STRUCTURE ELUCIDATION BY INTEGRATED SPECTROSCOPIC METHODS

Miscellaneous Methods UNIT 14 STRUCTURE ELUCIDATION BY INTEGRATED SPECTROSCOPIC METHODS Structure 14.1 Introduction Objectives 14.2 Molecular Formula and Index of Hydrogen Deficiency 14.3 Structural Information Available from Different Types of Spectra Mass Spectrum UV-VIS Spectrum IR Spectrum 1 H-NMR Spectrum 14.4 Structure Elucidation of Organic Molecules 14.5 Summary 14.6 Terminal Questions 14.7 Answers 14.1 INTRODUCTION This is the last Unit of this course. You have studied in details about the theoretical aspects and the applications of different types of atomic and molecular spectroscopic methods in the earlier Units. You have learnt that the molecular spectra contain important information about the structure of the molecule. In this Unit you will learn that how the information available from different types of molecular spectra can be used for the structure elucidation of the Organic molecules. In this context we will use the information available from UV-VIS, IR, 1 H-NMR and mass spectrometry. To facilitate this, we would first recapitulate the important structural information available from different types of spectra and then, with the help of worked examples, illustrate how to elucidate the structure of a molecule from its given spectral data. This Unit, thus, aims to integrate the information available from different types of spectral methods and apply this knowledge to arrive at the structure of a molecule. In other words, you will be able to correlate the spectral signals of a molecule to different structural units present in the molecule. Once you understand the correlation between different spectral features and the structure of the molecule, you can also predict the important spectral details for a given molecule from its structure. Objectives After studying this Unit, you should be able to: define and calculate the Index of Hydrogen Deficiency, discuss the importance of molecular formula and Index of Hydrogen Deficiency in the structure elucidation of an organic compound, list important structural information available from UV-VIS, IR, 1 H-NMR and mass spectra of a molecule, correlate the signals and their intensities in different types of spectra of a molecule to various structural units present in it, integrate the information available from different types of spectra to elucidate the structure of a molecule, and 56

predict the spectral data for a compound from its structural formula. 14.2 MOLECULAR FORMULA AND INDEX OF HYDROGEN DEFICIENCY We have mentioned that in this unit you will learn as to how the information available from different types of molecular spectra can be used to elucidate the structure of a molecule. It is interesting to know that in addition to spectra, the molecular formula can also provide important structural leads. It is, therefore, worthwhile to know about the kind of information available from the molecular formula. You know that the molecular formula indicates the elements constituting the molecule and the number of atoms of each element present in it. This can provide information about the unsaturation and/or the ring structures in the molecule. The information can be obtained by computing a parameter called Index of Hydrogen Deficiency (IHD). It is defined as the number of pairs of hydrogen atoms that must be removed from the corresponding saturated formula to give the molecular formula under consideration. Let us learn about it with the help of some simple examples. In case of ethane (C 2 H 4 ), there are two i.e., one pair of hydrogen atoms less than that required for the corresponding saturated compound having molecular formula C 2 H 6. Hence, its IHD will be 1. Similarly, for ethyne (C 2 H 2 ), the number of hydrogen atoms required for the saturated hydrocarbon would be C 2 H 6 C 2 H 2 = 4. Hence, the number of pairs of H atoms needed would be 4/2= 2. So, IHD of ethyne is 2. Let us now take the example of cyclopropane. What is its IHD? If we see the molecular formula C 3 H 6, we find that the numbers of hydrogen atoms are two less than that for the corresponding saturated aliphatic hydrocarbon, i.e. propane. Thus, according to our definition its IHD should be 1. You know, that this compound itself does not contain any multiple bond or unsaturation per se. Then how do we account for the IHD of 1? In this case, the IHD is due to the presence of a ring in the molecule. Thus, a ring contributes to IHD in the same way as a double bond. Similarly, we can consider the contributions of halogen atoms and nitrogen atoms on IHD. These considerations give the following formula for the computation of the index of hydrogen deficiency for a molecule from its molecular formula: Number of hydrogen atoms IHD = Number of carbon atoms 2 Number of halogen atoms Number of nitrogen atoms 2 2 You may note here that the divalent atoms such as oxygen and sulphur are not taken into account while calculating the IHD. Thus, for a compound with the molecular formula as C 4 H 10 O, the index of hydrogen deficiency would be zero as shown below: 1 Structure Elucidation by Integrated Spectroscopic Methods The index of hydrogen deficiency helps us to know about the number of multiple bonds (i.e. double or triple) or rings present in a molecule. The determination of IHD alongwith other tests/data helps us to choose between alternative structures proposed for an unknown compound. 10 IHD = 4 1 = 5 5 = 0 2 It implies that the molecule of C 4 H 10 O does not contain any unsaturation or a ring structure. Why don t you try to calculate the IHD for the compounds with molecular formulas as given in the following SAQ. SAQ 1 Calculate the IHD for the compounds having the following molecular formulae: 57

