Subject Chemistry Paper No and Title Module No and Title Module Tag 12: rganic Spectroscopy 29: Combined problem on UV, IR, 1 H NMR, 13 C NMR and Mass - Part I CHE_P12_M29
TABLE F CNTENTS 1. Learning utcomes 2. Introduction 3. Problems and their solutions
1. Learning utcomes After studying this module, you shall be able to Solve problem related with electronic transitions Learn how to differentiate molecule on the basis of IR spectroscopy Correlate spectra with structure of compound Interpret the spectroscopic data 2. Introduction The knowledge and concepts of UV-visible, IR, 1 H NMR, 13 C NMR and Mass help us in solving problems based on the experimental data. It will help us in analysing the experimental data to elucidate the structure of any organic compound. While analysing the data the following point must be kept in mind: In UV-visible spectroscopy; the types of bonds and electrons plays important role in understanding the electronic transitions. UV-visible spectroscopy gives information regarding the presence of conjugation, carbonyl group etc. The IR values gives information regarding the functional group present in the molecule The 1 H NMR tells us the number and environment of neighbouring hydrogens present. The 13 C NMR helps in getting the information about the type of carbon atom(s) present in the molecule. Mass spectral data gives information about the total mass and fragmentation pattern of the molecule. By combining all the information one can find the structure of the molecule.
3. Problems and their solutions Q.1 What are the sources for UV and Visible light in UV-visible spectrophotometer? Ans. The source of UV light in spectrophotometer: Hydrogen-deuterium discharge lamp The source of visible light in spectrophotometer: Tungsten Filament lamp Q.2 Calculate max for the given compound using Woodward-Fieser Rules. Ans. This is an example of heteroannular diene. There are four ring residues on the double bonds. Base value: 215 nm Ring residue = 5 4 = 20 nm Total value = 235 nm Q.3 What are the possible infra-red bands in case of 1-butyne? Ans. The chemical structure of 1-butyne is: The possible important absorption bands in infra-red spectroscopy are: C-H stretching: 3300 cm -1 C-H stretching in CH2 and CH3: 2850 2950 cm -1 CC stretching: 2100 cm -1 C-H bending: 1450 cm -1 Q.4 How will you distinguish o-hydroxybenzaldehyde and p-hydroxybenzaldehyde on the basis of IR spectroscopy?
Ans. In o-hydroxybenzaldehyde, there is a possibility of intramolecular hydrogen bonding. Due to the formation of hydrogen bonds between >C=..H--; the IR values will be shifted towards lower wave numbers. Also, concentration does not cause any further shift in the frequencies of >C= and H--. In p-hydroxybenzaldehyde, the possibility of intramolecular hydrogen bonding is not possible. These types of molecule undergo intermolecular hydrogen bonding at high concentration. n dilution with non-polar solvents, the frequencies of >C= and H--, will be shifted towards higher wave number. Q.5 A compound having molecular formula C6H102 shows three 1 H NMR signals with relative intensity 1:1:3 at 3.99, 2.60 and 1.00 ppm respectively. IR spectrum shows a significant peak at 1771 cm -1. Deduce the structure of the compound. Ans. Double bond equivalent (DBE) for C6H102 is 2. The presence of IR value at 1771 cm -1 indicates that the compound is a cyclic ester. This is also in accordance with DBE. The total number of proton in the molecule is 10 with intensity ratio 1:1:3, so the possibility is 2:2:6. Hence, the compound has two non-equivalent CH2 group and two equivalents CH3 groups. A possible structure of the compound is
Q.6 A compound having molecular formula C4H102 shows the presence of a peak at 3400 cm -1 in its IR spectrum. The 13 C NMR spectrum shows peaks at δ 66; δ 37 and δ 17. 1H NMR shows the presence of an octet for one proton and doublet for 3 protons apart from other peaks. Deduce the structure of the compound. Ans. Double bond equivalent (DBE) for C4H102 is 0. The peak at 3400 cm -1 is for -H group and hence the compound is an alcohol or carboxylic acid. The presence of carboxylic group is ruled out from DBE and 13 C NMR. The chemical shift in 13 C NMR at δ 66 indicates that the carbon is attached with oxygen through single bond. The splitting of 1H NMR shows the partial structure as: H 2 C H 3 C CH 2 H From the combined provided data, the possible structure is 66 H H 2 C H 3 C 17 H 37 CH 2 66 H Q.7 Deduce the structure of an organic compound having molecular mass 88 and 1 H NMR as follows: 1.22 (t, 3H); 2.00 (s, 3H) and 3.99 (q, 2H) Ans. The presence of a triplet for 3H and quartet for 2H at 1.22 and 3.99 ppm shows that the partial structure CH3-CH2- is a part of the final structure. The presence of singlet at 2.00 ppm for three protons may be due to CH3-N or CH3-C. The presence of one nitrogen is ruled out as the molecular mass is even, ie 88. The possible structure is CH3-CH2--C-CH3
Q8. Can you distinguish between inter and intra-molecular hydrogen bonding on the basis of 1 H NMR spectroscopy? Explain. Ans. Yes, we can distinguish between inter and intra-molecular hydrogen bonding on the basis of 1 H NMR spectroscopy. The intermolecular hydrogen bonding is concentration and temperature dependent. The hydrogen bonding is more in the case of concentrated solution. This results in the deshielding to proton and resonance occurs at higher value of. n the other hand, intramolecular hydrogen bonding does not get affected with dilution. Q9. Comment on the mass spectra of compounds containing one bromine atom. Ans. Bromine exists in two isotopic forms with equal abundance, ie 79 Br and 81 Br. A compound which contains one bromine atom exhibits a pair of peaks of almost equal intensity and the peaks are 2 mass units apart. Q10. What are the special features of amines that help them to identify in mass spectroscopy. Ans. (a) The amines are basic compounds and hence has tendency to capture protons. Thus, M+1 peak shows increase in intensity. (b) An odd molecular mass of the compound indicates the presence of nitrogen atom. (c) The salt form of amines does not volatilize rather decomposes to give a free amine and acid. For example, if acid component is HCl or HBr, then strong peaks at m/e 36 and 38 for HCl and m/e 80 and 82 for HBr are obtained.