13.24: Mass Spectrometry: molecular weight of the sample

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1 hapter 13: Spectroscopy Methods of structure determination Nuclear Magnetic Resonances (NMR) Spectroscopy (Sections ) Infrared (IR) Spectroscopy (Sections ) Ultraviolet-visible (UV-Vis) Spectroscopy (Section 13.23) Mass (MS) spectrometry (not really spectroscopy) (Section 13.24) Molecular Spectroscopy: the interaction of electromagnetic radiation (light) with matter (organic compounds). This interaction gives specific structural information : Mass Spectrometry: molecular weight of the sample formula The mass spectrometer gives the mass to charge ratio (m/z), therefore the sample (analyte) must be an ion. Mass spectrometry is a gas phase technique- the sample must be vaporized. Electron-impact ionization ionization chamber e _ Sample Inlet torr R- R- + (M + ) mass analyzer m/z electron beam 70 ev (6700 KJ/mol) proton neutron electron u u u 2 1

2 mass charge m = = B2 r 2 z 2V B= magnetic field strength r = radius of the analyzer tube V= voltage (accelerator plate) Ions of non-selected mass/charge ratio are not detected Ions of selected mass/charge ratio are detected Magnetic Field, B o Ionization chamber The Mass Spectrometer 3 Exact Masses of ommon Natural Isotopes Isotope mass natural abundance (1.11%) 14 N N (0.38%) (0.04%) (0.20%) Isotope mass natural abundance 19 F l l (32.5%) 79 Br Br (98%) 127 I

3 Molecular Ion (parent ion, M) = molecular mass of the analyte; sample minus an electron Base peak- largest (most abundant) peak in a mass spectra; arbitrarily assigned a relative abundance of 100%. m/z=78 (M + ) (100%) 6 6 m/z = m/z=79 (M+1) (~ 6.6% of M + ) 5 The radical cation (M + ) is unstable and will fragment into smaller ions Relative abundance (%) m/z=14 m/z=15 m/z=16 (M + ) m/z=17 (M+1) - e _ + m/z = 16 m/z = m/z = 15 charge neutral not detected charge neutral not detected Relative abundance (%) m/z=15 m/z=29 m/z m/z m/z=43 m/z=44 (M) m/z=45 (M+1) - e _ + - e _ charge neutral not detected m/z = 15 m/z = 44 + m/z = m/z = 29 charge neutral not detected 6 + charge neutral not detected 3

4 l m/z=112 (M + ) 35 l l (32.5%) m/z=77 m/z=113 (M + +1) m/z=114 (M + +2) m/z=115 (M + +3) m/z Br m/z=77 m/z=156 (M + ) m/z=158 (M + +2) 79 Br Br (98%) m/z=157 (M + +1) m/z=159 (M + +3) m/z 7 Mass spectra can be quite complicated and interpretation difficult. Some functional groups have characteristic fragmentation It is difficult to assign an entire structure based only on the mass spectra. owever, the mass spectra gives the mass and formula of the sample which is very important information. To obtain the formula, the molecular ion must be observed. Soft ionization techniques Methods have been developed to get large molecules such as polymers and biological macromolecules (proteins, peptides, nucleic acids) into the vapor phase 8 4

5 13.25: Molecular Formula as a lue to Structure Nitrogen rule: In general, small organic molecules with an odd mass must have an odd number of nitrogens. rganic molecules with an even mass have zero or an even number of nitrogens If the mass can be determined accurately enough, then the molecular formula can be determined (high-resolution mass spectrometry) Information can be obtained from the molecular formula: Degrees of unsaturation: the number of rings and/or π-bonds in a molecule (Index of ydrogen Deficiency) 9 Degrees of unsaturation saturated hydrocarbon n 2n+2 cycloalkane (1 ring) n 2n alkene (1 π-bond) n 2n alkyne (2 π-bonds) n 2n-2 For each ring or π-bond, -2 from the formula of the saturated alkane ydrogen Deficiency = 1 Degrees of Unsaturation =

