HC5_MAP_NMR2_JCR.pdf

Molecular Analytics Premaster L5 Nuclear Magnetic Resonance Spectroscopy Part 2: 13C-NMR Jesper C. Ruiter Sectie BioAnalytische Chemie Vrije Universiteit Amsterdam [email protected] Friday 6 October 2023 Previously on MAP • What is the property spin? How can you calculate the number of spin states...

Molecular Analytics Premaster L5 Nuclear Magnetic Resonance Spectroscopy Part 2: 13C-NMR Jesper C. Ruiter Sectie BioAnalytische Chemie Vrije Universiteit Amsterdam [email protected] Friday 6 October 2023 Previously on MAP • 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) 2 Where are we? L1 L4 L5 L2 T2 T1 L3 3 Check-in • Question: • Do you guys know what week it is? • Week 40 (fair) • Its…………….. • Nobel prize week ☺ 4 The Nobel Prize in Chemistry 2023 Thursday 5 October 2023 Quantum dots Herbert Fröhlich In 1937 predicted the effects 6 Into the quantum realm • text 7 How small are these quantum dots? • Example: 8 Quantum dots: a new class of materials Molecule (indigo) Allura Red AC Quantum dot Same structure Same material Different properties Different colour? New atoms in a new conformation Size-dependent properties • Optical • Catalytic • Magnetic • Thermal • Electric 9 Alexei Ekimov • Examined coloured glass CuCl2 Size I just witnessed a size-dependent quantum effect • Early 1980s succeeded in creating size-dependent quantum effects in coloured glass • Colour came from nanoparticles of CuCl2 10 Meanwhile 11 I also observed a sizedependent quantum effect but in liquids Luis Brus • First scientist in the world to prove size-dependent quantum effects in particles floating freely in a fluid Problem Quality was unpredictable Let sit on the workbench CdS CdS 12 Moungi Bawendi I figured out how to make high quality quantum dots Revolutionised the production of quantum dots Almost perfect particles This high quality was necessary for applications 13 Why do we care? • Questions you might have: • Why does it matter if a substance’s absorbance is slightly more towards blue? • Why is this so amazing? The optical changes revealed that the substance characteristics had completely changed A substance’s optical properties are governed by its electrons These same electrons also govern the substance’s other properties When scientists observed the changed absorption they understood that they were looking at an entirely new material (in principle) 14 Applications Biomedical imaging Quantum dot Catalysis 15 Learning objectives of today • Knowing why the resonances of • Become familiar with chemical shift ranges of carbon atoms in different functional groups • Understanding what effect electronegativity, hybridisations and anisotropy have on the 13C nuclei are more difficult to observe than those of protons chemical shift of carbon atoms • Knowing what the difference is between proton-coupled and proton-decoupled • Understanding what the nuclear Overhauser effect is and why it is beneficial for • Identifying different carbons in the different DEPT traces (45, 90 and 135) • Recognising how many 13C resonance peaks will be observed in a 13C 13C spectra 13C NMR NMR spectrum by identifying the chemical (non)equivalent carbons in molecules • Recognising the 13C patterns of mono and disubstituted benzene rings 16 How to solve 1H NMR spectra (1/4) Step 1: calculate the hydrogen deficiency index (HDI) The HDI is the number of π bonds and/or rings a molecule contains 17 Calculating the HDI (2𝐶 + 2 + 𝑁 − 𝐻 − 𝑋) 𝐻𝐷𝐼 = 2 • In this formula: • • • • • C = # of carbon atoms N = # of nitrogen atoms H = # of hydrogen atoms X = # of halogen atoms Oxygen atoms can be ignored (no change in the No. of H is required) Compounds with an index of one → must have one double bond or one ring Compounds with an index of two → Could have a triple bond or it could have two double bonds, two rings or one of each Benzene contains one ring and three double bonds → HDI of benzene is, therefore, four 18 How to solve 1H NMR spectra (1/4) Step 1 𝐻𝐷𝐼 = (2𝐶 + 2 + 𝑁 − 𝐻 − 𝑋) 2 × 7 + 2 − 14 ⇒ =1 2 2 HDI = 1, so a double bond or a ring (no benzene!) 