Chemistry and Biochemistry University of Massachusetts Dartmouth

Chemistry and Biochemistry University of Massachusetts Dartmouth
Chapter 13: NMR CARBON-13 (13C) NMR SPECTROSCOPY Developed much later than 1H NMR. Why? – Only 1% of carbons are 13C (NMR active), whereas almost 100% of hydrogens are 1H (NMR active) – So you need 100 times more sample (in principle) or a new methodology to see 13C signals….. – FT-NMR (Fourier Transform) is the new technique – look at entire spectrum many times and sum. – Old technique was Continuous Wave (CW). Scan through the spectrum once. Dr. Sivappa Rasapalli Chemistry and Biochemistry University of Massachusetts Dartmouth

Carbon-13 NMR Spectroscopy
H1 and C13 NMR Spectroscopy both give us information about the number of chemically nonequivalent nuclei (nonequivalent hydrogens or nonequivalent carbons) both give us information about the environment of the nuclei (hybridization state, attached atoms, etc.) 6

SPIN PROPERTIES OF ATOMIC NUCLEI
What is spin? The Simple explanation Spin is a fundamental property of nature like electrical charge or mass. Spin is a measure of angular momentum (rotation about an axis) hence the term Spin comes in multiples of 1/2 (0, 1/2, 1, 3/2, 2, 5/2…) and can be + or -. Protons, electrons, and neutrons possess spin. Individual unpaired electrons, protons, and neutrons each possesses a spin of 1/2 Atomic nuclei composed of neutrons and protons may also possess spin. The spin of an atomic nucleus is determined by the number of protons and neutrons in the nucleus. Atoms with an odd number of protons will have spin Atoms with an odd number of neutrons will have spin Atoms with an odd number of both protons and neutrons will have spin Atoms with an even number of both protons and neutrons will not have spin The value of nuclear spin is represented by the symbol I, the nuclear spin quantum number. (I = 0, 1/2, 1, 3/2, 2, 5/2….) A nucleus with spin of I can exist in (2I+1) spin states. The shell model for the nucleus tells us that nucleons (protons and neutrons), just like electrons, fill orbitals. When the number of protons or neutrons equals 2, 8, 20, 28, 50, 82, and 126, orbitals are filled. Because nucleons have spin, just like electrons do, their spin can pair up when the orbitals are being filled and cancel out. Odd numbers mean unfilled orbitals, that do not cancel out. What is spin? The fundamental explanation. The shell model for the nucleus tells us that nucleons (protons and neutrons), just like electrons, fill orbitals. When the number of protons or neutrons equals 2, 8, 20, 28, 50, 82, and 126, orbitals are filled. Because nucleons have spin, just like electrons do, their spin can pair up when the orbitals are being filled and cancel out. Odd numbers mean unfilled orbitals, that do not cancel out.

A Basic Concept in Electromagnetic Theory- Direct Application to NMR
A moving perpendicular external magnetic field will induce an electric current in a closed loop An electric current in a closed loop will create a perpendicular magnetic field

Where’s Waldo? 12C is the most abundant natural isotope of carbon, but has a nuclear spin I = 0, rendering it unobservable by NMR. Limited to the observation of the 13C nucleus which constitutes only 1.1% of naturally occurring carbon. 12C has no magnetic spin and produces no NMR signal. 13C accounts for only 1.1% of naturally occurring carbon C-13 NMR has δ 0 to 220 ppm (1HNMR d 0 to 12 ppm) No integration for C-13 spectra Since the 13C isotope of carbon is present in only 1.1% natural abundance, there is only a 1 in 10,000 chance that two 13C atoms will occur next to each other in a molecule

13C NMR Spectroscopy One carbon in 3 molecules of squalene is 13C
DE= gBo h 2 p Bo = external magnetic field strength = magnetogyric ratio 1H= 26,752 13C= 6.7 One carbon in 3 molecules of squalene is 13C Natural Abundance 1H 99.9% (I= 1/2) 12C 98.9% (I= 0) 13C 1.1% (I= 1/2) 13C is a much less sensitive nuclei than 1H for NMR spectroscopy New techniques (hardware and software) has made 13C NMR routine • Pulsed NMR techniques (FT or time domain NMR) • Signal averaging (improved signal to noise)

13C Transition Energy Nucleus  Field strength B0 (Tesla) Frequency 
(106 rad/Tesla  sec) Field strength B0 (Tesla) Frequency  (MHz) 1H 267.53 1.00 42.6 4.70 200. 7.05 300. 2H 41.1 6.5 13C 67.28 10.7 50.0 75.0 19F 251.7 40.0 The magnetogyric ratio, , for the 13C is 67.3 compared to for 1H. Remember the resonance condition for a nucleus is given by:  = (/2)B0 If the gyromagnetic ratio is lowered, the E is also lowered. Where a 1H spectrum using a 1.41 T magnet is observed at 60 MHz, a 13C spectrum is observed at 15 MHz – roughly 4 times less energetic. Boltzmann: Nupper/Nlower = e-E/kT = e-h/kT @ 298 K the ratio is 1,000,000 / 1,000,002 The combined effects of smaller excess populations in the lower energy state, low natural abundance, and slow relaxation rates result in a 13C signal that is typically 6000 times weaker than that observed for 1H. With FT instruments, this is not a problem – simply take more scans! (recall S/N increases as the square root of the number of scans). 16 scans on a 5-10 mg sample will give a good 1H spectrum, 512 scans on a 50 mg sample will give a good 13C spectrum.

