnetic nuclei are oriented randomly. When a sample containing these nuclei is placed between the poles of a strong magnet, however, the nuclei adopt specific orientations. A spinning 1H and 13C nucleus can orient so that its own tiny magnetic field is aligned either with (parallel to) or against (antiparallel to) the external field. The parallel orientation is slightly lower in energy by an amount that depends on the strength of the external field, making this spin state very slightly favored over the antiparallel orientation.

Nuclear Magnetic Resonance Spectroscopy When exposed to a certain frequency of electromagnetic radiation, oriented nuclei absorb energy which causes a spinflip from a state of lower energy to higher energy Nuclear magnetic resonance - Nuclei are in resonance with applied radiation Frequency that causes resonance depends on: Strength of external magnetic field Identity of the nucleus Electronic environment of the nucleus

Nuclear Magnetic Resonance Spectroscopy Larmor equation Relation between Resonance frequency of a nucleus Magnetic field and the magnetogyric ratio of the nucleus If a very strong magnetic field is applied, the energy difference between the two spin states is larger and higher-frequence (higher-energy) radiation is required for a spin-flip. If a weaker magnetic field is applied, less energy is required to effect the transition between nuclear spin states. The Larmor equation can be used to calculate the resonance frequency of a nucleus based on the strength of the magnetic field B0 and the magnetogyric ratio of the nucleus. The latter is a ratio of the isotope’s magnetic moment to its angular momentum.

Worked Example Calculate the amount of energy required to spin-flip a proton in a spectrometer operating at 300 MHz Analyze if the increase of spectrometer frequency from 200 MHz to 300 MHz increases or decreases the amount of energy necessary for resonance Remember E= hv and lambda x v = c (the speed of light)

Worked Example Solution: Increasing the spectrometer frequency from 200 MHz to 300 MHz increases the amount of energy needed for resonance Use E = h v h = 6.626 x 10-34Js Multiply by Avogadro’s number to get per mol.

The Nature of NMR Absorptions Absorption frequencies differ across 1H and 13C molecules Shielding: Opposing magnetic field produced by electrons surrounding nuclei to counteract the effects of an external magnetic field Effect on the nucleus is lesser than the applied magnetic field Beffective = Bapplied – Blocal Individual variances in the electronic environment of nuclei leads to different shielding intensities All nuclei in molecules are surrounded by electrons. When an external magnetic field is applied to a molecule, the electrons moving around nuclei set up tiny local magnetic fields of their own. These local magnetic fields act in opposition to the applied field so that the effective field actually felt by the nucleus is a bit weaker than the applied field. The nucleus is experience shielding from the full effect of the applied field by the surrounding electrons. Because each chemically distinct nucleus in a molecule is in a slighly different electronic environment, each nucleus is shielded to a slightly different extent and the effective magnetic field felt by each is slightly different. Thes tiny differences in the effective magnetic fields experienced by different nuclei can be detected, and we thus see a distinct NMR signal for each chemically distinct 13C or 1H nucleus in a molecule. As a result, an NMR spectrum effectively maps the carbon-hydrogen framework of an organic molecule.

NMR Spectrum of 1H and 13C The horizontal axis shows the effective field strength felt by the nuclei, and the veritcal axis indicates the intensity of absorption of rf energy. Each peak in the NMR spectrum corresponds to a chemicall distinct 1H or 13C nucleus in the molecule. Different amounts of energy are required to spin-flip different kinds of nuclei. As a result, 1H and 13C spectra cannot be observed simultaneously on the same spectrometer. The two spectra must be recorded separately. The 13C NMR of methyl acetate shows three peaks, one for each of the three chemically distinct carbon atoms in the molecule. The 1H NMR spectrum shows only two peaks, however, even though methyl acetate has six hydrogens. One peak is due to the CH3C=O hydrogens, and the other is due to the –OCH3 hydrogens. Because the three hydrogens in each methyl group have the same electronic environment, they are shielded to the same extent and are said to be equivalent. Chemically equivalent nuclei always show the a single absorption. The two methyl groups themselves, however, are nonequivalent, so the two sets of hydrogens absorb at different positions.