19 How to solve 1H NMR spectra (2/4) Step 2: Fill in the table ☺ δ Multiplicity Integral Assignment 0.85 ppm Doublet 5.85 = 6H CH3 next to CH (x2) 20 How to solve 1H NMR spectra (2/4) Step 2: Fill in the table ☺ δ Multiplicity Integral Assignment 0.85 ppm Doublet 5.85 = 6H CH3 next to CH (x2) 1.1 ppm Triplet 2.89 = 3H CH3 next to CH2 1.9 ppm Nonet 1.04 = 1H CH next to 2x CH3 and a CH2 2.3 ppm Quartet 1.92 = 2H CH2 next to a CH3 3.8 ppm Doublet 1.97 = 2H CH2 next to CH (attached to O, due to deshielding) HDI = 1, so a double bond or a ring (no benzene!) from the molecular formula (C7H14O2), we can rule out an acid or aldehyde (proton signals not there!) so lets try an ester 21 How to solve 1H NMR spectra (3/4) Step 3: Draw the different assignments you made and piece the ‘’puzzle’’ pieces together • bla 22 How to solve 1H NMR spectra (4/4) Step 4: Use the spin-spin splitting rule to assign the multiplicities of the protons in your proposed structure and see if they match the 1H NMR spectrum If the multiplicities match the 1H NMR spectrum, you probably have the correct structure ☺ If the multiplicities do not match, the 1H NMR spectrum, you probably have the incorrect structure  In the latter case, rearrange the puzzle pieces until the multiplicities match the 1H NMR spectrum 23 Your turn! Propose a structure (3 min) • bla 24 Answers • Step 1: HDI = 1 (same chemical formula) • Step 2: δ Multiplicity Integral Assignment 1.1 ppm Triplet 2.97 = 3H CH3 next to CH2 1.4 ppm Singlet 8.6 = 9H CH3 (no neighbours) x3 2.2 ppm Quartet 1.98 = 2H CH2 next to CH3 (slight deshielding) • Step 3 and 4: 25 Disclaimer! This is not the set-in-stone way to solve 1H NMR spectra!!! This is my personal way to solve these type of spectra You should use the approach that works for you If you struggle, you can try this approach and see if it works for you 26 The carbon-13 nucleus 13C has an odd mass and as a result has a nuclear spin, with I = The resonances of 13C 1 2 nuclei are more difficult to observe than those of protons (1H) for two reasons: 1: The natural abundance of 2: The magnetogyric ratio of a 13C 13C is very low (1.08%) nucleus is smaller than that of hydrogen, meaning that 13C nuclei always have resonance at a lower frequency compared to protons 27 13C • chemical shifts Similar to 1H NMR, TMS is used as a reference compound. Note that here the blacarbons of the methyl groups are used as reference, not the hydrogens The chemical shifts appear over a range (0–220 ppm) much larger than that observed for protons (0–12 ppm) Because of the large range of values, nearly every nonequivalent carbon atom gives rise to a peak with a different chemical shift 28 Correlation chart for 13C chemical shifts 29 Approximate 13C chemical shift ranges for selected types of carbon < 190 ppm Carboxylic acid > 190 ppm derivatives Ketones Aldehydes 30 Correlation chart for carbonyl and nitrile functional groups 31 Shifting the chemical shift (1/3) Electronegativity, hybridisation, and anisotropy all affect 13C chemical shifts in nearly the same way as they affect 1H chemical shifts • bla However, the 13C chemical shifts are about 20 times larger Question: Why? Answer: The electronegative atom is directly attached to the 13C atom Effect occurs through only a single bond 32 Electronegativity effect: distance In 1H NMR, the effect of an EN element on chemical shift diminishes with distance Always in the same direction (deshielding and downfield) • bla In 13C NMR, an EN element also causes a downfield shift in the α and β carbons Usually, it leads to a small upfield shift for the γ carbon 33 Shifting the chemical shift (2/3) Equal to 1H shifts, changes in hybridisation also produce larger shifts for the nucleus that is directly involved • bla In 13C 13C NMR the carbons of carbonyl groups have the largest chemical shifts, due to sp2 hybridisation and to the fact that an EN oxygen is directly attached to the carbonyl carbon, deshielding it even further 34 Shifting the chemical shift (3/3) Anisotropy is responsible for the large chemical shifts of the carbons in aromatic rings and alkenes • bla 35 Proton-coupled • 13C spectra (1/2) The probability of finding two 13C atoms in the same molecule is low blaThe probability of finding two 13C atoms next to each other is even lower Therefore, we rarely observe homonuclear (C-C) spin-spin splitting patterns where the