HC4_MAP_NMR1_JCR.pdf

Molecular Analytics Premaster L4 Nuclear Magnetic Resonance Spectroscopy Part 1: 1H-NMR Jesper C. Ruiter Sectie BioAnalytische Chemie Vrije Universiteit Amsterdam [email protected] Friday 29 September 2023 Previously on MAP • What do the following terms mean: molecular mass, mass number, nominal...

Molecular Analytics Premaster L4 Nuclear Magnetic Resonance Spectroscopy Part 1: 1H-NMR Jesper C. Ruiter Sectie BioAnalytische Chemie Vrije Universiteit Amsterdam [email protected] Friday 29 September 2023 Previously on MAP • What do the following terms mean: molecular mass, mass number, nominal mass, exact mass? • What is m/z? • What are the molecular ion and base peak? • What is mass resolution? What is mass accuracy and how do we calculate both? • What is a mass spectrometer? • What is a mass spectrum? • How does the ionisation method EI work? What is the general reaction equation of EI? • How do heterolytical and homolytical fragmentations look like? Can you draw the mechanisms for both for alkyl halides, alcohols, ethers and ketones? • How does McLafferty rearrangement look like? Can you draw the mechanism for a ketone compound? • What is a mass analyzer and how does the TOF mass analyzer work? • How are the formed and separated ions detected in MS? 2 Where are we? L1 L4 L5 L2 T2 T1 L3 3 Check-in 4 Check-in 5 Check-in 6 Learning objectives of today • Understanding what spin is and how you calculate the number of spin states from the nuclear spin quantum number • Understanding why spin states in an applied magnetic field have different energies • Understanding when the nuclear magnetic resonance phenomenon can occur • Understanding what shielding is and how electronegativity, hybridisation, and hydrogen bonding affect it • Knowing what chemical shift is and in what range different hydrogens of functional groups have their chemical shift • Understanding anisotropy and what the effects of it are on hydrogens • Knowing what the spin-spin splitting rule is and how to assign multiplet peaks to hydrogens in simple molecules • Knowing what a coupling constant is and how to calculate it • Identifying the structures of unknown compounds based on their 1H NMR spectrum 7 From the top: What is NMR spectroscopy? • • NMR is a spectroscopic technique to observe local magnetic fields around atomic nuclei Absorption of radio waves in the presence of magnetic fields is accomponied by a special type of nuclear transition, therefore, such type of spectroscopy is known as Nuclear Magnetic Resonance Spectroscopy NMR is a very powerful technique for structure elucidation 8 The electromagnetic spectrum in detail NMR region ~4 to 900 mHz 9 Nuclear spin states Many atomic nuclei have a property called spin: the nuclei behave as if they were spinning Any atomic nucleus that possesses either odd mass, odd atomic number, or both has a quantized spin angular momentum Common nuclei that possess spin include: 1𝐻, 2𝐻, 13𝐶, 14𝑁, 17𝑂, 𝑎𝑛𝑑 19𝐹 1 1 6 7 8 9 For each nucleus with spin, the number of allowed spin states it may adopt is quantized and determined by its nuclear spin quantum number I. There are 2I + 1 allowed spin states with integral differences ranging from +I to -I The individual spin states fit into the sequence: +I, (I-1), … (-I + 1), -I 10 Nuclear spin states of a proton A proton (hydrogen nucleus) has a spin quantum number I= 1 2 Question How many allowed spin states does a proton have? 11 Nuclear spin states of a proton A proton (hydrogen nucleus) has a spin quantum number I= 1 2 Question How many allowed spin states does a proton have? A proton has a spin quantum number of I = 1 2 1 Filling in the formula gives: 2 × 2 + 1 = 2 spin states Fitting the individual spin states into +I, (I-1), … (-I + 1), -I gives us 1 1 − 2 and + 2 12 Spin quantum numbers of common nuclei Element 1 1𝐻 2 1𝐻 12 6𝐶 13 6𝐶 Nuclear spin quantum number 1 2 1 0 1 2 1 Number of spin states 2 3 0 2 3 14 7𝑁 17 8𝑂 19 9𝐹 31 15𝑃 35 17𝐶𝑙 0 5 2 1 2 1 2 3 2 0 6 2 2 4 16 8𝑂 In the absence of an applied magnetic field: • All the spin states of a given nucleus are of equivalent energy (degenerate) • All of the spin states should be almost equally populated (in a collection of atoms) • With the same number of atoms having each of the allowed spins Question: What happens when we apply a magnetic field? 