Working of an NMR Spectrometer Organic sample dissolved in a suitable solvent is placed in a thin glass tube between the poles of a magnet 1H and 13C nuclei respond to the magnetic field by aligning themselves to one of the two possible orientations followed by rf irradiation Varying the strength of the applied field causes each nucleus to resonate at a slightly varied field strength Absorption of rf energy is monitored by a sensitive detector that displays signals as a peak Commonly deuterochloroform CDCl3 is used as a solvent because it contains no hydrogens; deuterium does not resonate in the NMR. If the frequency of the rf irradiation is held constant and the strength of the applied magnetic field is varied, each nucleus comes into resonance at a s slightly different field strength. A sensitive detector monitors the absorption of rf energy, and its electronic signal is then amplified and displayed as a peak.

Operation of a Basic NMR Spectrometer

NMR Spectrometer Time taken by IR spectroscopy is about 10–13 s Time taken by NMR spectroscopy is about 10–3 s Provides a blurring effect that is used in the measurement of rates and activation energies of vary fast processes The absorption of infrared energy by a molecule is nearly instantaneous. The absorbtion of radiofrequency energy by a hydrogen or carbon nucleus is considerably slower. As a result, the spectum recorded by an NMR is a time-averaged spectrum reflecting the average of any changes (such as a ring-flip) that might be occuring in a molecule. Because of this blurring effet, NMR spectroscopy can be used to measure the rates and activation energies of very fast processes. In cyclohexane, for example, a ring-flip occurs so rapidly at room temperature that axial and equatorial hydrogens can’t be distinguished by NMR: only a single averaged 1H absorption is seen for cyclohexane at 25oC. At – 90oC, however, the ring-flip is slowed down enough that two absorption peaks are visible, one for the six axial hydrogens and one for the six equatorial hydrogens. Knowing the temperature and the rate at which signal blurring begins to occur, it’s possible to calculate that the activation energy for the cyclohexane ring-flip is 45 kJ/mol.

Worked Example Explain why 2-chloropropene shows signals for three kinds of protons in its 1H NMR spectrum Solution: 2-Chloropropene has three kinds of protons Protons b and c differ because one is cis to the chlorine and the other is trans

The Chemical Shift The left segment of the chart is the downfield Nuclei absorbing on the downfield have less shielding as they require a lower field for resistance The right segment is the upfield Nuclei absorbing on the upfield have more shielding as they require a higher field strength for resistance NMR spectra are displayed on charts that show the applied field strength increasing from left to right. The left part of the chart is referred to as downfield and nuclei that absorb there are said to be deshielded relative to their upfield (right side of chart) shielded counterparts.

The NMR Chart To define the position of an absorption, the NMR chart is calibrated and a reference point is used. A small amount of tetramethylsilane (CH3)4Si is added to the sample so that a reference absorption peak is produced when a spectrum is run. TMS produces a single peak in both proton and carbon NMR spectra which occurs upfield of other absorptions commonly found in organic molecules.

The Chemical Shift Chemical shift is the position on the chart at which a nucleus absorbs The delta (δ) scale is used in calibration of the NMR chart 1 δ = 1 part-per-million of the spectrometer operating frequency The delta scale is used as the units of measurement can be used to compare values across other instruments The position on the chart at which a nucleus absorbs is called its chemical shift. The chemical shift of TMS is set as the zero point, and other absorptions normally occur downfield, to the left on the chart. NMR charts are calibrated using an arbitrary scale called the delta scale, where 1 delta equals 1 part-per-million (1 ppm) of the spectrometer operating frequency. By converting the absorption in Hz for a particular nucleus to a delta value, the chemical shift of the same nucleus will have the same chemical shift value on NMRs having differing field strengths. Almost all 1H absorptions occur from 0 to 10 delta downfield from the proton absorption of TMS, and almost all 13C absorptions occur from 1 to 220 delta downfield from the carbon absorption of TMS. The advantage of using an instrument with higher field strength (say, 500 MHz) rather than lower field strength (200 MHz) is that different NMR absorptions are more widely separated at the higher field strength. The chances that two signals will accidentally overlap are therefore lessened, and interpretation of spectra becomes easier.