Miscellaneous Methods (a) C 4 H 4 (b) C 5 H 10 O (c) C 6 H 12 O 6.... Having understood the importance of molecular formula in structure elucidation and the method of computing the index of hydrogen deficiency, let us now take up a comprehensive account of the information about the structural features/units of a molecule available from the different types of spectra exhibited by it. 14.3 STRUCTURAL INFORMATION AVAILABLE FROM DIFFERENT TYPES OF SPECTRA In this section, you will learn about important structured information about a molecule which can be obtained from its different types of spectra. Let us begin with the information available from the mass spectrum of a molecule. 14.3.1 Mass Spectrum i) The m/z value of the molecular ion, M gives the molecular mass and can also be used for generating the molecular formula (subsec. 13.4.1). i iv) The relative intensities of [M 1] and [M 2] peaks can be related to the number and nature of hetero atoms present in a molecule. You would recall from Unit 13 that a typical pattern of M1 and M2 peaks is observed if a chlorine or bromine atom is present in the molecule. The odd molecular mass is indicative of the presence of a nitrogen atom in the molecule. However, this has to be further confirmed by other means or by analysing the fragmentation patterns for the typical nitrogen containing functional groups. The characteristic peaks arising from typical fragmentation patterns of various classes of functional groups such as α cleavage, loss of small molecules such as H 2 O, C 2 H 4, etc. are quite useful. v) Certain peaks which may be attributed to the rearrangement of the molecular ion or its fragments ions also give significant structural leads. You are suggested to have a relook at the subsection 13.4.1 of Unit 13. Table 13.2 containing commonly lost fragments and stable fragment ions observed in the mass spectrum is being reproduced here so as to facilitate you in the interpretation of the mass spectra of the examples being taken up in the next section. Table 13.2: Some commonly lost fragments and the stable fragments in the mass spectrum Commonly lost fragments Fragment lost Peak obtained Fragment lost Peak obtained... CH M 15 OCH 3 3 M. - 31. OH M. 17. CN M. 26. Cl. CH 3 C O M. - 35 M. - 43 58

H 2 C CH 2. M 28. OCH 2 CH 3 M. - 45. CH 2 CH 3 M. 29. M. - 91 CH 2 Structure Elucidation by Integrated Spectroscopic Methods m/z values m/z = 43 m/z = 91 Common stable ions CH 3 C O. CH 2 Ion.. m/z = M - 1 R O. CH R C O 14.3.2 UV-VIS Spectrum i) An absorption in the UV-VIS region of the spectrum, i.e., the UV-VIS spectrum of a molecule indicates the presence of certain functional groups that have characteristic n π*, π π * transitions. You may refer to Table 2.1 of Unit 2. These transitions are so characteristic that the absence of a UV-VIS spectrum for a molecule eliminates the presence of a number of functional groups in the molecule. The λ max and the intensity of the absorption bands are indicative of the extent of conjugation in the molecule; larger the wavelength, greater the conjugation. You can refer back to Table 2.2 of Unit 2 for details of the characteristic UV absorptions of various structural units in the molecules. 14.3.3 IR Spectrum You would recall from Unit 3 of Block 1 that the region of IR spectrum ranges from 600-4000 cm 1 and the infrared spectrum can be broadly divided into two regions as explain below. i) The region spanning from 3600 to 1200 cm 1 is called the functional group region and includes the stretching vibrations or group frequencies of common functional groups. The absorptions in this region provide information about the presence of characteristic functional groups. i The region that includes all frequencies below 1200 cm 1 is called the fingerprint region and includes molecular vibrations, characteristic of the entire molecule or large fragments of the molecule. This region is useful for confirming the identity of a particular molecule by comparing with the spectral absorptions of its authentic sample. In addition to the presence (or even the absence) of a particular functional group, the analysis of IR spectrum gives the following information: a) Presence of hydrogen bonding in the molecule (3200-3600 cm 1 ) b) Geometry (cis or trans) in case of alkenes 59