13 Nuclear magnetic moments Spin states are not of equivalent energy in an applied magnetic field The nucleus is a charged particle, and (any) moving charge generates a magnetic field of its own As a result, the nucleus has a magnetic moment (µ) generated by its charge and spin 14 Nuclear magnetic moments: hydrogen 1 1 A hydrogen nucleus may have a clockwise (+ 2) or counter clockwise (− 2) spin In an applied magnetic field, all protons have their magnetic moments either aligned with field or opposed to it 15 What about the energy? • The + 2 spin state is of lower energy (= aligned with the field) • The − 2 spin state is of higher energy (= opposed to the field) 1 1 N S N S N N S − 1 2 against field β spin states + 1 2 with field α spin states E S B0 Aligned: Oppossed: Lower energy Higher energy No field Applied field 16 Absorption of energy The nuclear magnetic resonance phenomenon occurs when nuclei aligned with an applied field are induced to absorb energy and change their spin orientation Inducing transitions between α and β states = NMR spectroscopy 1 − 2 + hν 1 2 Quantized process Energy absorbed must equal the energy difference between the two states involved 1 − 2 + 1 2 Magnetic field direction 𝐸𝑎𝑏𝑠𝑜𝑟𝑏𝑒𝑑 = 𝐸−1𝑠𝑡𝑎𝑡𝑒 − 𝐸+1𝑠𝑡𝑎𝑡𝑒 = ℎν) 2 2 17 Strength of the magnetic field B0 • ΔE is a function of the strength of the applied magnetic field B0 ∆𝐸 = 𝑓(𝐵0 ) 18 The stronger the applied magnetic field, the greater the ΔE between the possible spin states Magnetogyric ratio γ (1/2) • The magnitude of the energy-level separation also depends on the nucleus involved • Each nucleus has a different ratio of magnetic moment to angular momentum • This ratio is called → magnetogyric ratio γ • γ is constant for each nucleus • Determines the energy dependence on the magnetic field ∆𝐸 = 𝑓 𝛾𝐵0 = ℎν 19 Magnetogyric ratio γ (2/2) • Angular momentum of the nucleus is quantized in units ℎ 2𝜋 ℎ ∆𝐸 = 𝛾 𝐵0 = ℎν 2𝜋 • Solving for frequency of the absorbed energy gives: 𝛾 ν= 𝐵0 2𝜋 • γ for 1H is 267.53 radians/Tesla, if our B0 = 1 Tesla then we find: ν= 267.53 × 1 = 42.6 𝑀𝐻𝑧 2𝜋 An (unshielded) proton should absorb radiation of frequency 42.6 MHz in a field strength of 1 Tesla 20 Table of frequencies and field strengths where selected nuclei have their nuclear resonances 21 The mechanism of absorption (resonance) • Protons absorb energy because they begin to precess in an applied magnetic field • Think of a child’s spinning top. The top begins to wobble (precess) about its axis • A spinning nucleus behaves similarly under the influence of an applied magnetic field 22 Resonance (1/2) When the magnetic field is applied, the nucleus begins to precess about its own axis of spin with angular frequency ω (Larmor frequency) The frequency at which a proton precesses is directly proportional to the strength of the applied magnetic field The stronger the applied field, the higher the angular frequency ω Precession generates an oscillating electric field of the same frequency (since the nucleus is charged) If radiofrequency waves of this frequency are supplied to the precessing proton, the energy can be absorbed 23 Resonance (2/2) When hν = hω the two fields can couple Energy can be transferred from the incoming radiation to the nucleus → causing a spin change This condition is called resonance The nucleus is said to have resonance with the incoming electromagnetic wave 24 Population densities of nuclear spin states • For a proton, if the applied magnetic field has a strength of 1.41 Tesla, resonance occurs ~60 MHz • Using ∆𝐸 = ℎν ⇒ 2.39 𝑘𝐽 ∙ 𝑚𝑜𝑙 (energy difference between the two spin states) • Thermal energy