Class Notes (811,169)
Canada (494,539)
Chemistry (407)
CHEM 3750 (7)


46 Pages
Unlock Document

University of Guelph
CHEM 3750
Adrian Schwan

1 - INTRODUCTION TO ORGANIC SPECTROSCOPY 1 1. H Nuclear Magnetic Resonance Spectroscopy ( SF10 9.1-9.9) Atomic nuclei with odd mass numbers have angular momentum and behave as though they were spinning on an axis, that is, they behave as tiny magnets. In the absence of an applied field, these tiny magnets are randomly oriented, but when they are put in the presence of an applied magnetic fiold (H ), they will align with or against that field. A slight excess will align with the field (α- spin) while the others will be anti parallel (β-spin). H o Alignment in the presence In the absence of an of a strong magnetic field applied magnetic field α-spin β-spin The β−spin nuclei are higher in energy and irradiation of light can induce the flipping of α-spins to β-spins. Such a process is called resonance. 1 13 19 These concepts apply to H, C, and F nuclei which have spin quantum numbers of ± / 2 i.e., two spin states. Many other atoms are also NMR active and they may have positive integers as spin states. Atoms with spin quantum numbers of zero are not NMR active. The two energetically different states afford the necessary condition for resonance. Energy differences can be quantitated to be CHEM*3750 SCHWANC OURSE N OTES F13 Chapter 1 Page 0 γH o ν = 2π H is the applied field strength at the nuclei o γ is a constant of the nucleus, the magnetogyric ratio. The frequency has units of Hertz (Hz) or cycles per second. One can irradiate the nuclei and achieve the 'flips' or resonance. The irradiation is in the radio frequency region of the electromagnetic spectrum. The radio frequency (rf) is varied and when the light energy matches the spin state energy difference, then the signal (a resonance) occurs. An important element of nuclear magnetic resonance is that the energy difference is proportional to the strength of the applied magnetic field. magnetic moment of nucleus is opposed toH o energy magnetic moment of nucleus is aligned wiHh external magnetic field strengto, a) Chemical Shift 1 Not all protons (for H NMR spectroscopy) in a given molecule absorb the same energy. The position of an NMR signal depends on the nucleus's electronic environment. In a covalently bound organic molecule, different hydrogens are under the influence of a number of electronic factors. The principal factors include the hybridization of the attached atom, the polarity of the bond and the presence of electron withdrawing or electron donating groups nearby. The applied magnetic field induces movement of the electrons in the bond, which in turn induces smaller magnetic fields. One may view the extremes of electronic environment as a hydrogen cation and a hydrogen anion. On the NMR scale the free proton would resonate at the extreme left while the free hydride would be observed at the extreme right. CHEM*3750 SCHWANC OURSE N OTES F13 Chapter 1 Page 1 + H H- downfield typical H region since proton upfield and hydride are the extremes Increasing applied magnetic fielo)(H , (rf constant) Increasing chemical shift scale Increasing radio frequency (rfo, (H constant) When hydrogens are surrounded by electron density, they shift upfield and are said to be shielded. Conversely, when their environment is lacking in electron density, the hydrogens are deshielded and move downfield. For example, consider the following trends based on electronegativity; CH 3 CH 3l CH 3r CH 3I δ 4.30 δ 3.05 δ 2.70 δ 2.10 fluoromethane chloromethane bromomethane iodomethane The δ scale is based on the resonance position of tetramethylsilan3 4(CH ) Si, TMS) which is observed as a singlet at 0.00 ppm. Any organic compounds that you will be seeing can be found downfield from TMS. Some organometallic compounds may have hydrogens that resonate upfield from TMS. Most hydrogens of organic compounds will be found between 0 and 10 on the δ scale, although the hydrogens of carboxylic acids are often observed farther downfield. The δ value can be calculated as follows: ν (sample)- ν (TMS) X 10 6 δ = ν of the instrumHno ( CHEM*3750 SCHWANC OURSE NOTES F13 Chapter 1 Page 2 A chemical shift expressed in δ has units of parts per million (ppm). All hydrogens that are not chemically equivalent will afford a unique resonance in the H NMR spectrum, since each is in a unique environment in the molecule. Table 9.1 on SF10 p. 387 gives a listing of typical peak positions. For typical chemical shifts, some important beginning points are 0.9 ppm for a methyl group, 5.25 ppm for a hydrogen on a double bond and 7.27 ppm for aromatic hydrogens. This table offers additional detail. Type of hydrogen Chemical shift Situation Typical (δ) range in ppm primary alkyl RCH 3 0.8 - 1.0 alkane and primary alkyl RCH 2’ 1.2 -1.4 alkane-like primary alkyl R3CH 1.4 – 1.7 hydrogen allylic (next to double 1.6 – 1.9 adjacent to bond), R 2=CR-CH 3 unsaturated benzylic ArCH 2 2.2 – 2.5 groups ketone RC(O)-CH 3 2.1 - 2.6 alkyne, RC≡CH 1.7 - 3.1 chloroalkane, RCH C2 3.6 - 3.8 bromoalkane, RCH Br2 3.4 - 3.6 iodoalkane, RCH I2 3.1 - 3.3 ether, RCH OR’ 3.3 - 3.9 2 alcohol, RCH 2H 3.3 – 4.0 adjacent to ester oxygen, 4.1 – 4.3 RCH O2(O)R’ electronegative ester oxygen, 3.6 – 3.8 groups CH 3C(O)R alkyl methyl ether, 3.3 – 3.5 CH 3R aryl methyl ether, 3.6 – 3.8 CH 3Ar terminal alkene, 4.6 - 5.0 R 2=CH 2 alkene (vinylic) internal alkene, 5.2 – 5.7 hydrogen R 2=CH-R aromatic, Ar-H 6.0 – 8.9 aldedyde, RC(O)- H 9.5 - 10.0 hydroxyl hydrogen, R- 0.5 – 5.0 variable and OH unreliable amine, RNH 2 0.5 – 5.0 Some examples: CHEM*3750 SCHWANC OURSE N OTES F13 Chapter 1 Page 3 1,4 Dimethoxybenzene 10 9 8 7 6 5 4 3 2 1 ppm CHEM*3750 SCHWANC OURSE NOTES F13 Chapter 1 Page 4 Methyl propiolate (Methyl propynoate) 10 9 8 7 6 5 4 3 2 1 ppm Succinimide O NH O 10 9 8 7 6 5 4 3 2 1 ppm CHEM*3750 SCHWANC OURSE N OTES F13 Chapter 1 Page 5 b) Signal Areas The area under the curve of a given signal is proportional to the number of hydrogens responsible for the signal. Hence if one can integrate the signal, one would have the relative number of hydrogens for each resonance. So to summarize, to this point, H NMR tells what sort of electronic environment that the individual hydrogen or groups of hydrogens are in, whereas the integration tells us the number of hydrogens at a given resonance. 1,4-Dimethoxybenzene 10 9 8 7 6 5 4 3 2 1 ppm 0.15 0.23 CHEM*3750 SCHWANC OURSE N OTES F13 Chapter 1 Page 6 Methyl Propiolate 10 9 8 7 6 5 4 3 2 1 ppm 0.38 0.12 Succinimide 10 9 8 7 6 5 4 3 2 1 ppm 0.14 0.68 CHEM*3750 SCHWANC OURSE N OTES F13 Chapter 1 Page 7 PRACTICE PROBLEMS 1 You are capable of doing Questions 9.1→9.6, 9.2SF10. 1. Calculate the chemical shift δ (in ppm) for a proton that has resonance 128 Hz downfield from TMS on a spectrometer that operates at 60 MHz. 2. Sketch the H NMR spectrum that you would HCl2C CHCl 2 expect for the accompanying molecule. Be sure to C address issues of coupling, integration and H Cl chemical shift. 