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by: Michelle Gulgowski DVM


Marketplace > University of Kentucky > Chemistry > CHE 230 > ORGANIC CHEMISTRY I
Michelle Gulgowski DVM
GPA 3.91

Robert Grossman

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Robert Grossman
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Date Created: 10/23/15
Grossman CHE 230 F99 Chapter 15 and 214 Spectroscopy 151 Introduction to Spectroscopy When we run a reaction in the laboratory or when we isolate a compound from nature one of our first tasks is to identify the compound that we have obtained There are a number of different analytical techniques that can be used to do this We can compare its physical properties melting point boiling point optical rotation physical appearance odor and taste to a known compound to see if they are identical This method is not terribly reliable We can burn the compound measure the amount of C02 H20 and other compounds that are produced and use the proportions of the combustion products to determine an empirical formula This method elemental analysis is the oldest method of organic structure determination Berzelius a Swedish chemist invented the technique about 1808 not long after the concept of atomic weights had been invented by the Englishman John Dalton He determined the molecular formulae of eight organic carboxylic acids and got all but one right Elemental analysis is still used today but it is a relatively crude method used more to determine the purity of a compound whose structure is already known than to identify an unknown compound The most widely used methods of organic compound identification today are those which measure the interactions of compounds with electromagnetic radiation of different wavelengths In order of increasing energy of the radiation these are as follows 0 Nuclear Magnetic Resonance NMR spectroscopy measures the absorption of radio waves by H or C atoms in a magnetic field Different kinds of H or C atoms absorb energy of different wavelengths Infrared IR spectroscopy measures the absorption of infrared radiation heat by organic compounds Different functional groups eg C20 O H absorb energy of different wavelengths Ultraviolet Visible UV vis spectroscopy measures the absorption of visible and ultraviolet light by at bonds in an organic compound Bonds of different types CC C20 and with different extents of conjugation CC vs CC CC vs aromatic absorb energy of different wavelengths There is a fourth kind of spectroscopy mass spectroscopy MS that does not involve the absorption of light In fact quotmass spectroscopyquot is a misnomer for this reason However it is the uni versally used name for this technique so we will use it too MS provides a means of measuring the mass of a compound and certain pieces fragments of that compound upon blasting it with high energy electrons This is the technique we will discuss first Make sure you remember how to calculate degrees of unsaturation Grossman CHE 230 F99 152 Mass Spectroscopy The MS experiment works as follows A compound is vaporized by putting it into a vacuum It is then bombarded with high energy electrons These electrons bounce into the compound knocking one or more electrons out of an orbital A radical cation a compound with one unpaired electron and a positive charge is left behind The radical cation then passes into a magnetic field The magnetic field causes the path of the positively charged compound to curve The extent of curvature is determined by the mass to charge ratio which is written as mz The detector at the other end of the magnetic eld measures mz by finding where the fragments emerge from the magnetic field Usually z 1 so the method provides a measure of the mass of the particle The quantity of fragments emerging at a particular mz ratio the intensity is also measured electrons magnetic field compound injected radical cation formed low mz gt high mZ Electrons in the highest energy occupied molecular orbital HOMO are most likely to be ejected They already have more energy than other electrons in the compound so it takes less energy to eject them The HOMO is usually a lone pair or a 313 bond but it can be a 0 bond if there s no other choice Note the order of decreasing energy Electrons from O N S and the halogens are usually ejected most readily Note that the incoming electron does not become part of the compound One goes in but two come out O HCH gt HC H After the radical cation is formed and before it travels through the detector it may fragment that is break up into smaller pieces Fragmentation is a characteristic reaction of radicals Fragmentation occurs to give a neutral radical and an even electron cation and it usually occurs to give the stabler cation When this happens the cation is deflected through the magnetic field and detected and the neutral fragment is lost 117 Grossman CHE 230 F99 Each compound has a characteristic fragmentation pattern These are predictable but we re not going to worry about that gt 7 gt CH2 detected not detected Me Me 67 Me Me 639 Me 6 I gt I gt C Me Me Me Me Me Me Me detected not detected The mz ratios and intensities of the fragments that are detected are usually presented as a bar graph with intensity on the y axis and the mz ratio on the x axis As I said usually only one electron is ejected from each molecule so the x axis is usually thought of as the mass of the fragments The most intense peak is called the base peak The intensities of other peaks are measured as a percentage of the base peak The heaviest peak that is observed is usually the molecular ion M which is the unfragmented compound If M is very unstable or can fragment into very stable pieces it may not be observed at all On the other hand if M can t fragment into more stable pieces it might be the base peak neopentane gs X 9 8 E I I I I I I I I I I I I I I I 0 20 40 W1 60 80 100 The simplest piece of information that the mass spectrum can give us is the molecular weight and hence formula of a compound This can easily be distinguished to a single atomic mass unit amu For example M for propane MezCHZ weighs 44 amu while that for dimethyl ether MeZO weighs 46 amu If we have a sample that is one or the other of these we can easily identify it Similarly if a compound has a M of 110 amu its molecular formula must be either C8H14 C7H100 C6H602 or C6H10N2 There are tables available that tabulate the possible formulas for a given molecular ion Grossman CHE 230 