resulting from room temperature is sufficient to populate both energy levels (since ∆𝐸 is small) • There is a slight excess of nuclei in the lower-energy spin state • The magnitude of this difference can be calculated using the Boltzmann distribution equations 25 Ratio of nuclear spins in the upper and lower levels • Boltzmann ratio of nuclear spins in the upper and lower levels: 𝑁𝑢𝑝𝑝𝑒𝑟 = 𝑒 −∆𝐸/𝑘𝑇 = 𝑒 −ℎν/𝑘𝑇 𝑁𝑙𝑜𝑤𝑒𝑟 • h = 6.626∙10-34 J∙s (Planck’s constant), k = 1.38∙10-23 J/K, ΔE = energy difference between the upper and lower energy states, T = absolute temperature in K Question Calculate the Boltzmann ratio for an NMR instrument operating at 60 MHz, T = 298 K 26 Ratio of nuclear spins in the upper and lower levels Question Calculate the Boltzmann ratio for an NMR instrument operating at 60 MHz, T = 298 K 𝑁𝑢𝑝𝑝𝑒𝑟 − −∆𝐸/𝑘𝑇 = 𝑒 =𝑒 𝑁𝑙𝑜𝑤𝑒𝑟 6.626∙10−34 ×60∙106 𝐻𝑧 1.38∙10−23 ×298 = 0.999990 1000000 1000000 = 0.999990 ⇒ 𝑁𝑙𝑜𝑤𝑒𝑟 = = 1000009.67 𝑁𝑙𝑜𝑤𝑒𝑟 0.999990 In other words, in ~2 million nuclei, there are only 9 more nuclei in the lower spin state. 27 Excess population The excess nuclei are the ones that allow us to observe resonance When the 60 MHz, from previous example, radiation is applied, it induces both upward and downward transitions If Nupper = Nlower no net signal is observed → saturation If the power of the radiofrequency signal is too high, saturation is achieved quickly The small excess of nuclei in the lower spin state is important for NMR spectroscopy Sensitive NMR instrumentation is required to detect the signal 28 Increasing the frequency increases the population in Nlower Increasing the operating frequency of the NMR instrument, increases the ΔE between the two states, causing an increase in excess This is also the reason why modern NMR instruments are designed with increasingly higher operating frequencies The sensitivity of the instrument is increased. Resonance signals are higher, because more nuclei can 100 MHz 400 MHz 800 MHz undergo transition at higher frequency 29 Shielding (1/2) • Not all protons in a molecule have resonance at exactly the same frequency • Protons in a molecule are surrounded by electrons • Exist in slightly different electronic (magnetic) environments from one another • The valence-shell electron densities vary between protons • The protons are shielded by the electrons that surround them 30 Shielding (2/2) In an applied magnetic field, the valence electrons of the proton are caused to circulate This circulation generates a counter magnetic field that opposes the applied magnetic field This effect is called diamagnetic shielding or diamagnetic anisotropy 31 Chemical shift (1/2) • Due to diamagnetic anisotropy, each proton in a molecule is shielded from the applied magnetic field to an extent that depends on the electron density surrounding it The greater the electron density around a nucleus, the greater the induced counter field that opposes the applied field The counter field that shields a nucleus diminishes the net applied magnetic field that the nucleus experiences → the nucleus precesses at a lower frequency This also means that it absorbs radiofrequency radiation at this lower frequency Each proton in a molecule is in a slightly different chemical environment and, therefore, has a slightly different amount of electronic shielding → results in slightly different resonance frequency 32 Chemical shift (2/2) These differences are very small (slightly >1 ppm). Therefore, a reference compound is used For 1H and 13C this compound is TetraMethylSilane = Si(CH3)4 or TMS Dimensionless chemical shift (δ) expressed in ppm: 𝛿= ν − ν𝑟𝑒𝑓 × 106 𝑝𝑝𝑚 ν𝑟𝑒𝑓 The chemical shift expresses the amount by which a proton resonance is shifted from TMS Chemical shifts are reported in delta (δ) units or ppm The resonance of the protons in TMS are (by definition) at exactly 0.00 ppm 33 Describing chemical shift B A δ More deshielded, downfield More shielded, high field Frequency Magnetic field B is deshielded