spectroscopy Practice #2 1. This question is about the three organic compounds i
ID: 1044546 • Letter: S
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spectroscopy Practice #2 1. This question is about the three organic compounds involved in the following reaction. (a) The infrared spectra of all three compounds showed several absorptions. Describe what happens on a molecular level when molecules absorb infrared radiation. (b) Use the following information about their infrared spectra to deduce which bonds are present in the three compounds. All three compounds showed an absorption close to 1200 cm-1 There were broad absorptions in both W and X. The one in W was centred around 3000 cm, and in X around 3400 cm- Compounds W and Y showed absorptions close to 1700 cm-1 bonds in W. bonds in X bonds in Y .. The 'H NMR spectra of the three compounds were available. State what can be deduced from each of the following. (e) () The presence of two peaks in the spectrum of w. The presence of a triplet and a quartet, with areas in the ratio 3:2, respectively, in the spectra of both X and Y (ii)Explanation / Answer
(a) Absorption of IR radiation is a typical of molecular species that have closely lying rotational and vibrational states. A criterion for IR absorption is there should be a net change in dipole moment in a molecule as it vibrates or rotates. As the molecule wobbles / vibrates , there is a fluctuation in its dipole moment; this causes a field that interacts with the electric field associated with radiation. If there is a match in frequency of the radiation and the natural vibration of the molecule then the resonance occurs , which compels the absorbtion and in turn tweaks the amplitude of the molecular vibration. Thus the choice of functional groups are detremential in the IR spectroscopy which shows appreciable fluctuations in the dipole moment during vibrations.
(b) THe molecule W is CH3-COOH.
The molecule X is Ethyl alcohol which is undergoing dehydration reaction leaving out a molecule of water.
W shoes broad peak near 3000cm -1 which is due to the OH in the carboxylic acid.
The broad band near 3400 cm-1 corresponds to the alcoholic OH in the ethyl alcohol (X).
Y is the ester dehydration product where the C=O stretching occurs near the 1735 cm-1.
(C) One peak corresponds to the three H of CH3 and the other labile OH proton appear downfield.
For both the ester and the alcohol the ratio of the Hs in CH2 and CH3 are similar. In the ester the CH3 H's are symmetrical and there is a CH2 which are in the intensity of 3:2.
The C2H5OH also has the CH3 and CH2 peak in the same ratio due to the presence of H in the ratio 3:2.
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