Miscellaneous Methods c) Substitution pattern of aromatic compounds (690-850 cm 1 ) Table 3.1 containing characteristic group frequencies of some important functional groups is being reproduced here for your reference. Table 3.1: Characteristic group frequencies of some important functional groups Functional group Alkane (C-H) Alcohol (O- H) Structure (IR absorbing bond in bold) R 3 C-H (R= H or C) R RO-H IR frequency range (cm 1 ) 3000-2800 3600-3200 (broad band) 3500-3300 (One band) Functional group Alkyne (C C) Nitrile (C N) Structure (IR absorbing bond in bold) RC CR (R=H or C) RC N (R=C) IR frequency range (cm 1 ) 2260-2100 2260-2200 Amines, Amide (N-H) R N ------ H 3500-3300 (two bands), 1650-1560 (bending) Alkene (C C) R 2 C CR 2 (R=H or C) 1600-1680 Carboxylic acid (O-H) Aldehydes or Ketones (C=O) O C R O O C R R H 3200-2500 (broad band) 1750-1705 Benzene ring (C=C) Alcohol or Ether (C-O) (C=C ring breathing ) --- (R=H or C) 1450-1600 2 to 3 bands 1300-1000 O Carboxylic acid (C=O) C R OH (R=H or C) O 1790-1680 Alkyne (C-H) RC C H (R=C) O 3300 Amide (C=O) Acid anhydride (C=O) C R NR 2 (R-H or C) O O C C R O R (R=H or C) 1850-1800 and 1790-1730 1850-1800 and 1790-1740 (two bands) Acid chloride (C=O) Nitro Compound (NO 2 ) R C Cl (R=H or C) O R N O (R = C) 1815-1790 1660-1500 (asymmetric stretch) 1390-1260 (symmetric stretch) 1 H-NMR Spectrum 60

You would recall from Unit 12, the following spectral features of the 1 H-NMR and the structural information available from them. Spectral data tells us about the following important aspects of a molecule. Structure Elucidation by Integrated Spectroscopic Methods i) The number of different signals in the 1 H-NMR spectrum indicates about the different types of protons present in the molecule. The position of the signals i.e. their chemical shift values, tells about the electronic environment of a particular proton. The chemical shifts of different types of protons are given in Fig. 14.1. i iv) Fig. 14.1: Range of Chemical shift values for different types of protons The area under the peaks obtained from the integrals for the signals of various types of protons provides information about the ratio of the numbers of different types of protons present in a molecule. The spin-spin splitting pattern of a particular signal gives information about the number of neighbouring protons present around the given type of protons. You can also refer back to Unit 12, particularly Sections 12.6, 12.7 and 12.10 for the details of the above aspects. Having learnt about the information available from different types of spectral data, we can now take up how to use it to arrive at the structure of a given compound. You can study about this in detail in the next section after achieving the following SAQ. SAQ 2 Which spectrum you will be referring to in order to look for the presence of the following in a molecule? i) Conjugation i Hydrogen bonding Hetero atom........ 14.4 STRUCTURE ELUCIDATION OF ORGANIC MOLECULES 61

Miscellaneous Methods In this section, we will discuss the structural elucidation by taking some organic compounds as examples. We will analyse the data available from various types of spectra for a given compound and integrate the information about various structural units present in the compound to arrive at its structure. Example 1 A compound A, having molecular formula C 3 H 6 O, shows the following spectra. Deduce its structure. Fig. 14.2: Mass spectrum of compound A Fig. 14.3: IR spectrum of compound A 62

Structure Elucidation by Integrated Spectroscopic Methods Fig. 14.4: 1 H-NMR spectrum of compound A Solution There is no unique way of solving a problem of structure elucidation. What we need to do is to pick up all the important information available from the spectra and then integrate them to arrive at the structure. Sometime one or two spectra give enough clues to the structure. In such cases we suggest a tentative structure and check that whether it supports the other data. Let us make a beginning. IHD: We start with calculation of IHD. Number of hydrogen atoms IHD = Number of carbon atoms 2 Number of halogen atoms Number of nitrogen atoms 2 2 Therefore, IHD=3 6/2 = 0 The IHD of zero clearly shows that the molecule does not have any unsaturation or a ring structure. IR: A strong absorption around 1700 cm 1 is indicative of a carbonyl group (>C = O); this may account for the oxygen in the molecule. NMR: The NMR spectrum provides two important bits of information as given below. i) A single signal indicates that only one type of protons is present in the molecule i.e., all the protons are equivalent. 1 The chemical shift of δ = 2.2 indicates that the protons are slightly deshielded; may be due to being in the neighbourhood of an anisotropic group (>C = O) Mass: The mass spectrum shows prominent peaks at m/z 15, 43, 58. The M ion appears at m/z 58 while the base peak is at m/z 43. The M peak is in accordance with the molecular formula and the base peak (M 15) is indicative of a stable C OCH 3 ion 63

