Paper 12: Organic Spectroscopy

Transcription

1 Subject hemistry Paper No and Title Module No and Title Module Tag Paper 12: Organic Spectroscopy 34: ombined problem on UV, IR, 1 H NMR, 13 NMR and Mass- Part 6 HE_P12_M34

2 TABLE OF ONTENTS 1. Learning Outcomes 2. Introduction 3. Problems and their solutions 5. Summary

3 1. Learning Outcomes After studying this module, you shall be able to Know how to interpret a spectra Learn to deduce the structure Identify the peaks in the spectra 2. Introduction If only the spectra are provided one should know how to collect information from them and assemble all the structural features obtained to a final structure. By looking at the spectra we will deduce the structure of the compounds. 3. Problems and their solutions Q.1 The following compound with the formula 4H8O2 is an ester. Give its structure and assign the chemical shift values. Solution DBE = [(2 x 4+2)-8]/2 =1 The DBE =1, and it is given that the compound is an ester, thus one double bond is accounted by O-=O (ester) group. The total integral (2+3+3 = 8) is equal to the number of hydrogen atoms in the molecular formula. Therefore, each integral represents the actual number of hydrogen atoms. The quartet at 4.2 ppm, for two hydrogens and triplet at 1.3 ppm for 3 hydrogens, indicate H 2H 3 substructure. The downfield position of H 2 at 4.2 ppm indicates its attachment with oxygen atom. Hence, it is OH 2H 3 moiety. A singlet at 2.1 ppm, indicates H 3 group. Thus, the structure of the compound is: H 3OOH 2H 3 (or Ethyl acetate) Q.2 A monosubstituted aromatic hydrocarbon has molecular formula 9H12. The 1 H NMR spectrum is given. Identify the structure of the compound.

4 Solution DBE = [2 x ]/2 = 4 The DBE = 4 may be due to an alkene or a aromatic ring. But as given it is an aromatic hydrocarbon, thus four double bonds are accounted by a phenyl ring. The total integral (5+1+6 = 12) is equal to the number of hydrogen atoms in the molecular formula. Therefore, each integral represents the actual number of hydrogen atoms. Septet at 2.9 ppm for 1 H and doublet at 1.3 ppm, indicate presence of iso-propyl linkage. Peak at 7.3 ppm for 5 H, is due to mono substituted aromatic ring. Thus, the compound is iso-propyl benzene. H 3 H H 3 Ques 3. A carboxylic acid 4H7O2Br shows peak at ppm. The 1 H NMR is shown. Identify the compound. Solution: The most downfield signal at ppm is due to the carboxylic proton, of the carboxylic acid. The one double bond by DBE is accounted by the =O of carboxylic acid group. DBE = [2 x ]/2 = 1 The total integral ( = 7) is equal to the number of hydrogen atoms in the molecular formula. Therefore, each integral represents the actual number of hydrogen atoms. Triplet at 4.2 ppm for 1H, represents -H 2H- linkage. Trriplet at 1.1 ppm for 3 H, indicate H 3H 2- linkage. Multiplet at 2.1 ppm, for 1H is due to H-. Thus the structure is

5 H 3 H 2 H Br O OH Q.4 The NMR spectra of a monosubstituted hydrocarbon is provided, its molecular formula is 10H14. Deduce the structure Quartet 2 1 Solution: DBE = [2 x ]/2 = 4 It is given that the compound is a monosubstituted hydrocarbon, the four double bond equivalents are accounted by the phenyl ring. A triplet for 3H indicates presence of H2H3 group. A doublet for 3H, indicates H3-H- linkage. A quartet for 2H, indicates H-H2-H 3 linkage. A sextet for 1H, indicate H 3-H-H 2 linkage. ombining these linkages, the side chain will be H 3-H-H 2-H 2-H 3 Thus the structure of the compound is H 3 H H 2 H 2 H 3 Ques 5. A compound with molecular formula 10H12O, shows band near 1715 and cm -1 in IR spectrum. The 1 H NMR is given, Deduce its structure. Solution: DBE = 2 x ]/2 = 5 Band near 1750 cm-1 in IR show presence of keto group (=O). The DBE is 5, one double bond is accounted by =O group, the other four double bonds can be due to aromatic ring. Which is further confirmed by peaks near 7 ppm, which is due due protons of benzene ring. The triplet and quartet are due to H 3H 2- linkage. A singlet near 3.7 ppm is due to, protons of the carbon attached to aromatic ring. Thus the structure is

6 H 2 O H 2 H 3 Ques 6. The Mass, UV, 1 H and 13 NMR of a compound are given. Identify the compound. Solution: Mass spectra: The mass spectra show two peaks of equal intensity, separated by 2 units at 198 and 200. This indicates presence of bromine. The presence of bromine is also confirmed from the molecular formula given, 9H 11Br. The m/e = 91, indicates Ph-H 2- group. UV spectrum:

