hapter 9 Nuclear Magnetic Resonance and Mass Spectrometry reated by Professor William Tam & Dr. Phillis hang 1. Introduction Spectroscopy the study of the interaction of light with matter Spectroscopy provides information about molecular structure. Methods we will explore: Nuclear Magnetic Resonance (NMR) Spectroscopy Mass Spectrometry (MS) Infrared (IR) Spectroscopy (Section 2.15) v Electromagnetic spectrum 2. Nuclear Magnetic Resonance (NMR) Spectroscopy cosmic & g-rays X-rays ultraviolet visible infrared microwave radiowave v The nuclei of some isotopes, such as 1 and 13, behave as magnets. l: 0.1nm 200nm 400nm 800nm 50mm X-Ray rystallography UV IR NMR 1Å = 10-10 m 1nm = 10-9 m 1mm = 10-6 m v v When 1 or 13 atoms are placed in a magnetic field and irradiated with electromagnetic energy, some frequencies are absorbed ( magnetic resonance ) A graph that shows the characteristic energy absorption frequencies and intensities for a sample in a magnetic field is called a nuclear magnetic resonance (NMR) spectrum 1
Example of an NMR spectrum 1. The number of signals in the spectrum tells us how many different sets of protons there are in the molecule 2. The position of the signals in the spectrum along the x-axis tells us about the magnetic environment of each set of protons arising largely from the electron density in their environment 3. The area under the signal tells us about how many protons there are in the set being measured 4. The multiplicity (or splitting pattern) of each signal tells us about the number of protons on atoms adjacent to the one whose signal is being measured 2A. hemical Shift v The position of a signal along the x-axis of an NMR spectrum is called its chemical shift v The chemical shift of each signal gives information about the structural environment of the nuclei producing that signal v ounting the number of signals in a 1 NMR spectrum indicates, at a first approximation, the number of distinct proton environments in a molecule 2
v Normal range of 1 NMR chemical shifts "upfield" (more shielded) "downfield" (deshielded) 15-10 (low field d ppm (high field strength) strength) hemical shift depends on a given nucleus s magnetic environment. Magnetic environment is affected by such factors as electron density. v Reference compound TMS = tetramethylsilane Me Me as a reference standard (0 ppm) Reasons for the choice of TMS as reference t Resonance position at higher field than most other organic compounds t Unreactive and stable t Volatile and easily removed from sample (B.P. = 28 o ) Si Me Me v NMR solvent Normal NMR solvents should not contain hydrogen ommon solvents t Dl 3 t 6 D 6 t t D 3 OD D 3 OD 3 (d 6 -acetone) 3
v The 300-Mz 1 NMR spectrum of 1,4-dimethylbenzene 2B. Integration of Signal Areas Integral Step eights v The area under each signal in a 1 NMR spectrum is proportional to the number of hydrogen atoms producing that signal v It is signal area (integration), not signal height, that gives information about the number of hydrogen atoms R a O a b b b 2 a 3 b a b 2. oupling (Signal Splitting) v oupling is caused by the magnetic effect of nonequivalent hydrogen atoms that are within 2 or 3 bonds of the hydrogens producing the signal v The n+1 rule Rule of Multiplicity: If a proton (or a set of magnetically equivalent nuclei) has n neighbors of magnetically equivalent protons, its multiplicity is n + 1 4
v Examples (1) b a b l b a a : multiplicity = 3 + 1 = 4 (a quartet) b : multiplicity = 2 + 1 = 3 (a triplet) (2) a b l l l b a : multiplicity = 2 + 1 = 3 (a triplet) b : multiplicity = 1 + 1 = 2 (a doublet) v (3) Examples b a b Br b b b b a : multiplicity = 6 + 1 = 7 (a septet) b : multiplicity = 1 + 1 = 2 (a doublet) Note: All b s are chemically and magnetically equivalent. v Pascal s Triangle Use to predict relative intensity of various peaks in multiplet Given by the coefficient of binomial expansion (a + b) n singlet (s) 1 doublet (d) 1 1 triplet (t) 1 2 1 quartet (q) 1 3 3 1 quintet 1 4 6 4 1 sextet 1 5 10 10 5 1 5
