EM 03 rganic Spectroscopy: a Primer INDEX A. Introduction B. Infrared (IR) Spectroscopy 3. Proton Nuclear Magnetic Resonance ( NMR) Spectroscopy
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 0 X-rays Å diffraction atomic coordinates 0-70 electron ----- fragmentation mass of molecule bombardment & of its fragments 6.-3. UV light 90-30 nm electronic nature of the π systems 3.-. VIS light 00-800 nm transitions present in the molecule 0.-0.0 IR light.-30.0 µm vibrational functional groups transitions present in the molecule 0 6-0 7 radiowaves 0.- m nuclear chemical environment transitions of and atoms Spectrum of a molecule: a plot of extent of absorption vs. energy extent of absorption 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
3 B. Infrared (IR) Spectroscopy Infrared (IR) spectroscopy: a technique that derives information about functional groups present in a molecule on the basis of how the molecule reacts to energy absorption in the infrared range Wavelength of IR light of interest in structural organic chemistry:. < λ < 0 µ ( µ = 0 6 m) Frequency of above IR radiation: 0 ν Gz ( Gz = 0 z) Spectrometer: the instrument required for the conduct of a spectroscopic experiment Principle: given the way IR spectrometers are built, it is most convenient to express the energy of IR radiation in terms of its frequency. owever, expressing frequencies is Gz is impractical because of the large numbers involved Wavenumber (!, or more simply!): a quantity measured in cm, and linearly related to frequency, that expresses the number of ondulations of the EM field in cm of propagation: example: if! =. µ, =. 0 6 m =. 0 cm, then " = (. 0 ) = 000 cm Wavenumber calibration of the energy axis of the IR spectrum of a molecule: 000 ν 00 Transmittance: the percent light intensity transmitted through a test sample example: if a sample absorbs none of the electromagnetic radiation of a given frequency, then the transmittance at that frequency is 00%. If a sample absorbs all of the electromagnetic radiation of a given frequency, then the transmittance at that frequency is 0%. Transmittance calibration of the absorption axis of the IR spectrum of a molecule Infrared spectrum of a molecule: a plot of transmittance % (00 0%) vs. energy expressed in wavenumbers, ν (000 ν 00) transmittance % 00% " =! 0% 000 00! (cm ) Principle: absorption of IR light causes a molecule to undergo bond stretching and bond bending vibrations. In EM 03 we are only interested in stretching vibrations observed between 000 and 600 cm
haracteristic Infrared Stretching Absorptions of ommon Functional Groups Functional Group Alcohol Bond Frequency Range (cm ) Functional Group 300 360 (s, broad) Nitrile 00 0 (s) arboxylic acid Ether 000 60 Amine N 3300 330 (m) Alkane Alkene Alkyne Bond N Frequency Range (cm ) 0 60 (w m) 00-300 (s, broad) = 700 70 (s) Ester = 70 70 (s) 80 90 (m s) Acyl halide = 770 80 (s) = 300 300 (m) Acid anhydride = 70 790 (s) = 60 680 (m) 800 80 (s) 370 3330 (s) Amide = 630 700 (s) 00 60 (w m) Aldehyde, ketone = 680 730 (s) The spectra of alkanes, e.g., pentane: sp3- stretch at 800 90 cm. This signal is not diagnostic, because virtually every organic compound will show absorptions in this range The spectra of alkenes, e.g., 3-ethyl--pentene: sp stretch at 3000 300 cm, and in many but not all cases, a sharp signal at ~ 60 cm for the = stretching absorption =
The spectra of molecules containing benzene rings, e.g.,,-dimethylbenzene (p-xylene): sp stretch: 3000 300 cm, just as in the case of alkenes learly, IR absorptions in the range 3000 300 cm are diagnostic for the presence of alkene and / or benzene subunits in a molecule The spectra of terminal alkynes, e.g., 3,3,-dimethyl--butyne: sharp, intense sp - stretching absorption at! 3300 cm (highly diagnostic for the presence of a terminal alkyne) and weak, but highly diagnostic, stretching absorption at!"00 cm The IR spectra of internal alkynes: these molecules produce spectra IR spectra that display no diagnostic features, because: a. there is no sp- bond, therefore no signal at 3300 cm b. the 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 electromagnetic 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. learly, the triple bond of an ordinary internal alkyne has no dipole moment, so it does not absorb IR radiation
6 Example: the spectrum of,,-dimethyl--pentyne: The spectra of alcohols, e.g., isopropanol: stretch: 300-3600 cm, broad signal often of "parabolic" shape. ighly diagnostic for the presence of alcohols (note: we are not interested in the signals that appear below 600 cm ) Primary, secondary, tertiary amines: R N R R 3 N R R N R a primary amine a secondary amine a tertiary amine R, R, R 3 = alkyl groups (Me, Et,...)