F99 The molecular weight also gives a clue as to whether N atoms may be present The nitrogen rule is the following A compound has an odd molecular weight if and only if there are an odd number of nitrogen atoms in its formula The nitrogen rule holds for compounds containing H C N O Si P S or any of the halogens ie for any compounds we would see in this class Every C containing ion has a small peak accompanying it that weighs one mass unit more The ratio of the intensities of the M and M1 peaks is directly proportional to the number of C atoms in the ion The peak is due to small amounts of naturally occurring 13C which has 11 natural abundance The intensity of the M1 peak allows one to determine exactly of how many C atoms a particular ion consists For example a C6 ion has a 66 probability of having one 13C atom in it so its M1 peak is 66 of the intensity of the M4r peak Other isotopes that are used to identify peaks include Cl 75 35Cl 25 37Cl and Br 50 79Br 50 81Br Natural H N O F and I consist of almost exclusively a single isotope deuterium and tritium have very low natural abundance so among elements the most commonly encountered in organic compounds we need only to worry about isotope peaks from C Cl and Br The molecular ion unfortunately is not always observed One can also gain information from the fragmentation pattern The fragmentation pattem constitutes a kind of quotfingerprintquot for a compound Different structural isomers and even stereoisomers give different fragmentation patterns so the fragmentation pattern can be used to distinguish different compounds that have the same molecular formula The fragmentation pattern also provides evidence for and against certain structural elements in the unidentified compound If the major fragments can be identified and if a reasonable mechanism for their formation from the molecular ion can be written then this constitutes evidence that the parent ion has been correctly identified For example when a fragment ion of M 91 is observed it is almost always the very stable benzyl cation PhCHf suggesting that a benzyl group is present in the parent compound Ethyl esters RCOZEt often show fragments M 45 for the fragment RC0 and M 73 for the fragment R I don t expect you to be able to analyze fragmentation patterns To summarize MS allows one to determine the molecular weight of an unknown compound if the M peak can be identified An odd molecular weight suggests that an odd number ofN atoms are present The presence and size of an M2 peak tells you whether Cl or Br atoms are present The size of the MI peak tells you how many C atoms are present The fragmentation pattern observed in MS provides clues to the structure of a compound HRMS allows one to determine the exact molecular formula of the M peak or any other fragment Problem for class The mass spectrum of a compound shows an M peak at 124 amu and no M2 peaks Determine some molecular formulas for this compound 1 If the molecular weight is odd you have an odd number of N atoms Subtract 14 from the molecular weight and write N in your formula In the present case this is not necessary 1 1 4 Grossman CHE 230 F99 2 It will be clear from the M2 peak how many C1 or Br atoms you have For each C1 or Br atom subtract 35 or 79 from the molecular weight and write C1 or Br in your formula In the present case no C1 or Br atoms are present 3 Determine the maximum number of C atoms that can fit into the remaining molecular weight In the present case the answer is C10 since 1012 120 If you are given the intensity of the M1 peak divide it by 11 to determine exactly how many C atoms you have 4 Add enough H s to make up the mass of the M peak In this case C10H4 5 Calculate the degrees of unsaturation in your formula You may have too few fewer than zero ie a negative number or too many degrees of unsaturation in your compound Too few degrees of unsaturation means that you have too many H atoms a If you have too few number of degrees of unsaturation you must add degrees of unsatura tion ie remove H s without changing the molecular weight There are a few ways to do this 1 Replace CH4 with O This has the effect of adding one degree of unsaturation 2 Replace C2H4 with N2 This also has the effect of adding one degree of unsaturation 3 Replace H16 with O This has the effect of adding eight degrees of unsaturation b If you have a very large number of degrees of unsaturation you may want to subtract degrees of unsaturation Most of the compounds I would have you identify have 0 5 degrees of unsaturation Replace C with H12 This has the effect of subtracting seven degrees of unsaturation without changing the molecular weight In the present case we have nine degrees of unsaturation in a C10 compound It is not impossible to draw a compound with this formula HCEC2CCCECH2 but it s just not a very likely formula We can subtract seven degrees of unsaturation by converting C10H4 into C9H16 Now we have just two degrees of unsaturation much more reasonable 6 You now have your first reasonable formula for your compound Once you generate a reasonable formula you may do the following replacements to generate new formulas with the same molecular weight a Replace CH4 with O Adds a degree of unsaturation b Replace C2H4 with N2 Adds a degree of unsaturation c Replace CO with N2 Leaves degrees of unsaturation unchanged Thus C9H16 gt C8H120 Three degrees of unsaturation C8H120 gt C7H12N2 Three degrees of unsaturation C8H120 gt C7H802 Four degrees of unsaturation C8H120 gt C6H8N20 Four degrees of unsaturation C6H8N20 gt C5H8N4 Four degrees of unsaturation Etc 7 Draw at least one structure for each formula It s useful to remember that a six membered ring with three double bonds eg a phenyl group C6H5 or one or more N atoms may replace CH groups in the ring uses up four degrees of unsaturation 11 Grossman CHE 230 F99 More problems to do in class 1 88 amu no M2 2 119 amu no M2 3 108 amu and there s an M2 peak that is 13 the size of M 153 Infrared Spectroscopy The energy of a photon of electromagnetic radiation is related to its wavelength and its frequency E hv h c l where v is the frequency of the light in units of S 1 or Hz l is the wavelength of the light in In h is Planck s constant 663 X 10 34 J s and c is the speed of light 300 X 108 ms The energy of electromagnetic radiation is directly proportional to its frequency and inversely proportional to its wavelength Infrared IR radiation is electromagnetic radiation with