Miscellaneous Methods which is obtained by the loss of C H 3 radical from the molecular ion. Similarly the peak at m/z =15 can be attributed to the loss of C OCH 3 (M 43) from the molecular ion. Thus, the fragmentation pattern in mass spectrum indicates presence of CH 3 and COCH 3 groups. On the basis of the above information i.e. the presence of >C = O group (from IR), COCH 3 and CH 3 (from MS) we may propose the structure of the compound A to be the following. O CH 3 C CH 3 Compound A This structure also explains the equivalence of all the protons observed in 1 H-NMR spectrum. Thus, we can say that the compound A is acetone having the structural formula as shown above. Example 2 A compound B, having molecular formula C 7 H 8, shows the following spectra. From the given spectra, we will now see what could be its possible structure. Fig. 14.5: Mass Spectrum of Compound B Fig. 14.6: IR Spectrum of Compound B 64

Structure Elucidation by Integrated Spectroscopic Methods Fig. 14.7: 1 H-NMR Spectrum of Compound B Solution IHD: Let us again start with the molecular formula and calculate its IHD, we get Number of hydrogen atoms IHD = Number of carbon atoms 2 Number of halogen atoms Number of nitrogen atoms 2 2 = 7 2 8 1 = 7 4 1 = 4 The IHD of four suggests that the compound does contain a ring and/or double bonds. 1 NMR: The NMR spectrum is quite useful in this case and provides the following information. i) There are two types of protons i The integration of the peaks shows that the number of the two types of protons is in the ratio of 5:3. The signal at δ =7.2 indicates the presence of benzene ring. iv) The other signal for three protons at (δ =2.2) can be attributed to a CH 3 group attached to a benzene ring. => the presence of benzene ring accounts for the IHD of 4 (a ring and three double bonds) IR: There are very important pieces of information available from the IR spectrum. i) The bands in the region of 1400 1600 cm 1 point to the presence of a benzene ring; the signal at 1605 cm 1 is quite characteristic of the same. 65

Miscellaneous Methods Two strong peaks near 690 cm 1 and 750 cm 1 indicate monosubstituted benzene ring. These are due to out of plane bending of aromatic = C H. The overtones of bands are observed between 2000 1667 cm 1. i The signal at about 3050 cm 1 can be ascribed to sp 2 C H stretching, while at about 2900 cm 1 is due to sp 3 C H stretching. Thus, the IR spectrum clearly points to a monosubstituted benzene. Mass: The mass spectrum given in Fig. 14.5 shows the M peak at m/z 92 and the base peak at m/z 91. Looking at the table we can say that this peak could be due to benzyl carbocation, as shown in the following structure. CH 2 Benzyl carbocation The spectral data evidences indicate that compound B to have the following structure CH 3 And, hence the compound B is toluene. Toluene Example 3 A compound C, having molecular formula C 7 H 6 O, exhibits the following spectra: Fig. 14.8: Mass spectrum of compound C 66

Structure Elucidation by Integrated Spectroscopic Methods Fig. 14.9: IR spectrum of compound C Fig. 14.10: 1 H NMR spectrum of compound C Solution To arrive at its structure, we start with the determination of IHD from its molecular formula. We get, IHD = 7 2 6 1 = 7 3 1 = 5 Again, there may be a possibility that the ring and/or the double or triple bonds may be present in the structure of the molecule. IR: Similar to the previous compound the IR spectrum, suggests for a monosubstituted benzene. i) The aromatic C = C stretch is observed between 1600 1450 cm 1. 67