Worked Example The 1H NMR peak of CHCl3 was recorded on a spectrometer operating at 200 MHz providing the value of 1454 Hz Convert 1454 Hz into δ units Solution:

Chemical Shifts in 1H NMR Spectroscopy Chemical shifts are due to the varied electromagnetic fields produced by electrons surrounding nuclei Protons bonded to saturated, sp3-hybridized carbons absorb at higher fields Protons bonded to sp2-hybridized carbons absorb at lower fields Protons bonded to electronegative atoms absorb at lower fields

Regions of the 1H NMR Spectrum Nuclei that are more strongly shielded by electrons require a higher applied field to bring them into resonance and therefore absorb on the right side of the NMR chart. Nuclei that are less strongly shielded need a lower applied field for resonance and therefore absorb on the left side of the NMR chart.

Representative Chemical Shifts Note the slight difference in chemical shift positions between aliphatic (alkyl) methyl, methylene, and methine groups. Similarly note how the methyl group is shifted downfield when it is attached to an aromatic ring. Lastly, note that aldehydic and acid protons are deshielded downfield to a greater extent than most other absorptions.

Worked Example CH2Cl2 has a single 1H NMR peak Solution: Determine the location of absorption Solution: For CH2Cl2 , δ = 5.30 The location of absorption are the protons adjacent to the two halogens Based on the chart on the previous slide, we know that a hydrogen on a carbon to which is attached a chlorine will resonate from 2.5-4.5 ppm. Since two electronegative chlorines are attached to the hydrogen bearing the carbon, the resonance of the hydrogen will be shifted downfield a little further.

Integration of 1H NMR Absorptions: Proton Counting In the figure, the peak caused by (CH3)3C–protons is larger than the peak caused by –OCH3 protons Integration of the area under the peak can be used to quantify the different kinds of protons in a molecule The area under each peak is proportional to the number of protons causing that peak. By electronically measuring, or integrating, the area under each peak, it’s possible to measure the relative number of the different kinds of protons in a molecule. The heights of the vertical “integration” lines on the two peaks reflect the relative differences in the number of hydrogens that account for the two signals.

Worked Example Mention the number of peaks in the 1H NMR spectrum of 1,4-dimethyl-benzene (para-xylene or p-xylene) Mention the ratio of peak areas possible on integration of the spectrum

Worked Example Solution: There are two absorptions in the 1H NMR spectrum of p-xylene The four ring protons absorb at 7.05 δ and the six methyl-groups absorb at 2.23 δ The peak ratio of methyl protons:ring protons is 3:2

Worked Example

Worked Example How many distinct signals are present in the 13C NMR of p-xylene? Mention the ratio of peak areas possible on integration of the spectrum

Worked Example This is a C-13 spectrum

Spin-Spin Splitting in 1H NMR Spectra Multiplet: Absorption of a proton that splits into multiple peaks The phenomenon is called spin-spin splitting Caused by coupling of neighboring spins The –CH2Br protons appear as four peaks (a quartet) centered at 3.42 ppm and the –CH3 protons appear as three peaks (a triplet) centered at 1.68 ppm. Multiple absorptions of a nucleus are caused by the interaction, or coupling, of the spins of nearby nuclei. In other words, the tiny magnetic field produced by one nucleus affects the magnetic field felt by a neighboring nucleus.