wavelengths from 750 nm to 100 pm although IR spectroscopy is mostly concerned with the region from 25 pm to 25 pm IR radiation is less energetic than visible light which has wavelengths of 400 750 nm When IR radiation is absorbed by molecules it causes the atoms and bonds to bend twist and stretch like balls on springs deforming bond lengths and angles This molecular motion is measured as temperature so we perceive the radiation that causes it as heat IR spectroscopy measures a compound s absorption of IR radiation as a function of the frequency of the radiation A compound absorbs IR radiation of certain wavelengths and not others because its motion is quantized That is a molecule can vibrate only at certain frequencies and not at others The frequencies can be calculated using quantum mechanics equations but we usually use empirical methods to determine which frequencies correspond to which molecular motions The intensity of absorption of IR radiation of a particular wavelength is related to the change in dipole moment of the molecule upon undergoing the particular motion with which the wavelength is correlated Very polar bonds show large absorptions while nonpolar bonds show none at all The IR spectrum is usually presented as a graph of transmittance versus wavenumber Wavenumber is defined as the reciprocal of the wavelength or 1 l and is measured in cm l It is another unit of energy The transmittance of a compound is defined as the amount of light of a particular energy which is not absorbed by the compound and passes through it 100 Transmittance means that the compound does not absorb any of the light of that energy while 0 transmittance means that it absorbs it all When you are looking at an IR spectrum the troughs represent low transmittance and high absorption while the peaks represent high transmittance and low absorption We are interested in the troughs ie high absorbance regions Grossman CHE 230 F99 The motions of all but the simplest molecules are extremely complex so we divide the IR spectrum up into the interpretable region 4000 1500 cm l and the ngerprint region 1500 800 or so cm l The fingerprint region is usually not interpreted although some real IR jocks do so Usually it s just used to identify a known compound by comparing it to a reference IR spectrum In the interpretable region the most important absorbances and their intensities are as follows O H about 3500 cm l strong and broad N H about 3300 cm l strong C H 2850 3300 cm l strong to moderate Csp H 3300 cm 1 Csp2 H 3000 3 100 cm 1 Csp3 H 2850 3000 cm 1 CEN about 2250 cm l moderate CEC about 2200 cm l moderate RCECH to non existent RCECR CO 1650 1780 cm l very strong aldehydes 1730 cm 1 acyclic ketones 1715 cm 1 cyclohexanone 1715 cm 1 strained cyclic ketones 1750 1780 cm 1 esters 1735 cm 1 amides 1670 cm 1 subtract 20 25 cm 1 for being adjacent to a 313 bond CC CO CC 1500 1650 cm l intensity dependent on polarity of substituents These absorbances especially the carbonyl stretch tend to stand out from the background and from each other so they are diagnostic for these functional groups If the absorbances are present the functional groups are present if they are absent the functional groups are absent except for CC and CEC In later chapters you will learn more about the IR characteristics of particular functional groups For now I want you to know the ranges above To summarize IR spectra allow one to determine the presence or absence of certain mctional groups in a compound It is especially useful for identifying OH NH CEX CC and C20 groups 154 Physical Basis of NMR Spectroscopy Today the most widely used method for determining the structure of organic compounds is nuclear magnetic resonance spectroscopy or NMR spectrosocopy NMR spectroscopy involves putting a 117 Grossman CHE 230 F99 compound into a magnetic field and measuring the absorption of radio waves by the 1H 13C 19F 31F or other nuclei Each nucleus in a different environment absorbs radio waves of a different energy For example if one looks at the 1H NMR spectrum for a compound like CH3CHZOH one sees one absorption for the 0 H 1H one for the CH2 1H s and one for the CH3 1H s The 13C NMR spectrum shows one absorption for each C atom in the compound The NMR spectra thus give direct information about the nature of the chemical environment of each magnetically active nucleus in the molecule The physical basis of NMR spectroscopy is as follows Many nuclear isotopes have a magnetic moment called the nuclear spin This number is a multiple of 12 The 1H has a spin of 12 as do 13C 19F and 31F but 12C and 160 have spins of 0 and 14N and 2H have spins of 1 Only nuclei with spins of gt0 can be detected by NMR spectroscopy Because the nucleus has a charge and because of its quotspinquot nuclei act like tiny magnets If one applies an external magnetic field H0 to the nuclei the tiny magnets of the nuclei align themselves with the eld some parallel and some anti parallel The parallel arrangement has a slightly lower energy than the anti parallel arrangement The difference in energy is directly proportional to the strength of the field The proportionality constant is dependent on the nucleus Because the two alignments are different in energy there is a slightly higher population of nuclei in the lower energy state than in the higher energy state Radio waves are absorbed by the nuclei with a parallel alignment causing them to ip to the anti parallel alignment Note that radio waves are very low energy radiation There s not much difference in energy between the spin states even with a very strong magnetic field antiparallel alignment absorption of radio wave causes spin flip magnetic field strength Energy of nucleus parallel alignment Why should different 1H nuclei in a compound absorb radio waves of different energies Don t they experience the same external magnetic field They do and they don t The external magnetic field is indeed identical for all the nuclei However every nucleus is surrounded by electrons The electrons are charged so when they experience a magnetic field they circulate in such a way as to create an opposing magnetic field This is called shielding A 1H nucleus surrounded by a large number of electrons attached to electropositive elements will experience a much smaller magnetic field than a 1H nucleus surrounded by HR Grossman CHE 230 F99 a small number of electrons attached to electronegative elements The 1H attached to an electronegative