Magnetic alignment = g h / 4p Bo
RANDOM ORIENTATION = g h / 4p Bo For our purposes, consider the behavior of a 1H atom. The magnetic moment  is a vector quantity, that is it has both magnitude and direction. In the absence of an external magnetic field, the magnetic moments of a collection of a large number of hydrogen atoms orient themselves in a random fashion. That is, no particular orientation is preferred. How ever, If two magnets are brought near each other they will exert a force on each other and will try to align themselves. For simple bar magnets, the favored alignment is parallel (north pole of one magnet faces the south pole of the second). Similarly, when a magnetic nucleus (I>0) is placed between the poles of an external magnet, it too will try to align itself with respect to this externally applied magnetic field (Bo). In the absence of external field, each nuclei is energetically degenerate Add a strong external field (Bo) and the nuclear magnetic moment: aligns with (low energy) against (high-energy)

Nuclear Spin (cont.) 1. At zero external magnetic field, spins are degenerate! 2. Apply an external magnetic field – spins states will differ in energy depending upon relative orientation with respect to external field. -nuclei with I = ½ will adopt two specific orientations with respect to an externally-applied magnetic field... external magnetic field Bo -spin state +1/2 (lower energy) -spin state -1/2 (higher energy) radiofrequency energy source A spinning charged particle generates a magnetic field. A nucleus with a spin angular momentum will generate a magnetic moment (μ). If these tiny magnets are placed in an applied magnetic field (Bo), they will adopt two different states - one aligned with the field and one aligned against the field. The energy difference between these two states is what we are observing with NMR. In the macroscopic world, two magnets can be aligned in an infinite number of orientations . At the atomic level, these alignments are quantized. There are only a finite number of alignments a nucleus can take against an external magnetic field. This number depends on the value of its spin number I. Each possible alignment is assigned a value called Iz which ranges from -I to +I in steps of 1. These orientations are referred to as spin states. The diagram illustrates the possible spin states for a spin 1/2 nucleus. to convert lower energy spin state into higher energy spin state, require external energy source irradiate with radiofrequency (rf) radiation!

Spins Orientation in a Magnetic Field (Energy Levels)
1. At zero external magnetic field, spins are degenerate! 2. Apply an external magnetic field – spins states will differ in energy depending upon relative orientation with respect to external field. -nuclei with I = ½ will adopt two specific orientations with respect to an externally-applied magnetic field... radiofrequency required depends on E E depends on strength of Bo h  E apply magnetic field increasing Bo -spin: +1/2 lower energy -spin: -1/2 higher energy no external magnetic field A spinning charged particle generates a magnetic field. A nucleus with a spin angular momentum will generate a magnetic moment (μ). If these tiny magnets are placed in an applied magnetic field (Bo), they will adopt two different states - one aligned with the field and one aligned against the field. The energy difference between these two states is what we are observing with NMR. In the macroscopic world, two magnets can be aligned in an infinite number of orientations . At the atomic level, these alignments are quantized. There are only a finite number of alignments a nucleus can take against an external magnetic field. This number depends on the value of its spin number I. Each possible alignment is assigned a value called Iz which ranges from -I to +I in steps of 1. These orientations are referred to as spin states. The diagram illustrates the possible spin states for a spin 1/2 nucleus. In quantum mechanical terms, the nuclear magnetic moment of a nucleus can align with an externally applied magnetic field of strength Bo in only 2I + 1 ways, either parallel or opposing Bo. The energetically preferred orientation has the magnetic moment aligned parallel with the applied field (spin +1/2) and is often given the notation a, whereas the higher energy anti-parallel orientation (spin -1/2) is referred to as b. to convert lower energy spin state into higher energy spin state, require external energy source irradiate with radiofrequency (rf) radiation!

Difference in energy between the two states is given by: DE = g h Bo / 2p (DE = g hn) where: Bo = external magnetic field h = Planck’s constant; g – gyromagnetic ratio radiofrequency required depends on E E depends on strength of Bo h  E apply magnetic field increasing H0 -spin: +1/2 lower energy -spin: -1/2 higher energy no external magnetic field Bo =0 Bo >0

