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C Proton Nuclear Magnetic Resonance (1H NMR) Spectroscopy The IR spectra of acid anhydrides, e g , acetic anhydride: two C=O stretching absorptions,

[PDF] Hydrolysis of Acetic Anhydride with Heavy Water - Thermo Scientific

We also take advantage of isotopic substitution to suppress an otherwise large proton signal in the NMR spectrum originating from the reactant/solvent H2O

acetic anhydride - PNAS

9 mar 2009 · larized [1-13C] acetic anhydride and rapid chemical reactions to provide high SNR NMR spectra of amino acid derivatives and

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reactions of 1 with acetic anhydride and halo-substituted acetic anhydrides anhydride was performed in a NMR tube, the 1H-NMR spectrum of the solution

[PDF] Online Monitoring of the N-Acetylation Reaction of L - Magritek

N-Acetylation of L-Phenylalanine with acetic anhydride (Fig progress of the reaction, the NMR spectra also provide information about the addition of reactants

1 CHEM 203 Organic Spectroscopy: a Primer INDEX A. Introduction 2 B. Infrared (IR) Spectroscopy 3 C. Proton Nuclear Magnetic Resonance (1H NMR) Spectroscopy 14

2 A. Introduction The problem of determining the structure of organic molecules Spectroscopy: a technique to obtain structural information by studying how molecules interact with appropriate energy probes Energy sources utilized in modern organic structure determination: energy (eV) source wavelength (λ) effect information obtained ≈ 104 X-rays ≈ 1 Å diffraction atomic coordinates 10-70 electron ----- fragmentation mass of molecule bombardment & of its fragments 6.5-3.5 UV light 190-350 nm electronic nature of the π systems 3.1-1.5 VIS light 400-800 nm transitions present in the molecule 0.5-0.04 IR light 2.5-30.0 µm vibrational functional groups transitions present in the molecule 10-6-10-7 radiowaves 0.5-15 m nuclear chemical environment transitions of H and C atoms Spectrum of a molecule: a plot of extent of absorption vs. energy Energy of electromagnetic ("EM") radiation as a function of its frequency or of its wavelength: • relationship between energy (E) and frequency (ν) of EM radiation: E = h ν, where h = Planck's constant • relationship between wavelength (λ), frequency and speed (c ≈ 300,000 km/sec) of EM radiation: λ • ν = c • relationship between energy and wavelength (λ) of EM radiation: E = h c λ -1 energy

extent of absorption -1 , and linearly related to frequency, that expresses the number of ondulations of the EM field in 1 cm of propagation: example: if ! = 2.5 µ, = 2.5•10 -6 m = 2.5•10 -4 cm, then " = (2.5•10 -4 -1 = 4000 cm -1 -1 transmittance %! (cm-1)0%100%4000500

4 Characteristic Infrared Stretching Absorptions of Common Functional Groups Functional Group Bond Frequency Range (cm-1) Functional Group Bond Frequency Range (cm-1) Alcohol O-H 3400 - 3650 (s, broad) Nitrile 2210 - 2260 (w - m) C-O 1050 - 1150 (s) Carboxylic acid O-H 2500-3100 (s, broad) Ether C-O 1000 - 1260 C=O 1700 - 1720 (s) Amine N-H 3300 - 3350 (m) Ester C=O 1710 - 1750 (s) Alkane C-H 2850 - 2950 (m - s) Acyl halide C=O 1770 - 1820 (s) Alkene =C-H 3020 - 3100 (m) Acid anhydride C=O 1740 - 1790 (s) C=C 1640 - 1680 (m) 1800 - 1850 (s) Alkyne 3270 - 3330 (s) Amide C=O 1630 - 1700 (s) 2100 - 2260 (w - m) Aldehyde, ketone C=O 1680 - 1730 (s) The spectra of alkanes, e.g., pentane: sp3-C-H stretch at 2800 - 2950 cm-1. This signal is not diagnostic, because virtually every organic compound will show absorptions in this range The spectra of alkenes, e.g., 3-ethyl-1-pentene: sp2 C-H stretch at 3000 - 3100 cm-1, and in many - but not all - cases, a sharp signal at ~ 1650 cm-1 for the C=C stretching absorption CNCHCC

