Subject Chemistry Paper No and Title Module No and Title Module Tag Paper 12: Organic Spectroscopy Module 20: Applications of PMR in Structural Elucidation of Simple and Complex Compounds and 2-D NMR spectroscopy CHE_P12_M20_e-Text
TABLE OF CONTENTS 1. Learning Outcomes 2. Introduction 3. Applications of PMR in Structural Elucidation of Simple and Complex Compounds 3.1 Identification of structural isomers 3.2 Detection of H-bonding 3.3 Detection of Cis-trans isomerism and Conformers 3.4 Quantitative analysis 3.5 Use in medicine (Magnetic resonance imaging) 3.6 Structure determination of unknown compounds 4. Two-dimensional 4.1 COSY 4.2 HETCOR 5. Summary
1. Learning Outcomes After studying this module, you shall be able to To understand the Nuclear Magnetic Resonance (NMR) Spectroscopy To understand the Structural Information through To understand the Number of Signals of An Unknown Compound To know about the Spin-Spin Coupling/Splitting: To understand the Proton NMR Chemical Shifts for Common Functional Groups To know how elucidate an unknown structure of compound NMR Spectroscopy Applications (discuss some problems) 2. Introduction Today, NMR has become a sophisticated and powerful analytical tool that has found a variety of applications in many disciplines of scientific research, medicine, and various industries. In the past two decades with the developments of new methodology and instrumentation technique, NMR has become one of the most powerful and versatile spectroscopic techniques for the analysis of bio-macromolecules such as proteins, nucleic acid. Together with X-ray crystallography, is one of the two leading technologies for the structure determination of bio-macromolecules at atomic resolution. The following are some important applications of NMR Spectroscopy as described below used by an organic chemist. 3. Applications of PMR in Structural Elucidation of Simple and Complex Compounds 3.1 Identification of structure isomers Ethanol (I) and dimethyl ether (II) are functional group isomers and they can be identified by 1 H. In ethanol, there are three types of protons and hence three signals will be observed in the 1 H NMR, whereas in dimethyl ether, all the six protons are equivalent and hence only one signal will be observed. Similarly we can distinguish between n-propyl bromide (III) and isopropyl bromide (IV) by 1 H NMR. The 1 H NMR spectrum of n-propyl bromide will show three signals, whereas only two signals will be observed in the 1H NMR spectrum of isopropyl bromide. In a similar way a large number of structural isomers can easily be distinguished on the basis of 1 H.
3.2 Detection of H-bonding2.2 Detection of H-bonding In general hydrogen bonded protons show higher value of chemical shift than the non-hydrogen bonded protons. The downfield shift of the absorption depends on the strength of the hydrogen bonding. Greater is the degree of hydrogen bonding of proton, greater the downfield shift of a proton. The intermolecular and intramolecular hydrogen bonding can be distinguished by the use of 1 H. The intermolecular hydrogen bonding depends on the concentration of the sample and decreases on dilution with a non-polar solvent and also with increasing temperature. The pure ethanol show signals near 5 ppm but on dilution with non hydrogen-bonding solvents (CCl 4, CDCl 3, C 6 D 6 ), the OH signal shift upfield due to the breaking of intermolecular hydrogen bonds. The intramolecular hydrogen bonding does not depends on the concentration of the sample and hence show no change in their absorption position on dilution. 3.3 Detection of cis-trans isomerism and conformers 1 H can easily distinguish cis and trans isomers because the concerned protons have different values of chemical shifts as well as coupling constants. For isomeric olefins, J trans is always greater than the J cis. Thus it is possible to determine the configuration of geometrical isomers of a disubstituted olefin. Similarly the axial and equatorial protons can be distinguished on the basis of different values of their coupling constants and chemical shifts. The equatorial protons in a conformationally rigid cyclohexane ring appear more downfield than the axial protons on the same carbon. 3.4 Quantitative analysis2.4 Quantitative analysis 1 H has been used to determine the relative amount of keto and enol form in a tautomeric mixture. In the 1 H NMR spectrum of acetylacetone, the integration value for the methylene protons was found to be 10 mm and that for the methine (=CH ) proton is 22 mm.
The two methylene protons of keto form have integration of 10 mm One methine proton of the enol form has integration of 22 mm and hence 2H integration will be equal to 44 mm % of the keto form = 10/(44+10) 100 = 18.5% % of the enol form = 44/(44+10) 100 = 81.5% 3.5 Use in medicine (Magnetic resonance imaging) resonance imaging) The important application of 1 H is magnetic resonance imaging for medical diagnosis. Magnetic resonance imaging (MRI) is used to take the image of the any part of the human body, especially soft tissue, in all possible planes. The brightness of the various regions of an image is related to the number of protons in the tissue at a particular place and the relaxation time of the protons In the body, the hydrogens of water molecules attached to the surface of biological molecules relax faster than those in the free liquid. Furthermore the relaxation time of protons of water molecules is different in different types of tissues, especially in diseased tissues and non-diseased tissues. Specific pulse sequences detect these differences in relaxation time for the protons of water molecules in the tissue being examined. When the results of the scan are processed, the image produced shows different densities of signals depending upon the difference in relaxation times. Thus the MRI image can give idea about any abnormality in the different types of tissues. Unlike, other diagnostic techniques, such as computed tomography (CT), MRI is non-invasive and painless and it does not use ionized radiation and hence is very safe. 3.6 Structure determination of unknown compounds unknown compounds is used to determine the structure of known and novel compounds, and as such, is usually required by scientific journals for identity confirmation of synthesized new compounds. As we all know that the 1 H spectra have three features, chemical shift, signal intensity, and multiplicity and each one provide important information. Here we will see how to use these features together to assign the structure of an organic compound. 1. If the molecular formula is given, first determine the double bond equivalence of the compound which cab give important clues about the structure of the compound, whether the compounds is acyclic, cyclic, saturated or unsaturated or aromatic. Zero value of DBI indicates that the compound does not have any ring or π bonds. The DBI value of 1 indicates that the compound has either one ring or one π bond. The DBI value of 2 indicates that the compound has either one ring and one double bond or two double bonds or two rings. The DBI value of 3 indicates that the compound has either three double bonds or two double bonds and one ring or two rings and one double bond. The DBI value of 4 or more indicates that the compound is most likely to have an aromatic ring. (DBE) = [(2a+2)-(b-d)]/2
Where: a = No of Carbon atom; b = Hydrogen or monovalent atom; d = Nitrogen. 2. Consider the number of signals and integration of each signal which gives clues about the symmetry of the molecule. 3. Analyze carefully chemical shift, integration, and multiplicity of each signal, and then draw fragments relating each signal.. 4. Finally attach all the fragments to get the structure of the compound. Note that there may be more than one way of joining the fragments together. Spin-spin coupling data may help you to make a definite choice between different possibilities. The following exercise will give you hand on practice to solve such kind of problems Problem 1: An organic compound having molecular formula C 5 H 12 O gave the following spectral data: IR: 3400 cm -1 ; 1 H NMR (CDCl 3 ): 0.95 (d, 6H), 1.8 (m, 3H), 2.7 (brs, 1H), 3.9 (t, 2H). Find out structure of the compound. Solution: First calculate DBE. (DBE) = [(2 5+2) (12-0)]/2 = 0 DBE of given compound, C 5 H 12 O is zero that means there is no unsaturation in the compound. IR of given compound shows a peak at 3400 cm -1 which is a stretching frequency for OH group. 1 H NMR gives four signals, indicates the presence of four groups of non-equivalent protons. Now let s assign the hydrogens for each of these signals. A doublet at 0.95 ppm of six protons indicates the presence of one (CH 3 ) 2 CH group (Remember that the multiplicity depends on the number of protons present at the neighboring carbon atom). Presence of a triplet at 3.9 ppm of two protons indicates the presence of CH 2 CH 2 -Ofunctionality, because you can only observe a triplet if the next carbon have two hydrogens (Remember peak around 3.5-4.3 is typical for OCH functionality). In addition to this we have one oxygen atom in our molecule and it is present in the form of hydroxyl (OH) as confirmed by its IR.
Presence of a broad singlet at 2.7 for one proton is due to the hydrogen of OH group, which is further confirmed by IR. Finally now we have two fragments and after assembling these fragments the structure of the given compound would be (CH 3 ) 2 CHCH 2 CH 2 OH Problem 2: An organic compound of molecular formula C 9 H 12 showed following spectral data. 1 H NMR: 1.2 (d, J = 8 Hz, 6H), 2.87 (septet, 1H), 7.2 (s, 5H). Determine the structure of given compound. Solution: MF of the given compound is C 9 H 12 ; Hence DBE of given compound = [(2 9+2) (12 0)]/2 = 4. DBE of 4 indicates that the given compound may be aromatic. 1 H NMR gives three signals, indicates the presence of three sets of non-equivalent protons. A doublets at 1.2 ppm of six protons indicates the presence of (CH 3 ) 2 CH. This also confirms the presence of a septet of one proton at 2.87 ppm. Presence of singlet of five protons at 7.2 ppm indicates the presence of mono substituted aromatic ring.
