16.1 Introduction to NMR Spectroscopy. Spectroscopy. Spectroscopy. Spectroscopy. Spectroscopy. Spectroscopy 4/11/2013

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1 What is spectroscopy? NUCLEAR MAGNETIC RESONANCE (NMR) spectroscopy may be the most powerful method of gaining structural information about organic compounds. NMR involves an interaction between electromagnetic radiation (light) and the NUCLEUS of an atom. We will focus on C and H nuclei. WHY? The structure (connectivity) of a molecule affects how the radiation interacts with each nucleus in the molecule Protons and neutrons in a nucleus behave as if they are spinning. If the total number of neutrons and protons is an ODD number, the atoms will have net nuclear spin. Examples: The spinning charge in the nucleus creates a MAGNETIC MOMENT. We saw in Chapter 15 how a dipole moment creates an electric field. What does a magnetic moment create? 16-2 Like a bar magnet, a MAGNETIC MOMENT exists perpendicular to the axis of nuclear spin. If the normally disordered magnetic moments of atoms are exposed to an external magnetic field, their magnetic moments will align WHAT if the total number of neutrons and protons is an EVEN number? 16-4 The aligned magnetic moments can be either WITH or AGAINST the external magnetic field. The α and β spin states are not equal in energy. WHY? When an atom with an α spin state is exposed to radio waves of just the right energy, it can be promoted to a β spin state. The stronger the magnetic field, the greater the energy gap

2 The magnetic moment of the electrons generally reduces the effect of the external field. The more SHIELDED a nucleus is with electron density, the smaller the α β energy gap. WHY? The amount of radio wave energy necessary for the α β energy transition depends on the electronic environment for the atom. When the α spins are flipped to β spins, the atoms are said to be in RESONANCE. The use of the term resonance here is totally different from when we are talking about electrons in molecular orbitals. How does NMR spectroscopy tell us about molecular structure? Acquiring a 1 H NMR Spectrum NMR requires a strong magnetic field and radio wave energy. The strength of the magnetic field affects the energy gap Acquiring a 1 H NMR Spectrum The strong magnetic field is created when a high current is passed through a superconducting material at extremely low temperature ( 4 Kelvin). The greater the current, the greater the magnetic field. InmostcurrentNMRinstruments instruments, a brief pulse ofradio energy (all relevant wavelengths) is used to excite the sample. Each of the atoms is excited and then relaxes, emitting energy. The emitted energy is recorded as a FREE INDUCTION DECAY (FID) Acquiring a 1 H NMR Spectrum The FID contains all of the information for each atom. A mathematical treatment called a FOURIER TRANSFORM separates the signals so an individual signal can be observed for each atom that was excited. Suchan instrument is called an FT NMR. Often multiple FIDs are taken and averaged together. Before analysis, NMR samples must be prepared neat or in a liquid solution and placed in a small NMR tube. The sample is placed into the magnetic field and the tube is spun at a high rate to average magnetic field variations or tube imperfections Acquiring a 1 H NMR Spectrum Solvents are used such as chloroform d. WHY? The magnet is super cooled, but the sample is generally at room temperature

3 16.3 Characteristics of a 1 H NMR Spectrum NMR spectra contain a lot of structural information: Number of signals Signal location shift Signal area integration Signal shape splitting i pattern Protons with different electronic environments will give different signals. Protons that are HOMOTOPIC will have perfectly overlapping signals. Protons are HOMOTOPIC if the molecule has an axis of rotational symmetry that allows one proton to be rotated onto the other without changing the molecule. Find the rotational axis of symmetry in each molecule below Another test for homotopic protons is to replace the protons one at a time with another atom. If the resulting compounds are identical, then the protons that you replaced are homotopic. Protons that are ENANTIOTOPIC will also have perfectly overlapping signals. Protons are ENANTIOTOPIC if the molecule has a plane of reflection that makes one proton the mirror image of the other. Find the mirror plane that splits each molecule below The replacement test is universal. It will work to identify any equivalents protons whether they are homotopic or enantiotopic. If the protons are neither homotopic nor enantiotopic, then the are NOT chemically equivalent. Perform the replacement test on the protons shown in the molecule below. If the resulting compounds are enantiomers, then the protons that you replaced are enantiotopic How would you describe the relationship between the protons shown? Practice with SKILLBUILDER