C=C

5 The spectra of molecules containing benzene rings, e.g., 1,4-dimethylbenzene (p-xylene): sp2 C-H stretch: 3000 - 3100 cm-1, just as in the case of alkenes Clearly, IR absorptions in the range 3000 - 3100 cm-1 are diagnostic for the presence of alkene and / or benzene subunits in a molecule The IR spectra of internal alkynes: these molecules produce spectra IR spectra that display no diagnostic features, because: a. there is no sp-C-H bond, therefore no signal at 3300 cm-1 b. the CC triple bond stretch is too weak to be seen. Why? because: a bond must have a permanent electric dipole to absorb IR light. IR light is electro- magnetic radiation (=it has an electric component and a magnetic component) and it can only interact with objects that have either a permanent electrostatic dipole or a permanent magnetic dipole. Clearly, the CC triple bond of an ordinary internal alkyne has no dipole moment, so it does not absorb IR radiation

The spectra of terminal alkynes, e.g., 3,3,-dimethyl-1-butyne: sharp, intense sp -C-H stretching absorption at ! 3300 cm-1 (highly diagnostic for the presence of a terminal alkyne) and weak, but highly diagnostic, stretching absorption at !"2100 cm-1CC

CC - H

6 Example: the spectrum of 4,4,-dimethyl-2-pentyne: The spectra of alcohols, e.g., isopropanol: O-H stretch: 3400-3600 cm-1, broad signal often of "parabolic" shape. Highly diagnostic for the presence of alcohols (note: we are not interested in the signals that appear below 1600 cm-1) Primary, secondary, tertiary amines:

R 1 NH 2 R 1 H NR 2 R 1 NR 2 R 3 a primary amine a secondary amine a tertiary amine R 1 , R 2 , R 3 = alkyl groups (Me, Et, ...)

7 The IR spectra of primary and secondary amines, e.g., butylamine and dipropylamine: N-H stretch at ~ 3300 cm-1, significantly weaker than an O-H absorption. Primary amines often show a pair of N-H stretching absorption bands. Note: C-N stretching absorption (ca. 1100 cm-1) are not diagnostic. The IR spectra of tertiary amines: these molecules produce IR spectra that display no diagnostic features, because there is no N-H bond, therefore no signal at 3300 cm-1

a pair of N-H stretching absorptions is typical of primary amines

8 Example: the spectrum of triethylamine: Principle: molecules incorporating multiple functional groups generate IR spectra in which each functional group independently produces its own characteristic absorptions. example: the spectrum of 2-propyn-1-ol

C C CH-C O-H

The IR spectra of nitriles, e.g., acetonitrile: stretching absorption at ! 2200 cm-1. This signal is often weak, but it is highly diagnostic for the presence of a nitrile:CN

9 Principle: IR spectroscopy is especially useful to detect the presence of carbonyl groups, which absorb strongly between 1850 and 1650 cm-1. The IR spectra of aldehydes, e.g., butyraldehyde: C=O stretch at ca. 1720 cm-1: The IR spectra of ketones, e.g., acetone: C=O stretch at ca. 1715 cm-1:

10 The IR spectra of esters, e.g., ethyl acetate: C=O stretch at ca. 1740 cm-1: The IR spectra of carboxylic acids, e.g., acetic acid: C=O stretch ~ 1700 cm-1. Very broad O-H stretch starting at ~ 3400 cm-1, extending down to ~ 2400 cm-1, and centered around 3100-3000 cm-1:

11 The IR spectra of acid chlorides, e.g., lauroyl chloride: C=O stretch at unusually high frequency, typically at ca. 1800 cm-1: The IR spectra of acid anhydrides, e.g., acetic anhydride: two C=O stretching absorptions, one at very high frequency, ca. 1840 cm-1 (antisymmetric C=O / C=O stetch) the other at ca. 1750 cm-1 (symmetric C=O / C=O stetch)

12 Primary, secondary, tertiary amides: The IR spectra of primary and secondary amides, e.g., formamide and N-methylacetamide: Strong N-H stretch at ~ 3300 cm-1 and strong C=O stretch at ~ 1650-1700 cm-1 NH

