O que é MMN?

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Chapter 3 Suggested solutions for Chapter 3 11 the force constant f (more or less equals bond strength), and the reduced mass (µ) is given by equation. = 1 Suggested solution The reduced mass (µ) of a vibrating bond is given by this equation. m₂ the reduced mass of OH is 16/17 or about 0.94. When you double the mass of H, the reduced of OD becomes 32/18 or about 1.78 which is roughly double that for OH. But when you the mass of O, the reduced mass for SH is 32/33 or about 0.97 - hardly changed from OH! The change in reduced mass is enough to account for the changed frequency of the OD bond - it by about - but cannot account for the change from OH to SH as the two reduced are about the same. The only explanation of this can be that the SH bond is weaker than the OH bond by a factor of about 2. There is an important principle to be deduced from this problem. Very roughly, the reduced masses bonds involving heavier atoms (C, N, 0, S, etc.) are about the same and the differences in IR frequency are mostly due to changes in bond strength. This is most dramatic in comparing double, and triple bonds. Only with bonds involving hydrogen does the reduced mass become the more important factor, though it is also significant in comparing, say, C-O with C-Cl. Problem 5 Three compounds, each having the formula C₃H₅NO, have the IR spectra summarized here. are their structures? Without NMR data, it may be easier to tackle this problem by writing down all the possible structures for C₃H₅NO. In what specific ways would data help? a One sharp band above 3000 cm one strong band at about 1700 cm⁻¹ Two sharp bands above 3000 cm⁻¹; two bands between 1600 and 1700 cm⁻¹ : One strong broad band above 3000 cm a band at about 2200 cm⁻¹ Purpose of the problem show that IR alone has some usefulness in the identification of molecules but that NMR is even with very simple molecules. In answers to examination questions of this type it is to show how you interpret the data as well as to give a structure. If you get the answer right, interpretation is not so important, but, if you get the answer wrong, you should still get some credit for your interpretation. solution sharp band above 3000 cm⁻¹ must be an N-H and one strong band at about 1700 cm⁻¹ is probably a That leaves C₂H₄ and so we might have one of these (there are other less likely structures). ¹³C NMR would help because a(i) would have a carbonyl group and two signals for saturated C while a(ii) would also have a C=O but only one signal for saturated carbon as the compound is symmetrical. HN a(i) HN 0