The value, , is the magnetogyric ratio
Magnets, when brought together, will align exactly parallel to each other and will maintain this alignment in a static fashion. Magnetic nuclei, due to restrictions described by quantum mechanics, do not align exactly parallel to or against the external magnetic field but rather, they align at an angle. This has an important consequence that can be illustrated by considering a gyroscope A spinning gyroscope, when placed in a specific orientation, will tend to hold that orientation despite the effects of external forces like gravity. In a vertical gravity field (gravity pulling straight down) a gyroscope placed vertically will maintain this orientation motionlessly. If a force is applied to the gyroscope perpendicular to the gravity field, it will rotate about an axis parallel to the gravity field demonstrating something call precession. The frequency of this precession depends on two factors, the force exerted by the gravity field, and the force exerted by the gyroscope Just as a gyroscope will precess in a gravitational field, the magnetic moment μ associated with a spinning spherical charge will precess in an external magnetic field. In the following illustration, the spinning nucleus has been placed at the origin of a cartesian coordinate system, and the external field is oriented along the z-axis. The angular frequency at which this precession occurs is given by  = Bo/2 and is called the Larmor frequency. The value, , is the magnetogyric ratio and is characteristic for each type of nucleus. It relates to the strength of the nucleus' magnetic field. B is the strength of the externally applied magnetic field. For example, a 1H atom in a magnetic field B=1.41 Tesla has a Larmor frequency of 60 megahertz (MHz).  = Bo/2 The value, , is the magnetogyric ratio

Transition from the low energy to high energy spin state occurs through an absorption of a photon of radio-frequency (RF) energy RF The orientations a magnetic nucleus can take against an external magnetic are not of equal energy. Spin states which are oriented parallel to the external field are lower in energy than in the absence of an external field. In contrast, spin states whose orientations more nearly oppose the external field are higher in energy than in the absence of an external field. The difference in energy between the two spin states is dependent on the external magnetic field strength, and is always very small. The above and the following diagrams illustrate that the two spin states have the same energy when the external field is zero, but diverge as the field increases. Frequency of absorption: n = g Bo / 2p

Nuclear Spin (cont.) ENERGY OF A PHOTON E = h SPIN STATE ENRGY DIFFERENCE E = hB0/2 WHEN E = E, SPIN FLIP OCCURS h hB0/2 THE NECESSARY FREQUENCY IS:  B0/2 Difference in energy between the two nuclear spin states: depends on strength of external magnetic field For nucleus of H atom (proton), spin energy differences: -1/2  (MHz) Strong magnetic fields are necessary for nmr spectroscopy. The international unit for magnetic flux is the tesla (T). The earth's magnetic field is approximately 10-4 T at ground level. For nmr purposes, this small energy difference (ΔE) is usually given as a frequency in units of MHz (106 Hz), ranging from 20 to 900 Mz, depending on the magnetic field strength. Irradiation of a sample with radio frequency (rf) energy corresponding exactly to the spin state separation of a specific set of nuclei will cause excitation of those nuclei in the +1/2 state to the higher -1/2 spin state. Some important relationships in NMR The frequency of absorbed Units electromagnetic radiation Hz is proportional to the energy difference between two nuclear spin states kJ/mol which is proportional to the applied magnetic field tesla (T) 100 200 300 360 500 +1/2 2.34 4.73 6.35 8.46 11.75 H0 (Tesla) Thus, at H0 = 4.7 T (Tesla), use rf radiation of 200 MHz for 1H Nuclei

Nuclear Spin (cont.) ENERGY OF A PHOTON E = h SPIN STATE ENRGY DIFFERENCE E = hB0/2 WHEN E = E, SPIN FLIP OCCURS h hB0/2 THE NECESSARY FREQUENCY IS:  B0/2 Difference in energy between the two nuclear spin states: depends on strength of external magnetic field For nucleus of H atom (proton), spin energy differences: -1/2  (MHz) Strong magnetic fields are necessary for nmr spectroscopy. The international unit for magnetic flux is the tesla (T). The earth's magnetic field is approximately 10-4 T at ground level. For nmr purposes, this small energy difference (ΔE) is usually given as a frequency in units of MHz (106 Hz), ranging from 20 to 900 Mz, depending on the magnetic field strength. Irradiation of a sample with radio frequency (rf) energy corresponding exactly to the spin state separation of a specific set of nuclei will cause excitation of those nuclei in the +1/2 state to the higher -1/2 spin state. Some important relationships in NMR The frequency of absorbed Units electromagnetic radiation Hz is proportional to the energy difference between two nuclear spin states kJ/mol which is proportional to the applied magnetic field tesla (T) 25 50 75 90 125 +1/2 2.34 4.73 6.35 8.46 11.75 H0 (Tesla) Thus, at B0 = 4.7 T (Tesla), use rf radiation of 50 MHz etc for 13C nuclei.

Fourier Transform NMR 13C-spectra of CH3CH2CH2CH2CH2OH
Radio-frequency pulse given. Nuclei absorb energy and precess (spin) like little tops. Signal-averaging: pulsed NMR allows for many FID’s (NMR spectra) to be accumulated over time. These FID’s are added together and averaged. Signals (resonances) build up while the “noise” is random and cancels out during the averaging. Enhanced signal to noise ratio and allows for NMR spectra to be collected on insensitive nuclei such as 13C and small samples. low 13C abundance; a single molecule will have at most only one 13C atom; however, we are sampling a very large number of molecules, even in a 50 mg sample! thus our sampling will ‘see’ a 13C at every C position in the molecule! 13C-spectra of CH3CH2CH2CH2CH2OH after one scan average of 200 scans

Features of 13C NMR Spectra
Each unique C in a structure gives a single peak in the spectrum; there is rarely any overlap. The C NMR spectrum spans over 200 ppm; chemical shifts only ppm apart can be distinguished; this allows for over 2x105 possible chemical shifts for carbon. The intensity (size) of each peak is NOT directly related to the number of that type of carbon. Other factors contribute to the size of a peak: Peaks from carbon atoms that have attached hydrogen atoms are bigger than those that don’t have hydrogens attached. Carbon chemical shifts are usually reported as downfield from the carbon signal of tetramethylsilane (TMS).