2 N H R 1 N R 2 R 1 a primary amide a secondary amide a tertiary amide R, R 1 , R 2 = alkyl groups O R O R O R

13 The IR spectra of tertiary amides, e.g., N,N-dimethylacetamide: strong C=O stretch at ~ 1650-1700 cm-1

14 C. Proton Nuclear Magnetic Resonance (1H NMR) Spectroscopy Nuclear Magnetic Resonanc e (NMR) spectroscopy: a techni que that derives structural information on the basis of how a molecule reacts to energy absorption in the radiowave range. Absorption of radiowaves induces transitions of magnetic nuclei such as 1H (= a proton: the nucleus of H atoms) and 13C (the nucleus of a less common, non-radioactive, isotope of carbon that constitutes about 1% of total carbon), which reveal a great deal about the chemic al surroundings of the corresponding atoms NMR spectroscopy as one of the most powerful tools available for organic structural studies The basis of NMR: quantum behavior of magnetic nuclei, such as 1H and 13C, in a magnetic field • certain nuclei, such as 1H and 13C, have a permanent magnetic dipole, i.e., they behave like tiny "compass needles" • if one takes a population of compass needles and places them in a magnetic field (e.g., the magnetic field of the Earth), the needles all orient themselves in such a way that their own magnetic field opposes the external one: their S pole would point to the magnetic N pole, and their N pole would point to the magnetic S pole. This corresponds to an energetically favorable situation • Atomic nuclei, however, are subatomic particles, and as such they are subject to the laws of quantum mechanics. Quantum theory teaches that if a population of magnetic nuclei (1H, 13C, ...) is placed in a magnetic field (e.g., inside a powerful magnet), they may orient themselves in such a way that their magnetic field either opposes the external one (= more energetically favorable situation), or reinforces it (less energetically favorable situation)! Moreover, only a very small excess population (of the order of parts per million) will position itself in the more energetically favorable state: • Subjecting the population of magnetic nuclei thus distributed between the two energy states to an external source of energy exactly equal to ΔE will induce transitions of nuclei between the two states. Some nuclei will jump a population of magnetic nuclei (

1 H, 13

C, ...)

in the absence of a strong external field. Their magnetic moments are oriented at random.

Bold: the N pole of the nuclear magnetic field

Plain: the S pole of the nuclear magnetic field

apply a magnetic field, B B

S pole

N pole

Energy

In the presence of a magnetic field, B, the

nuclei populate two energy states that differ by !E. Nuclei in the less energetic state orient their field against B (like compass needles would do); those in the higher state, along B. A small excess population in present in the less energetic state.

15 from the lower state to the higher one; some will decay from the higher state to the lower one. • The energy corresponding to ΔE is readily available in the form of radiowaves. Modern NMR spectrometers employ radiofrequencies of the order of 100- 900 MHz • Nuclei jumping from the lower to the higher state will absorb an energy equal to ΔE; those decaying from the higher to the lower state will emit an energy equal to ΔE. • Quantum mechanics teaches us that the probability that a nucleus in the lower state will undergo an "up" transition is equal to the probability that a nucleus in the upper state will undergo a "down" one. Since there is a slight excess population in the lower energy state, the radiowaves will cause a greater number of "up" transitions relative to "down" ones. Overall, the system will absorb energy. • The magnitude of ΔE is a function of the external magnetic field, B, plus the combined effect of all magnetic fields generated by other magnetic subatomic particles (mostly electrons) in the vicinity of the nucleus undergoing such transitions • The energy corresponding to ΔE is readily available in the form of radiowaves. • The magnitude of ΔE is a function of the external magnetic field, B0, plus the combined effect of all magnetic fields generated by other magnetic subatomic particles (mostly electrons, but also other magnetic nuclei) in the vicinity of the nucleus undergoing such transitions • Net effect of electrons: generation of a secondary B field that opposes (=diminishes the intensity of) the external one, B0, thereby "shielding" the nuclei from B0 B

S pole

N pole

Energy

!E a population of magnetic nuclei ( 1 H, 13

C, ..) in

the presence of an external magnetic field, B 0

The nuclei populate two energy states that differ

by !E. A small excess population in present in the less energetic state. Nuclear transitions between the two states will occur if one supplies the system with an energy equal to !E. Overall, the system will absorb energy, due to the slight excess population in the lower state.

Bold: the N pole of the nuclear magnetic field

16 Different protons in an organic molecule are embedded in regions of the molecule possessing distinct electronic densities; i.e., they find themselves in distinct chemical environments that shield them to a greater or lesser extent from B0 Each proton in a given chemical environment will absorb at a specific frequency, giving rise to a characteristic absorption signal. Problem: given a molecule containing a certain number of protons, how many signals can we expect to observe in the 1H NMR spectrum of that molecule? Two or more protons in a molecule that happen to be in the same chemical environment, and that therefore will absorb at the same frequency, are said to be chemically equivalent. To determine whether 2 protons, Ha and Hb, in a molecule are chemically equivalent, one needs to carry out a so-called substitution test as follows: • one imagines replacing Ha with a generic atom X, thereby creating a new molecule, A • one imagines replacing Hb also with atom X, thereby creating a new molecule, B • one now compares molecules A and B: - If A and B are the same molecule or if they are enantiomers: Ha and Hb are chemically equivalent and they will absorb at the same frequency - If A and B are diastereomers or if they are constitutional isomers: Ha and Hb are not chemically equivalent and they will absorb at different frequencies example: the question of chemical equivalence of the 1H's in MeI example: the question of equivalence of the 1H's in dichloroacetaldehyde: H

a C I H b H c

We want to predict the number of

H NMR absorption

signals produced by the molecule of MeI. To do so, we need to determine whether H a , H b , and H c are chemically equivalent or not. Consider H a and H b first:

Replacement of H

a with X gives:

Replacement of H

b with X gives: X C I H H H C I X H A B compare A and B: they are the same molecule: H a and H b are equivalent • repeating the exercise for H c leads to the conclusion that all H's are chemically equivalent • the molecule will produce one signal in the 1

H NMR spectrum

ClCClHaCOHbWe want to predict the number of 1H NMR signalsproduced by this molecule, so we need to know whether Ha and Hb are chemically equivalent

17 the molecule will produce two signals in the 1H NMR spectrum: one for Ha and one for Hb Principle: each group of chemically equivalent protons within a molecule will produce its own characteristic absorption signal The difference between the external field B0 and the effective field Beff felt by a proton embedded in a molecule is called the chemical shift Chemical shifts are very small: of the order of parts per million (ppm) of the external field. However, chemical shifts can be determined with great precision by measuring the frequency of the radio waves absorbed by the nuclei undergoing the above transitions Expressing chemical shifts as frequencies creates practical problems that are nicely circumvented when one expresses them as a fraction of the external field B0; i.e., in ppm of B0. In this way, the chemical shift unit becomes independent of external factors such as B0 itself To construct a scale of chemical shifts, it is best to agree on a reference compound and measure the chemical shifts of other molecules relative to the chemical shift of the nuclei in the reference compound Reference compound in NMR spectroscopy: tetramethylsilane [(CH3)4Si, "TMS"]. The elec tron density surrounding the protons in the molecul e of TMS is unusually high. Consequently, these protons are highly "shielded" from B0. The protons in most other organic compounds are much less shielded (= they are more "deshielded"). Notice that all 12 H's in TMS are chemically e quivalent , so they all abso rb at the same frequency The 1H NMR chart: Replacement of Ha with X gives:Replacement of Hb with X gives:ClCClXCOHbClCClHaCOXABcompare A and B: they are constitutional isomers. So,Ha and Hb are not equivalent

18 Typical proton chemical shifts: Characteristic Proton (1H) NMR Chemical Shifts Type of Hydrogen Structure Chemical Shift δ (ppm) Type of Hydrogen Structure Chemical Shift δ (ppm) Reference (CH3)4Si 0.00 Amines 2.3 - 3.0 Alkane, primary -CH3 0.7 - 1.3 Alcohol, ether 3.3 - 4.0 Alkane, secondary -CH2- 1.2 - 1.4 Ester 3.7 - 4.2 Alkane, tertiary 1.4 - 1.7 Olefinic C=C-H 5.0 - 6.5 Allylic, primary C=C-CH3 1.6 - 1.9 Aromatic Ar-H 6.5 - 8.0 Methyl carbonyl 2.1 - 2.4 Aldehyde 9.7 - 10.0 Aromatic methyl Ar-CH3 2.5 - 2.7 Amine -NH2 1 - 5, variable Alkyne 2.5 - 2.7 Alcohol -OH 1 - 5, variable Alkyl halide (X = Cl, Br, I) 2.5 - 4.0 Carboxylic acid -COOH 11.0 - 12.0 0

1 234
5 6 7 8910
intensity of absoption chemical shift in ppm of B 0 relative to TMS set at 0.0 (arbitrary value) signal of the 1

H's in TMS

H's in most organic molecules absorb

between 0 and 10 ppm relative to TMS protons that are more shielded produce signals closer to that of the TMS protons that are less shielded produce signals farther away from TMS shielding deshielding 0 1 234
5 6 7 8910
ppm alkane- like C-H's X C C

X = C, O

H HC-Z

Z = N, I, S

HC-Z

Z = O, Cl, Br

olefinic C-H HZ-C Z'