3. Tell precisely how you would use H NMR to distinguish between the following pairs of compounds. a) 1-bromopropane vs. 2-bromopropane b) CH 3H 2 C C H vs.CH 3 C C CH 3 c) O O CH 3 vs. H CH3 CH 3 4. a) Determine the ratios of the peaks areas in the following three NMR spectra. b) Then use this information, together with the chemical shifts to match each spectrum to the appropriate compound on the list on the next page. c) Assign the peaks in each spectrum to the protons they represent in the molecule. SPECTRUM A CHEM*3750 SCHWANC OURSE NOTES F13 Chapter 1 Page 8 SPECTRUM B SPECTRUM C LIST OF MOLECULES CH3 CH3 CH3O OCH3 C CH HO Br CH 3 CH C C CH BrCH2 C 3 CH CH CH CH CH 3 CH3 3 2 3 Br CHEM*3750 SCHWANC OURSEN OTESF13 Chapter 1 Page 9 SOLUTIONS TO PRACTICE PROBLEMS 1 Q1. 6 (observed shift from TMS) X 10 δ = (operating frequency [in Hz]) 6 128 X 10 128 = = 60,000,000 60 = 2.13 ppm Q2 Cl2HC CHCl 2 C H Cl integrates to 2 integrates to 1 about 6 ppm about 4 ppm Q3. a) CH CH CH Br 3 2 2 -3 signals: a sextet, a triplet and another triple CH 3 CHBr CH 3 -2 signals: a septet and doublet b) CH 3CH2 C C H -3 signals: will contain the expected look for an ethyl grou and also a singlet CH 3 C C CH 3 -will have a lone singlet CHEM*3750 SCHWANC OURSE N OTES F13 Chapter 1 Page 10 c) O -3 signals: will contain a triplet for the CH group CH3 3 H -will contain a pentet f2r the CH group and also a triplet near 10 ppm O -will have a lone singlet CH3 CH3 Q4. SPECTRUM A 2.3 & 2.5 ppm (can't tell HO which is which) CH C C CH 3 CH 3 1.5 ppm -integration from left to right is 1:1:6 SPECTRUM B -aromatic H's at 6.8 ppm CH 3 OCH 3 -CH 3groups at 3.7 ppm -integration from left to right is 2:3 SPECTRUM C CH3 1.9 ppm BrCH 2 C CH 3 Br 3.9 ppm -integration from left to right is 1:3 _______________________________________________________ CHEM*3750 SCHWANC OURSE NOTES F13 Chapter 1 Page 11 c) Signal Splitting (Coupling) When hydrogens are close to one another in a molecule, they usually exert a slight chemical shift influence on one another. Consider two non- equivalent hydrogens as shown below: H a H b C C The chemical shift of Ha is influenced by whether the spin of H is aligned b with or against the applied field. If the spinbof H is aligned with the applied field, it adds to the applied field and a smaller external strength is required than that necessary for H an the absence of any perturbations. Hence a peak at lower field is observed. However this absorption size is only half of the Ha protons since only half of the bprotons are in the α spin state. The other half ofbH are in the β spin state and since those are aligned against the external field, the local field strength around Ha in this case is diminished. To achieve resonance, Ho has to be increased and an upfield shift is observed. The resonance of Ha is said to be split into a doublet. Note that one hydrogen has induced the appearance of a doublet and also note that the doublet peaks are of equal height. The same arguments apply to the influence of Ha on the appearance of the Hb resonance are applicable. Indeed the coupling constant, J , is defined as the distance between the two split peaks. It is the same for bothbHs effect on a and H a's effect on b . δ Hawithout the influence of H b Ha is split into a double H bs aligned Hb is aligned with the against the applied applied field field CHEM*3750 SCHWANC OURSE NOTES F13 Chapter 1 Page 12 H H a b J Jab ab spin of b :α β spin of Ha: α β Jab =coupling constant When Ha is near two H bs (equivalent H's) then thebH resonance will again be a doublet as it is affected by a lone hydrogen. However, Ha will now feel the effect of two hydrogen's spin states. The various combinations will afford a triplet (three peaks) in a 1:2:1 ratio. Note that two hydrogens have induced a triplet. 1 2 1 a b c Similarly ifaHis near a methyl group, i.e., three equivalbnt H 's,athen H will appear as a quartet due to 8 possible spin orientations and their relative ratio is 1:3:3:1. Note that three hydrogens have induced splitting and the appearance of a quartet. The number of lines that appear for a given resonance is the multiplicity. 