element is said to be deshielded The same is true of other nuclei If we keep a magnetic field constant and vary the radio wave frequency different nuclei will resonate at different frequencies We can measure this resonance and plot it as a function of radio wave frequency In practice though it s easier to keep the radio wave frequency constant and to vary the magnetic field The most commonly used 1H NMR spectrometers use radio waves with frequencies of 60 MHz to 500 Mhz while the frequencies are one fourth of this for 13C NMR experiments We do the NMR experiment by dissolving the compound to be analyzed in a solvent that lacks any 1H nuclei CCl4 or a deuterated solvent like CDC13 or D20 The solution is placed in a tube which is placed in a magnet The sample is irradiated with radio waves of constant frequency as the magnetic field is slowly varied At different magnetic field strengths different absorbances are measured When we vary the magnetic field we only need to change its strength by a few millionths to observe all the different resonances of all the different atoms in the compound This is why we measure the resonance of nuclei in ppm Most 1H nuclei resonate within a range of 10 ppm of each other Most 13C nuclei on the other hand resonate in a range of about 220 ppm of each other The radio wave frequencies used for 13C NMR spectroscopy are one fourth of those used for 1H NMR spectroscopy We use tetramethylsilane or TMS which has twelve identical 1H nuclei as a standard for 1H and 13C resonance We arbitrarily define the magnetic field strength required for TMS to resonate with a given radio wave energy as 0 ppm The resonances of other kinds of 1H s are then measured in ppm with respect to TMS The resonance of a particular kind of 1H is called its chemical shift and it is often written as 6 In the case of ethanol three resonances are observed at about 6 55 38 and 12 ppm down eld more deshielded of TMS As I said the 1HS in most organic compounds resonate in the region 0 10 ppm although there are exceptions to this rule 13C nuclei usually resonate in the range 0 220 ppm TMS was chosen as a standard because most 1H and 13C nuclei resonate downfield of the same nuclei in TMS When an NMR spectrum is plotted 0 ppm is placed on the right with increased deshielding as we move to the left The right of the spectrum is called upfield while the left of the spectrum is called down eld The chemical shift values of the nuclei in a particular compound are independent of the radio wave frequency used to measure them We can use a 60 MHz or 400 MHz instrument to measure the chemical shifts of the 1H s in ethanol but the values are identical The magnetic field strengths required for resonance are of course different but the ppm change in magnetic field strength from TMS is not Grossman CHE 230 F99 155 Interpreting NMR Spectra Symmetry Let s look start to look at specific examples The first thing to predict is how many resonances a compound will display Just because a compound has eight 1H s doesn t mean that eight resonances will be observed Symmetry will often reduce the number of resonances Let s look at methyl acetate CH30COCH3 We have three C atoms each of which is a different kind the carbonyl C the methyl C attached to O and the methyl C attached to C so we expect to see three different 13C resonances We have six H atoms but some of these have identical environments The three H s in the methyl group attached to C have the same connections and exchange their environments constantly by rapid rotation about the C C 0 bond so they experience the same deshieldin g so they resonate at the same frequency They are said to be chemically equivalent Likewise for the three H s in the methyl group attached to O But the three H s in one methyl group have a different environment from the three H s in the other methyl group different connections In the end we expect to see two 1H resonances for this compound Atoms with different connectivities are in principle expected to have different resonances Atoms with identical connectivities may or may not be chemically equivalent For example in cyclopentanol the two H atoms on C2 are not equivalent One is cis to the OH and the other is trans to the OH Atoms that have the same connectivity but don t have identical resonances are said to be chemically inequivalent or diastereotopic from tonoljg the Greek word for place so meaning diastereomeric places In this class we will label chemically equivalent atoms with identical letters and chemically inequivalent atoms with different letters So all three H s attached to C1 in methyl acetate will be labelled Ha and the three attached to C3 will be labelled Hb Here s a method for determining whether or not two atoms with the same connectivities are chemically equivalent Draw the compound twice In one of the structures replace one of the atoms in question with a different test atom In the other replace the other atom in question with the test atom If the two structures are identical or enantiomers then the atoms in the original structure are chemically equivalent if the two structures are diastereomers then the atoms in the original structure are chemically inequivalent In any compound containing a stereocenter that has a group XCHZY where X and Y are different the H atoms in the CH2 group are always chemically inequivalent This is because replacing one H atom in XCHZY with a test atom generates a new stereocenter so replacing one or the other H atom always generates different diastereomeIs of the compound our positive test for chemically inequivalent atoms A compound doesn t have to have a stereocenter to give a positive test for chemically inequivalent atoms though see cyclopentanol 1110 Grossman CHE 230 F99 Cyclohexane has axial and equatorial H s so we might predict it would show two resonances This would be somewhat incorrect The amount of time required to record for a compound to interact with radio waves of 100 MHz frequency about 10 ns is much longer than the amount of time required for cyclohexane to ip back and forth many times As a result every H spends time in both the axial and equatorial positions during the time scale of the experiment and the axial and equatorial H s can t be distinguished At room temperature cyclohexane shows just one resonance for all the 1H s However at very low temperatures 90 C where ring ipping is greatly slowed two resonances are seen In practice for spectra at or around