6 orrection for other elements: For Group VII elements (halogens): subtract 1 from the -deficiency for each halogen, For Group VI elements ( and S): No correction is needed For Group V elements (N and P): add 1 to the -deficiency for each N or P N l : Principles of molecular spectroscopy: Electromagnetic radiation organic molecule (ground state) light hν organic relaxation organic molecule + hν molecule (excited state) (ground state) Electromagnetic radiation has the properties of a particle (photon) and a wave. λ = distance of one wave ν = frequency: waves per unit time (sec -1, z) c = speed of light (3.0 x 10 8 m sec -1 ) h = Plank s constant (6.63 x J sec) 12 6

7 Quantum: the energy of a photon E = hν ν = c λ E = h c λ E α ν E α λ 1 ν α λ 1!-rays x-rays UV Vis IR microwaves radiowaves λ (cm) short high high Wavelength (λ) Frequency (ν) Energy (E) long low low : Principles of molecular spectroscopy: Quantized Energy Levels molecules have discrete energy levels (no continuum between levels) A molecule absorbs electromagnetic radiation when the energy of photon corresponds to the difference in energy between two states 14 7

8 organic molecule (ground state) light hν organic relaxation organic molecule + hν molecule (excited state) (ground state) UV-Vis: valance electron transitions - gives information about π-bonds and conjugated systems Infrared: molecular vibrations (stretches, bends) - identify functional groups Radiowaves: nuclear spin in a magnetic field (NMR) - gives a map of the and framework Ultraviolet-Visible (UV-Vis) Spectroscopy UV Vis λ nm Recall bonding of a π-bond from hapter

9 π-molecular orbitals of butadiene Ψ 4 : 3 Nodes 0 bonding interactions 3 antibonding interactions ANTIBNDING M Ψ 3 : 2 Nodes 1 bonding interactions 2 antibonding interactions ANTIBNDING M Ψ 2 : 1 Nodes 2 bonding interactions 1 antibonding interactions BNDING M Ψ 1 : 0 Nodes 3 bonding interactions 0 antibonding interactions BNDING M ψ 2 is the ighest ccupied Molecular rbital (M) ψ 3 is the Lowest Unoccupied Molecular rbital (LUM) 17 UV-Vis light causes electrons in lower energy molecular orbitals to be promoted to higher energy molecular orbitals. M LUM hromophore: light absorbing portion of a molecule Butadiene Butadiene 18 9

10 Molecular orbitals of conjugated polyenes Antibonding Energy Bonding nm 217 nm 258 nm 290 nm 19 Molecules with extended conjugation move toward the visible region 380 nm 400 nm 450 nm 500 nm 550 nm 600 nm 700 nm 780 nm violet-indigo blue green yellow orange red olor of absorbed light λ olor observed violet 400 nm yellow blue 450 orange blue-green 500 red yellow-green 530 red-violet yellow 550 violet orange 600 blue-green red 700 green 20 10

11 Many natural pigments have conjugated systems N N Mg N N 2 3 hlorophyll + anthocyanin β-carotene lycopene 21 hromophore: light absorbing portion of a molecule Beer s Law: There is a linear relationship between absorbance and concentration A = ε c l A = absorbance c = concentration (M, mol/l) l = sample path length (cm) ε = molar absorptivity (extinction coefficient) a proportionality constant for a specific absorbance of a substance 22 11

12 13.20: Introduction to Infrared Spectroscopy Vis Near IR Infrared (IR) Far IR microwave x 10-4 cm 1.6 x 10-3 cm µm 16 µm _ 4000! 600! _ λ (cm) E α 1 λ _ λ is expressed as ν (wavenumber), reciprocal cm (cm -1 ) _ ν = 1 therefore E α _ ν λ IR radiation causes changes in a molecular vibrations 23 Stretch: change in bond length Symmetric stretch Antisymmetric stretch Bend: change in bond angle > > > > > > > > > > > > > > scissoring rocking wagging twisting in-plane bend out-of-plane bend > > Animation of bond streches and bends:

13 Bond Stretch: ooke s Law X Y _! = 1 2 " c f m x m y m x + m y 1 2 _! = vibrational frequency c = speed of light m x = mass of X m y = mass of Y m x m y m x + m y = reduced mass (µ) _ E α ν α f f = spring constant; type of bond between X and Y (single, double or triple) ooke s law simulation: 25 Interpretation of an Infrared Spectra: organic molecules contain many atoms. As a result, there are many stretching and bending modes- IR spectra have many absorption bands Four distinct regions of an IR spectra X- single bond region triple bond region double bond region fingerprint region 4000 cm cm cm cm cm N- N = = 26 13

14 Fingerprint region ( cm -1 )- low energy single bond stretching and bending modes. The fingerprint region is unique for any given organic compound. owever, there are few diagnostic absorptions. Double-bond regions ( cm -1 ) = cm -1 = cm -1 Triple-bond region: ( cm -1 ) cm -1 (weak, often not observed) N cm -1 X- Single-bond region ( cm -1 ) cm -1 (broad) cm -1 (very broad) N cm cm -1 sp cm -1 sp 2 = cm -1 sp cm haracteristic Absorption Frequencies Table 13.4, p. 554 Alkenes = cm -1 medium - strong = cm -1 medium Aromatic = cm -1 strong = cm -1 strong Alkynes cm -1 strong cm -1 weak - medium Alcohols cm -1 strong cm -1 strong and broad Amines -N cm -1 medium N cm -1 medium arbonyl = cm -1 strong arboxylic acids cm -1 strong and very broad Nitrile N cm -1 weak-medium 28 14

15 % transmittance % transmittance - - hexane cm % transmittance % transmittance =- - = - cm cm -1 cm ( 2 ) 4 2 N - 3 ( 2 ) 4 2 N ( 2 ) 4 2 N 2 3 ( 2 ) 4 2 N- 3 N- N

16 (3 ) ( 2 ) = 1705 cm -1 = 1710 cm -1 3 ( 2 ) ( 2 ) = 1730 cm -1 - = 1715 cm Typical IR Absorptions for Funtional Groups X- Regon ( cm -1 ) - - N- Triple Bond Region ( ) N Double Bond Region ( ) N Fingerprint Region ( cm -1 ) difficult to interpret Functional Group Alkanes, Alkyl groups - cm -1 Functional Group cm Amines N Alkenes =- = Alkynes!-! Amide N- arbonyl Group = Alkyl alides -l -Br -I Alcohols arboxylic Acids - Nitriles!N Nitro Group -N Aromatics = ,

17 13.3: Introduction to 1 NMR direct observation of the s of a molecules Nuclei are positively charged and spin on an axis; they create a tiny magnetic field + + Not all nuclei are suitable for NMR. 1 and 13 are the most important NMR active nuclei in organic chemistry Natural Abundance % % % (not NMR active) 33 (a) Normally the nuclear magnetic fields are randomly oriented (b) When placed in an external magnetic field (B o ), the nuclear magnetic field will either aligned with (lower energy) or oppose (higher energy) the external magnetic field Fig 13.3, p

18 The energy difference between aligned and opposed to the external magnetic field (B o ) is generally small and is dependant upon B o Applied EM radiation (radio waves) causes the spin to flip and the nuclei are said to be in resonance with B o ΔE = h ν Note that h 2π ΔE = γb o h 2 π B o = external magnetic field strength γ = gyromagnetic ratio 1 = 26, = 6.7 is a constant and is sometimes denoted as h 35 NMR Active Nuclei: nuclear spin quantum number (I) atomic mass and atomic number Number of spin states = 2I + 1 (number of possible energy levels) Even mass nuclei that have even number of neutron have I = 0 (NMR inactive) Even mass nuclei that have odd number of neutrons have an integer spin quantum number (I = 1, 2, 3, etc) dd mass nuclei have half-integer spin quantum number (I = 1/2, 3/2, 5/2, etc) I= 1/2: 1, 13, 19 F, 31 P I= 1: 2, 14 N I= 3/2: 15 N I= 0: 12,

