OC307-11-Solving NMR - Organic Chemistry 307 Solving NMR...

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Organic Chemistry 307 Solving NMR Problems – H. D. Roth A Guide to Solving NMR Problems NMR spectroscopy is a great tool for determining structures of organic compounds. As you know 1 H spectra have three features, chemical shift, signal intensity, and multiplicity, each providing helpful information. In this document we show how you use these features together to assign structures from 1 H and 13 C spectra. Use this approach. I Begin with a general examination of the spectrum: a) determine how many signals (clusters) there are in the 1 H spectrum and note their chemical shifts, and count the signals in the 13 C spectrum; b) determine how many 1 H nuclei are present in each cluster (look at the integration); c) determine the multiplicity of each signal cluster (count the peaks or read the label attached to the peak); d) determine what additional information is available (often the molecular formula is given; that gives you a chance to subtract groups you’ve identified and to check what is left to identify); e) compare the number of 1 H signals with those in the 13 C spectrum; this may tell you whether all carbon atoms bear 1 H nuclei or whether there are carbons without 1 H nuclei attached. II Now proceed to the interpretation of the spectrum: f) for each peak match the number of 1 H nuclei and its multiplicity with a corresponding peak and combine them to groups (coupling is mutual); g) record each group you recognize (that’ll give you partial credit J ); h) when you have assigned all 1 H groups that you recognize, add their formulas up and compare with the molecular formula; the difference will be a group or groups that you haven’t identified as yet or cannot see in the 1 H spectrum; step e) above will give you similar information. i) enter the “difference” and combine all fragments. III Check your work: make sure that you have accounted for all signals and for every atom of the molecular structure and that your combined structure satisfies all chemical shifts. That’s all, folks.
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Solving NMR Problems – H. D. Roth Lets begin by looking at a really quite simple compound; it has a molecular formula of C 4 H 6 Cl 2 . The 1 H spectrum has 3 signals and the 13 C spectrum shows 4 signals; make a note that there must be a C without any H attached. 0 2 4 6 8 10 PPM tri 3 H q 2 H s 1 H For your interpretation of the individual peaks, it’s best to start at one of the edges, low field or high, left or right. At low field (left) our spectrum has a singlet (1 H) at 9.7 ppm; that chemical shift is unmistakable: it can only indicate an aldehyde. Because this signal is a singlet (n + 1 = 1; n = 0), there cannot be any 1 H nuclei on the adjacent carbon. You can enter the aldehyde fragment without neighbor in the “box provided” for your first piece of partial credit. (I place groups with low-field signals on the left, but that is not crucial). H
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This note was uploaded on 09/30/2011 for the course ORGANIC CH 01:160:307 taught by Professor Babiarz during the Fall '11 term at Rutgers.

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OC307-11-Solving NMR - Organic Chemistry 307 Solving NMR...

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