room temperature we can look at the at structure to determine which atoms are inequivalent just like we did to determine whether the compound was chiral In the case of cyclohexane all H atoms are labelled Ha and all C atoms are Ca Problems for home 1 Identify the number of 1H and 13C resonances each of these compounds will display a cis 13 Dimethylcyclohexane b Benzene c Toluene methylbenzene d Diethyl ether ethoxyethane e 2 Methylpentane f 3 Ethylpentane 156 Interpreting 1H NMR Spectra The 1H NMR spectrum gives us information about the number of chemically different H atoms the chemical environment of each atom the number of H atoms giving rise to each resonance and the number of nearby magnetic nuclei usually other H atoms We ll talk in detail about each of these points Chemically inequivalent H s resonate at different eld strengths while chemically equivalent H s resonate at the same eld strength We have talked in detail about how you can tell whether H atoms are chemically 1111 Grossman CHE 230 F99 identical or different For example in 11 dimethylcyclohexane we expect to see four sets of resonances one from the Me H s one from C2 and C6 one from C3 and C5 and one from C4 The resonances of 1H s tend to fall into the range 05 11 ppm downfield of TMS The range can be divided into six regions 1H nuclei in certain kinds of chemical environments resonate in certain regions while others resonate in other regions Saturated alkyl groups resonate from 05 to 15 ppm H atoms on C s adjacent to C20 or CC groups and RCECHresonate between 15 and 25 ppm H atoms attached to C atoms bearing one heteroatom H C X X 0 N S Hal resonate between 25 and 45 ppm Alkenyl H atoms R2C2CHR and H atoms attached to C atoms bearing two heteroatoms CXZR resonate between 5 and 65 ppm Aryl H atoms resonate between 65 and 80 ppm Finally H atoms in XCHO groups X C N O resonate between 78 and 105 ppm These ranges are approximate H atoms in compounds with unusual structures can resonate at unusual frequencies Carboxylic acid H s resonate way downfield while alcohol and amine H s may resonate anywhere Type of H atom Chemical Shift 6 saturated alkyl 0 5 15 CC C H OC C H CEC H 15 25 X C H X20 N S Hal 25 45 CC H RXzC H 45 65 aryl H 6 5 85 OC H 78 105 alkyl O H varies alkyl N H varies RCOZH 110 140 A H atom near two deshielding groups falls further downfield than if either one of the deshielding groups were present alone For example MeOCHzMe has 6 about 34 ppm PhCHZMe has 6 about 23 ppm but MeOCHzPh has 6 about 55 Within each region methyl H s resonate slightly more upfield smaller 6 than methylene or methine H s and electronegative atoms or groups that are one C removed tend to shift the resonance a little more downfield larger 6 The chemical shifts of H atoms attached to heteroatoms are not constant even for a given compound and vary tremendously depending on concentration solvent and temperature If one measures the areas under the resonances in a 1H NMR spectrum one can determine the number of H atoms contributing to each resonance This process is called integration It used to be done by cutting the peaks out of the paper and weighing them but today we do it by computer Let s look at the 1H NMR spectrum of methyl formate HC02CH3 We expect to see two resonances one at about 6 40 ppm for the methyl group and one at about 6 80 ppm for the carbonyl H If we integrate these two peaks the ratio of 1 1 17 Grossman CHE 230 F99 the 6 80 peak to the 6 40 peak will be approximately 13 since one H atom contributed to the downfield peak while three contributed to the upfield peak Note that we obtain only the ratio of H atoms contributing to the different resonances not necessarily the total number of H atoms in that compound For example if we integrate the spectrum of methyl pivalate MegCCOzMe which has two resonances at 6 37 and 12 ppm we find a 13 ratio of peaks the same as in methyl formate We can t tell a priori whether we have four eight twelve or sixteen H atoms in the compound just that we have two kinds of H atoms in a 31 ratio The most complex aspect of 1H NMR spectra is spinspin coupling Consider CHBrzCHClz Ha will experience a magnetic field of a particular strength and we expect to see one resonance from it However there is a non equivalent magnetic atom next door on C2 Hb Hb has about a 50 probability of aligning with the external magnetic field and about a 50 probability of aligning against it When Hb aligns with the field Ha experiences a slightly stronger magnetic field and when Hb aligns against the field Ha experiences a slightly weaker magnetic field We therefore expect that the resonance due to Ha would be split into two resonances of equal height centered around the quottruequot resonance frequency This expectation is true The Ha resonance is observed to be a doublet Likewise the Hb resonance is also a doublet due to the effect of Ha on the magnetic field experienced by Hb 1H 1H coupling is often observed between chemically inequivalent 1 H s separated by two or three bonds This is called twobond and threebond coupling Three bond coupling ie coupling between 1H atoms on adjacent C atoms H C C H is observed most frequently Two bond coupling is observed between diastereotopic H atoms Four and five bond coupling is observed only occasionally in compounds with rigid geometries Coupling between equivalent 1 H s is not observed We have seen that in CHBrzCHClz Ha splits the resonance of Hb into two signals The extent of splitting is called the coupling constant and is commonly abbreviated J We have also seen that Hb splits the resonance of Ha Hb and Ha split each other with the same coupling constant Coupling constants are usually measured in Hz not in ppm because they are independent of spectrometer frequency when expressed in Hz Three bond coupling constants usually range between 0 and 12 Hz while two bond coupling constants may range up to 20 Hz The H C C H three bond coupling constant is usually about 70 Hz when there is free rotation about the C C bond When there is not free rotation as in cyclic compounds or in double bonds the coupling constants may be very large or very small In cyclohexanes that are conformationally locked into a chair the coupling between adjacent axial H s is usually large between equatorial and an adjacent axial is medium and between adjacent equatorial H s is usually very small The coupling between H s that are trans on a double bond is usually quite large 15 Hz for cis it is