H1 and C13 NMR Spectroscopy
Number of peaks Chemical shifts Integration Spin-Spin Splitting Number of peaks Chemical shifts Integration Spin-Spin Splitting

13C NMR 13C spectra are typically recorded from 0 – 220 ppm; with the zero being the methyl carbon in TMS (much wider range than 1H spectra!) 13C nuclei are shielded or deshielded (CHEMICAL SHIFT) due to the same factors as for 1H NMR. 1. Electron withdrawing ability (by inductance or resonance) of nearby groups. 2. Hybridization. 3. Electron current effects. 1. Chemical shift (ppm scale) – Functional group information. 2. Intensity information (vertical scale) –(# of equivalent carbons). 3. Peak Multiplicity: Nearest neighbour analysis – more difficult for 13C NMR. • Same trends apply as for 1H NMR , but the chemical shift range is about 20 times greater! Reduces overlap of peaks. Also… can see carbonyls! • TMS is still the reference

13C NMR 13C has only about 1.1% natural abundance (of carbon atoms). As a result, C is about 50 times less sensitive than H nucleus to the NMR phenomena. Hundreds of spectra are acquired and then averaged to give a spectrum with good signal-to-noise ratio. Chemical shift range is normally ppm (1H is 0-15 ppm) Carbon atoms with more hydrogens absorb more strongly. The number of signals in a 13C spectrum gives the number of different types of carbon atoms in a molecule. It is unlikely that a 13C would be adjacent to another 13C, so splitting by carbon is negligible. Because 13C NMR signals are not split, the number of signals equals the number of lines in the 13C spectrum. All signals are singlets In contrast to the 1H NMR situation, peak intensity is not proportional to the number of absorbing carbons, so 13C NMR signals are not integrated. Instrument’s Resonance frequency is ~ one-fourth, 75 MHz instead of 300 MHz.

Predicting 13C Spectra

Predicting 13C NMR Each type of non-equivalent C gives a unique signal
Equivalent C’s have the same environment, hence they give the same signal

Symmetry in C-13 NMR Each unique carbon in a molecule gives rise to a 13C NMR signal. Therefore, if there are fewer signals in the spectrum than carbon atoms in the compound, the molecule must possess symmetry. Examples:

How many signals would you expect?

Enantiotopic vs Diastereotopic carbons’s
* * * * *

Predicting 13C NMR Determine the number of signals in the proton-decoupled C-13 NMR spectrum of each of the following compounds:

1H and 13C NMR compared: 13C signals are spread over a much wider range than 1H signals making it easier to identify and count individual nuclei The following slides show the 1H NMR and the 13C spectrum of 1-chloropentane. It is much easier to identify the compound as 1-chloropentane by its 13C spectrum than by its 1H spectrum. 6

1H Proton Spectrum Chemical shift (, ppm) ClCH2 CH3 ClCH2CH2CH2CH2CH3
1.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0 9.0 10.0 Chemical shift (, ppm) 1

Carbon Spectrum 13C ClCH2CH2CH2CH2CH3 a separate, distinct peak appears for each of the 5 carbons CDCl3 20 40 60 80 100 120 140 160 180 200 Chemical shift (, ppm) 1

Intensity of 13C NMR signals
128.0 128.5 Chemical shifts give an idea of the chemical and electronic environment of the 13C nuclei due to shielding and deshielding effects range: ppm from TMS 13C NMR spectra will give a map of the carbon framework. The number of resonances equals the number of non-equivalent carbons. Intensity information is qualitative for 13C NMR (unlike 1H NMR): Intensity depends on number of equivalent carbons, NOE and motion. 132.8 128.0 137.1 128.5 132.8 17.8 13.9 40.5 TMS CDCl3 200.3 137.1

13C-NMR: Integration 1H-NMR: Integration reveals relative number of hydrogens per signal 13C-NMR: Integration reveals relative number of carbons per signal Rarely useful due to slow relaxation time for 13C time for nucleus to relax from excited spin state to ground state

13C Chemical shifts are most affected by:
electronegativity of groups attached to carbon hybridization state of carbon 6