Miscellaneous Methods Two strong peaks near 690 cm 1 and 750 cm 1 are indicative of out of plane i iv) = C H bending for monosubstituted benzene. In addition, the strong signal at 1700 cm 1 suggests for the carbonyl group. The doublet, between 2850 cm 1 and 2750 cm 1 is characteristic of C H stretch of aldehydes. This further qualifies the nature of carbonyl group. Thus, an IHD of 4 out of 5 is accounted for by the presence of benzene unit (which contains one ring and three double bonds) and the fifth by the carbonyl group. Thus, you may see here that the IR spectrum has provided enough leads to suggest a plausible structure. In the molecule, an aldehyde ( CHO) group is linked to the monosubstituted benzene ring. Hence, the compound has the structure C 6 H 5 CHO. Let us look into other spectra and see do they support the assignment. NMR: The NMR spectrum also supports the proposed structure. Let us see the available information. i) The multiplet between δ 7.4 7.9 is suggestive of aromatic protons. The one proton signal at δ 9.8 is quite useful as it is suggestive of an aldehydic proton. The complex multiplet between δ 7.4 7.9 when resolved under high resolution shows that there are three different types of aromatic protons and these are in the ratio of 2:1:2. The multiplet when observed at higher resolution (at 300 MHz) shows a doublet (2H) and two triplets. The doublet at δ =7.5 is due to two ortho protons. The triplet due to (1H) at δ 7.6 is due to para proton and the other triplet is due to 2 meta protons appears at about δ 7.9. Thus, the 1 H-NMR spectral data also supports that the compound is benzaldehyde. Let us see what does the mass spectrum say. Mass: The mass spectrum also supports the suggested structure. Let us analyse the evidence. i) The molecular ion gives an intense M peak at m/z = 106. i iv) An equally intense M 1 peak at m/z 105 is suggestive of the loss of one hydrogen via the α -cleavage process a very favourable fragmentation mechanism. The characteristic peak at m/z 77 can be due to the loss of CHO to give C 6 H 5 ion. A further loss of HC = CH from C 6 H 5 ion yields C 4 H 3 ion; the peak for which is observed at m/z 51 in the mass spectrum. Thus, all the above spectral data confirm that the compound C is benzaldehyde and it has the following structure. CHO Example 4 Benzaldehyde A compound D having molecular formula C 3 H 6 O 2 shows the following spectral data. Find out the structure. 68

Structure Elucidation by Integrated Spectroscopic Methods Fig. 14.11: Mass spectrum of compound D Fig. 14.12: IR spectrum of compound D Fig. 14.13: 1 H NMR spectrum of compound D Solution You can compute the IHD for the molecule to be 1 suggesting a single double bond or a ring. IR: The IR spectrum of this molecule shows a feature that we have not observed so far. Let us see what information do we get from IR. i) The intense peak close to 1700 cm 1 is characteristic of the carbonyl group. The broad band around 3000 cm 1 is characteristic of OH stretching; probably hydrogen bonded OH group. 69

Miscellaneous Methods These two pieces of information indicate towards the presence of a COOH functional group. Let us see other spectra for more information. NMR: We can arrive at the following observations from its NMR spectrum. i) The NMR spectrum gives three signals indicative of three different types of protons. The small peak at a highly downfield δ = 10.5 which is likely to be due to the acidic proton of the COOH. i The typical pattern of a two proton quartet and three proton triplet at about δ 1.2 and δ 2.3 respectively are suggestive of CH 2 CH 3 group. On the basis of the two spectra, the tentative structure of the molecule can be proposed to be CH 3 CH 2 COOH. Let us check whether the mass spectrum supports it or not. Mass: The mass spectrum contains the following information. i) The molecular ion peak M is at m/z 74. i iv) A M 17 peak at m/z 57 is indicating the loss of OH. The intense peak at m/z 29 could be due to the loss of COOH. The peak at m/z 45 probably arises because of COOH ion. As the mass spectrum clearly accounts for the COOH group we can say that the proposed structure is correct. The given compound D is propanoic acid with the following structure. CH 3 CH 2 COOH (Propanoic acid) Example 5 A compound E having molecular formula C 5 H 9 O 2 Cl exhibits the following spectra. Suggest its structure. Fig. 14.14: Mass spectrum of compound E 70

Structure Elucidation by Integrated Spectroscopic Methods Fig. 14.15: IR spectrum of compound E Fig. 14.16: 1 H-NMR spectrum of compound E IHD: As usual we first calculate the IHD from the molecular formula. IHD = 5 2 9 2 1 1 = 5 5 1 = 1 So the molecule has a double bond or a ring. Let us see what do the spectra suggest? IR: The following observations from IR spectrum are given below. i) The strong band in the region 1750 1735 cm 1 indicates C = O stretch. This along with two strong stretching absorptions in the range 1300 1000 cm 1 due to C O indicate the presence of ester function (-O-C=O) 71