7 The ultraviolet spectrum shows a typical benzenoid absorption without further conjugation or auxochromes. This is onsistent with the presence of Ph-H2- group. From the molecular formula DBE can be calculated: DBE = 4 Thus, either there are four double bonds or an aromatic ring in the compound. 1 H NMR From the 1 H NMR the number of hydrogens for each peak can be found: hemical shift 7.2 ppm Multiplicity multiplet triplet triplet quintet Integral Relative no of H 19/8= 2.37 (multiply by 2) 8/8=1 (multiply by 2) 8/8=1 (multiply by 2) 8/8=1 (multiply by 2) No of H 4.74 ~ 5H Thus, total = 11 protons are present. In the 1 H NMR, 5 protons near 7.2 ppm indicates, a monosubstituted benzene ring, which supports presence of Ph-H 2- group. The triplet at 3.3 ppm and 2.8 ppm for 2H each, indicate a -H 2H2- group, the peak at 3.3 ppm is relatively downfield thus this -H2- group is attached with electron withdrawing Bromine. Thus it is H 2H2Br A quintet (or pentet) at 2.2 ppm for 2H, suggest presence of -H 2H2H 2- moiety. 13 NMR and DEPT spectrum: 13 spectrum shows total 7 peaks, out of this 4 peaks for aromatic or olefinic carbons, 3 peaks for aliphatic carbons. The molecular formula has 9 carbons. The 13 DEPT spectrum shows 3 H peaks in the aromatic/olefinic range and 3 H 2 peaks in the aliphatic chemical shift range There is one peak in the 13 NMR spectrum, which is absent in 13 DEPT spectrum. This indicates one quaternary (non-protonated) carbon. This is for the substituted carbon of the aromatic ring. There are 4 resonances in the aromatic region, 3 x H and 1 x quaternary carbon, which is typical of a monosubstituted benzene ring. Thus the sub-structures present are: PhH 2 -H 2-H 2-H 2-Br Since the total number of carbons is 9 the compound is 1-bromo-3-phenylpropane.

8 Ques 7. From the IR, UV, NMR spectrums given, identify the compound.

9 Solution: The molecular formula is given as 9H 11NO 2. DBE = 5. Double bond equivalent indicates one aromatic ring and one double bond. Mass spectra The mass spectrum has difference of 28 units between 165 (M) and 137 which suggests loss of ethylene (H2=H2) or O. UV Spectrum alculate the extinction coefficient from the UV spectrum: In the UV spectrum, the presence of extensive conjugation is apparent from the large extinction coefficient (17,267). IR spectrum Infra red spectrum, shows a strong absorption at 1680 cm -1 and may be due to a =O stretch at an unusually low frequency (such as an amide or strongly conjugated ketone). 1 H NMR spectrum The numbers of protons in different environments are: hemical shift 7.9 ppm 6.6 ppm 4.3 ppm 4.0 ppm 1.4 ppm Integral Relative no of H No of H Thus, there are a total of = 11 protons which is consistent with the molecular formula provided. The triplet and quartet pattern is due to an ethyl group and the downfield shift of the H 2 at 4.3 ppm indicates that it must be attached to a heteroatom so this is possibly an -O-H 2-H 3 group.

10 13 NMR spectrum From the 13 spectrum there are 7 types of carbon present in the molecule. 4 carbons are in the aromatic/olefinic chemical shift range. 2 carbons in the aliphatic chemical shift range and 1 carbon at low field (167 ppm) characteristic of a carbonyl carbon. But, the molecular formula 9H 11NO 2 has 9 carbon atoms. So there must be some symmetry in the molecule. From the 13 off-resonance decoupled spectrum there are 2 H resonances in the aromatic/olefinic chemical shift range, one H 2 and one H 3 carbon in the aliphatic chemical shift range. The presence of 4 proton symmetrical pattern in the aromatic region near 7.9 and 6.6 ppm indicates a para disubstituted benzene ring. This is also, confirmed by the presence of two quaternary 13 resonances at 152 and 119 ppm in the 13 spectrum and two H 13 resonances at 131 and 113 ppm. The 4 resonances in the aromatic region (2 x H and 2 x quaternary carbons) confirm the presence of a p- disubstituted benzene ring. The signals at 14 ppm (H 3) and 60 ppm (H 2) in the 13 NMR spectrum confirm the presence of the ethoxy group. The quaternary carbon signal at 167 ppm in the 13 NMR spectrum indicates an ester or an amide carbonyl group. Thus the following structural are present in the molecule: The presence of an -NH 2 group is confirmed by the exchangeable signal at 4.0 in the 1 H NMR spectrum and the characteristic N-H stretching vibrations at cm -1 in the IR spectrum. The presence of one aromatic ring plus the double bond in the carbonyl group is consistent with the calculated degree of unsaturation there can be no other rings or multiple bonds in the structure. Thus it has the following structure (ethyl 4-aminobenzoate): The amine -NH 2 group is confirmed as it is exchangeable with D 2O. The 1 H chemical shift of the -O-H 2- at 4.37 ppm is due to the ester structure. The 13 chemical shifts of the quaternary carbons in the aromatic ring aromatic ring are at approximately 152 and 119 ppm.

11 The fragmentation pattern in the mass spectrum at m/e 137, 120 and 92 can be rationalised as follows: