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# Nuclear Magnetic Resonance

Example of how to solve an 1H-NMR Spectroscopy problem Formula: C6H14O H‐NMR: 3.3 ppm (triplet; integral = 1)  1.6 ppm (sextet; integral = 1)  0.90 ppm (triplet; integral = 1.5) 1. The first step is to calculate the DBE (double bond equivalency) value, which will tell us how  many pi bonds are present or if there is a potential benzene ring present in the structure. DBE = # of carbons – (# of hydrogens ÷ 2) + (# of nitrogens ÷ 2)  + 1         = 6 – (14/2) + (0/2) + 1          = 0 (this means there are no pi bonds within the structure) 2. Knowing that the integral is proportional to the relative number of equivalent hydrogen  atoms present within the structure, we can use the integral information given in the  problem and the number of hydrogens, also given in the problem through the molecular  formula, to find the factor in which to multiply the integral by so as to determine how many  hydrogen atoms are present within each signal. The first signal had an integral = 1 The second signal had an integral = 1 The third signal had an integral = 1.5  If you multiply each of these numbers by a factor of four, then you get: 4 hydrogen atoms for the first signal 4 hydrogen atoms for the second signal 6 hydrogen atoms for the third signal 3. Now you can determine the implications for each integral and splitting pattern. ‐ Before you find each implication, you need to find the number of neighbors each  signal has.  This is revealed through the splitting pattern given in the problem. The first signal is a triplet.  From the general rule stated above, the signal for a  proton with n neighbors is split into n+1 lines.  Because there are 3 splits in the  signal, this means that there are 3‐1 neighbors, which equals two neighbors.   The second signal is a sextet.  This means that there are 6‐1 neighbors, which  equals five neighbors.