Spin-Spin Splitting in 1H NMR Spectra Alignment of both –CH2Br proton spins with the applied field will result in: Slightly larger total effective field and slight reduction in the applied field to achieve resonance There is no effect if one of the –CH2Br proton spins aligns with the applied field and the other aligns against it Alignment of both –CH2Br proton spins against the applied field results in: Smaller effective field and an increased applied field to achieve resonance Consider the –CH3 protons in bromoethane. The three equivalent –CH3 protons are neighbored by two other magnetic nuclei – the two protons on the adjacent –CH2Br group. Each of the neighboring –CH2Br protons has its own nuclear spin, which can align either with or against the applied field, producing a tiny effect that is felt by the –CH3 protons. There are three ways in which the spins of the two –CH2Br protons can align. If both protons align with the applied field, the total effective field felt by the neighboring –CH3 protons is slightly larger than it would be otherwise. Consequently, the applied field necessary to cause resonance is slightly reduced. Alternatively, if one of the –CH2Br proton spins aligns with the field and one aligns against the field, there is no effect on the neighboring –CH3 protons. (This arrangement can occur in two ways.) Finally, if both –CH2Br proton spins align against the applied field, the effective field felt by the –CH3 protons is slightly smaller than it would be otherwise, and the applied field needed for resonance is slightly increased. Any given molecule has only one of the three possible alignments of –CH2Br spins, but in a large collection of molecules, all three spin states are represented in a 1:2:1 statistical rtio. We therefore find that the neighboring –CH3 protons come into resonance at three slightly different values of the applied field, and we see a 1:2:1 triplet in the NMR spectrum.

The Origin of Spin-Spin Splitting in Bromoethane In the same way that the –CH3 absorption is split into a triplet, the –CH2Br absorption is split into a quartet. The three spins of the neighboring –CH3 protons can align in four possible combinations: all three with the applied field, two with and one against (three ways), one with and two against (three ways), or all three against. Thus, four peaks are produced for the –CH2Br protons in a 1:3:3:1 ratio.

Spin-Spin Splitting in 1H NMR Spectra n + 1 rule: Protons that exhibit n + 1 peaks in the NMR spectrum possess n = number of equivalent neighboring protons Coupling constant is the distance between peaks in a multiplet As a general rule, protons that have n equivalent neighboring protons show n+1 peaks in their NMR spectrum. The distance between peaks in a multiplet is called the coupling constant and is denoted J. Coupling constants are measured in hertz and generally fall in the range 0 to 118 Hz. The exact value of the coupling constant depends on the geometry of the molecule.Typically for open chain alkanes, J = 6-8 Hz. The same coupling constant is shared by both groups of hydrogens whos spins are coupled and is independent of spectrometer field strength.

Spin-Spin Splitting in 1H NMR Spectra It is possible to identify multiplets in a complex NMR that are related Multiplets that have the same coupling constant can be related Multiplet-causing protons are situated adjacent to each other in the molecule Because coupling is a reciprocal interaction between two adjacent groups of protons, it’s sometimes possible to tell which multiplets in a complex NMR spectrum are related to each other. If two multiplets have the same coupling constant, they are probably related, and the protons causing those multiplets are therefore adjacent in the molecule.

Rules of Spin-Spin Splitting Chemically equivalent protons do not show spin-spin splitting The signal of a proton with n equivalent neighboring protons is split into a multiplet of n + 1 peaks with a coupling constant Two groups of protons coupled together have the same coupling constant, J 1) Chemically equivalent protons may be on the same carbon or on different carbons, but their signals don’t split. 2) Protons that are farther than two carbon atoms apart don’t usually couple, although they sometimes show weak coupling when they are separated by a pi bond.