19 ontinuous wave (W) NMR Pulsed (FT) NMR : Nuclear Shielding and 1 hemical Shift Different nuclei absorb EM radiation at different wavelength (energy required to bring about resonance) Nuclei of a given type, will resonate at different energies depending on their chemical and electronic environment. The position (chemical shift, δ) and pattern (splitting or multiplicity) of the NMR signals gives important information about the chemical environment of the nuclei. The integration of the signal is proportional to the number of nuclei giving rise to that signal 38 19

20 hemical shift: the exact field strength (in ppm) that a nuclei comes into resonance relative to a reference standard (TMS) Electron clouds shield nuclei from the external magnetic field causing them to resonate at slightly higher energy Shielding: influence of neighboring functional groups on the electronic structure around a nuclei and consequently the chemical shift of their resonance. l l l 3 3 Si 3 3 Tetramethylsilane (TMS); Reference standard δ = 0 for 1 NMR downfield lower magnetic field less shielded (deshielded) l 3 δ = 7.28 ppm upfield higher magnetic field more shielded TMS hemical shift (δ, ppm) 39 downfield lower magnetic field less shielded (deshielded) N 2 3 δ= 4.20 ppm 2 δ= 3.50 ppm 3 upfield higher magnetic field more shielded TMS hemical shift (δ, ppm) Vertical scale= intensity of the signal orizontal scale= chemical shift (δ), dependent upon the field strength of the external magnetic field; for 1, δ is usually from 1-10 ppm ν- ν δ= TMS chemical shift in z = operating frequency in Mz ν ο 14,100 gauss: 60 Mz for 1 (60 million hertz) ppm= 60 z 15 Mz for ,000 gauss: 600 Mz for 1 ppm = 600 z 150 Mz for

21 13.5: Effect of Molecular Structure on 1 hemical Shift Electronegative substituents deshield nearby protons less shielded more shielded δ 4.3 δ 3.2 δ 2.2 δ 0.9 δ 0.0 The deshielding effect of a group drops off quickly with distance (number of bonds between the substituent and the proton) 3 -F ( 3 ) 3 -N 3-3 ( 3 ) 4 -Si δ= The influence of neighboring groups (deshielding) on 1 chemical shifts is cumulative l l l l l l δ = ppm l 3 2 l l δ = ppm 42 21

22 Typical 1 NMR chemical shifts ranges; additional substitution can move the resonances out of the range NMR Shift Ranges N R R 2 aromatics Ph R vinyl R 2 X R R 2 R 2 Ar R 2 X= F, l, Br, I R 2 N R 2 X R R 2 X=, R 2 N -R 2 R!- R 2 N- R sat. alkanes R " (ppm) also see Table 13.1 (p. 526) 43 Protons attached to sp 2 and sp hybridize carbons are deshielded relative to protons attached to sp 3 hybridized carbons 3 3 δ = ppm Please read about ring current effects of π-bonds (Figs , p ) δ = ppm 44 22

23 13.6: Interpreting 1 NMR Spectra Equivalence (chemical-shift equivalence): chemically and magnetically equivalent nuclei resonate at the same energy and give a single signal or pattern N δ= 4.20 ppm 2 δ= 3.50 ppm 3 TMS

24 Test of Equivalence: 1. Do a mental substitution of the nuclei you are testing with an arbitrary label 2. Ask what is the relationship of the compounds with the arbitrary label 3. If the labeled compounds are identical (or enantiomers), then the original nuclei are chemically equivalent and do not normally give rise to a separate resonance in the NMR spectra If the labeled compounds are not identical (and not enantiomers), then the original nuclei are not chemically equivalent and can give rise to different resonances in the NMR spectra Identical, so the protons are equivalent Identical, so the methyl groups are equivalent These are geometric isomers (not identical and not enantiomers). The three methyl groups are therefore not chemically equivalent and can give rise to different resonances