interaction occurs between two But, the spins of protons attached (directly) to 13C 13C atoms atoms do interact with the spin of carbon and cause the carbon signal to be split according to the n + 1 rule This is called heteronuclear coupling, since it involves two different types of atoms (C-H) 36 The effect of attached protons on 13C Note that you are not seeing protons directly You are observing only the effect of protons on 13C resonances 37 atoms Proton-coupled 13C spectra (2/2) Spectra that show the spin-spin splitting, or coupling, between 13C and the protons directly attached to it are called proton-coupled spectra Proton-coupled spectra for large molecules are often difficult to interpret, due to overlap of the multiplets of different carbons 38 Proton-coupled 13C spectra (2/2) Spectra that show the spin-spin splitting, or coupling, between 13C and the protons directly attached to it are called proton-coupled spectra Proton-coupled spectra for large molecules are often difficult to interpret, due to overlap of the multiplets of different carbons 39 Proton-decoupled The majority of 13C 13C spectra (1/2) NMR spectra are obtained as proton-decoupled spectra The decoupling technique obliterates all interactions between protons and Result: only singlets are observed in a decoupled 13C 13C nuclei NMR spectrum Advantage: this technique simplifies the spectrum ☺ Disadvantage: the information on attached hydrogens is lost  40 Proton-decoupled 13C spectra (2/2) Proton decoupling is achieved in the process of determining a 13C NMR spectrum by simultaneously irradiating all the protons in the molecule with a broad spectrum of frequencies in the proper range Most modern NMR spectrometers provide a second tunable radiofrequency generator, the decoupler, to achieve this The irradiation causes the protons to become saturated, and they undergo rapid upward and downward transitions These rapid transitions decouple any spin-spin interactions between the hydrogens and the 13C nuclei being observed The carbon nucleus ‘’senses’’ only one average spin state for the attached hydrogens 41 Proton-decoupled 13C spectrum example (1/2) 42 Proton-decoupled 13C spectrum example (2/2) 43 If there are no equivalent carbon atoms in a molecule, a 13C peak will be observed for each carbon Coffee time☺ 44 Nuclear Overhauser Enhancement (NOE) (1/4) When a proton-decoupled 13C spectrum is obtained, the intensities of many of the carbon • bla resonances increase significantly above those observed in a proton-coupled experiment C atoms with H atoms directly attached are enhanced the most The enhancement increases (not always linearly) as more hydrogens are attached This effect is known as the nuclear Overhauser effect The degree of increase in the signal is called the nuclear Overhauser enhancement The effect can be positive or negative In the case of 13C interacting with 1H, the effect is positive 45 Nuclear Overhauser Enhancement (NOE) (2/4) • • The maximum enhancement that can be observed is given by the relationship: 𝑁𝑂𝐸𝑚𝑎𝑥 Where: 1 𝛾𝑖𝑟𝑟 = 2 𝛾𝑜𝑏𝑠 • γirr is the magnetogyric ratio of the nucleus being irradiated • γobs is the magnetogyric ratio of the nucleus being observed Remember: the NOEmax is the enhancement of the signal, and it must be added to the original signal strength: Total predicted intensity (max) = 1 + NOEmax • For a proton-decoupled 𝑁𝑂𝐸𝑚𝑎𝑥 13C spectrum, you would calculate: 46 1 267.5 Indicating that the 13C signals can be enhanced up to = = 1.988 200% (theoretically) by irradiation of the hydrogens 2 67.28 Nuclear Overhauser Enhancement (NOE) (3/4) The NOE is a definite bonus received in the determination of proton-decoupled spectra There are a lot of hydrogens, and 13C, 13C with its low abundance, commonly produces weak signals Because NOE increases the intensity of the carbon signals, it substantially increases the sensitivity In a proton-decoupled 13C spectrum, the total NOE for a given carbon increases as the number of nearby hydrogens increases Typically, we find that the intensities of the signals in a 13C spectrum assume the order: CH3> CH2 > CH >> C 47 Nuclear Overhauser Enhancement (NOE) (4/4) • Although the hydrogens producing the NOE effect influence carbon atoms