PAPER No.12 :Organic Spectroscopy MODULE No.30: Combined problem on UV, IR, 1 H NMR, 13 C NMR and Mass - Part II

Transcription

1 Subject Chemistry Paper No and Title Module No and Title Module Tag 12 : rganic Spectroscopy 30: Combined problem on UV, IR, 1 H NMR, 13 C NMR and Mass Part-II CHE_P12_M30

2 TABLE F CNTENTS 1. Learning utcomes 2. Introduction 3. Problems and their solutions

3 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.

4 3. Problems and their solutions Q.1 List all possible electronic transitions (i) CH4 (ii) CH3Br (iii) CH3CCH3 and (iv) CH3CH Ans. S. No. Compound Possible electronic transitions (i) CH4 * (ii) CH3Br * & n * (iii) CH3CCH3 n * ; * & n * (iv) CH3CH n * ; * & n * Q.2 The UV spectrum of acetone shows peaks at 280 nm (15) and 190 nm (100). (i) Identify the electronic transition for each. (ii) Which is more intense. Ans. (i) Peaks at 280 nm (15) corresponds to n *, whereas peaks at 190 nm (100) corresponds to *. (ii) * transition is more intense having equal to 100. Q.3 Why oxygen (2), nitrogen (N2), hydrogen (H2) and chlorine (Cl2) are IR inactive? Ans. xygen (2), nitrogen (N2), hydrogen (H2) and chlorine (Cl2) are linear in nature, having (3N-5 = = 1) fundamental vibrational mode of freedom, which is symmetric stretch. Since during symmetric stretch vibration, the dipole moment of the molecule does not change, hence this mode of vibration is IR inactive or these molecules are IR inactive.

5 Q.4 How could you distinguish between ethanol and acetone using IR spectroscopy? Ans. Ethanol (CH3CH2H) Acetone (CH3CH3) (i) Shows broad absorption band at (i) Broad absorption band at cm -1 due to -H stretching in cm -1 due to -H stretching in intermolecular hydrogen-bonded alcohol molecules. (ii) Shows strong absorption band at cm -1 (usually at 1050 cm -1 ) due to C- stretching. intermolecular hydrogen-bonded alcohol molecules is absent. (ii) Shows strong absorption band at cm -1 (usually at 1100 cm -1 ) due to C- -C stretching. Q.5 Arrange the following in the order of increasing frequency of carbonyl absorption, giving reasons. (I) (II) (III) Ans. Incorporation of the carbonyl group in a small ring (5, 4 or 3-membered), raises the stretching frequency due to the ring strain in the molecule (larger the ring strain in the molecule, larger will be the carbonyl stretching (C=) frequency). The carbon atom of a carbonyl group is sp 2 hybridized, which implies that the bond angles will be 120 and the C- sigma bond has 33% s-character. If this group is incorporated in a small ring, the C- C-C bond angle is reduced to 108 (5-membered ring), 90 (4-membered ring) or 60 (3-membered ring). When this happens, the C-C bonds of the ring assume greater p- character and the C- sigma bond has correspondingly greater s-character. The double bond of the carbonyl group is therefore shorter and stronger, and exhibits a larger stretching frequency. Cyclopentanone (I) appears at 1748 cm -1, having lesser ring strain in comparison to cyclobutanone (1783 cm -1 (III)) and cyclopropane (1850 cm -1 (II) having maximum ring strain).

6 1748 cm -1 (I) 1783 cm -1 (III) 1850 cm -1 (II) Q.6 How many signals are possible in cyclooctatetraene? Suggest a possible value in ppm. Ans: Since all the carbons are chemically and magnetically equivalent, hence cyclooctatetraene will have one signal in 13 C NMR spectrum. The possible value is 132 ppm. Cyclooctatetraene 132 ppm Q.7 What are the possible mass fragmentation of 4-n-butyltoluene? 4-n-butyltoluene Ans. The benzylic bond cleavage expels the fragment of mass 43 (C3H7. ; as a radical).

7 Due to McLafferty rearrangement, a neutral molecule is expelled; propene (C3H6) with mass number 42. Q8. A compound having molecular formula C3H6 absorbs in carbonyl reason in IR spectrum. How can 1 H NMR spectra establish that the compound is a aldehyde or a ketone. Ans. The molecule having molecular formula C3H6 exist in two different structural formulas: Propanal 2-Propanone In propanal, three signals will be obtained whereas in 2-propanone (acetone) only one signal will be obtained. Q9. How many signals are possible in the 13 C NMR spectrum of 1-butene? Ans. 1-Butene has molecular formula C4H8 and the structure given below All the four carbons are in different chemical environments and hence this molecule will show 4 signals in the 13 C NMR spectrum. The possible values in ppm is given in the structure. Q10. Predict the types of functional group may be inferred from the given IR spectrum.

8 Ans.