v Pascal s Triangle For For a b Br Br l l a b l Br l Br Due to symmetry, a and b are identical Þ a singlet a b Þ two doublets 3. ow to Interpret Proton NMR Spectra 1. ount the number of signals to determine how many distinct proton environments are in the molecule (neglecting, for the time being, the possibility of overlapping signals) 2. Use chemical shift tables or charts to correlate chemical shifts with possible structural environments 2014 by John Wiley & Sons, Inc. All rights reserved. 3. Determine the relative area of each signal, as compared with the area of other signals, as an indication of the relative number of protons producing the signal 4. Interpret the splitting pattern for each signal to determine how many hydrogen atoms are present on carbon atoms adjacent to those producing the signal and sketch possible molecular fragments 5. Join the fragments to make a molecule in a fashion that is consistent with the data v Example: 1 NMR (300 Mz) of an unknown compound with molecular formula 3 7 Br 6
v Three distinct signals at ~ d3.4, 1.8 and 1.1 ppm Þ d3.4 ppm: likely to be near an electronegative group (Br) d (ppm): 3.4 1.8 1.1 Integral: 2 2 3 d (ppm): 3.4 1.8 1.1 Multiplicity: triplet sextet triplet 2 's on adjacent 5 's on adjacent 2 's on adjacent omplete structure: most downfield signal Br 2 's from integration triplet 2 2 3 2 's from integration sextet most upfield signal 3 's from integration triplet 7
4. Shielding and Deshielding of Protons: More About hemical Shift Protons absorb at different NMR frequencies depending on the electron density around them and the effects of local induced magnetic fields. The magnetic field associated with a spinning proton The spinning proton resembles a tiny bar magnet 2014 by John Wiley & Sons, Inc. All rights reserved. 2014 by John Wiley & Sons, Inc. All rights reserved. 2014 by John Wiley & Sons, Inc. All rights reserved. 8
2014 by John Wiley & Sons, Inc. All rights reserved. Protons of ydrogen Atoms in Alkyl - Groups The applied magnetic field causes s electrons to circulate in a way that induces a local magnetic field. The hydrogen of a - bond experiences a net smaller magnetic field than the applied field. The proton is said to be shielded from the applied magnetic field. The chemical shift for hydrogens of unsubstituted alkanes is typically in the range of d 0.8 1.8. 2014 by John Wiley & Sons, Inc. All rights reserved. 9
Protons of ydrogens Near Electronegative Groups Electronegative groups draw electron density away from nearby hydrogen atoms. Electronegative groups diminish the shielding of protons by circulating s electrons. The proton is said to be deshielded from the applied magnetic field. The chemical shift for hydrogens bonded to a carbon bearing an oxygen or halogen atom is typically in the range of d 3.1 4.0. 2014 by John Wiley & Sons, Inc. All rights reserved. 2014 by John Wiley & Sons, Inc. All rights reserved. 2014 by John Wiley & Sons, Inc. All rights reserved. 10
2014 by John Wiley & Sons, Inc. All rights reserved. 2014 by John Wiley & Sons, Inc. All rights reserved. 11
Electronegativity omplete structure: most downfield signal Br 2 2 3 most upfield signal 2 's from integration triplet 2 's from integration sextet 2014 by John Wiley & Sons, Inc. All rights reserved. 3 's from integration triplet If we were to consider only the relative electronegativities of carbon in its three hybridization states, we might expect the following order of protons attached to each type of carbon: In fact, protons of terminal alkynes absorb between d 2.0 and d 3.0, and the order is (higher frequency) sp 2 < sp < sp 3 (lower frequency) (higher frequency) sp < sp2 < sp 3 (lower frequency) 2014 by John Wiley & Sons, Inc. All rights reserved. 2014 by John Wiley & Sons, Inc. All rights reserved. 12