13C NMR Chemical Shifts Several functionalities appear directly on 13C NMR which are not ‘visible’ in 1H NMR: - Quaternary carbons - ipso carbons - Carbonyl carbons sp3 carbon sp3-EWG alkyne carbons In contrast to the small range of chemical shifts in 1H NMR (1-10 ppm usually), 13C NMR absorptions occur over a much broader range (0-220 ppm). The chemical shifts of carbon atoms in 13C NMR depend on the same effects as the chemical shifts of protons in 1H NMR. The 13C chemical shift is dependent both on the presence of electronegative groups and on the steric environment. 1) Simple interior (2o & 3o=methylenes and methines ) = 25 – 45 ppm. Methyl groups which terminate unbranched alkyl chains, 5-15 ppm. 2) Carbons adjacent to halogens tend to have chemical shifts in the: ppm 3) The presence of an electronegative atom such as oxygen tends to move the chemical shift of the adjacent carbon down into the region: ppm. 4) sp2 carbons are less shielded than sp3 carbons Alkenes = 110 – 140 ppm Conjugation has little effect on the shift. Conjugation with an oxygen, however, has a dramatic shielding effect, which is attributed to contributions from resonance. 5) Alkynyl carbons = ppm, and carbons which are immediately adjacent are upfield. 6) Aromatic carbons have chemical shifts in the range: and are shifted within this range by the nature of the attached substituent. alkene carbons aromatic carbons carbonyl carbons 220 200 180 160 140 120 100 80 60 40 20 0.0 downfield d (ppm) upfield deshielded shielded higher DE lower DE

13C NMR

Chemical Shift - 13C-NMR Trends RCH3 < R2CH2 < R3CH
Electronegative atoms cause downfield shift Pi bonds cause downfield shift C=O ppm

Chemical Shift - 13C-NMR Chemical Shift Range of 13C
7) Carbonyls are the most highly deshielded carbons typically observed. Their intensity is usually weaker since they are doubly bonded and have no attached hydrogens. 8) Typical chemical shifts occur in the region ppm with esters, carboxylic acids and amides at the low end, and simple ketones and aldehydes at the high end of the range. Note the carbonyl range

Carbonyl Carbon Chemical Shifts
anhydrides nitriles acid chlorides amides esters carboxylic acids aldehydes conj. Ketones ketones 220 210 200 190 180 170 160 150 140 130 120 110

Alkane: 2-methylpentane

Alcohol: 2-hexanol

Alkyl Halide: 3-bromopentane

Alkene: 1-hexene

Aromatic Ring: eugenol

Carboxylic Acid: pentanoic acid

Ester: ethyl valerate

Amide: pentanamide

Ketone: 3-methyl-2-pentanone

Aldehyde: 2-methylpentanal

Carbon-13 NMR Spectrum of Geraniol
ppm Carbon # 8 9

Spin-Spin Coupling in 13C NMR
Homonuclear coupling of 13C-13C is possible in theory. However, due to the low natural abundance of 13C, it is rare to find two 13C’s in the same molecule, let alone adjacent to one another. No need to consider 13C-13C coupling except for enrichment studies! Heteronuclear coupling between 13C and the 1H atoms attached to them is observed (1H abundance ~99%). Because the 1H atoms are directly attached, the coupling constants (1J)are large, typically Hz. When such spectra are observed, they are referred to as proton coupled spectra (or non-decoupled spectra). The lack of splitting in a 13C spectrum is a consequence of the low natural abundance of 13C Recall that splitting occurs when two NMR active nuclei—like two protons—are close to each other. Because of the low natural abundance of 13C nuclei (1.1%), the chance of two 13C nuclei being bonded to each other is very small (0.01%), and so no carbon-carbon splitting is observed A 13C NMR signal can also be split by nearby protons. This 1H-13C splitting is usually eliminated from the spectrum by using an instrumental technique that decouples the proton-carbon interactions, so that every peak in a 13C NMR spectrum appears as a singlet The two features of a 13C NMR spectrum that provide the most structural information are the number of signals observed and the chemical shifts of those signals

Where’s Waldo? 12C is the most abundant natural isotope of carbon, but has a nuclear spin I = 0, rendering it unobservable by NMR. Limited to the observation of the 13C nucleus which constitutes only 1.1% of naturally occurring carbon. 12C has no magnetic spin and produces no NMR signal. 13C accounts for only 1.1% of naturally occurring carbon C-13 NMR has δ 0 to 220 ppm (1HNMR d 0 to 12 ppm) No integration for C-13 spectra Since the 13C isotope of carbon is present in only 1.1% natural abundance, there is only a 1 in 10,000 chance that two 13C atoms will occur next to each other in a molecule

Spin-Spin Coupling in 13C NMR
Homonuclear coupling of 13C-13C is possible in theory. However, due to the low natural abundance of 13C, it is rare to find two 13C’s in the same molecule, let alone adjacent to one another. No need to consider 13C-13C coupling except for enrichment studies! Heteronuclear coupling between 13C and the 1H atoms attached to them is observed (1H abundance ~99%). Because the 1H atoms are directly attached, the coupling constants (1J)are large, typically Hz. When such spectra are observed, they are referred to as proton coupled spectra (or non-decoupled spectra). The lack of splitting in a 13C spectrum is a consequence of the low natural abundance of 13C Recall that splitting occurs when two NMR active nuclei—like two protons—are close to each other. Because of the low natural abundance of 13C nuclei (1.1%), the chance of two 13C nuclei being bonded to each other is very small (0.01%), and so no carbon-carbon splitting is observed A 13C NMR signal can also be split by nearby protons. This 1H-13C splitting is usually eliminated from the spectrum by using an instrumental technique that decouples the proton-carbon interactions, so that every peak in a 13C NMR spectrum appears as a singlet The two features of a 13C NMR spectrum that provide the most structural information are the number of signals observed and the chemical shifts of those signals