7 The IR spectra of primary and secondary amines, e.g., butylamine and dipropylamine: N stretch at ~ 3300 cm, significantly weaker than an absorption. Primary amines often show a pair of N stretching absorption bands. Note: N stretching absorption (ca. 00 cm ) are not diagnostic. a pair of N stretching absorptions is typical of primary amines The IR spectra of tertiary amines: these molecules produce IR spectra that display no diagnostic features, because there is no N bond, therefore no signal at 3300 cm
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 -propyn--ol The IR spectra of nitriles, e.g., acetonitrile: N stretching absorption at! 00 cm. This signal is often weak, but it is highly diagnostic for the presence of a nitrile:
9 N Principle: IR spectroscopy is especially useful to detect the presence of carbonyl groups, which absorb strongly between 80 and 60 cm. The IR spectra of aldehydes, e.g., butyraldehyde: = stretch at ca. 70 cm : The IR spectra of ketones, e.g., acetone: = stretch at ca. 7 cm :
0 The IR spectra of esters, e.g., ethyl acetate: = stretch at ca. 70 cm : The IR spectra of carboxylic acids, e.g., acetic acid: = stretch ~ 700 cm. Very broad stretch starting at ~ 300 cm, extending down to ~ 00 cm, and centered around 300-3000 cm :
The IR spectra of acid chlorides, e.g., lauroyl chloride: = stretch at unusually high frequency, typically at ca. 800 cm : The IR spectra of acid anhydrides, e.g., acetic anhydride: two = stretching absorptions, one at very high frequency, ca. 80 cm (antisymmetric = / = stetch) the other at ca. 70 cm (symmetric = / = stetch)
Primary, secondary, tertiary amides: R R N R N R N a primary amide a secondary amide a tertiary amide R, R, R = alkyl groups The IR spectra of primary and secondary amides, e.g., formamide and N-methylacetamide: Strong N stretch at ~ 3300 cm and strong = stretch at ~ 60-700 cm R R
The IR spectra of tertiary amides, e.g., N,N-dimethylacetamide: strong = stretch at ~ 60-700 cm 3
. Proton Nuclear Magnetic Resonance ( NMR) Spectroscopy Nuclear Magnetic Resonance (NMR) spectroscopy: a technique 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 (= a proton: the nucleus of atoms) and 3 (the nucleus of a less common, non-radioactive, isotope of carbon that constitutes about % of total carbon), which reveal a great deal about the chemical 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 and 3, in a magnetic field certain nuclei, such as and 3, 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 (, 3,...) 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: apply a magnetic field, B N pole Energy!E a population of magnetic nuclei (, 3,...) 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 B S pole 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. 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
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 00-900 Mz 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. verall, 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 B 0 N pole S pole Energy!E a population of magnetic nuclei (, 3,..) 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. verall, the system will absorb energy, due to the slight excess population in the lower state. Bold: the N pole of the nuclear magnetic field 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, B 0, 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, B 0, thereby "shielding" the nuclei from B 0
6 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 B 0 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 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 protons, a and b, in a molecule are chemically equivalent, one needs to carry out a so-called substitution test as follows: one imagines replacing a with a generic atom X, thereby creating a new molecule, A one imagines replacing b 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: a and b are chemically equivalent and they will absorb at the same frequency If A and B are diastereomers or if they are constitutional isomers: a and b are not chemically equivalent and they will absorb at different frequencies example: the question of chemical equivalence of the 's in MeI c b a I Replacement of a with X gives: Replacement of b with X gives: We want to predict the number of NMR absorption signals produced by the molecule of MeI. To do so, we need to determine whether a, b, and c are chemically equivalent or not. onsider a and b first: X X I I A B compare A and B: they are the same molecule: a and b are equivalent repeating the exercise for c leads to the conclusion that all 's are chemically equivalent the molecule will produce one signal in the NMR spectrum example: the question of equivalence of the 's in dichloroacetaldehyde: l l a b We want to predict the number of NMR signals produced by this molecule, so we need to know whether a and b are chemically equivalent
7 Replacement of a with X gives: Replacement of b with X gives: l A l B l X b l a X compare A and B: they are constitutional isomers. So, a and b are not equivalent the molecule will produce two signals in the NMR spectrum: one for a and one for b Principle: each group of chemically equivalent protons within a molecule will produce its own characteristic absorption signal The difference between the external field B 0 and the effective field B eff felt by a proton embedded in a molecule is called the chemical shift hemical shifts are very small: of the order of parts per million () of the external field. owever, 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 B 0 ; i.e., in of B 0. In this way, the chemical shift unit becomes independent of external factors such as B 0 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 [( 3 ) Si, "TMS"]. The electron density surrounding the protons in the molecule of TMS is unusually high. onsequently, these protons are highly "shielded" from B 0. The protons in most other organic compounds are much less shielded (= they are more "deshielded"). Notice that all 's in TMS are chemically equivalent, so they all absorb at the same frequency The NMR chart:
8 intensity of absoption 's in most organic molecules absorb between 0 and 0 relative to TMS signal of the 's in TMS protons that are less shielded produce signals farther away from TMS protons that are more shielded produce signals closer to that of the TMS 0 deshielding 9 8 7 6 3 shielding 0 chemical shift in of B 0 relative to TMS set at 0.0 (arbitrary value) Typical proton chemical shifts: aromatic olefinic Z Z' Z, Z' = heteroatoms Z Z =, l, X X =, Z Z = N, I, S alkanelike 's 0 9 8 7 6 3 0 haracteristic Proton ( ) NMR hemical Shifts Type of ydrogen Structure hemical Shift δ () Type of ydrogen Structure hemical Shift δ () Reference ( 3) Si 0.00 Amines.3 3.0 N Alkane, primary - 3 0.7.3 Alcohol, ether 3.3.0 Alkane, secondary - -.. Ester 3.7. Alkane, tertiary..7 lefinic =-.0 6. Allylic, primary =- 3.6.9 Aromatic Ar 6. 8.0 Methyl carbonyl.. Aldehyde 9.7 0.0 3 Aromatic methyl Ar 3..7 Amine N, variable Alkyne..7 Alcohol, variable Alkyl halide (X = l,, I) X..0 arboxylic acid.0.0
9 The proton NMR spectrum of 3 I. 3 -I: the three chemically equivalent protons absorb at. relative to TMS = they have the same chemical shift of. relative to TMS TMS 0 9 8 7 6 3 0 The NMR spectrum of methyl acetate integral = area under the signal one group of chem. equiv. protons another group of chem. equiv. protons 3 3 's on one methyl are not chem. equiv. to those on the other: they will have different chemical shifts 3.6 3.9 3 TMS 0 9 8 7 6 3 0 Integration of NMR spectra Noteworthy feature of the above spectrum: the areas under the two signals are identical Principle: in a proton NMR spectrum, the ratio of the areas under two or more 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. is produced by as many protons as those that generate the signal at 3.6
0 The NMR spectrum of methyl pivalate: one group of another group chem. equiv. of chem. equiv. protons protons 3 3 3 3 notice the integral ratio of : 3. ( 3 ) 3 not chemically equiv.: different chemical shifts 3.7 3 TMS 0 9 8 7 6 3 0 Integration of the spectrum of methyl pivalate indicates that the signal at. is produced by three times as many protons as those that generate the signal at 3.7 The effect of neighboring protons on the appearance of the NMR spectrum of a given proton: signal splitting The NMR spectrum of Me 3 J AB = J BA = 6 z 3 3 9 3 3 3 3. 6.3. TMS 0 9 8 7 6 3 0 Why are the signals of and of "doubled up"?