Based on these fragments, final structure of the compound would be Problem 3: Molecular formula of an organic compound is C 7 H 12 O 3 exhibit the following 1H NMR and IR data. IR: 1710, 1750 cm -1 ; 1 H NMR (CDCl 3 ): 1.25 (t, 3H), 2.2 (s, 3H), 2.55 (t, 2H), 2.7 (t, 2H), 4.15 (q, 2H). Determine the structure of the compound. Solution: DBE = [(2 7+2) (12-0)]/2 = 2 DBE of given compound is 2 that means there is unsaturation in the compound. However the 1H NMR data does not show any peak in the region 4.5-6.8. IR of given compound shows peaks at 1710 cm -1 and 1750 cm -1 which indicates the presence of carbonyl group. The peak at lower frequency (1710 cm -1 ) is characteristic peak for keto group, whereas, peak at 1750 cm -1 is due to ester functional group. 1 H NMR gives five signals, indicates the presence of five sets of non-equivalent protons. A doublet at 1.25 ppm of three protons indicates the presence of one CH 3 CH 2 group. A singlet at 2.2 ppm of three protons indicates the presence of CH 3 CO- group (Remember that CH 3 protons in CH 3 COCH 3 resonate at 2.15 ppm). A quartet at 4.15 ppm of two protons indicates the presence of one OCH 2 CH 3 group (Remember peak around 3.5-4.3 is typical for OCH functionality). Two triplets at 2.55 and 2.7 ppm each of two protons indicate the presence of one -CH 2 CH 2 - group. Since the methylene protons show coupling due to each other only and with no other proton that indicates that both the methylene carbons are attached to the CO functionality. Based on the above fragments, the structure of the given compound would be CH 3 COCH 2 CH 2 COOCH 2 CH 3. 4. Two-dimensional
Since the development of pulsed, a number of advanced two dimensional NMR techniques have been devised. There are various types of 2D NMR experiments such as Correlation spectroscopy (COSY), Heteronuclear Correlation spectroscopy (HETCOR), Exclusive correlation spectroscopy (ECOSY), Total correlation spectroscopy (TOCSY)., Incredible natural-abundance double-quantum transfer experiment (INADEQUATE), Heteronuclear single-quantum correlation spectroscopy (HSQC), Heteronuclear multiple-bond correlation spectroscopy (HMBC), Nuclear Overhauser effect spectroscopy (NOESY), Rotating frame nuclear Overhauser effect spectroscopy (ROESY). They are especially valuable when the structure of complicated natural products and the conformations of bio-molecules has to be determined. Out of these several experiments, two find the most frequent applications. These two are 1 H- 1 H correlation spectroscopy, abbreviated as COSY (Correlated spectroscopy) and 13 C- 1 H correlations spectroscopy (Heteronuclear correlation; HETCOR). In one-dimensional 1 H NMR spectrum, there are two axes, one represents the chemical shift and the other intensity of the signal. In 2D NMR spectra there are three axes, out of these, two axes represent the chemical shifts and the third axis represents the intensity of the signal. In COSY experiments both the axes have chemical shifts of the protons, whereas in HETCOR, the chemical shift range of the proton is plotted on one axis and the chemical shift range of 13 C is plotted on the second axis. Thus COSY identifies pairs of protons which are coupled to each other. HETCOR is used to identify carbon atoms and the protons directly attach to it. 4.1 COSY To see what type of information can we get from a COSY spectrum, let s consider the COSY spectrum of 2-hexanone. Both the X- and Y-axes have the one dimensional 1 H NMR of the 2- hexanone, each axes is calibrated according to the chemical shift values in ppm. You can observe that there are many contours lying along the diagonal extending from the lower left corner to the upper right corner. This diagonal bisects the 2D NMR into two parts which are mirror-image of each other. The off-diagonal contours are called cross peaks and contain the connectivity information that we require. The contours lying on the diagonal serve only as reference points. Each cross peak has X and Y coordinates. One coordinate corresponds to the chemical shift of a proton, the other to the chemical shift of a proton to which it is coupled. Because the diagonal
splits the COSY spectrum in half, each cross peak is duplicated on the other side of the other diagonal with the same coordinates, except in reverse order. Therefore, we only need to examine only half of the cross peaks. Now to examine the COSY, we just start with a cross peak and draw two perpendicular lines (dashed lines) connecting to the diagonal. These lines are parallel to each spectrum axes. The peaks which are intersected on the diagonal by these lines are said to be coupled to each other. Figure 1: COSY spectrum of 2-hexanone Let s start analysing the COSY spectrum of 2-hexanone. First consider the off diagonal spot at 1.6 (Figure 1). Now draw a horizontal line from this point. This line intersects the spot on the diagonal that corresponds to the H 2 proton. Again draw a vertical line from the same off diagonal spot, this line intersect the spot on the diagonal corresponding to the H 3 proton. This relation shows that the H 2 proton is coupled to the H 3 proton. Similarly draw a horizontal and a vertical line from the off diagonal spot at 1.3. These two lines intersect the spots on the diagonal corresponding to the H 3 and H 4 protons, respectively. This confirms that the H 3 proton is coupled to the H 4 proton. Similarly we can show the coupling between H 4 and H 5 protons. It is important to note here that the diagonal peak at 2.1 (due to H 1
proton) has no cross peaks, which shows that H 1 proton is not coupled to any other proton in the molecule. 4.2 HETCOR3.2 HETCOR HETCOR (heteronuclear chemical shift correlation) is also a type of COSY in which the two axes belong to the chemical shifts for different nuclei i.e., one axis for 1 H and the other for 13 C. By HETCOR spectrum, we can determine that which hydrogen is attached to which carbon. There are no diagonal peaks in HETCOR and is composed of only cross peaks, each relating a carbon to its directly attached proton. Let s look at the HETCOR spectrum of 2-hexanone (Figure 2) to explain the correlation between proton and carbon atoms. The 13 C NMR spectrum is shown on the X-axis and the NMR spectrum is shown on the Y-axis. Interpreting the spectrum is straightforward. If you draw a vertical line and a horizontal line from a spot designated as A, it indicates that the hydrogens that show a signal at 0.9 ppm in the 1 H NMR spectrum are bonded to the carbon that shows a signal at 14 ppm in the 13 C NMR spectrum. Similarly, a cross peak B shows that the hydrogens that show a signal at 1.3 ppm are bonded to the carbon that shows a signal at 22 ppm. In a similar manner you can identify the attachment pattern of other carbon and hydrogen atoms. If a line does not have cross peak, this means that this carbon atoms has no attached proton for example a quaternary carbon). Here the chemical shift of the carbonyl carbon is not displayed in the 13 C spectrum because it has no attached hydrogens. Note: Although the one dimensional 1 H and 13 C spectra of 2-hexanone are simple enough to be interpreted directly, however these techniques are more frequently used in case of many complicated molecules where signals can be assigned only with the help of 2-D NMR spectroscopy.
Figure 2: HETCOR spectrum of 2-hexanone 5. Summary In the past two decades, NMR being a sophisticated and powerful analytical tool, has become one of the most powerful and versatile spectroscopic techniques for the analysis of biomacromolecules such as proteins, nucleic acids. NMR is extremely useful in the detection of functional isomers, H-bonding, cis-trans isomerism, conformers. 1 H has been used to determine the relative amount of keto and enol form in a tautomeric mixture. Magnetic resonance imaging (MRI) is one of the most powerful tools in modern clinical diagnosis, which is based on the topological representation of NMR parameters like proton density and transverse relaxation times and longitudinal relaxation times. Differences in these parameters allow impressive anatomical discrimination to be made, and make it possible to distinguish pathological from healthy tissues. is used to determine the structure of known and novel compounds, and as such, is usually required by scientific journals for identity confirmation of synthesized new compounds.
Two-dimensional nuclear magnetic resonance spectroscopy (2D NMR) is a set of nuclear magnetic resonance spectroscopy (NMR) techniques which give data plotted in a space defined by two frequency axes rather than one. Many 2D NMR techniques are there like HETCOR, COSY, EECOSY, NOESY, HMBC, HMQC, ROESY, HSQC, INADEQUATE, TOCSY etc., which are very useful for detecting various structural aspects of an unknown compound. COSY AND HETCOR have been described in this module in detail.