1H NMR (Proton with Carbon-13 coupling)

13C NMR Spectrum – Carbon 13 and Proton-Coupled
• Signals due to methyl groups become quartets - coupled to 3 hydrogens. • ….methylene groups become triplets - coupled to 2 hydrogens. • ….methine groups become doublets - coupled to 1 hydrogen. • ….quaternary or carbonyl carbons remain singlets. Proton-Coupled

1H – 13C Splitting The splitting follows the simple N+1 rule:
The multiplet analysis gives useful information, but there are two major limitations: 1) If the 13C signal is weak (common) the outer peaks of the multiplet may be lost in the noise of the spectrum. 2) Due to the large J-constants, the multiplets quickly begin to overlap and become congested. quaternary singlet methine doublet methylene triplet quaternary quartet Can we observe 13C nuclei independent of 1H nuclei? • Yes, because 13C nuclei absorb at a different frequency than 1H nuclei. Simply use a different Rf generator tuned to a different frequency

Three equal intensity lines at 77 ppm
Effect of Coupling Three equal intensity lines at 77 ppm CDCl3 solvent 13C- 2D coupling Coupling can cause 13C NMR spectra to become very complicated (convoluted) quite easily. 1H Coupled Since only 1% of carbon is 13C, the chances of seeing 13C-13C coupling is negligible: = .01 x .01 = • Need 2-dimensional techniques to do nearest neighbour analysis. (beyond the scope of this course) • To simplify spectrum remove coupling with 1H, so run “1H decoupled” spectrum (all signals are singlets).

1H Decoupling To simplify the 13C spectrum, and to increase the intensity of the observed signals, a decoupler is used to remove the spin effects of the 1H nucleus. A second RF generator irradiates at the 1H resonance frequency causing the saturation – effectively averaging all their spin states to zero 1H channel- 13C channel 13C n pulse 13C FID

Decoupling Proton-decoupled mode,
a sample is irradiated with two different radiofrequencies. One to excite all 13C nuclei, a second to cause all protons in the molecule to undergo rapid transitions between their nuclear spin states. On the time scale of a 13C-NMR spectrum, each proton is in an average or effectively constant nuclear spin state, with the result that 1H-13C spin-spin interactions are not observed and they are decoupled.

13C Proton Decoupled Spectrum
13C{1H}

13C Off-resonance & Broadband decoupled spectra

Effect of Decoupling 1H Coupled 1H Decoupled
Due to signal enhancement and spectral simplification, 13C spectra are usually reported as 1H decoupled. Each chemically unique carbon in the molecule gives rise to a single peak. Of course chemically equivalent carbons contribute to the same peak! The number of different signals (peaks) indicates the number of different kinds of carbon. The location (chemical shift) indicates the type of functional group. Peak areas (~heights) are NOT proportional to number of carbons. Carbon atoms with more hydrogens give stronger signals, due to more efficient relaxation (transfer of spin to the hydrogens). However, peak areas (~heights) can be compared within the same type of carbons (e.g. methyls) 1H Decoupled

H1 Decoupling Techniques
J values for C-H are typically Hz (C-C-H and C-C-C-H are 0-60Hz) . Thus a CH3 group would appear as a quartet, CH2-triplet CH-doublet etc. The H1 nuclei are irradiated with a broadband Rf to remove coupling to Carbon.

Attached Protons Affect T1 and Signal Intensity
7 carbons give 7 signals, but intensities are not equal CH3 OH 20 40 60 80 100 120 140 160 180 200 Chemical shift (, ppm) 1

Example: Ethanol 1H-13C spin-spin coupling: spin-spin coupling tells how many protons are attached to the 13C nuclei. (i.e., primary, secondary tertiary, or quaternary carbon) 13C spectra are usually collected with the 1H-13C coupling “turned off” (broad band decoupled). In this mode all 13C resonances appear as singlets.

2-methylbutane (CH3)2CHCH2CH3
Other isomers of C5H12 pentane CH3CH2CH2CH2CH3 3 peaks 2,3-dimethylpropane (CH3)4C 2 peaks chemically equivalent carbon atoms There are four chemically different carbon atoms in the molecule so there are four peaks in the C-13 nmr spectrum. H  C  C  C  C  H H CH3 NO SPLITTING WITH C-13 ONLY ONE PEAK FOR EACH CARBON

Example: cyclohexane

Example: cyclohexene

Example: 1,3-cyclohexadiene

Example: 1,4-cyclohexadiene

Example: m-nitrotoluene
6 5 4 2 3 1 7 2 4 6 5 7 1 3

C6H12O2

HCCCBr H H Br H HCCCH H H H Isomers of C3H7Br 3 peaks 2 peaks
all three carbons are different the two outer carbons are similar HCCCBr H H Br H HCCCH H H H

Can we tell the isomers apart?