more bla distant than the ones to which they are attached, their effectiveness drops off quickly with distance The interaction of the spin-spin dipoles operates through space (not bonds) and its magnitude decreases as a function of the inverse of r3 (r = radial distance from the hydrogen of origin) r 𝐶→𝐻 1 𝑁𝑂𝐸 = 𝑓 3 𝑟 Thus, the nuclei must be rather close together in the molecule in order to exhibit the NOE effect 48 A quick dip into DEPT (1/2) Distortionless Enhancement by Polarization Transfer (DEPT) In the DEPT technique, a sample is irradiated with a complex sequence of pulses in both the 13C Result: the 13C and 1H channels signals for the carbon atoms in the molecule will exhibit different phases Depending on the number of hydrogens attached to each carbon Each type of carbon will behave slightly differently, depending on the duration of the pulses These differences can be detected, and spectra produced in each experiment can be plotted 49 A quick dip into DEPT (2/2) One method of presenting the results of a DEPT experiment is to plot four different subspectra, which each give different information Example Isopentyl acetate DEPT results 50 DEPT results isopentyl acetate DEPT 135 Signals arising from CH or CH3 give positive peaks Signals arising from CH2 will give negative peaks DEPT 90 Only those carbons that bear a single hydrogen are seen DEPT 45 Only signals that arise from protonated carbons are detected (carbon 6 is not seen) 51 Equivalent carbons Question How many peaks are obsereved in the 13C NMR spectrum? 52 Equivalent carbons Answer 4 peaks! The 13C atoms that are equivalent appear at the same chemical shift 53 Equivalent carbons 54 Test: How many 13C resonances? 55 Test: How many 7 4 13C resonances? 4 7 2 3 56 Nonequivalent carbon atoms (1/2) Often, methyl groups are nonequivalent and will appear at different chemical shifts in the spectrum • bla Several cases where nonequivalent methyl groups may appear include: Groups located near a stereocenter 57 Nonequivalent carbon atoms (2/2) Often, methyl groups are nonequivalent and will appear at different chemical shifts in the spectrum • bla Several cases where nonequivalent methyl groups may appear include: Compounds that are conformationally rigid 58 Compounds with aromatic rings (1/2) Compounds with C-C double bonds or aromatic rings give rise to chemical shifts in the range from 100–175 ppm • bla Relatively few other peaks appear in this range → a great deal of useful information is available when peaks appear here A monosubstituted benzene ring shows four peaks in the aromatic carbon area of a protondecoupled 13C spectrum Ortho and meta carbons are doubled by symmetry 59 Compounds with aromatic rings (1/2) Often the carbon with no protons attached (ipso), has a very weak peak (weak NOE) Note that carbons c and d are not easy to assign by inspection of the spectrum 60 The 13C NMR spectrum of toluene 61 Disubstituted benzene rings (1/2) Depending on the mode of substitution, a symmetrically disubstituted benzene ring • blacan show two, three or four peaks in the proton-decoupled 13C spectrum Ortho Meta Para 62 Disubstituted benzene rings (2/2) Ortho Meta Para 63 Take home message You know the most important points of this lecture if you can answer the following questions: • What are the two reasons that cause the resonances of 13C nuclei more difficult to observe than protons? • Can you (roughly) give the chemical shift ranges of carbon atoms presented in slides (18-20)? • How do electronegativity, hybridisation and anisotropy affect shielding in 13C • What is the difference between proton-coupled and proton-decoupled NMR spectra? • What is the nuclear Overhauser effect and why is it useful in • What types of carbons can be found in the DEPT-45, DEPT-90 and DEPT-135 trace, 13C 13C NMR? NMR? respectively? • Why are there sometimes less peaks observed in a 13C NMR spectrum than carbons present in a molecule? • Can you recognise the different patterns for disubstituted benzene rings? How many peaks are observed for ortho, meta, and para-substituted benzene rings? 64 Next lecture Friday 13 October at 13:30–15:15 ‘’Spectral interpretation’’ NU-4C07 Groups on Canvas + assignment 65 End of today Questions? 66