v Shielding and deshielding by circulation Protons of ydrogen Atoms Near p Electrons of p electrons The p electrons in alkenes, alkynes, and aromatic rings also circulate as to generate an induced, local magnetic field Whether shielding or deshielding occurs depends on the location of the protons in the induced magnetic field. 2014 by John Wiley & Sons, Inc. All rights reserved. Protons of ydrogen Atoms Near p Electrons The hydrogens of benzene absorb at d 7.27. ydrogens bonded to substituted benzene rings have chemical shifts in the range of d 6.0 8.5. The chemical shift of alkene hydrogens is typically in the range of d 4.0 6.0. The chemical shift of an alkyne hydrogen is typically in the range of of d 2.5 3.1. 2014 by John Wiley & Sons, Inc. All rights reserved. 13
e.g. d c Aldehydes b a d (ppm) a & b : 7.9 & 7.4 (deshielded) c & d : 0.91 1.2 (shielded) R Electronegativity effect + Anisotropy effect Þ d = 8.5 10 ppm (deshielded) O 2014 by John Wiley & Sons, Inc. All rights reserved. 2014 by John Wiley & Sons, Inc. All rights reserved. 5. hemical Shift Equivalent and Nonequivalent Protons v Two or more protons that are in identical environments have the same chemical shift and, therefore, give only one 1 NMR signal v hemically equivalent protons are chemical shift equivalent in 1 NMR spectra 5A. omotopic and eterotopic Atoms v If replacing the hydrogens by a different atom gives the same compound, the hydrogens are said to be homotopic v omotopic hydrogens have identical environments and will have the same chemical shift. They are said to be chemical shift equivalent 14
same compounds Br Br Br Ethane v The six hydrogens of ethane are homotopic and are, therefore, chemical shift equivalent v Ethane, consequently, gives only one signal in its 1 NMR spectrum Br Br Br same compounds v If replacing hydrogens by a different atom gives different compounds, the hydrogens are said to be heterotopic v eterotopic atoms have different chemical shifts and are not chemical shift equivalent same compounds Þ these 3 s of the 3 group are homotopic Þ the 3 group gives only one 1 NMR signal l l l Br Br Br l Br Br Br l different compounds Þ heterotopic These 2 s are also homotopic to each other Br v 3 2 Br two sets of hydrogens that are heterotopic with respect to each other two 1 NMR signals 15
v Other examples v Other examples 3 (1) 3 Þ 2 1 NMR signals (3) 3 3 3 Þ 3 1 NMR signals (2) Þ 4 1 NMR signals 3 v Application to 13 NMR spectroscopy Examples (3) Þ 5 13 NMR signals (1) 3 3 Þ 1 13 NMR signal O O (2) 3 3 Þ 4 13 NMR signals (4) O O Þ 4 13 NMR signals 16
5B. Enantiotopic and Diastereotopic ydrogen Atoms v If replacement of each of two hydrogen atoms by the same group yields compounds that are enantiomers, the two hydrogen atoms are said to be enantiotopic v Enantiotopic hydrogen atoms have the same chemical shift and give only one 1 NMR signal: enantiotopic 3 Br 3 G Br enantiomers 3 G Br v If replacement of each of two hydrogen atoms by the same group yields compounds that are diastereomers, the two hydrogen atoms are said to be diastereotopic O chirality 3 center 3 b G O 3 3 b a O diastereotopic 3 3 G a v Except for accidental coincidence, diastereotopic protons do not have the same chemical shift and give rise to different 1 NMR signals. (the difference in chemical shift may be small) diastereomers 17
Neuman projections help us see that diastereotopic experience different environments. In every conformation a and b experience different environments. They give rise to signals with different chemical shifts. G Br b a Br b a diastereotopic Br G diastereomers 6. Spin Spin oupling: More About Signal Splitting and Nonequivalent or Equivalent Protons Signal splitting arises due to spinspin coupling. Spin-spin coupling effects are transferred primarily through bonding electrons and lead to spin-spin splitting. 2014 by John Wiley & Sons, Inc. All rights reserved. 18