Worked Example The integrated 1H NMR spectrum of a compound of formula C4H10O is shown below Propose a structure

Worked Example Solution: The molecular formula (C4H10O) indicates that the compound has no multiple bonds or rings The 1H NMR spectrum shows two signals, corresponding to two types of hydrogens in the ratio 1.50:1.00, or 3:2 Since the unknown contains 10 hydrogens, four protons are of one type and six are of the other type The upfield signal at 1.22 δ is due to saturated primary protons

Worked Example The downfield signal at 3.49 δ is due to protons on carbon adjacent to an electronegative atom - in this case, oxygen The signal at 1.23 δ is a triplet, indicating two neighboring protons The signal at 3.49 δ is a quartet, indicating three neighboring protons This splitting pattern is characteristic of an ethyl group The compound is diethyl ether, CH3CH2OCH2CH3

1H NMR Spectroscopy and Proton Equivalence Proton NMR is much more sensitive than 13C and the active nucleus (1H) is nearly 100 % of the natural abundance Proton NMR shows how many kinds of nonequivalent hydrogens are in a compound Theoretical equivalence can be predicted by comparing structures formed by replacing each H with X and determining the number of different compounds Equivalent H’s have the same signal while nonequivalent H’s give different signals There are degrees of nonequivalence

1H NMR Spectroscopy and Proton Equivalence One use of 1H NMR is to ascertain the number of electronically non-equivalent hydrogens present in a molecule In relatively small molecules, a brief look at the structure can help determine the kinds of protons present and the number of possible NMR absorptions Equivalence or nonequivalence of two protons can be determined by comparison of structures formed if each hydrogen were replaced by an X group

1H NMR Spectroscopy and Proton Equivalence Possibilities If the protons are chemically unrelated and non-equivalent, the products formed by substitution would be different constitutional isomers In other words, substitute two hydrogens that you think might be identical with an X group. If the two molecules that are created are the same molecule, then the H’s are chemically equivalent.

1H NMR Spectroscopy and Proton Equivalence If the protons are chemically identical, the same product would be formed despite the substitution Protons that are equivalent are said to be homotopic. Homotopic show an identical NMR absorption.

1H NMR Spectroscopy and Proton Equivalence If the hydrogens are homotopic but not identical, substitution will form a new chirality center Hydrogens that lead to formation of enantiomers upon substitution with X are called enantiotopic Protons in two structures whose substitution by X would lead to two different enantiomers are said to be enantiotopic. Enantiotopic hydrogens, even though not identical, are nevertheless electronically equivalent and thus have the same NMR absorption.

1H NMR Spectroscopy and Proton Equivalence If the hydrogens are neither homotopic nor enantiotopic, substitution of a hydrogen at C3 would form a second chirality center Hydrogens, whose substitution by X leads to different diastereomers, are said to be diastereotopic. Diastereotopic hydrogens are neither chemically nor electronically equivalent. These hydrogens are completely different and would likely show different NMR absorptions.

Worked Example How many absorptions will (S)-malate, an intermediate in carbohydrate metabolism have in its 1H NMR spectrum? Explain

Worked Example Solution: Because (S)-malate already has a chirality center(starred), the two protons next to it are diastereotopic and absorb at different values The 1H NMR spectrum of (S)-malate has four absorptions

More Complex Spin-Spin Splitting Patterns Some hydrogens in a molecule possess accidentally overlapping signals In the spectrum of toluene (methylbenzene), the five aromatic ring protons produce a complex, overlapping pattern though they are not equivalent

More Complex Spin-Spin Splitting Patterns Splitting of a signal by two or more nonequivalent kinds of protons causes a complication in 1H NMR spectroscopy The aldehydic proton at 9.69 is split into a doublet by the hydrogen at C-2. The black tracings represent the phenyl hydrogens. The green signals correspond to the vinyl proton at C-3. Its signal is shifted downfield because the carbon to which it is attached is not only a vinyl carbon but is also next to an aromatic ring (think electronegative) which pulls the proton signal further downfield than would otherwise be expected for a vinyl proton. The blue tracings represent the splitting that is observed for the vinyl proton at C-2. Perfunctorally, one might expect this signal to be split into a triplet because the carbon to which the proton is attached is flanked by a total of two protons – one on the carbon of the aldehyde and one at C-3. BUT, the protons at C-1 and at C-3 are not equivalent so they do not split the proton at C-2 equally. Instead, the proton at C-3 splits the signal for the proton at C-2 into a doublet with a characteristic coupling constant. THEN, each of these two peaks in the doublet are split by the non-equivalent proton at C-1 into a doublet. Since the coupling constant of this splitting is different the spectrum reveals a doublet of doublets for the blue signal. Note: The right most peak of the green signal and the left most peak of the blue signal are higher than their same colored partners. This can be used to identify the signal to which that absorption is coupled. The higher peak directs the interpreter to look to the right (in the case of the green signal) and to the left (in the case of the blue signal) to find the signal corresponding to the coupled proton.