relative to A or the shielding of A is bigger 34 Chemical environment & chemical shift Different protons have different chemical shifts with characteristic values It is very useful to know these ranges of chemical shifts for these common type of protons 35 How to obtain an NMR spectrum? • Early days: Continuous-Wave (CW) instruments • Modern days: Pulsed Fourier Transform (FT) Instrument FT Excite all transitions of a When the pulse is stopped, the FT to get the frequency particular nucleus at once excited nuclei lose their spectrum with a short burst of energy excitation energy (relax) and → pulse return to their original spin state 36 Coffee time☺ 37 Chemical equivalence (1/2) All protons found in chemically identical environments within a molecule are chemically equivalent (same chemical shift) Molecules give rise to one NMR absorption peak All protons are chemically equivalent Molecules give rise to two NMR absorption peaks Two different sets of chemically equivalent protons 38 Chemical equivalence (2/2) 39 Chemical equivalence (2/2) Molecules give rise to three NMR absorption peaks Three different sets of chemically equivalent protons 40 Integrals and integration The NMR spectrum reveals how many of each type of protons are contained within the molecule The area under each peak = proportional to the number of hydrogens generating that peak The NMR spectrometer can electronically integrate the area under each peak A vertical rising line, integral, is traced over each peak by the NMR spectrometer 41 Example: integrated spectrum of benzyl acetate 42 Local diamagnetic shielding (1/3) Electronegativity effects The chemical shift increases as the electronegativity of the attached element increases Compound CH3X CH3F CH3OH CH3Cl CH3Br CH3I CH4 Si(CH3)4 Element X F O Cl Br I H Si Electronegativity of X 4.0 3.5 3.1 2.8 2.5 2.1 1.8 Chemical shift (δ) 4.26 3.40 3.05 2.68 2.16 0.23 0 Multiple substituents have a stronger effect than a single substituent. Influence of the substituent decreases rapidly with distance (very little effect on protons that are more than three carbons distant) No. Substituents/ distance CHCl3 CH2Cl2 CH3Cl -CH2Br -CH2-CH2Br -CH2-CH2CH2Br Chemical shift (δ) 7.27 5.30 3.05 3.30 1.69 1.25 43 Local diamagnetic shielding (2/3): hybridization • Hydrogens attached to purely sp3 carbon (C-CH3) atoms have resonance in the range from 0 – 2 ppm • Methyl, methylene and methine hydrogens follow this relationship 44 Local diamagnetic shielding (2/3): hybridization 2 • Vinyl hydrogens (-C=C-H) have resonance in the range from 4.5 to 7 ppm • In a sp2 1s C-H bond, the C atom has more s character → makes it more electronegative than an sp3 C atom. • Results in less shielding for the H nucleus than in an sp3 1s bond • Thus vinyl hydrogens have a greater chemical shift (5 to 6 ppm) • Aromatic hydrogens appear further downfield (7 to 8 ppm) 45 Local diamagnetic shielding (2/3): hybridization 3 • Acetylenic hydrogens (C-H, sp-1s) appear around 2 tot 3 ppm due to anisotropy • It would be expected that acetylenic hydrogens have a greater chemical shift than vinyl hydrogens, but this is not observed 46 Local diamagnetic shielding (3/3) Hydrogen bonding and exchangeable hydrogens Protons than can exhibit hydrogen bonding (e.g., hydroxyl or amino protons) show extremely variable absorption positions over a wide range The more hydrogen bonding that takes place, the more deshielded a proton becomes Amount of hydrogen bonding is often a function of concentration and temperature The more concentrated the solution, the more molecules can come into contact with each other and hydrogen bond At high dilution (no H bonding) hydroxyl protons absorb near 0.5 – 1 ppm In concentrated solution, their absorption is closer to 4 – 5 ppm 47 Anisotropy…? 