Miscellaneous Methods Let us look into other spectra to identify the groups attached to the ester function. NMR i) The 1 H-NMR spectrum shows the presence of four different types of protons. i The two quartets at δ = 4.23 and at δ = 4.39 are for one proton and two protons respectively. The other two signals, a doublet at δ = 1.68 and a triplet at δ = 1.31 are for two protons each. iv) One triplet and one quartet could be attributed to ethyl group ( CH 2 CH 3 ). v) The other quartet and a doublet may be due to CH 3 CH(Cl) unit. We are still not very sure but the tentative structure at this stage is given below. Cl O CH 3 CH C O CH 2 CH 3 let us see how does the mass spectrum help us. Mass: The interpretation of the mass spectra is given below. i) The M peak at m/z 136 is very-very small. The base peak at m/z = 29 suggests for the presence of CH 2 CH 3. i iv) The peak at m/z = 91 (M 45) is suggestive of the loss of OCH 2 CH 3 fragment; the peak at m/z =45 corroborates this. The reasonably intense peak at m/z = 63 does go into support the proposed structure as this may correspond to CH 3 CH(Cl) moiety which may be obtained from the loss of CO from the m/z = 91 peak of CH 3 CH(Cl)CO v) The peak at m/z = 26 supports this. Thus, we can say that the structure proposed above is correct and the given compound has the following structure Cl O Example 6 CH 3 CH C O CH 2 CH 3 Ethylchloropropionate A compound F has a molecular formula C 4 H 11 N and it exhibits the following spectra. Determine its structure on the basis of these spectra. 72

Fig. 14.17: Mass spectrum of compound F Structure Elucidation by Integrated Spectroscopic Methods Fig. 14.18: IR spectrum of compound F Solution Fig. 14.19: 1 H-NMR spectrum of compound F If we calculate the IHD from the molecular formula of compound F by the formula given below, No.of hydrogen atoms No.of nitrogen atoms IHD = No.of carbon atoms 1 2 2 We get, 11 1 IHD = 4 1 2 2 10 = 5 2 = 5 5 = 0 73

Miscellaneous Methods Thus, the compound has no unsaturation or double/triple bond or rings. Mass: The compound contains nitrogen as is evident from its molecular formula. The molecular ion M peak is at m/z at 73 which is in accordance with the nitrogen rule. The base peak at m/z 30 indicates that the compounds could be a primary amine as primary amines gives this characteristic peak due to α -cleavage as shown below.. [R CH 2 NH 2 ] R CH 2 = N H 2 m/z = 30 IR: The interpretation of the IR signal is as follows. i) The two bands in the range of 3500 3300 cm 1 are supportive of the presence of a primary amine as these bands could be arising due to the N H stretching. i The other amines are ruled out as the secondary amines show only one band in this region and tertairy amines do not show any band here. The presence of amino group is supported by the broad band near 800 cm 1 due to the out of plane N H bending and the C N stretching band around 1090 cm 1. NMR: The 1 H-NMR spectrum shows 4 sets of signals at δ= 0.92 (triplet, 3H), δ= 1.16 singlet (2H, broad), δ= 1.3 1.5 (multiplet, 4H) and δ= 2.7 (triplet, 2H).Let us now try to identify the structural units present in the compound on the basis of these signals and their splitting patterns. i) The three proton triplet at δ 0.92 could be due to the protons of a methyl group, next to a CH 2 group, i.e. CH 3 CH 2. The broad singlet due to two H at δ 1.16 could be due to NH 2 protons which do not show coupling with neighbouring protons. i The two proton triplet at δ= 2.7 is quite downfield and is likely to be for a CH 2 group attached to NH 2 group. The splitting into triplet, could be due to the presence of a neighbouring CH 2 group. This yields the following structural unit. CH 2 CH 2 NH 2 iv) The remaining 4 protons appear as a multiplet between δ 1.3-1.5. If we put together this unit and the CH 3 CH 2 unit, we get the structure indicated below: CH 3 CH 2 CH 2 CH 2 NH 2 Butanamine Thus, on the basis of the spectra the compound is found to be butanamine with the structure given above. 14.5 SUMMARY The determination of molecular formula is an important aspect in the structure elucidation of an organic compound. From the molecular formula, the Index of Hydrogen Deficiency can be calculated. The knowledge of IHD helps to know about the possibility of presence of unsaturation and/or rings in a molecule. Different kind of information is available from various spectra of a molecule. The mass spectral data helps us to know the molecular mass of a compound and the fragment ions 74

of the molecule. An analysis of the fragment ion peaks indicates the type of functional groups or structural units that may be present in the molecule. The UV spectral data indicates about the extent of conjugation present in a compound. The IR spectral data tells about different types of functional groups present in a compound. It also gives information about the presence/absence of hydrogen bonding, substituent pattern of aromatic rings etc. The 1 H-NMR reveals the number of different types of protons present in a molecule, their number and the electronic environment, etc. Structure Elucidation by Integrated Spectroscopic Methods Together, the variety of information available from different types of spectra leads us to the elucidation of structure of an unknown compound. The structure-spectra correlation also helps us to predict the spectral data for a given compound. The spectral studies, thus, help a lot in structure determination and identification of compounds. 14.6 TERMINAL QUESTIONS 1. Calculate IHD for the following molecules: (a) C 6 H 6 (b) C 4 H 8 O 2 O (c) (d) 2. A compound G having molecular formula C 6 H 6 O exhibits the following spectra as given in Fig. 14.20 to 14.22. Fig. 14.20: Mass Spectrum of compound G Fig. 14.21: IR Spectrum of compound G 75