Tree Diagram for the C2 proton of trans-cinnamaldehyde The two different protons at C-1 and C-3 split the proton at C-2 into a doublet of doublets each with a different coupling constant.

Worked Example 3-Bromo-1-phenyl-1-propene shows a complex NMR spectrum in which the vinylic proton at C2 is coupled with both the C1 vinylic proton (J = 16 Hz) and the C3 methylene protons (J = 8 Hz) Draw a tree diagram for the C2 proton signal and account for the fact that a five-line multiplet is observed

Worked Example Solution: C2 proton couples with vinylic proton (J = 16) Hz C2 proton’s signal is split into a doublet C2 proton also couples with the two C3 protons (J = 8 Hz) Each leg of the C2 proton doublet is split into a triplet to produce a total of six lines

Worked Example

Uses of 1H NMR Spectroscopy The technique is used to identify likely products in the laboratory quickly and easily NMR can help prove that hydroboration-oxidation of alkenes occurs with non-Markovnikov regiochemistry to yield the less highly substituted alcohol

1H NMR Spectra of Cyclohexylmethanol If Markovnikov addition occurred, a methyl group absorbing around 1 ppm would be observed in the spectrum as is evident in the bottom tracing. Similarly, there would be no CH2O resonance between 3 and 4 ppm. The top tracing reveals just such a signal and demonstrates the absence of a singlet at 1 ppm indicating the absence of a methyl group.

Worked Example Mention how 1H NMR is used to determine the regiochemistry of electrophilic addition to alkenes Determine whether addition of HCl to 1-methylcyclohexene yields 1-chloro-1-methylcyclohexane or 1-chloro-2-methylcyclohexane

Worked Example Solution: Referring to 1H NMR methyl group absorption The unsplit methyl group in the left appears as a doublet in the product on the right Bonding of a proton to a carbon that is also bonded to an electronegative atom causes a downfield absorption in the 2.5–4.0 region 1H NMR spectrum of the product would confirm the product to be 1-chloro-1-methylcyclohexane

13C NMR Spectroscopy: Signal Averaging and FT–NMR Carbon-13 is the only naturally occurring carbon isotope that possesses a nuclear spin, but its natural abundance is 1.1% Signal averaging and Fourier-transform NMR (FT–NMR) help in detecting carbon 13 Due to the excess random electronic background noise present in 13C NMR, an average is taken from hundreds or thousands of individual NMR spectra FT NMR involves the sample being irradiated with a broad spectrum of radiofrequencies at the same time. The complex signal is then manipulated using Fourier Transform mathematics to get a usable signal. This saves time over the traditional method of varying the magnetic field. At the same time, there is a lot of noise in a CMR spectrum because of the low natural abundance of C-13. By doing many scans, the noise averages out whereas the signals resulting from individual carbons do not.

Carbon-13 NMR Spectra of 1-Pentanol

13C NMR Spectroscopy: Signal Averaging and FT–NMR Spin-spin splitting is observed only in 1H NMR The low natural abundance of 13C nucleus is the reason that coupling with adjacent carbons is highly unlikely Due to the broadband decoupling method used to record 13C spectra, hydrogen coupling is not seen The lack of spin-spin splitting is because it is rare that two 13C nuclei are next to each other. Broadband decoupling involves irradiating all of the protons so they don’t affect the splitting of carbon nuclei. This allows for singlets for each carbon in the CMR spectrum.