25 l 3 3 omotopic: equivalent Enantiotopic: equivalent Diastereotopic: non-equivalent 50 25

26 Integration of 1 NMR resonances: The area under an NMR resonance is proportional to the number of equivalent nuclei that give rise to that resonance. δ= 4.20, 2 δ= 3.50, δ3 N TMS The relative area under the resonances at δ= 4.20 and 3.50 is 2: : Spin-spin splitting in 1 NMR spectroscopy protons on adjacent carbons will interact and split each others resonances into multiple peaks (multiplets) n + 1 rule: equivalent protons that have n equivalent protons on the adjacent carbon will be split into n + 1 peaks. δ= δ= δ= Resonances always split each other. The resonance at δ= 4.1 splits the resonance at δ = 1.2; therefore, the resonance at δ = 1.2 must split the resonance at δ =

27 The multiplicity is defined by the number of peaks and the pattern (see Table 13.2 for common multiplicities patterns and relative intensities) δ= 2.0 s, 3 δ= 1.2 t, 3 δ= 4.1 q, : 2 : 1 1 : 3 : 3 : 1 53 The resonance of a proton with n equivalent protons on the adjacent carbon will be split into n + 1 peaks with a coupling constant J. oupling constant: distance between peaks of a split pattern; J is expressed in z. Protons coupled to each other have the same coupling constant J. δ= 2.0 s, 3 δ= 1.2 t, J=7.2 z, 3 δ= 4.1 q, J=7.2 z, 2 3 J 3 ab J 3 ab J 3 3 ab J ab J ab 54 27

28 13.8: Splitting Patterns: The Ethyl Group Two equivalent protons on an adjacent carbon will split a proton a triplet (t), three peaks of 1:2:1 relative intensity Three equivalent protons on an adjacent carbon will split a proton into a quartet (q), four peaks of 1:3:3:1 relative intensity δ= 1.2, t J= 7.0, 3 δ= 1.2, t J= 7.2, 3 δ= 3.0, q J= 7.2, 2 δ= , m, 5 δ= 2.6, q, J= 7.0, 2 δ= 8.0, 2 δ= , m, : Splitting Patterns: The Isopropyl Group ne proton on an adjacent carbon will split a proton into a doublet (d), two peaks of 1:1 relative intensity Six equivalent protons on an adjacent carbon will split a proton into a septet (s), seven peaks of 1:6:15:20:15:6:1 relative intensity 3 3 δ= , m, 5 δ= 2.9, s, J= 6.9, 1 δ= 1.2, d J= 6.9, 6 δ= 8.1, d, J= 6.1 z, 2 2 N 3 3 δ= 7.4, d J= 6.1 z, 2 δ= 3.0, s, J= 6.9, 1 δ= 1.2, d J= 6.9,

29 13.10: Splitting Patterns: Pairs of Doublets 3 δ= 1.2, s, δ= 8.0, d, J= δ= 7.4, d, J= δ= 3.9, s, 3 fig 13.19, p : omplex Splitting Patterns: when a protons is adjacent to more than one set of non-equivalent protons, they will split independently J 1-2 = 7.0 J 2-3 = 16.0 J 1-2 = 7.0 J 2-3 = splits 3 into a doublet with coupling constant J 2-3 = splits 1 into a doublet with coupling constant J 1-2 = splits 2 into a doublet with a coupling constants J 1-2 =7.0; 3 further splits 2 into a doublet (doublet of doublets) with coupling constants J 2-3 =