Miscellaneous Methods Fig.14.22: 1 H-NMR Spectrum of compound G Arrive at its structure and correlate different signals with structural units present in the molecule. 3. Predict the important spectral signals which you would expect in the IR, 1 H-NMR and mass spectra of the following compounds: i) Ethyl ethanoate Benzoic acid 4. Which type of spectral data will you use to differentiate between isomeric compounds having molecular formula C 3 H 6 O? 14.7 ANSWERS Self Assessment Questions 1. (a) 3 (b) 1 (c) 1 2. (i) UV ( IR (i Mass Terminal Questions 1. The IHD can be calculated with the help of following formula. Number of hydrogen atoms IHD = Number of carbon atoms 2 Number of halogen atoms Number of nitrogen atoms 1 2 2 i) IHD= (6-6/2) 1 = 31 = 4 IHD= (4-8/2) 1 = 1 i The molecular formula for the given compound is C 6 H 8. The corresponding IHD is 76

iv) IHD= (6-8/2) 1= 21=3 The molecular formula for the given compound is C 4 H 6 O. The corresponding IHD is IHD= (4-6/2) 1= 11=2 Structure Elucidation by Integrated Spectroscopic Methods 2. The structure of the compound G on the basis of the given spectra is benzyl acetate 3. The most significant spectral signals expected in the IR, 1 H-NMR and mass spectra are as follows: i) Ethylethanoate, CH 3 COOC 2 H 5 Mass : The molecular ion peak would appear at m/z = 88. The fragment CH 3 CO is very stable and is likely to give an intense signal at m/z=43. accordingly, we would expect a signal at M-43 i.e., at m/z =45. Other important fragments are expected at m/z =29 and 15 (for ethyl group and methyl group). IR : The most important signal in the IR spectrum of ethyl eethanoate is for the carbonyl absorption. As this is a part of the ester linkage it is expected to appear at around 1740 to 1750 cm -1. NMR: The NMR spectrum is expected to be quite simple. The ethyl part of the molecule would give a combination of a triplet ( for CH 3 ) and a quartet (for CH 2 ). The methyl group triplet is expected to be around δ = 1-1.5 and the CH 2 quartet is likely to go downfield to about 4 ppm. As the other methyl group protons cannot have any coupling interactions, it is expected to appear as a singlet around 2 ppm (due to the deshielding effect of the carbonyl group). Benzoic acid, C 6 H 5 COOH Mass i) The molecular ion M peak is expected at m/z = 122. The loss of OH is quite favorable and is expected to give a peak at m/z=105 77

Miscellaneous Methods i Similarly, the loss of COOH is also favourable and is expected to give a peak at m/z=77; the expected ion being 6 H 5 iv) A further loss of HC = CH from 6 H 5 4 H 3 ion; the peak for which may be observed at m/z 51. IR In benzoic acid we can expect 3-4 important signals/ features in the IR spectrum. These are as follows: i) Characteristic bands in the range of 1600 cm -1 for the benzene ring The carbonyl group at around 1700-1750 cm -1 A broad signal about 3000 cm -1 for hydrogen bonding i Characteristic pattern in the range of 700-750 cm -1 for the mono substituted benzene. NMR The aromatic protons are expected as a multiplet between δ 7.4 7.9 due to the difference in chemical shift positions of the ortho, meta and para protons and the coupling between them. The acidic proton of the COOH group is expected to be quite downfield around 11-12 ppm. 4. Let us first write down the structural formulae for the possible isomers of the compound with the molecular formula, C 3 H 6 O. O i) CH 3 C CH 3 H 3 C CH 2 C H O i CH 2 CH CH 2 OH iv) C H 3 C CH 2 OH These compounds can be differentiated by using UV-VIS, IR, NMR or Mass spectrometry; NMR being the most convenient. The NMR of the compound i) would give just one signal in NMR whereas the compounds and iv) would give three different signals each. On the other hand the compound i would give as many as four signals. The compounds and iv) can be further differentiated by the location of the signals and the splitting pattern. 78