Characteristics of 13C NMR Spectroscopy 13C NMR provides a count of the different carbon atoms in a molecule 13C resonances are 0 to 220 ppm downfield from TMS The exact chemical shift of each 13C resonance is dependent on that carbon’s electronic environment within the molecule. Carbons bonded to oxygen, nitrogen, or halogen absorb downfield of typical alkane carbons. Sp3 hybridized carbons generally absorb from 0 to 90 ppm, while sp2 carbons absorb from 110 to 220 ppm. Carbonyl carbons absorb between 160 to 220 ppm.

Characteristics of 13C NMR Spectroscopy General factors that determine chemical shifts The electronegativity of nearby atoms The diamagnetic anisotropy of pi systems The absorption of sp3-hybridized carbons and sp2 carbons Diamagnetic anisotrophy refers to the density of electrons in a pi system which causes shielding or deshielding of an adjacent or nearby signal depending on where the proton or carbon is relative to the pi cloud.

Carbon-13 Spectra of 2-butanone and para-bromoacetophenone Carbon nuclei relax at differing rates following being irradiated with RF energy. As a result some signals are smaller than other signals. Typically, those carbon signals to which a hydrogen is attached will show up more strongly than signals corresponding to carbons to which no hydrogen is attached. Also evident in the lower spectrum is the combination of resonances for carbons 4 and 4’ and 5 and 5’. These carbons are chemically equivalent because of symmetry in the molecule.

Worked Example Classify the resonances in the 13C spectrum of methyl propanoate, CH3CH2CO2CH3

Worked Example Solution: Methyl propanoate has four unique carbons that individually absorb in specific regions of the 13C spectrum Methyl group carbons typically absorb upfield relative to other carbons in the molecule. In the case of carbon 1, it is attached directly to an electronegative oxygen which causes the signal to move considerably downfield. The methylene at C-3 is downfield relative to the methyl group C-4. It is not directly attached to an oxygen; as a result it is more upfield than the methyl group at C-1.

DEPT 13C NMR Spectroscopy DEPT-NMR (distortionless enhancement by polarization transfer) Stages of a DEPT experiment Run a broadband-decoupled spectrum Run a DEPT-90 Run a DEPT-135 The DEPT experiment manipulates the nuclear spins of carbon nuclei DEPT CMR makes it possible to distinguish between signals due to CH3, CH2, CH, and quaternary carbons. Run in three stages the first stage involves a broadband-decoupled experiment to locate the positions of the chemical signals. The DEPT 90 experiment follows which locates only those carbons to which one H is attached. Thirdly a DEPT-135 experiment is run in which CH3 and CH resonances appear as positive signals, CH2 carbon resonances appear as negative signals and quaternary carbon signals are absent.

DEPT-NMR Spectra for 6-methyl-5-hepten-2-ol Putting together all of the information it is possible to tell how many hydrogens are attached to each carbon. The top spectrum shows all of the signals. The middle spectrum shows only the CH carbons. The lower spectrum reveals positive and negative peaks. The positive peaks that were not present in the middle experiment are CH3 resonances. The negative peaks are CH2 resonances. Those carbons in the top spectrum which are not present in the bottom spectrum are quarternary carbons.

Uses of 13C NMR Spectroscopy Helps in determining molecular structures Provides a count of non-equivalent carbons Provides information on the electronic environment of each carbon and the number of attached protons Provides answers on molecule structure that IR spectrometry or mass spectrometry cannot provide

13C NMR Spectrum of 1-methylcyclohexane This is the product obtained from the dehydration of 1-chloro-1-methylcyclohexene from the reaction on the previous slide. If methylenecyclohexane were the product we would see only five signals in the CMR. The fact that we see seven signals strongly suggests that, in accordance with Zaitsev’s rule, 1-methylcyclohexene is the product obtained.