medium sized ca 8 Hz and for terminal CH2 groups the coupling between the two H s is usually very small ca 1 Hz 1113 Grossman CHE 230 F99 One can calculate coupling constants in Hz by measuring the chemical shifts of the two peaks that are split and using the formula coupling constant Hz 2 A6 ppm spectrometer frequency MHz where A6 is the difference in the chemical shift of the two peaks Now consider CHBrzCHZCl There are one Ha and two Hb Consider the environment of Ha It experiences a magnetic field strength that is affected by the spins of its two neighbors Hb Before when there was one Hb we saw that Ha could experience two different magnetic fields depending on whether the spin of Hb was up or down Now Ha might experience three different magnetic field strengths depending on whether the Hb s are up up up down down up or down down Down up is experienced the same way as up down since the sum of the effect on the magentic field is 0 The resonance for Ha will be split into three resonances a triplet with a 121 ratio What about Hb in CHBrzCHZCl It has one inequivalent neighbor Ha Its resonance is a doublet The distance between two lines of the triplet for Ha the coupling constant is the same as the distance between the two lines of the doublet for Hb Also the integration of the triplet due to Ha and the doublet due to Hb will give a 12 ratio for the two peaks In other words the splitting doesn t change the total area under the resonances Now look at CHBT2CH3 Ha has three neighbors Hb These might be up up up up up down up down up down up up down down up down up down up down down or down down down As a result Ha might experience four different magnetic field strengths with a probability of 133 1 The signal for Ha is a quartet in the ratio 133 1 The signal for Hb remains a doublet The number of peaks into which a resonance is split is called its multiplicity The multiplicity of a 1H with n equivalent neighbors is n1 with the relative intensities of the resonances given by the polynomial expansion We call the multiplicities singlets doublets triplets quartets quintets sextets etc Other magnetically active nuclei such as 31F 2H and 19F can also split the resonance of a 1H To summarize The 1H NMR spectrum detects a single resonance for every non equivalent 1H in the compound The chemical shift of the resonance which is independent of spectrometer frequency when it is expressed in ppm gives information about the chemical environment of the 1H s The resonances can be integrated to obtain the number of H atoms contributing to each The 39 39 quot 39 of a tells you how many neighbors the 1H s contributing to that resonance have The coupling constant J is independent of spectrometer frequency when expressed in Hz The value of the coupling constant gives you information on the nature of the spatial relationship between the atoms that are coupled two bond or three bond fixed or varying orientation etc 1114 Grossmdn CHE 230 F99 Let s look at some specific examples Dichloroacetaldehyde ClzCHCHO has two inequivalent H atoms H1 and H2 We expect to see two resonances one at around 6 105 ppm and one around 6 55 ppm The H2 resonance is so far downfield because there are two Cl atoms attached to C2 One H atom contributes to one resonance and one H atom contributes to the other resonance so we expect them to integrate to a 1 1 ratio The 1H resonating at 10 ppm H1 has one magnetically active neighbor H2 These atoms are separated by three bonds so we expect to see a coupling constant The resonance at 6 10 ppm is split into two peaks of equal intensity a doublet Likewise H2 has one magnetically active neighbor so its resonance is split into a doublet too The spectrum consists of two doublets There is one H1 H2 coupling constant so the J values for one resonance obtained by measuring the distance between the peaks in ppm and multiplying by the spectrometer frequency is the same as the J value for the other resonance Chloroacetaldehyde ClCHzCHO has two kinds of inequivalent H atoms one H1 and two H2 We expect to see two resonances one at around 6 100 ppm and one around 6 45 ppm One H atom contributes to the downfield resonance and two H atoms contribute to the upfield resonance so we expect them to integrate to a 12 ratio This time H1 has two equivalent magnetically active neighbors so the downfield resonance is split into a triplet with 121 intensity One the other hand each H2 has one magnetically active and inequivalent neighbor so the upfield resonance is split into a doublet of 1 1 intensity Acetaldehyde CH3CHO has two kinds of inequivalent H atoms one H1 and three H2 We expect to see two resonances one at around 6 95 ppm and one around 6 22 ppm One H atom contributes to the downfield resonance and three H atoms contribute to the upfield resonance so we expect them to integrate to a 13 ratio H1 has three equivalent magnetically active neighbors so the downfield resonance is split into a quartet with 1331 intensity One the other hand each H2 has one magnetically active and inequivalent neighbor so the upfield resonance is split into a doublet of 1 1 intensity Bromoethane CH3CH2Br has two kinds of inequivalent H atoms one H1 and two H2 We expect to see two resonances one at around 6 34 ppm and one around 6 18 ppm The CH3 peak is a little more downfield of the normal saturated alkyl range because of the effect of the electronegative Br atom Two H atoms contribute to the downfield resonance and three H atoms contribute to the upfield resonance so we expect them to integrate to a 23 ratio Each H1 has three equivalent magnetically active and inequivalent neighbors so the downfield resonance is split into a quartet with 1331 intensity One the other hand each H2 has two magnetically active and inequivalent neighbors so the upfield resonance is split into a triplet of 121 intensity A downfield quartet and an upfield triplet with identical coupling constants and in a 23 ratio is diagnostic for the presence of an ethyl group attached to an electronegative atom Isopropanol CH3CHOHCH3 has three kinds of inequivalent H atoms six H1H3 one H2 and one OH We expect to see three resonances at 6 36 12 and anywhere between 1 and 6 in 161 ratio One tends not to see coupling between 1H s attached to heteroatoms and C atoms so we expect to see a doublet for 1 1 1 5 Grossman CHE 230 F99 the H1H3 resonance one magnetically active inequivalent neighbor a septet for the H2 resonance six neighbors not counting the OH and