SOME USEFUL BOOKS AND WEBSITES 1. Robert M. Silverstein, Francis X. Webster and David Kiemle (2005). Spectrometric Identification of Organic Compounds, 7 th edition. Wiley. Structure Elucidation by Integrated Spectroscopic Methods 2. William Kemp (1991). Organic Spectroscopy, 3 rd edition. W.H. Freeman & Company. 3. Douglas A. Skoog, F. James Holler and Timothy A. Nieman. (1998) Principles of Instrumental Analysis, 5th edition. Brooks/Cole. 4. S.M. Khopkar (2008) Basic Concepts in Analytical Chemistry, 3rd edition. New Age International publishers. http://www.cem.msu.edu/~reusch/virtualtext/spectrpy/nmr/nmr2.htm#nmr11 http://www.chemistry.nmsu.edu/studntres/chem435/lab8/intro.html http://www.chem.uic.edu/web1/ocol/spec/ms.htm http://base-peak.wiley.com/ http://www.chem.ucla.edu/~webspectra/solvingspectralproblems.html 79

Miscellaneous Methods Absorption signals 26 Acetone 26, 63 Analysers 43 Anisotropic 19, 21, 31 Applications of mass spectrometry Qualitative applications 46 Quantitative applications 51 Applications of NMR spectroscopy Qualitative applications 28 Ethanol 28 Methanol 30 Benzyl alcohol 30 Phenol 31 Base peak 39 Benzaldehyde 67, 68 Boltzmann constant 9 Boltzmann distribution law 9, 12 Broad or wide line 27 Butanamine 73 Characteristics of mass spectrum 38 Chemical exchange 30, 32 Chemical ionisation 42 Continuous wave spectrometry 12 Coupling constant 14, 23 Deshielding 15, 16, 17 Desorption sources 41 Diamagnetic effect 16, 17 Direct insertion probe 41 Double focussing analysers 44 Electromagnets 25, 43 Electron ionisation 42 Ethychloropropionate 71 Factors affecting chemical shift Anisotropy of chemical bonds19, 20, 33 Electronegativity 19 Hydrogen bonds 21 Field ionisation 42 Field sweep method 12 Fourier transform spectroscopy 14 Fourier transformation 14 Fragment ions 39 Fragmentation by cleavage at a single bond 48 by α-cleavage in molecules with heteroatoms 48 due to intramolecular rearrangement 48, 49 Free induction decay 14, 32 Frequency sweep method 12 Gas phase sources 41 Index of hydrogen deficiency 57, 58 Induced magnetic field 15, 16, 19 Instrumentation for mass spectrometry Inlet device 40, 41 Ionisation chamber 40 Analyser 40, 43 Detector 40, 44, 45 Processing and output devices 40, 45 Instrumentation for NMR spectroscopy Magnet 25 Sample probe 25 Detector system 25 INDEX Sample handling 26 Integral modes 26 Integration trace 26 Isotopic peaks 39 Larmor frequency 11 Larmor precession 10, 12 Longitudinal relaxation 13 Magnetic moment 7, 8, 9 Magnetic sector analyzer 43 Magnitude of coupling constants 23 Mass spectrum 38, 40 McLafferty rearrangement 48 Mechanism of resonance 11 Molecular ion 38 Multiplicity of spin-spin interactions 23 Natural abundance 39 Nitrogen rule 46 Nuclear g-factor 7 Paramagnetic effect 17 Pascal s triangle 22, 23 Permanent magnets 25 Precessional motion 10, 11, 12, 34 Propanoic acid 69 Quantisation 8 Quantitative applications Analysis of multicomponent mixtures 27 Elemental analysis 27 Organic functional group analysis 28 Radical cation 38 Radical ion 38 Relaxation mechanisms 12, 13 Representation of NMR 26 Resolution or sharp line spectrum 27 Scan 14, 15 Shielding 15 Shielding constant 16 Shielding mechanism 16 Spin angular momentum quantum number 6, 7, 13, 32 Spin down 9 Spin flip 12 Spin up 9 Spin-lattice relaxation 13, 32 Spin-spin coupling 21, 23 Spin-spin relaxation 13 Standard for chemical shift 17 Structure elucidation of organic compounds 61 Superconducting magnets 25 Tetramethylsilane 17, 21, 26 Theory of mass spectrometry 38 Theory of NMR spectroscopy 6 Toluene 65 Transducer 41 Transverse relaxation 13 Types of nuclei 6 Unit of chemical shift 17 80