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Unformatted text preview: you thinking about the positions of IR bands in terms of the two main influences: reduced mass and bond strength. Suggested solution Using the equation on page 64 of the textbook we find that the reduced mass of OH is 16/17 or about 0.94. When you double the mass of H, the reduced mass of OD becomes 32/18 or about 1.78 – nearly double that of OH. But when you double the mass of O, the reduced mass of SH is 32/33 or about 0.97 – hardly changed from OH! The change in the reduced mass from OH to OD is enough to account for the change in stretching frequency – a change of about √2. But the change in reduced mass from OH to SH cannot account for the change in frequency. The explanation is that the S–H bond is weaker than the O–H bond by a factor of about 2. So both both O–D and S–H absorb at about the same frequency. There is an important principle to be deduced from this problem. Very roughly, all the reduced masses of all bonds involving the heavier elements (C, N, O, S etc) differ by relatively small amounts and the differences in stretching frequency are mainly due to changes in bond strength, though it can be significant in comparing, say, C–O with C–Cl. Solutions for Chapter 3 – Determining Organic Structures 5 With bonds involving hydrogen does the reduced mass becomes by far the important factor. PROBLEM 6 Three compounds, each having the formula C3H5NO, have the IR data summarized here. What are their structures? Without 13C NMR data it might be easier to draw some or all possible structures befo...
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This document was uploaded on 02/10/2014.
 Spring '14

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