a singlet possibly broad for the OH resonance Other problems to work 3 Chloropentane diastereotopic H s 1 chloropentane We need to discuss one more subject Suppose we have a compound like transcinnamaldehyde PhCHCHCHO H1 gives us a doublet at 6 967 with J 7 H2 H3 gives us a doublet at 6 742 with J 15 Hz It is shifted downfield from the normal vinyl region because it is next to the aryl ring The coupling constant for trans H atoms on an alkene is always particularly large What kind of pattern can we expect to see for H2 It has two neighbors but they are inequivalent The first neighbor H3 splits the resonance for H2 into a doublet with J 15 Hz The second neighbor H1 splits each of the doublet peaks into another doublet with J 7 Hz The pattern that is seen is called a doublet of doublets If the H1 H2 and H2 H3 coupling constants had been the same we would have seen an apparent triplet H2 H3 J 15Hz H1 J7Hz Problem to work in class What kind of 1H NMR spectrum would you expect to see for trans1 chloropropene Often many signals overlap each other and clean spectra aren t observed This is especially true when you have two sets of 1HS with similar chemical shifts and that are coupled to one another For example toluene We expect to see a singlet for the CH3 group and we do In the aryl region we expect to see three resonances in a 122 integrated ratio for the three kinds of aryl H s but what we see instead is either a singlet low frequency spectrometer or an uninterpretable mess high frequency spectrometer that integrates to 5 with respect to the Me resonance s 3 The reason is that the aryl H s all have similar chemical shifts and they are all coupled to one another so the splitting patterns are very complex so called firstorder coupling Higher frequency spectrometers tend to simplify coupling patterns because the J values in ppm are made smaller with respect to the chemical shift differences between peaks Problem for home 2 Predict the approximate chemical shift integration and multiplicity of each resonance in the 1H NMR spectrum of each of the compounds in problem 1 Assume that each set of H s resonates at a different field strength so no overlap between signals is observed 1116 Grossman CHE 230 F99 157 Interpreting 13C NMR Spectra If you stop to think about it the fact that we can measure 13C NMR spectra at all is quite remarkable Only 11 out of 1000 C atoms is 13C In contrast almost 100 of H atoms are 1H Thus if we have 10 mmol of say CHC13 in solution we have 10 mmol of 1H atoms but only 100 ymol of 13C atoms When we do 13C NMR spectroscopy we try to make our solutions as concentrated as possible and we also take many spectra and add them up to increase our signal to noise ratio When we take a 13C NMR spectrum we obtain a certain number of resonances corresponding to the number of inequivalent C atoms in the compound The chemical shifts of each of the resonances gives us information about the nature of each of the C atoms especially the hybridization and the groups to which they are attached Finally spin spin coupling gives us information as to the number of 1H atoms attached to each 13C atom 13C nuclei resonate at a far wider range 220 ppm of frequencies than do protons ca 10 ppm but like 1H s 13C nuclei with certain hybridizations tend to resonate in certain regions Csp3 atoms resonate in the region between 0 and 80 ppm Csp between 65 and 120 ppm aryl and alkenyl Csp2 between 100 and 160 ppm and C20 between 165 and 220 ppm There is much more overlap of ranges in 13C NMR than there is in 1H NMR and it s harder to predict exactly where particular resonances will be found in each range In the alkyl range Csp3 O tends to resonate further downfield than alkyl Csp3 and alkyl Csp3 tends to resonate further upf1eld when it is 1 and further down eld when it is 3 or 4 In the carbonyl region ketones and aldehydes resonate further downfield than esters Type of C atom Chemical Shift 6 Csp3 0 50 Csp3 X 40 80 Csp 65 120 alkenyl aryl Csp2 100 160 C20 165 21 0 Problem to work in class Methyl acetate shows peaks at 6 170 50 and 20 ppm Identify the C atom corresponding to each peak Integration of 13C NMR spectra is rarely done The area under 13C resonances is not proportional to the number of atoms contributing to that resonance unless special experimental techniques are used However it is true that 4 C atoms tend to give signals of much weaker intensity than 1 2 and 3 C atoms 1117 Grossman CHE 230 F99 One more feature of 13C NMR spectra needs to be discussed If we look at a compound like chloroform CHC13 we notice that the C atom is attached to a magnetic atom the 1H This atom should affect the magnetic environment of the 13C nucleus splitting it into a doublet Actually in normal 13C NMR operating mode called spindecoupled mode we blast the 1H s with radio waves at the same time that we measure the 13C resonances The radio waves that excite 1H nuclei are of different energy from those that excite 13C nuclei so the two pulses don t interfere with one another The purpose of blasting the 1H s is to cause them to ip their spins so quickly that the 13C nuclei can t tell whether the 1H s are aligned with or against the magnetic field and so no splitting is observed It is possible though to run the 13C spectra in spincoupled mode Under these conditions each C atom experiences the magnetic field splitting caused by the magnetic nuclei directly attached to it In practice one only needs to worry about splitting from 1H 2H 19F and 13F Since 13C has only 11 natural abundance the probability of finding a 13C nucleus adjacent to another 13C nucleus is very small so 13C 13C splitting is not normally observed Just as we saw in 1H spectra a C atom attached to n H atoms has a multiplicity of n1 with intensities determined by the coefficients of the polynomial expansion In 13C NMR spectra coupling patterns are usually quite simple because one needs only worry about 1H s attached directly to the 13C atom In other words one bond coupling only is seen The resonances of 3 C atoms are split into doublets those of 2 C atoms split into triplets those of 1 C atoms split into quartets and those of 4 C atoms are not split at all C H coupling constants in a particular compound are independent of the spectrometer frequency when they are expressed in Hz C H coupling constants are usually not measured or reported unless there is a particular reason to do so Problem to work in class Predict the number of resonances