4 There are some shortcuts you can take to identify how many signals you should see in the 1 H NMR: 1. The two protons on a CH 2 group will be equivalent if there are NO chirality centers in the molecule. 2. The two protons on a CH 2 group will NOT be equivalent if there is a chirality center in the molecule. There are some shortcuts you can take to identify how many signals you should see in the 1 H NMR: 3. The three protons on any methyl group will always be equivalent to each other. 4. Multiple protons are equivalent if they can be interchanged through either a rotation or mirror plane. Practice with SKILLBUILDER Identify all the groups of equivalent protons in the molecules below and describe their relationships. Recall that cyclohexane chairs have six equitorial and six axial protons. Do the axial and equitorial protons have different electronic environments? Try the replacement test for an axial/equatorial pair. How many signals should you see for the molecule in the 1 H NMR? At room temperature, the chair interconversion occurs rapidly. The NMR is not fast enough to see the individual structures, so the average is observed (one signal). What might you expect to see if the temperature of the NMR sample were brought down to 100 C? Tetramethylsilane (TMS) is used as the standard for NMR chemical shift. In many NMR solvents, 1% TMS is added as an internal standard. The shift for a proton signal is calculated as a comparison to TMS: For benzene on a 300 MHz instrument:

5 The shift for a proton signal is calculated as a comparison to TMS: The shift for a proton signal is calculated as a comparison to TMS: For benzene on a 60 MHz instrument: The Hz of the signal is different in different instruments, but the shift relative to TMS (δ) is constant The shift relative e to TMS (δ) is a dimensionless number because the Hz units cancel out. Units for δ are often given as ppm (parts per million), which simply indicates that signals are reported as a fraction of the operating frequency of the spectrometer. Most 1 H signals appear between 0 and 10 ppm Early NMRs analyzed samples at a constant energy over a range of magnetic field strengths from low field strength (DOWNFIELD) to high field strength (UPFIELD). Shielded protons required a stronger external magnetic field to be excited at the same energy as deshielded protons. WHY? Current NMRs analyze samples at a constant field strength over a range of energies. Shielded protons have a smaller magnetic force acting on them, so they have smaller energy gaps and absorb lower energy radio waves. Higher Energy Lower Energy Alkane protons generally give signals around 1 2 ppm. Protons can be shifted downfield when nearby electronegative atoms cause deshielding. HOW? To predict chemical shifts, start with the standard ppm for the type of proton (methyl, methylene, or methine). Use Table 16.1 to adjust the ppm depending on proximity to certain function groups

6 16-31 Handbooks can be used for functional groups beyond Table Practice with SKILLBUILDER Predict chemical shifts for all of the protons in the molecule below. When the electrons in a pi system are subjected to an external magnetic field, they circulate a great deal causing DIAMAGNETIC ANISOTROPY DIAMAGNETIC ANISOTROPY means that different regions in space will have different magnetic strengths. Why are some regions more shielded than others? The result of the DIAMAGNETIC ANISOTROPY effect is similar to deshielding for aromatic protons. What about the other protons? Why does it appear that only one signal is given for all of the aromatic protons?

7 The result of the DIAMAGNETIC ANISOTROPY effect is similar to shielding for protons that extend into the pi system. Explain all of the shifts in Table Some of the protons in [14]annulene appear at 8 ppm while others appear at 1 ppm. Which are which? Practice CONCEPTUAL CHECKPOINTs Integration The INTEGRATION or area under the peak quantifies the relative number of protons giving rise to a signal. A computer will calculate the area of each peak representing that area with a STEP CURVE Integration The computer operator sets one of the peaks to a whole number to let it represent a number of protons. The computer uses the integration ratios to set the values for the other peaks The curve height represents the integration Integrations represent numbers of protons, so you must adjust the values to whole numbers. If the integration of the first peak is doubled, the computer will adjust the others according to the ratio Integration Integration The INTEGRATION are relative quantities rather than an absolute count of the number of protons. Predict the 1 H shifts and integrations for tert butyl methyl ether. Symmetry can also affect integrations. Predict the 1 H shifts and integrations for 3 pentanone. Practice with SKILLBUILDER