48 Anisotropy (1/2) Caused by the presence of an unsaturated system (one with π electrons) in the vicinity of protons Example: benzene When placed in a magnetic field, the π electrons in the aromatic ring are induced to circulate around the ring → ring current The moving electrons generate a magnetic field, which has a large enough spatial volume that it influences the shielding of benzene hydrogens 49 Anisotropy (2/2) Anisotropy is caused by the presence of π electrons. Below some common multiple bond sytems Anisotropy is the reason why the acetylenic hydrogens have a lower chemical shift than expected based on hybridisation effects. The hydrogens are shielded 50 Spin-spin splitting (n+1) rule (1/2) Question How many peaks will we observe in the NMR spectrum? 51 Spin-spin splitting (n+1) rule (1/2) Question How many peaks will we observe in the NMR spectrum? 5 peaks! 52 Spin-spin splitting (n+1) rule (2/2) Each proton ‘’senses’’ the number of equivalent protons (n) on the carbon atoms next to the one to which it is bonded Its resonance peak is split into (n+1) components Let us look at 1,1,2 trichloroethane The methine hydrogen (Ha) is next to a carbon bearing two methylene protons → two equivalent neighbours so (n = 2) and it is split into (n+1) = 3 peaks (triplet) The methylene hydrogens (Hb and Hc) are next to a carbon bearing one methine hydrogen → one neighbour so (n = 1) and it is split into (n+1) = 2 peaks (doublet) 53 NMR spectrum of 1,1,2-trichloroethane 54 Test: splitting of 2-nitropropane The spin-spin splitting gives a new type of structural information Question How many peaks will we observe in the NMR spectrum? It reveals how many hydrogens are adjacent to each type of hydrogen Rule of thumb: Hydrogens further than 3 bonds from each other do not couple with each other 55 Common splitting patterns The intensity ratios of multiplets derived from the n + 1 rule follow the entries in the mathematical mnemonic device called Pascal’s triangle 56 The origin of spin-spin splitting (1/3) Spin-spin splitting arises because hydrogens on adjacent carbon atoms can ‘’sense’’ one another Let’s look at an example: A hydrogen on carbon A can sense the spin direction of the hydrogen on carbon B 1 2 In some molecules the hydrogen on carbon B has spin + (X-type molecules) 1 2 In some molecules the hydrogen on carbon B has spin − (Y-type molecules) 57 The origin of spin-spin splitting (2/3) The chemical shift of proton A is influenced by the direction of the spin in proton B Proton A is said to be coupled to proton B 1 2 1 2 Its magnetic environment is affected by whether proton B has a + or − spin state Proton A absorbs at a slightly different chemical shift value in X-type molecules than in Ytype molecules X-type molecules: Proton A is slightly deshielded because the field of proton B is aligned with the magnetic field Y-type molecules: Proton A is slightly shielded because the field of proton B is aligned against the magnetic field 58 The origin of spin-spin splitting (3/3) ~ equal numbers of X- and Y-type molecules at any given time → two absorptions of nearly equal intensity are observed for proton A The resonance of proton A is said to have been split by proton B Phenomenon is called spin-spin splitting Two doublets will be observed in any situation of this type Except one in which protons A and B are identical by symmetry Shows only one peak 59 The coupling constant (1/2) • The distance between the peaks in a simple multiplet is called the coupling constant J • It is a measure of how strongly a nucleus is affected by the spin states of its neighbour • The spacing between the multiplet peaks is measured on the same scale as the chemical shift • Expressed in Hz 60 The coupling constant (2/2) For the interaction of most aliphatic protons in acyclic systems, the magnitudes of coupling constants are always near 7.5 Hz Two hydrogens on adjacent carbon atoms can be described as a three-bond interaction Usually abbreviated as 3J 2J, 4J, 5J and sometimes even further are possible, but these are mostly outside of the scope of this course The coupling constants of the groups of protons that split one another must be identical 