6A. Vicinal oupling v Vicinal coupling is coupling between hydrogen atoms on adjacent carbons (vicinal hydrogens), where separation between the hydrogens is by three s bonds v Vicinal coupling between heterotopic protons generally follows the n + 1 rule. v Exceptions to the n + 1 rule can occur when diastereotopic hydrogens or conformationally restricted systems are involved a b 3 J or vicinal coupling These hydrogens are more than 3 bonds away from any other hydrogens. These hydrogens are more than 3 bonds away from any other hydrogens. n = 0 n + 1 = 1 (singlet) Recall These hydrogens are spin-spin coupled to two protons. n = 2 n + 1 = 3 (triplet) These hydrogens are spin-spin coupled to two protons. n = 2 n + 1 = 3 (triplet) These hydrogens are spin-spin coupled to five protons. n = 0 n + 1 = 1 (singlet) n = 5 n + 1 = 6 (sextet) 19
TE EMIAL SIFT OF PROTON A IS AFFETED BY TE SPIN OF ITS NEIGBORS aligned with B o opposed to B o 50 % of +1/2-1/2 molecules A A 50 % of molecules B o downfield neighbor aligned upfield neighbor opposed At any given time about half of the molecules in solution will have spin +1/2 and the other half will have spin -1/2. 6B. oupling onstants Recognizing Splitting Patterns Protons that are coupled share a coupling constant (J). oupling constants are determined by measuring the distance (in hertz) between each peak of a signal. A vicinal coupling constant is 6-8 hertz. 20
X a a b b b If we measure the separation of peaks for two spin-spin coupled hydrogens, they have the same coupling constant (J ab ). This is called reciprocity of coupling constants. 6. The Dependence of oupling onstants on Dihedral Angle v The magnitude of a coupling constant is related to the dihedral angle (f) between coupled protons. f v Karplus curve v Karplus curve examples v f ~0 o or 180 o Þ Maximum value v f ~90 o Þ ~0 z a b (axial, axial) b a f = 180º J a,b = 10-14 z b a (equatorial, equatorial) a b f = 60º J a,b = 4-5 z 21
v Karplus curve examples b a (equatorial, axial) b a f = 60º J a,b = 4-5 z 7. Proton NMR Spectra and Rate Processes Think of NMR like a camera with a slow shutter speed. Just as a photo will blur for an object moving rapidly, the NMR spectrum of a fast molecular process will be blurred. Examples of rapid processes that blur NMR spectra hemical Exchange of Protons onformational hanges hemical Exchange auses Spin Decoupling We do not typically observe splitting between protons of the O and 2 group in ethanol. 22
hemical Exchange auses Spin Decoupling hemical Exchange auses Spin Decoupling Protons attached to electronegative atoms (such as O) with lone pairs of electrons can undergo rapid chemical exchange. These exchangable protons can be transferred rapidly from one molecule to another. The exchange is so rapid that the hydroxyl proton does not couple with - protons. Rapid exchange causes spin decoupling. Spin decoupling is seen in the 1 NMR spectra of alcohols, amines, and carboxylic acids. The signals of N and O protons are normally unsplit and broad. hemical Exchange auses Spin Decoupling Protons that undergo rapid chemical exchange will exchange with D from D 2 O. v Why don t we see coupling with the O proton, e.g. 2 O (triplet?) Because the acidic protons are exchangeable about 10 5 protons per second (residence time 10-5 sec), but the NMR experiment requires a time of 10-2 10-3 sec. to take a spectrum, usually we just see an average. Thus, O protons are usually a broad singlet. If placed in D 2 O, the 1 NMR signal from these protons will disappear. This is an easy method for identifying 1 NMR signals produced by exchangeable protons. 2014 by John Wiley & Sons, Inc. All rights reserved. 23