B 0 N pole in the absence of neighboring 's in the presence of neighboring 's S pole under observation senses a magnetic field of intensity B eff < B 0 observed senses a field B tot = B eff + B : deshielded J (z) observed senses a field B tot = B eff B : shielded chemical shift of the proton under observation chemical shift of the proton under observation 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) oupling constant, J (measured in z): the separation between the two signals of the doublet (with a few provisos not covered in EM 03) The chemical shift of the proton under observation as the mid-point of the doublet (with a few provisos not covered in EM 03) Principle: quantum mechanics reveals that: spin-spin coupling is observed only between chemically non-equivalent protons: (chemically equivalent do couple, but coupling cannot be observed) if two protons, and, are coupled, then J AB = J BA Principle: the signal of a proton under observation,, will be independently split into a doublet by every other proton,,,..., that spin interacts with Splitting of the signal of 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 /, /, /8, etc., of that of the unsplit signal of Appearance of n signals for a proton that spin-interacts with n other protons
Principle: coupling is generally seen only between protons separated by or 3 bonds The NMR spectrum of cinnamaldehyde: : doublet (9.8 ), because it spin-interacts with only : doublet of doublets (6.7 ): it spin-interacts with and : doublet (7. ), because it spin-interacts with only 's on the benzene ring ("aromatic 's"): ~ 7. note is separated from by 3 bonds: coupling is observed is separated from by bonds: no coupling is seen is separated from or by 3 bonds: coupling is observed is separated from the aromatic 's by or more bonds: no coupling If a proton spin interacts with or more chemically equivalent protons,,,... n, then J A-B = J A-B =... = J A-Bn (with a few provisos not covered in EM 03). For geometric reasons, the n signals then collapse into n+ lines. Example: J = J = 7 z J = 7 z J = z if J becomes equal to J, the inner lines will overlap intens. ratio: ::: intens. ratio: :: Triplets, quartets, quintets, sextets, septets... multiplets: signals composed of 3,,, 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 : : a doublet of doublets of doublets (ddd) becomes a quartet with intensity ratios : 3 : 3 : a dddd becomes a quintet with intensity ratios : : 6 : : and so on The NMR spectrum of - 3
3 : quartet (3. ), because it spin-interacts with 3 chemically equivalent 's : triplet (.7 ): because it spin-interacts with chemically equivalent 's The n+ rule of spin-spin coupling as apparent from the spectra of:,,-trichloroethane l l l triplet, because it spininteracts w/ chem. equivalent 's ( ) doublet, because of spininteraction w/ ( ) l l l TMS 0 9 8 7 6 3 0,-dibromoethane doublet: interaction with ( ) quartet: interaction w/ 3 chem.eq. 's TMS 0 9 8 7 6 3 0 bromoethane triplet: interaction w/ chem. eq. 's quartet: interaction w/ 3 chem.eq. 's TMS 0 9 8 7 6 3 0
-bromopropane doublet: interaction with septet: interaction w/ 6 chem.eq. 's TMS 0 9 8 7 6 3 0 para-methoxypropiophenone D E E D E singlet: no spin interaction doublet: interaction w/ doublet: interaction w/ Me quartet: interaction w/ 3 chem. eq. E 's D D triplet: interaction w/ chem. eq. D 's E E E TMS 0 9 8 7 6 3 0 note: the difference in chemical shift between the aromatic (=attached to a benzene ring) protons and is readily explainable on the basis of resonance interactions: Et the carbonyl group draws electrons away from the atom bearing : is deshielded and it absorbs downfield (= away from TMS) Et Me Me the methoxy group donates electrons toward the atom bearing : is shielded and it resonates upfield (= toward TMS)
-Propanol sextet: interaction w/ 's & 3 's in such a manner that accidentally J BA = J B triplet: interaction w/ chem.eq. 's 3 triplet: interaction w/ chem.eq. 's TMS 0 9 8 7 6 3 0 Noteworthy features of the spectrum of -propanol: The coupling constants J BA and J B should be different, because and are not chemically equivalent. owever, in this case (and in other similar cases) all J's happen to be the same. The signal of, which would otherwise appear as a triplet of quartets (splitting by 's and 3 's), collapses into a sextet of intensity ratio ::0:0:: No coupling is apparent between the proton and. This is because protons normally exchange (= they "jump" from molecule to molecule) very rapidly. As a consequence, protons reside on a given molecule for a period of time that is too short to modulate the field of neighboring 's; that is, to spin couple. The same is true for N protons in amines. note: spin coupling between protons and neighboring hydrogens may be seen when the rate of exchange becomes sufficiently slow. Exchanging the in and N bonds with deuterium by treatment of the sample with D D exchange causes the signal of and N protons to disappear from the spectrum useful technique to recognize the signals and N protons D exchange may improve the resolution (= the "sharpness") of NMR spectra of alcohols and amines