61 Some 3J coupling constants 62 Alkanes Alkanes can have three different types of hydrogens (methyl, methylene, and methine) each of which appears in its own region of the NMR spectrum • R-CH3 0.7–1.3 ppm • R-CH2-R 1.2–1.4 ppm • R3CH 1.4–1.7 ppm 63 Alkenes Alkenes have two types of hydrogens: vinyl (attached directly to the double bond) and allylic hydrogens (attached to the α carbon) • C=C-H 4.5–6.5 ppm • C=C-CH2 1.6–2.6 ppm 64 Aromatic compounds Aromatic compounds have two characteristic types of hydrogens: aromatic ring hydrogens (benzene ring hydrogens) and benzylic hydrogens (attached to an adjacent carbon atom) 6.5–8.0 ppm 2.3–2.7 ppm 65 Alkynes Terminal alkynes (those with a triple bond at the end of a chain) will show acetylenic hydrogens, as well as the α hydrogens found on the carbon atom next to the triple bond • C≡C-H 1.7–2.7 ppm • C≡C-CH 1.6–2.6 ppm 66 Alkyl halides In alkyl halides, the α hydrogen (the one attached to the same carbon as the halogen) will be deshielded • -CH-I 2.0–4.0 ppm • -CH-Br 2.7–4.1 ppm • -CH-Cl 3.1–4.1 ppm • -CH-F 4.2–4.8 ppm 67 Alcohols In alcohols, both the hydroxyl proton and the α hydrogens (those on the same carbon as the OH) have characteristic chemical shifts • C-OH 0.5–5.0 ppm • CH-O-H 3.2–3.8 ppm 68 Ethers In ethers, the α hydrogens (those attached to the α carbon) are highly deshielded • R-O-CH 3.2–3.8 ppm 69 Amines Two characteristic types of hydrogens are found in amines: those attached to the nitrogen and those attached to the α carbon • R-N-H 0.5–4.0 ppm • -CH-N2.2–2.9 ppm 3.0 – 5.0 ppm 70 Nitriles In nitriles, only the α hydrogens (those attached to the same carbon as the cyano group) have a characteristic chemical shift • -CH-C≡N 2.1–3.0 ppm 71 Aldehydes Two types of hydrogens are found in aldehydes: the aldehyde hydrogen and the α hydrogens (those hydrogens attached to the same carbon as the aldehyde group • R-CHO 9.0–10.0 ppm • R-CH-CH=O 2.1–2.4 ppm 72 Ketones Ketones have only one distinct type of hydrogen atom: those attached to the α carbon 2.1 – 2.4 ppm 73 Esters Two distinct types of hydrogens are found in esters: those on the carbon atom attached to the oxygen atom in the alcohol part of the ester and those on the α carbon in the acid part of the ester 74 Carboxylic acids Carboxylic acids have the acidic proton (the one attached to the COOH group) and the α hydrogens (those attached to the same carbon as the carboxyl group) • R-COOH 11.0–12.0 ppm • -CH-COOH 2.1–2.5 ppm Hydrogen bonding and exchange may cause the peak to become broadened 75 Amides • • • R(CO)-N-H -CH-CONHR(CO)-N-CH 5.0–9.0 ppm 2.1–2.5 ppm 2.2–2.9 ppm 76 Nitroalkanes In nitroalkanes, α hydrogens, those hydrogen atoms that are attached to the same carbon atom to which the nitro group is attached, have a characteristically large chemical shift • -CH-NO2 4.1–4.4 ppm 77 Take home message You know the most important points of this lecture if you can answer the following questions: • What is the property spin? How can you calculate the number of spin states, provided you are given the nuclear spin quantum number? • Why have spin states different energies when an external magnetic field is applied? • When can the nuclear magnetic resonance phenomenon occur? • What is meant by shielding? What are the effects of electronegativity, hybridisation and hydrogen bonding on shielding? • What is chemical shift? Can you roughly sum up the chemical shifts of the common type of protons shown in slide 35? • What is anisotropy? What causes it? And what effect does it have on hydrogens in the vicinity? • What is the spin-spin splitting rule? Can you draw the multiplets of the ethyl (-CH2CH3) group? • What is a coupling constant? How do you find the value of this coupling constant in an 1H NMR spectrum? • How do you identify structures of unknown compounds based on their 1H NMR spectrum? (practise in the tutorial) 78 Next lecture Friday 6 October at 13:30–15:15 ‘’NMR part 2’’ NU-4C07 79 End of today Questions? 80