+ 50 Proton NMR Spectra and Rate Processes v Protons of alcohols (RO) and amines (RN 2 ) may appear over a wide range from 0.5 5.0 ppm ydrogen-bonding is the reason for this range - 30 2014 by John Wiley & Sons, Inc. All rights reserved. onformational hanges At temperatures near room temperature, - single bonds can rotate rapidly. When we measure the 1 NMR spectra of compounds with single bonds that allow rotation, the spectra we obtain often reflects the hydrogen atoms in their average environment. onformational hanges We already said these protons are in the same environment But, imagine for a moment that - bonds didn t rotate If the - bond is frozen, these three hydrogens are not equivalent! 24
onformational hanges owever, the - bond rotates ~1 million times per second. Thus, they are all in the same average environment. onformational hanges At room temperature, cyclohexane chair conformations interconvert so fast that we only see one 1 NMR signal. onformational hanges At cold temperatures, cyclohexane chair conformations are slow. onformational hanges At 100 o, undecadeuteriocyclohexane gives only two 1 NMR signals of equal intensity. The 1 NMR spectrum becomes very complicated due to complex spin-spin couplings between non-equivalent axial and equatorial protons. Interconversions of the two conformations is slow at this temperature. (The deuterons are invisible to the 1 NMR) 25
8. arbon-13 NMR Spectroscopy 8A. Interpretation of 13 NMR Spectra v Unlike 1 with natural abundance ~99.98%, only 1.1% of carbon, namely 13, is NMR active 8B. One Peak for Each Magnetically Distinct arbon Atom Each distinct carbon produces one signal in a 13 spectrum. Splitting of 13 signals into multiple peaks is not observed in routine 13 spectra. The odds of two 13 atoms (~1% abundance) being next to each other, to split each other, is very low! Although 13-13 signal splitting does not occur in 13 NMR, 1 atoms attached can split 13 NMR signals To simplify the 13 NMR spectrum, such 1-13 splitting is instrumentally eliminated. A spectrum with 1-13 splitting eliminated is called broadband proton decoupled. v Example: 2-Butanol Proton-coupled 13 NMR spectrum 3 2 3 O 26
v Example: 2-Butanol Proton-decoupled 13 NMR spectrum 3 2 3 O In a boradband proton-decoupled 13 NMR spectrum, each carbon atom in a distinct environment gives a signal consisting of only one peak. Most 13 NMR spectra are obtained in the simplified broadband decoupled mode first and then in modes that provide information from the 1-13 couplings. 8. 13 hemical Shifts v Decreased electron density around an atom deshields the atom from the magnetic field and causes its signal to occur further downfield (higher ppm, to the left) in the NMR spectrum v Relatively higher electron density around an atom shields the atom from the magnetic field and causes the signal to occur upfield (lower ppm, to the right) in the NMR spectrum Electronegative groups deshield the carbons to which they are attached. The lower the electron density in the vicinity of a given carbon, the less the carbon will be shielded. (a) (b) (c) l 2 3 O 1-hloro-2-propanol (b) (a) (c) 27
The Dl 3 solvent peaks at d 77 should be disregarded. (a) (b) (c) l 2 3 O 1-hloro-2-propanol solvent 2014 by John Wiley & Sons, Inc. All rights reserved. BROMOYLOEXANE 1,2-DILOROBENZENE a b c l a b c l 28
8D. DEPT 13 Spectra v Distortionless Enhancement by Polarization Transfer v DEPT 13 NMR spectra indicate: how many hydrogen atoms are bonded to each carbon the chemical shift information contained in a broadband proton-decoupled 13 NMR spectrum. v The carbon signals in a DEPT spectrum are classified as 3, 2,, or accordingly 29
(a) (b) (c) l 2 3 (a) (b) (c) v The broadband proton-decoupled 13 NMR spectrum of methyl methacrylate O 1-hloro-2-propanol (a) (b) (c) l 2 3 O 1-hloro-2-propanol (a) (b) (c) 9. Two-Dimensional (2D) NMR Techniques v OSY 1 1 correlation spectroscopy v ETOR eteronuclear correlation spectroscopy 2014 by John Wiley & Sons, Inc. All rights reserved. 30
9A. 1 1 OSY ross-peak orrelations 9B. 1 13 eteronuclear orrelation ross-peak orrelations 2 3 1 4 v OSY of 2-chlorobutane 2 3 1 4 v ETOR of 2-chlorobutane 4 1 4 3 1 3 2 2 Quiz 1 Quiz 2 2014 by John Wiley & Sons, Inc. All rights reserved. 2014 by John Wiley & Sons, Inc. All rights reserved. 31