Edited C13 NMR Edited C13 NMRs APT= Attached Proton Test “APT” DEPT= Distortionless Enhancement by Polarization Transfer

Multiplicity detection
DEPT : CH, CH3 CH2 APT : CH, CH3 C , CH2 Normal C13

Attached Proton Test “APT”
Acquiring a 13C after 1/J seconds: methine and methyl C’s produce negative peaks (odd number of attached H’s) methylene and quaternary C’s produce positive peaks (even number of attached H’s) Remember 1JCH is essentially the same for all tetrahedral carbons Thus acquiring the C signal after a pre-determined time can give positive peaks, negative peaks, or even no peaks at all, depending on how many H’s are attached. APT has been superseded by DEPT Distortionless Enhancement by Polarization Transfer Complex pulse sequence allowing selective reception of signals from different C types: -C, -CH, -CH2, -CH3 Attached Proton Test “APT” Acquiring a 13C after 1/J seconds: methine and methyl C’s produce negative peaks (odd number of attached H’s) methylene and quaternary C’s produce positive peaks (even number of attached H’s) Remember 1JCH is essentially the same for all tetrahedral carbons Thus acquiring the C signal after a pre-determined time can give positive peaks, negative peaks, or even no peaks at all, depending on how many H’s are attached. APT has been superseded by DEPT Distortionless Enhancement by Polarization Transfer Complex pulse sequence allowing selective reception of signals from different C types: -C, -CH, -CH2, -CH3 DEPT spectra (Distortionless Enhancement by Polarization Transfer) a modern 13C NMR spectra that allows you to determine the number of attached hydrogens.

Measuring a 13C NMR spectrum involves
In DEPT, a second transmitter irradiates 1H during the sequence, which affects the appearance of the 13C spectrum. some 13C signals stay the same some 13C signals disappear some 13C signals are inverted 9

13C NMR - DEPT Distortionless enhancement by polarization transfer (DEPT) spectra permit identification of CH3, CH2, and CH carbon atoms. DEPT 45 shows 1o, 2o,and 3o carbons. DEPT 90 shows only 3o carbons. DEPT 135 shows 1o and 3o carbons as positive peaks and 2o carbons as negative peaks.

Using DEPT to Count Hydrogens Attached to 13C

Proton Decoupled Spectrum
CCH2CH2CH2CH3 O CH CH CH2 CH CH2 O CH2 CH3 C C 20 40 60 80 100 120 140 160 180 200 Chemical shift (, ppm) 1

DEPT 135 Spectrum CCH2CH2CH2CH3 O CH CH CH3 CH CH2 CH2 CH2
CH and CH3 unaffected C and C=O nulled CH2 inverted CH2 CH2 CH2 20 40 60 80 100 120 140 160 180 200 Chemical shift (, ppm) 1

DEPT 135 Spectrum Broad-band decoupled DEPT
7 8 2 4 Broad-band decoupled 5 1 6 CH3 DEPT CH CH3 CH CH3 CH2 CH2 CH2’s give negative resonances CH’s and CH3’s give positive resonances Quaternary carbon (no attached H’s) are not observed

DEPT DEPT – Distortionless Enhancement by Polarization Transfer
Allows us to observe the number of hydrogens attached to a particular carbon. DEPT Pulse Sequence methyl methylene methine quaternary DEPT-45 Positive peak Not observed DEPT-90 No obs. peak DEPT-135 Negative Peak Positive Peak

13C Citronellol

DEPT-135 Ipsenol DEPT-135 CH3 positive CH2 negative CH positive
C not observed

DEPT - Ipsenol 7 4 2 8 10 5 3 6 9,1 CH3, CH2 (none), CH (+), C (none)
CH3 (+), CH2 (-), CH (+), C (none) DEPT-135 8 10 5 3 6 9,1

DEPT Spectra of Codeine

DEPT Spectra of 1-phenyl-1-butanone
CH3 (+), CH2 (-), CH (+), C (none) DEPT 90 CH3, CH2 (none), CH (+), C (none) DEPT 45 CH3, CH2, CH (+), C (none)

DEPT Spectra of CH3 (+), CH2 (-), CH (+), C (none)
CH3, CH2 (none), CH (+), C (none)

Summary of Edited 13C NMR

Summary Number of signals indicates the number of types of carbon in the sample. (Is symmetry present?) Chemical shifts show what types of carbons are in the sample. Quaternary/ipso carbons will be smaller than carbons with protons attached. DEPT differentiates between primary, secondary, and tertiary carbons.