8 When a signal is observed in the 1 H NMR, often it is split into multiple peaks. Multiplicity or a splitting patterns results. Multiplicity results from magnetic effects that protons have on each other. Consider protons H a and H b We already saw that protons align with or against the external magnetic field. H b will be aligned with the magnetic field in some molecules. Other molecules in the sample will have H b aligned against the magnetic field. Some H b atoms have a slight shielding affect on H a and others have a slight deshielding effect The resulting multiplicity or splitting pattern for H a is a doublet. Consider an example where there are two protons on the adjacent carbon. There are three possible affects the H b protons have on H a. A doublet generally results when a proton is split by only one other proton on an adjacent carbon Half of the H a atoms will not be affected by the H b atoms. WHY? ¼ of the H a atoms will be shielded and ¼ deshielded. H a appears as a triplet. WHY? The three peaks in the triplet have an integration ratio of 1:2:1. WHY?

9 Consider a scenario where H a has three equivalent H b atoms splitting it. Explain how the magnetic fields cause shielding or deshielding. H a appears as a quartet. What should the integration ratios be for the four peaks of the quartet? Table 16.3 shows how the multiplicity trend continues. By analyzing the splitting pattern of a signal in the 1 H NMR, you can determine the number of equivalent protons on adjacent carbons. The trend in Table 16.3 also allows us to predict splitting patterns. Explain how the n+1 rule is used Remember three key rules: 1. Equivalent protons cannot split one another. Predict the splitting patterns observed for 1,2 dichloroethane. 2. To split each other, protons must be within a two or three bond distance. Remember three key rules: 3. The n+1 rule only applies to protons that are all equivalent. The splitting pattern observed for the proton shown below will be more complex than a simple triplet. Complex splitting will be discussed later in this section. Practice with SKILLBUILDER

10 Predict splitting patterns for all of the protons in the molecule below The degree to which a neighboring proton will shield or deshield its neighbor is called a COUPLING CONSTANT. The coupling constant or J value is the distance between peaks of a splitting pattern measured in units of Hz. When protons split each other, their coupling constants will be equal. J ab = J ba The COUPLING CONSTANT will be constant even if an NMR instrument with a stronger or weaker magnetic field is used. Higher field strength instruments will give better resolution between peak because the coupling constant is a smaller percentage of the overall Hz available. Sometimes recognizable splitting patterns will stand out in a spectrum. An isolated ethyl group gives a triplet and a quartet. Note the integrations. The triplet and quartet must have the same coupling constant if they are splitting each other A peak with an integration equal to 9 suggests the presence of a tert butyl group. An isolated disopropyl group gives a doublet and a septet. Note the integrations. Practice with CONCEPTUAL CHECKPOINT Complex splitting results when a proton is split by NONEQUIVALENT neighboring protons. In themoleculeshown shown, H b is split into a quartet by H a and into a triplet by H c. If J ab is much greater than J bc, the signal will appear as a quartet of triplets

11 Complex splitting results when a proton is split by NONEQUIVALENT neighboring protons. Complex splitting results when a proton is split by NONEQUIVALENT neighboring protons. If J bc is much greater than J ab, the signal will appear as a triplet of quartets. If J bc is similar to J ab, the signal will appear as a multiplet If J bc is equal to J ab, what type of patterns will be observed? Predict the splitting patterns for (S) pent 2 en 4 ol. Practice with CONCEPTUAL CHECKPOINT Splitting is not observed for some protons. Consider ethanol: The protons bonded to carbon split each other, but the hydroxyl proton is not split The hydroxyl proton and other labile or exchangeable protons undergo rapid exchange with trace amounts of acid. Show a reasonable mechanism. Such exchange blurs the shielding/deshielding effect of the neighboring ihb i protons giving ii a singlet that is often broadened. If ethanol is rigorously purified to remove traces of acid, then hydroxyl proton splitting is generally observed. Aldehyde protons also often appear as singlet because their coupling constants are sometimes too small to cause observable splitting Signals for exchangeable protons such as those shown below disappear completely when the 1 H NMR sample is prepared for analysis in a deuterated solvent such as chloroform d. WHY? 16.8 Predicting Expected 1 H Spectra for a Compound Predict the chemical shift, integration, and splitting patterns for all of the protons in the following molecule. Draw a spectrum for the molecule. O O Protic compounds have exchangeable protons. Practice with SKILLBUILDER