Quiz 3 Quiz 4 2014 by John Wiley & Sons, Inc. All rights reserved. 2014 by John Wiley & Sons, Inc. All rights reserved. 10. An Introduction to Mass Spectrometry Mass spectrometry (MS) involves formation of ions separation of the ions Based on detection of the ions mass and charge Relative abundance of each detected ion 3 2 3 Represents the formula weight of the detected ions. m/z is the mass (m) to charge (z) ratio. base peak is the tallest peak molecular ion Small peaks with m/z +1 or +2 due to heavy isotopes z usually = +1 32
11. Formation of Ions: Electron Impact Ionization v In the mass spectrometer, a molecule in the gaseous phase under low pressure is bombarded with a beam of high-energy electrons (70 ev or ~ 1600 kcal/mol) EI v This beam can dislodge an electron from a molecule to give a radical cation which is called the molecular ion, M + or more accurately M 70 ev e - M radical cation with net +1 charge 12. Depicting the Molecular Ion 3 2 3 Radical cations from ionization of nonbonding on p electron Sometimes the choice of where to locate the radical cation is arbitrary 3 2 3 When a molecule contains O, N, or a p bond, we place the odd electron and charge there. 3 O 3 N 3 3 2 2 3 Methanol Trimethylamine 1-Butene 3 O 3 N 3 3 2 2 3 Methanol Trimethylamine 1-Butene 33
v Ionization potentials of selected molecules ompound Ionization Potential (ev) 3 ( 2 ) 3 N 2 8.7 6 6 (benzene) 9.2 2 4 10.5 3 O 10.8 2 6 11.5 4 12.7 34
13A. Fragmentation by leavage at a Single Bond v When a molecular ion fragments, it will yield a neutral radical (not detected) and a carbocation (detected) with an even number of electrons v The fragmentation will be dictated to some extent by the stability of the carbocation generated: Ar 2 + > 2 = 2 + > 3 o > 2 o > 1 o > 3 + v e.g. This fragmentation pathway is more predominant More stable carbocation Less stable carbocation 35
v As the carbon skeleton becomes more highly branched, the intensity of the molecular ion peak decreases v Butane vs. isobutane 70eV a a (43) + 3 Fragmentation e - b M + (58) b + 2 3 (29) 70eV e - M + (58) (43) + 3 2014 by John Wiley & Sons, Inc. All rights reserved. m/z = 29 13B. Fragmentation of Longer hain and Branched Alkanes m/z = 44 Larger fragment peak m/z = 15 v Octane vs. isooctane (85) + (71) + Smaller fragment peak Molecular ion peak M + (114) (57) + (43) + M + (114) (57) + 36
v Masses are usually rounded off to whole numbers assuming: = 1, = 12, N = 14, O = 16, F = 19 etc. [ 8 18 ] (M +, 114) fragmentation - 3 2 (29) - 3 2 2 (29+14) Molecular ion (parent peak) [ 6 13 ] (85) [ 5 11 ] (71) Daughter ions v Partial MS of octane ( 8 18, M = 114) 57 14 ( 2 ) 29 ( 3 2 ) 85 71 114 M + 2014 by John Wiley & Sons, Inc. All rights reserved. 2014 by John Wiley & Sons, Inc. All rights reserved. 13. Fragmentation to Form Resonance-Stabilized ations Alkenes ionize and undergo fragmentation to yield resonance-stabilized allylic carbocations. 2014 by John Wiley & Sons, Inc. All rights reserved. 37
16. Fragmentation to Form Resonance-Stabilized ations v Alkenes Important fragmentation of terminal alkenes t Allyl carbocation (m/e = 41) Ÿ Alkene Fragmentation Fairly prominent M + Fragment ions of n2n + and n2n-1 + Terminal alkenes lose allyl cation if possible to form resonance-stabilized allylic cations R - R R R + (41) 2014 by John Wiley & Sons, Inc. All rights reserved. + [ 2 = 2 2 3 ] + 2 = 2 + 2 3 arbon-carbon bonds next to an atom with a lone pair of electrons readily break because the resulting carbocation is resonancestabilized. arbon-carbon bonds next to the carbonyl group of an aldehyde or ketone break to form a resonance-stabilized acylium ion. acylium ion 38