1 3 3 1 a b c d CHEM*3750 SCHWANC OURSE NOTES F13 Chapter 1 Page 13 The lines that appear are all equally spaced, by a distance that is abe J coupling constant. The coupling constant may vary and is on the order of 0 - 20 Hz. In freely rotating systems, H-C-C-H, the three bond coupling constant is ca. 6-8 Hz. There are other molecular subunits for which the coupling constant may be known or accurately predicted. Jab Jab Jab Jab Jab As an example n-propyl chloride with three different types of hydrogens would afford the following types of peaks. CH 3 ca. 1.1 ppm (triplet, J = ca. 7 Hz, 3H CH 2 ca. 1.7 ppm (sextet, J ca. 7 Hz, 2H CH 2l ca. 3.5 ppm (triplet, J = ca. 7 Hz, 2H Note that the central methylene group is coupled to two sets of different hydrogens. The resonance appears as a sextet because the coupling constants are essentially the same to each of the other hydrogens. That is the central methylene group is coupled to five hydrogens with J = 7 Hz. Many coupling constants can be predicted (or expected) base on a significant compilation of data from millions of compounds. Typical values are listed in a table on the next page. This information is a useful reference. CHEM*3750 SCHWANC OURSE NOTES F13 Chapter 1 Page 14 Representative Coupling Constants Coupling Coupling Coupling arrangement J (Hz) arrangement J (Hz) arrangement J (Hz) H H H H a,a: 8 to 14 6 to 8 H 6 to 10 a,e: 0 to 7 C C (ortho) H e,e: 0 to 5 (usually 7-8) H H H 11 to 18 8 to 11 cis: 6 to 12 trans: 4 to 8 H H H H H 6 to 15 5 to 7 cis: 2 to 5 O trans: 1 to 3 H H H H H H H 4 to10 0 to 5 H C 0 to 1.5 (often 0) (4 bond!) CH R R' H There are typical and recognizable coupling patterns that are indicative and rather diagnostic of certain fragments of molecules. See the table on the following page and appreciate that integration of the sets of coupling resonances will also be diagnostic. There are times when a given hydrogen is surrounded by a number of different hydrogens and they all couple with a different coupling constant. Then there can be a series of splittings that are superpositioned on one another. This type of resonance is sometimes simply called a multiplet. CHEM*3750 SCHWANC OURSENOTES F13 Chapter 1 Page 15 Commonly Observed Splitting Patterns appearance situation appearance CH CH X Y X ≠ Y CH 2H X CH CH Y 2 2 X ≠ Y CH CH 3 CH CH 3 2 CH 3 CH CH 3 Some examples: trans-3-phenylpropenal CHEM*3750 SCHWANC OURSEN OTESF13 Chapter 1 Page 16 3-methyl-1-butanol 1,1,2-trichloroethane ⇓ expansions ⇓ CHEM*3750 SCHWANC OURSE N OTESF13 Chapter 1 Page 17 1,2-dibromo-3,3-dimethylbutane borneol Me Me Me OH 3.5 3.0 2.5 2.0 1.5 1.0 0.5 ppm 0.10 0.33 0.32 0.34 0.00 CHEM*3750 SCHWANC OURSE N OTES F13 Chapter 1 Page 18 13 2. C NMR Spectroscopy ( SF10 9.11A-D) The major isotope of carbon (12) does not does not have magnetic spin and cannot produce NMR signals. However, carbon-13 which is 1.1% naturally abundant has a spin number of ½. Using a pulsed NMR instrument so that numerous scans can be accumulated, 13C NMR resonances can be observed and are very useful. Some of the initial points to be mentioned are that coupling is usually not observed in 13C NMR spectra. First of all, having two13C atoms side by side is statistically rare and so most oftenC atom is beside a 12C atom and there is no 1 13 coupling. H- C coupling can be readily observed but in most cases, the NMR acquisition is carried out so that no coupling is observed and theC signals appear simply as single lines. The H- C coupling is eliminated by broad band 1 proton decoupling, a