How many peaks would you expect there to be in the carbon-13 spectrum of…
butane CH3CH2CH2CH3 2-methylpropane CH3CH(CH3)CH3 butanal CH3CH2CH2CHO butanone CH3COCH2CH3 pentan-2-one CH3COCH2CH2CH3 pentan-3-one CH3CH2COCH2CH3 cyclohexane C6H12

How many peaks would you expect there to be in the carbon-13 spectrum of…
butane CH3CH2CH2CH3 2 2-methylpropane CH3CH(CH3)CH3 2 butanal CH3CH2CH2CHO 4 butanone CH3COCH2CH3 4 pentan-2-one CH3COCH2CH2CH3 5 pentan-3-one CH3CH2COCH2CH3 3 cyclohexane C6H 19

Identify the isomers of C4H8O

Identify the isomers of C4H8O
A butanal B butanone C 2-methylpropanal

C13 NMR-Important points
The 13C nucleus is present in only 1.08% natural abundance. Therefore, acquisition of a spectrum usually takes much longer than in 1H NMR. The magnetogyric ratio of the 13C nucleus is about 1/4 that of the 1H nucleus. Therefore, the resonance frequency in 13C NMR is much lower than in 1H NMR. (75 MHz for 13C as opposed to 300 MHz for 1H in a 7.04 Tesla field). At these lower frequencies, the excess population of nuclei in the lower spin state is reduced, which, in turn, reduces the sensitivity of NMR detection. Unlike 1H NMR, the area of a peak is not proportional to the number of carbons giving rise to the signal. Therefore, integrations are usually not done. Each unique carbon in a molecule gives rise to a 13C NMR signal. Therefore, if there are fewer signals in the spectrum than carbon atoms in the compound, the molecule must possess symmetry. When running a spectrum, the protons are usually decoupled from their respective carbons to give a singlet for each carbon atom. This is called a proton-decoupled spectrum.

2D NMR: COSY AND HETCOR 1

1D NMR = 1 frequency axis 2D NMR = 2 frequency axes
2D NMR Terminology 1D NMR = 1 frequency axis 2D NMR = 2 frequency axes COSY = Correlated Spectroscopy 1H-1H COSY provides connectivity information by allowing one to identify spin-coupled protons. x,y-coordinates of cross peaks are spin-coupled protons 9

1H-1H COSY CH3CCH2CH2CH2CH3 O 1H 1H 9

HETCOR 1H and 13C spectra plotted separately on two frequency axes Coordinates of cross peak connect signal of carbon to protons that are bonded to it. 9

1H-13C HETCOR CH3CCH2CH2CH2CH3 O 13C 1H 9

Solving Combined Spectra Problems:
Mass Spectra: Molecular Formula Nitrogen Rule  # of nitrogen atoms in the molecule M+1 peak  # of carbons Degrees of Unsaturation: # of rings and/or -bonds Infrared Spectra: Functional Groups C=O O-H C=C N-H CC CO-OH CN 1H NMR: Chemical Shift ()  chemical environment of the H's Integration  # of H's giving rise to the resonance Spin-Spin Coupling (multiplicity)  # of non-equivalent H's on the adjacent carbons (vicinal coupling). 13C NMR: # of resonances  symmetry of carbon framework Type of Carbonyl Each piece of evidence gives a fragment (puzzle piece) of the structure. Piece the puzzle together to give a proposed structure. The proposed structure should be consistent with all the evidence.

13C NMR: 7.9 35.5 212.1 C5H10O

= 2.61 (d, 3H) = 2.61 (t, 3H) = 2.61 (pentet, J=7, 2H) = 2.61
C10H14 127.0 31.2 128.2 125.7 41.7 21.8 12.3 147.6 = 2.61 (d, 3H) = 2.61 (t, 3H) = 2.61 (pentet, J=7, 2H) = 2.61 (sextet, J=7, 1H) = (m, 5H)

% T cm-1 Infrared (IR): Characteristic O–H stretching absorption at
3300 to 3600 cm1 Sharp absorption near 3600 cm-1 except if H-bonded: then broad absorption 3300 to 3400 cm1 range Strong C–O stretching absorption near 1050 cm1 % T O-H C-O cm-1

1H NMR: protons attached to the carbon bearing the hydroxyl
group are deshielded by the electron-withdrawing nature of the oxygen,  3.3 to 4.7 = 0.9, d, 3H = 1.5, q, 2H = 1.7, m, 1H = 3.65, t, 2H = 2.25, br s, 1H 22.6 61.2 41.7 24.7 CDCl3 O-H C-O

Usually no spin-spin coupling between the O–H proton and
neighboring protons on carbon due to exchange reaction The chemical shift of the -OH proton occurs over a large range ( ppm). It chemical shift is dependent upon the sample concentration and temperature. This proton is often observed as a broad singlet (br s). Exchangable protons are often not to be observed at all.

13C NMR: The oxygen of an alcohol will deshield the carbon
it is attached to. The chemical shift range is ppm  14  19  35  62 CH3 — CH2 — CH2 — CH2 — OH DMSO-d6 (solvent)

13C: 129.4 128.4 126.3 68.8 45.8 22.7 13C: 128.8 127.8 126.5 76.3 32.3 10.6

Magnetic Resonance Imaging (MRI)
MRI uses the principles of nuclear magnetic resonance to image tissue MRI normally uses the magnetic resonance of protons on water and very sophisticated computer methods to obtain images. Other nuclei within the tissue can also be used (31P) or a imaging (contrast) agent can be administered

Alzheimer’s Disease 78 years old Normal 25 years old Normal
MRI images Alzheimer’s Disease 78 years old Normal 25 years old Normal 86 years old fMRI:

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