30 J 1,2 = J 2,3 δ= 0.9, t, J = 7.4, 3 J 1,2 δ= 2.6, 2 δ= 1.6, 2 J 2, : 1 NMR Spectra of Alcohols Usually no spin-spin coupling between the proton and neighboring protons on carbon due to exchange reaction A + A The chemical shift of the - proton occurs over a large range ( ppm). This proton usually appears as a broad singlet. It is not uncommon for this proton not to be observed : NMR and onformation (please read) NMR is like a camera with a slow shutter speed. What is observed is a weighted time average of all conformations present

31 Summary of 1-1 spin-spin coupling chemically equivalent protons do not exhibit spin-spin coupling to each other. the resonance of a proton that has n equivalent protons on the adjacent carbon is split into n+1 peaks (multiplicity) with a coupling constant J. protons that are coupled to each other have the same coupling constant non-equivalent protons will split a common proton independently: complex coupling. Spin-spin coupling is normally observed between nuclei that are one, two and three bonds away. Four-bond coupling can be observed in certain situations (i.e., aromatic rings), but is not common. 61 Summary of 1 -NMR Spectroscopy the number of proton resonances equals the number of non-equivalent protons the chemical shift (δ, ppm) of a proton is diagnostic of the chemical environment (shielding and deshielding) Integration: number of equivalent protons giving rise to a resonance spin-spin coupling is dependent upon the number of equivalent protons on the adjacent carbon(s) 62 31

32 13 NMR Spectroscopy: Natural Abundance % (I= 1/2) % (I= 0) % (I= 1/2) % 1.1 % 1.1 % ΔE= γb o h 2 π B o = external magnetic field strength γ = magnetogyric ratio 1 = 26, = is a much less sensitive nuclei than 1 for NMR spectroscopy New techniques (hardware and software) has made 13 NMR routine Pulsed NMR techniques (FT or time domain NMR) Signal averaging (improved signal to noise) Animation: 63 Fourier Transform (FT) deconvolutes all of the FID s and gives an NMR spectra. time 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 13 and small samples. 13 -spectra of after one scan average of 200 scans 64 32

33 hemical shifts give an idea of the chemical and electronic environment of the 13 nuclei due to shielding and deshielding effects range: ppm from TMS 13 NMR spectra will give a map of the carbon framework. The number of resonances equals the number of non-equivalent carbons Dl 3 TMS 65 hemical Shift Range of Typical 13 NMR Shift Ranges R 3 -Br R 3 -F R 3 -l R 3 -I R 2 N-R 3 nitriles R 3 R R 3 aromatics R R 3 Ar-R 3 ketones & aldehydes carbonyls esters, amides & acids vinyl alkyne saturated alkanes ! (PPM) Note the carbonyl range 66 33

34 13.18: Using DEPT to ount ydrogens Attached to spin-spin coupling: spin-spin coupling tells how many protons are attached to the 13 nuclei. (i.e., primary, secondary tertiary, or quaternary carbon) 13 spectra are usually collected with the 1-13 coupling turned off (broad band decoupled). In this mode all 13 resonances appear as singlets. DEPT spectra (Distortionless Enhancement by Polarization Transfer) a modern 13 NMR spectra that allows you to determine the number of attached hydrogens Broad-band decoupled DEPT s give negative resonances s and 3 s give positive resonances Quaternary carbon (no attached s) are not observed 13.19: 2D NMR: SY and ETR (please read) 68 34

35 Solving ombined 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 π-bonds Infrared Spectra: Functional Groups = - = N- - N 1 NMR: hemical Shift (δ) chemical environment of the 's Integration # of 's giving rise to the resonance Spin-Spin oupling (multiplicity) # of non-equivalent 's on the adjacent carbons (vicinal coupling). 13 NMR: # of resonances symmetry of carbon framework Type of arbonyl 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 (Fig (p. 572) 70 35

36 Problem (Fig , p. 573) (Fig , p. 574): NMR:

37 δ= 2.61 (d, 3) δ= 2.61 (t, 3) δ= 2.61 (sextet, J=7, 1) δ= 2.61 (pentet, J=7, 2) δ= (m, 5) 73 37