process that is achieved by continuously irradiating thH NMR frequency range with another radio frequency source. The result is that α - spin and β-spin energy levels do not hold any protons very long and thus they 13 cannot be observed by the C atoms. This makes the protons essentially invisible. A broad band 1H decoupled 1C NMR spectrum will show a number of vertical lines. There will be one line for each carbon in the molecule, once symmetry considerations have been taken into account. Solvent peaks are also evident. a) Symmetry Gaining information from a13C NMR spectrum is often simply an exercise in symmetry. The spectra of two isomeric trimethylcyclohexanes on the following page illustrate this point. b)13C Chemical Shifts The 13C NMR scale is much wider than that for the proton. A typical ppm range is δ -15 to +240 ppm, with most peaks occurring between -5 and +215 ppm. Again the 0.0 point on the scale is defined by the position of the peak of 13 TMS, but in this instance, it is a resonance. TMS has only one peak because all the methyl carbons in the molecule are identical due to symmetry. 13 There can be less predictability in theC NMR spectrum, but there are nevertheless expected ranges of chemical shifts for functional groups. OnSF10 p. 418, a listing is provided. The trends iC chemical shifts essentially follow 1 those of H chemical shifts. That is, electronegative atoms push the chemical shift downfield, and the more of those atoms directly attached, the farther the resonance is shifted. Additional detail is shown on the following table and chart. CHEM*3750 SCHWANC OURSE N OTES F13 Chapter 1 Page 19 200 150 100 50 0 R -CH 3 8 30 Saturaedc abon s 3 - o e ec o eat ve e e e s- R-CH 2R' 20 45 l g i l R C H,R C 30 60 3 4 C O 50 0 Saturaedc abon s 3 -elec o e ati i effc s - g y C Cl 40 65 C-Br U ns uratd carbon(sp2) - C C 100 1 5 C Arom aicringcarbo s 107-160 N ( rO ) A cid A ids - C O E s e s Anh d d es 155 185 y H (orC ) Aldehyd es C O K e o es 185 220 200 150 100 50 0 13 Additional C NMR chemical detail for unsaturated carbon containing functional groups. Type of carbon Chemical shift (δ) Typical range in ppm nitrile, C≡N 109 – 125 117 – 123 acid anhydride, 154 – 175 RC(O)OC(O)R acid chloride, 160 – 180 165 – 175 RC(O)Cl amide, RC(O)NR 2 155 – 180 ester, RC(O)OR 160 – 185 167 – 174 carboxylic acid, 165 – 186 RC(O)OH aldehyde, RC(O)H 190 – 208 unsaturated ketone, 190 – 210 C=CC(O)C ketone, RC(O)C 200 – 220 CHEM*3750 SCHWANC OURSE N OTES F13 Chapter 1 Page 20 Two 1C NMR spectra for trimethylcyclohexanes 80 60 40 20 0 ppm 80 60 40 20 0 ppm On the next page the 13C NMR spectra of three isomeric dichlorobenzenes can be solved on the basis of symmetry and intensity of signals. CHEM*3750 SCHWANC OURSE NOTES F13 Chapter 1 Page 21 o-dichlorobenzene p-dichlorobenzene m-dichlorobenzene 140 130 120 140 130 120 140 130 120 Some examples: Succinimide 220 200 180 160 140 120 100 80 60 40 20 0 ppm CHEM*3750 SCHWANC OURSE N OTES F13 Chapter 1 Page 22 Ethyl oxirane (ethyl epoxide, 1,2 epoxybutane) 180 160 140 120 100 80 60 40 20 0 ppm 1-decene 160 140 120 100 80 60 40 20 0 ppm CHEM*3750 SCHWANC OURSE NOTES F13 Ch
More Less

Related notes for CHEM 3750

Log In


Don't have an account?

Join OneClass

Access over 10 million pages of study
documents for 1.3 million courses.

Sign up

Join to view


By registering, I agree to the Terms and Privacy Policies
Already have an account?
Just a few more details

So we can recommend you notes for your school.

Reset Password

Please enter below the email address you registered with and we will send you a link to reset your password.

Add your courses

Get notes from the top students in your class.