Z, Z' =

heteroatoms aromatic C-H C O O H C O H

CHNCHO

CHOC O CH CCH 3 O HC OCH CX H

19 The proton NMR spectrum of CH3I The 1H NMR spectrum of methyl acetate Integration of 1H NMR spectra Noteworthy feature of the above spectrum: the areas under the two signals are identical Principle: in a proton NMR spect rum, the ra tio of t he area s under two or m ore signals is proportional to the ratio of the number of protons producing such signals. Integration of the spectrum of methyl acetate indicates that the signal at 1.2 ppm is produced by as many protons as those that generate the signal at 3.6 ppm 0

1 234
5 6 7 8910
ppm TMS

2.2 ppm

CH 3 -I: the three chemically equivalent protons absorb at 2.2 ppm relative to

TMS = they have the same chemical

shift of 2.2 ppm relative to TMS 0 1 234
5 6 7 8910
ppm TMS

1.9 ppm

3.6 ppm

H 3 C O O CH 3 one group of chem. equiv. protons another group of chem. equiv. protons

H's on one methyl are

not chem. equiv. to those on the other: they will have different chemical shifts O-CH 3 H 3 C O integral = area under the signal

20 The 1H NMR spectrum of methyl pivalate: Integration of the spectrum of methyl pivalate indicates that the signal at 1.2 ppm is produced by three times as many protons as those that generate the signal at 3.7 ppm The effect of neighboring protons on the appearance of the NMR spectrum of a given proton: signal splitting The 1H NMR spectrum of Br2CH-CHBr-CMe3 Why are the signals of HA and of HB "doubled up"? 0

1 234
5 6 7 8910
ppm TMS

1.2 ppm

3.7 ppm

C O O CH 3 one group of chem. equiv. protons another group of chem. equiv. protons O-CH 3 CH 3 CH 3 CH 3 C(CH 3 3 notice the integral ratio of 1 : 3 not chemically equiv.: different chemical shifts 0 1 234
5 6 7 8910
ppm TMS

1.2 ppm

6.3 ppm

BrC Br C Br C CH 3 CH 3 CH 3 H A H B 1 1 9 BrC Br H A C Br H B C CH 3 CH 3 CH 3

4.4 ppm

J AB = J BA = 6 Hz H A H B

21 Doublet: a system composed of two signals of essentially equal intensity, and arising from a proton that senses the magnetic field of a neighboring proton Spin-spin coupling (or spin interaction) Coupling constant, J (measured in Hz): the separation between the two signals of the doublet (with a few provisos not covered in CHEM 203) The chemical shift of the proton under observation as the mid-point of the doublet (with a few provisos not covered in CHEM 203) Principle: quantum mechanics reveals that: • spin-spin coupling is observed only between chemically non-equivalent protons: (chemically equivalent 1H do couple, but coupling cannot be observed) • if two protons, HA and HB, are coupled, then JAB = JBA Principle: the signal of a proton under observation, HA, will be independently split into a doublet by every other proton, HB, HC, ...., that spin interacts with HA Splitting of the signal of HA into a doublet (d), a doublet of doublets (dd), a doublet of doublets of doublets (ddd) ... , in which each line has an intensity equal to 1/2, 1/4, 1/8, etc., of that of the unsplit signal of HA Appearance of 2n signals for a proton that spin-interacts with n other protons B

S pole

N pole

H under observation

senses a magnetic field of intensity B eff < B 0 in the absence of neighboring 1 H's in the presence of neighboring 1 H's observed 1

H senses a field

B tot = B eff - B H : shielded observed 1

H senses a field

B tot = B eff + B H : deshielded ppm chemical shift of the proton under observation ppm

J (Hz)

chemical shift of the proton under observation

22 Principle: coupling is generally seen only between protons separated by 2 or 3 bonds The 1H NMR spectrum of cinnamaldehyde: If a proton HA spin interacts with 2 or more chemically equivalent protons, HB1, HB2, ... HBn, then JA-B1 = JA-B2 = .... = JA-Bn (with a few provisos not covered in CHEM 203). For geometric reasons, the 2n signals then collapse into n+1 lines. Example: Triplets, quartets, quintets, sextets, septets ...... multiplets: signals composed of 3, 4, 5, 6, 7, .... multiple lines Principle: the intensity ratios of a triplet, quartet, quintet, etc., produced as detailed above are given by the values of the corresponding row of a Pascal triangle a doublet of doublets (dd) becomes a triplet with intensity ratios 1 : 2 : 1 a doublet of doublets of doublets (ddd) becomes a quartet with intensity ratios 1 : 3 : 3 : 1 a dddd becomes a quintet with intensity ratios 1 : 4 : 6 : 4 : 1 and so on The NMR spectrum of Br-CH2-CH3 H

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