their chemical shifts and their multiplicities in the 13C NMR spectrum of 4 ethylisopropoxybenzene To summarize The spindecoupled normal 13C NMR spectrum detects a single resonance for every non equivalent C atom in the compound The chemical shift of the resonance is usually expressed in ppm which is independent of spectrometer frequency The chemical environment of the C atoms in a compound can be deduced by examining the chemical shifts of the resonances In the spincoupled 13C NMR spectrum each resonance may be split into two to four peaks depending on the number of H atoms attached to the C in question The splitting distance called the coupling constant or J is usually expressed in Hz which is independent of spectrometer frequency I want you to be able to predictthe 1H and 13C NMR spectra of a given compound to the extent demonstrated above I also want you to be able to deduce the structure of a compound if you are given a 1H NMR spectrum and perhaps other data such as IR MS and 13C NMR spectra 111R Grossman CHE 230 F99 Problem for home 3 Predict the approximate chemical shift and multiplicity of each resonance in the 13C NMR spectrum of each of the compounds in problem 1 158 NMR Spectroscopy in Biology Magnetic resonance imaging or MRI is a new medical diagnostic technique based on the principles of NMR spectroscopy The only reason MRI isn t called NMR is that the public is afraid of the word quotnuclearquot In MRI a person is placed inside a giant magnet and an NMR spectrum of the H20 in his or her body is obtained The basis of MRI is that different body tissues have different chemical environments so the water in say a liver tumor has a different resonant frequency from the water in the rest of the liver This allows the doctor to quotlookquot into a person s body in a non invasive manner and look for abnormal biological structures NMR spectroscopy is unique among diagnostic techniques in that it provides information about the environment of specific atoms in a molecule in solution This is important when trying to elucidate the interactions of proteins with one another and with the small molecules to which they bind Answers to problems assigned in class 1 Identify the number of 1H and 13C resonances each of these compounds will display a cis13 Dimethylcyclohexane There is a plane of symmetry containing C2 and C5 that relates the two CH3 s to one another The plane relates C1 to C3 and C4 to C6 There is no plane that relates the H s on top of the ring to those on the bottom of the ring Five 13C and eight 1H resonances will be seen b Benzene Benzene is a flat molecule Its C atoms are spZ hybridized so all the atoms in benzene are in a single plane Benzene has a six fold axis of symmetry through the center of the ring Rotating benzene about 1119 Grossman CHE 230 F99 this axis maps every C and H onto the next C and H T he double bonds don t lower the symmetry in benzene because two equally good resonance structures for benzene can be drawn where the double bonds are drawn in different places As a result all the C C bonds in benzene are equivalent to one another Often a quottoilet bowlquot resonance structure is drawn for benzene for this reason Benzene also has seven planes of symmetry Every C is equivalent to the next C and likewise for the H s One 13C and one 1H resonance will be seen a six fold axis of sym etry c Toluene methylbenzene Toluene no longer has the six fold axis of symmetry of benzene but it has a plane of symmetry containing the Me group and two C s of the ring Five 13C and four 1H resonances will be seen A d Diethyl ether ethoxyethane In its hi ghest symmetry conformation diethyl ether has two planes of symmetry It also has a two fold axis of symmetry through O in the plane of the paper One relates the left half of the compound to the right half the other containing the CS and O relate the HS on each C to each other Two 13C and two 1H resonances will be seen 12 b an HH bun ng L a 2L A H3C 0 CH3 H3C 0 CH3 e 2 Methylpentane 1 1 70 Grossman CHE 230 F99 This compound can be drawn in a conformation in which it has one plane of symmetry the plane of the paper This relates the two Me groups on the one C to each other and it realtes the two H s in each CH2 to one another Five 13C and five 1H resonances will be seen A A a a d d H39 CH3 L1H H39 CH3gH AA 1 C E 1 e H B D CH3 1 H CH3 Hii cHiic f 3 Ethylpentane This compound can be thought of as three ethyl groups hanging off of one C It has a three fold axis of symmetry containing the central C and its H It also has several planes of symmetry each conatins the central CH and one of the Et groups and it relates the two H s of the contained Et group to one another and the two uncontained Et groups to one another Three 13C and three 1H resonances will be seen 2 Predict the approximate chemical shift integration and multiplicity of each resonance in the 1H NMR spectrum of each of the compounds in problem 1 Assume that each set of H s resonates at a different field strength so no overlap between signals is observed a cis13 Dimethylcyclohexane In reality many of these resonances will overlap with one another and give very complex splitting pattems Set 6 I Integration Multiplicity a 09 6 d b 15 2 ddddq c 12 1 dt d 12 1 dt e 12 2 dddd 1171 Grossman CHE 230 F99 f 12 2 dddd g 12 1 dtt h 12 1 dtt b Benzene Set I 6 I Integration I Multiplicity a I 73 s c Toluene rnethylbenzene Set I 6 I Integration I Multiplicity a 23 3 s b 71 2 d c 73 2 dd 1 72 1 t d Diethyl ether ethoxyethane Set I 6 I Integration I Multiplicity a 13 3 t b 34 2 q e 2 Methylpentane Again many of these resonances will overlap with one another Set 6 Integration Multiplicity a 09 6 d b 15 1 t of septets c 12 2 dt d 11 2 tq e 10 3 t f 3 Ethylpentane Set 6 Integration Multiplicity 1 1 77 Grossman CHE 230 F99 a 15 1 septet b 12 6 dq c 09 9 t 3 Predict the approximate chemical shift and multiplicity of each resonance in the 13C NMR spectrum of each of the compounds in problem 1 a cis13 Dimethylcyclohexane Set 6 Multiplicity A 10 q B 30 d C 25 t D 20 t E 23 t b Benzene Set I 6 I Multiplicity A I 128 I d c Toluene methylbenzene Set I 6 I Multiplicity A 21 q B 140 s C 129 d D 130 d E 126 d d Diethyl ether ethoxyethane Set I 6 I Multiplicity A 1 5 q B 60 t e 2 Methylpentane 11 77 Grossman CHE 230 F99 Set 6 Multiplicity A 1 5 q B 30 d C 25 t D 28 t E 12 q f 3 Ethylpentane Set 6 Multiplicity A 35 d B 25 t C 12 q 11 74


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