Worked Example Propose a structure for an aromatic hydrocarbon, C11H16, that has the following 13C NMR spectral data: Broadband decoupled: 29.5, 31.8, 50.2, 125.5, 127.5, 130.3, 139.8 δ DEPT-90: 125.5, 127.5, 130.3 δ DEPT-135: positive peaks at 29.5, 125.5, 127.5, 130.3 δ; negative peak at 50.2 δ Broadband shows there are seven resonances which means four carbons likely have identical environments in the molecule The DEPT 90 experiment suggests signals at 125.5, 127.5, and 130.3 are CH resonances. The positive peaks in the DEPT-135 experiment show that the resonance at 29.5 is a CH3 group while the rest are CH resonances. The negative peak at 50.2 corresponds to a CH2 group. Therefore, the peaks unaccounted for in the DEPT experiment that are present in the Broadband experiment, 31.8 and 139.8 are quarternary carbons. Summing up, we have two quarternary carbons, three CH resonances, one methyl resonance, and one methylene resonance, not accounting for duplication of four signals due to symmetry.

Worked Example Solution: Calculate the degree of unsaturation of the unknown compound C11H16 has 4 degrees of unsaturation Look for elements of symmetry 7 peaks appearing in the 13C NMR spectrum indicate a plane of symmetry According to the DEPT-90 spectrum, 3 of the kinds of carbons in the aromatic ring are CH carbons Since the molecule has four degrees of unsaturation, and CH resonances occur between 120 and 130 ppm we can infer an aromatic (benzene ring). With three such resonances, a monosubstituted benzene ring is likely. This takes into account all of the CH- resonances and the 4 degrees of unsaturation. The signal at 139.8 ppm is undoubtedly the aromatic quaternary carbon, but there is one more in the molecule, presumably an aliphatic one because of its chemical shift of 31.8 ppm. At this point, the aromatic ring eliminates C6H5 from the chemical formula leaving C4H11 to be accounted for. The presence of at least one methylene group at 50.2 suggests it could be connected to the benzene ring because the last remaining quaternary carbon resonates at 31.8 which is further downfield than the methylene signal. This then suggest a quaternary carbon attached to the benzylic methlene. Since C3H9 remains to be accounted for and these resonances seem symmetrical based on the presence of only seven carbon based signals, it is likely there are three methyl groups on a quaternary carbon.

Worked Example The unknown structure is a monosubstituted benzene ring with a substituent containing CH2 and CH3 carbons

How Many Signals? Each set of chemically equivalent protons give a signal in the 1H NMR spectrum.

How Many Signals?

How Many Signals?

Relative Positions of the Signals Protons in electron-poor environments show signals at high frequencies. Electron withdrawal causes NMR signals to appear at a higher frequency (at a larger d value).

Relative Positions of the Signals The closer the electronegative the atom (or group), the more it deshields the protons.

Where 1H NMR Signals Appear

Where They Show a Signal Methine protons appear at higher frequency than methylene protons, which appear at a higher frequency than methyl protons.

60-MHz Versus 300-MHz a 60-MHz 1H NMR spectrum a 300-MHz

Where 13C NMR Signals Appear

(Distinguishes CH3, CH2, and CH Groups) A DEPT 13C NMR Spectrum (Distinguishes CH3, CH2, and CH Groups)

Two-Dimensional NMR Spectroscopy (A COSY Spectrum) Cross peaks indicate pairs of protons that are coupled.

A COSY Spectrum of 1-Nitropropane

Ethyl Isopropyl Ketone A HETCOR Spectrum of Ethyl Isopropyl Ketone A HETCOR spectrum of indicates coupling between protons and the carbon to which they are attached.

Nuclear Magnetic Resonance (NMR) Magnetic Resonance Imaging (MRI) an MRI scanner

An MRI The white region is a brain lesion. The spectrum indicates an abscess.