12 16.9 Using 1 H Spectra to Distinguish Between Compounds The three molecules below might be difficult to distinguish by IR of MS. WHY? 16.9 Using 1 H Spectra to Distinguish Between Compounds Explain how 1 H NMR could be used to distinguish between the two molecules below. Explain how 1 H NMR could distinguish between them. Practice with SKILLBUILDER Analyzing a 1 H NMR Spectrum With a given formula and 1 H NMR spectrum, you can determine a molecule s structure by a four step process: 1. Calculate the degree or unsaturation or hydrogen deficiency index (HDI). What does the HDI tell you? 2. Considerthenumberof number of NMR signals andintegration to look for symmetry in the molecule. 3. Analyze each signal, and draw molecular fragments that match the shift, integration, and multiplicity. 4. Assemble the fragments into a complete structure like puzzle pieces. Practice with SKILLBUILDER Analyzing a 1 H NMR Spectrum Consider the data below, and propose a structure for the molecule. The formula is C 7 H 13 Cl. 1 H NMR data: δ 5.3 (dq 1H); 5.1 (d 1H); 3.4 (s 2H); 2.0 (d 3H); 1.0 (s 6H) 1. Calculate 2. Consider the number of NMR signals and integration to look for symmetry in the molecule Analyzing a 1 H NMR Spectrum Consider the data below, and propose a structure for the molecule The formula is C 7 H 13 Cl. 1 H NMR data: δ 5.3 (dq 1H); 5.1 (d 1H); 3.4 (s 2H); 2.0 (d 3H); 1.0 (s 6H) 3. Analyze each signal, and draw molecular fragments that match the shift, integration, and multiplicity. 4. Assemble the fragments into a complete structure like puzzle pieces Acquiring a 13 C NMR Spectrum Because 1 H is by far the most abundant isotope of hydrogen, 1 H NMR signals are generally strong. 13 C only accounts for about 1% of carbon atoms in nature, so a sensitive receiver coil and/or a concentrated NMR sample is needed. d In 1 H NMR, shift, splitting, and integration are important. In 13 C NMR, only the number of signals and the shift will be considered

13 16.11 Acquiring a 13 C NMR Spectrum In 13 C NMR, the 1 H 13 C splitting is often so complex that the spectrum is unreadable. To elucidate the 13 C spectrum and make it easier to determine the total number of 13 C signals, 13 C NMR are generally DECOUPLED. In the vast majority of 13 C spectra, all of the signals are singlets Chemical Shifts in 13 C NMR Spectra Compared to 1 H, 13 C atoms require a different frequency of energy to excite (resonate). Compared to the standard TMS, 13 C NMR signals generally appear between 220 and 0 ppm. Each signal on the 13 C spectra represents a carbon with a unique electronic environment. Planes and axes of symmetry can cause carbon signals to overlap if their electronic surroundings are equivalent Chemical Shifts in 13 C NMR Spectra Chemical Shifts in 13 C NMR Spectra Like 1 H signals, chemical shifts for 13 C signals are affected by shielding or deshielding. Note how symmetry affects the number of signals for the molecules above. How many 13 C signals should be observed for the molecule below? Practice with SKILLBUILDER Chemical Shifts in 13 C NMR Spectra Predict the number of signals and chemical shifts in the 13 C NMR spectrum for the molecule below. O O Application DEPT 13 C NMR Spectra 13 C spectra generally give singlets that do not provide information about the number of hydrogen atoms attached to each carbon. DISTORTIONLESS ENHANCEMENT BY POLARIZATION TRANSFER (DEPT) 13 C NMR provides information i the number of hydrogen atoms attached to each carbon. Full decoupled 13 C spectrum shows all carbon peaks. In a DEPT 90, only CH signals appear. In a DEPT 135, CH 3 and CH give (+) signals, and CH 2 give ( ) signals

14 16.13 Application DEPT 13 C NMR Spectra Full decoupled 13 C spectrum: shows all carbon peaks. DEPT 90: Only CH signals appear. DEPT 135: CH 3 and CH = (+) signals, CH 2 = ( ) signals Application DEPT 13 C NMR Spectra Explain how DEPT 13 C spectra could be used to distinguish between the two molecules below. Practice with SKILLBUILDER Application Medically Speaking MRI (magnetic resonance imaging) instruments are essentially 1 H NMR spectrometers. The body is analyzed rather than a sample in an NMR tube. Different tissues have different concentrations of protons. WHY? The MRI gives a 3D image of different tissues. HOW? Are there any known side effects from exposure to either radio waves or a magnetic field?