Alkyl-substituted benzenes undergo loss of a hydrogen atom or methyl group to yield the tropylium ion Monosubstituted benzenes with other than alkyl groups lose their substituent to yield a phenyl cation. tropylium ion m/z = 91 Y = halogen, NO 2, etc. phenyl ion m/z = 77 13D. Fragmentation by leavage of Two Bonds 1,2-elimination: O + 2 O v Alcohols frequently show a prominent + peak at M - 18. This corresponds to the loss of a molecule of water M (M - 18) May lose 2 O by 1,2- or 1,4- elimination 1,4-elimination: M O O + 3 2 (M - 18) + 2 O 39
v Alcohols Most common fragmentation: loss of alkyl groups a O b M + (74) b 3 2 + O O (m/e = 45) 2014 by John Wiley & Sons, Inc. All rights reserved. 2014 by John Wiley & Sons, Inc. All rights reserved. v Aldehydes M + peak usually observed but may be fairly weak hexanal ommon fragmentation pattern t a-cleavage R O R + R O acylium ion + O (m/e = 29) 2014 by John Wiley & Sons, Inc. All rights reserved. 40
Mc Lafferty Rearrangement v Ketones a-cleavage a b O a b 2014 by John Wiley & Sons, Inc. All rights reserved. 41
O O ii O i i 2º radical (m/e = 86) observed ii O O 1º radical (m/e = 114) NOT observed v haracteristics of McLafferty rearrangements 1. No alkyl migrations to =O, only migrates O R O R v haracteristics of McLafferty rearrangements 2. 2 o is preferred over 1 o O ii i O 2º radical X R O not O 1º radical 42
v Aromatic hydrocarbons very intense M + peaks characteristic fragmentation pattern (when an alkyl group attached to the benzene ring): tropylium cation 2 3 3 + rearrangement benzyl cation tropylium cation (m/e = 91) 2014 by John Wiley & Sons, Inc. All rights reserved. 14. Isotopes in Mass Spectra v 13 and 12 About 1.1% of all carbon atoms are the 13 isotope v About 98.9% of the methane molecules in the sample will contain 12, and the other 1.1% will contain 13 43
v Example onsider 100 molecules of 4 1 1 12 1 1 1 1 13 1 1 1 1 12 2 1 1 1 12 1 1 M : 16 1 1 13 1 1 M + 1 = 17 1 1 12 2 1 M : 16 M + 1 = 17 12 : 100 13 : 1.11 1.11 molecules contain a 13 atom 4x0.016 = 0.064 molecules contain a 2 atom 1 : 100 2 : 0.016 Intensity of M + 1 peak: + 1.11+0.064=1.174% of the M peak + relative ion abundance M 100 m/z + + M +1 1.17 v Some elements that are common in organic molecules have isotopes that differ by two atomic mass units. v These include 16 O and 18 O, 32 S and 34 S, 35 l and 37 l, and 79 Br and 81 Br. v It is particularly easy to identify the presence of chlorine or bromine using mass spectrometry because multiple isotopes of chlorine and bromine are relatively abundant 44
The natural abundance of 35 l is 75.5% and that of 37 l is 24.5% In the mass spectrum for a sample containing chlorine, we would expect to find peaks separated by two mass units, in an approximately 3 : 1 (75.5% : 24.5%) ratio for the molecular ion or any fragments that contain chlorine The natural abundance of 79 Br is 51.5%, and that of 81 Br is 49.5% In the mass spectrum for a sample containing bromine we would expect to find peaks separated by two mass units in an approximately 1 : 1 ratio (49.5% : 51.5%) 45
14A. igh-resolution Mass Spectrometry Low-resolution mass spectrometers measure m/z to the nearest whole number unit. igh-resolution mass spectrometers an measure m/z values to three or four decimal places. Because the masses of the actual masses of individual atoms are not integers, a high-resolution mass spectrum can be used to determine molecular formulas. v Example 1 O 2, N 2 4 and 3 O all have M.W. of 32 (by MS), but accurate masses are different t O 2 = 2(15.9949) = 31.9898 t N 2 4 = 2(14.0031) + 4(1.00783) = 32.0375 t 4 O = 12.00000 + 4(1.00783) + 15.9949 = 32.0262 46
v Example 2 Both 3 8 O and 2 4 O 2 have M.W. of 60 (by low-res MS), but their accurate masses are different. t 3 8 O = 60.05754 t 2 4 O 2 = 60.02112 Low Resolution Mass 47
igh Resolution Mass 48
49
50
Identify which one of the following isomers of 6 14 has the -13 NMR below. A) 3 2 2 2 2 3 ) ( 3) 2 ( 3) 2 B) 3 2 2( 3) 2 D) 3 2( 3) 3 A) 3 2 2 2 2 3 ) ( 3) 2( 3) 2 B) 3 2 2( 3) 2 D) 3 2( 3) 3 51
52
O 72 m/e = 72? 53