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C,H14O 13C NMR: 1Η ΝMR: 29.7, t 29.8, q (3) 46.4, s 60.0, t 0.91, s, 9H 1.53, t, J = 7.3 Hz, 2H 2.13, bs, IH (exchanges) 3.70, t, J = 7.3 Hz, 2H |

C,H14O 13C NMR: 1Η ΝMR: 29.7, t 29.8, q (3) 46.4, s 60.0, t 0.91, s, 9H 1.53, t, J = 7.3 Hz, 2H 2.13, bs, IH (exchanges) 3.70, t, J = 7.3 Hz, 2H |

Chemistry
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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Determine the structure. Note that for the 13C data, multiplicity is given.  So a CH3 group would be a quartet (n+1) and a CH2 group would be a triplet, etc

 

 

 

CH140 13C NMR: 1Η ΝMR: 29.7, t 29.8, q (3) 46.4, s 60.0, t 0.91, s, 9H 1.53, t, J = 7.3 Hz, 2H 2.13, bs, lH (exchanges) 3.70, t, J = 7.3 Hz, 2H CH100 13C NMR: 'H NMR: 23.5, t 26.7, t (2) 68.9, t (2) 1.7, m, 6H 3.72, t, J = 7.4 Hz, 4H lim PPha A C12H1402 13C NMR: 'H NMR: 2.27, s, 3H 27.2, q 68.8, t 72.9, t 4.20, d, J = 4.5 Hz, 2H 4.57, s, 2H 127.7, d (2) 127.9, d 6.35, d, J = 16.1 Hz, IH 6.80, dt, J= 16.1, 4.5 Hz, IH 7.3, m, 5H 128.5, d (2) 130.4, d 137.6, s 142.9, d 198.0, s
Transcribed Image Text:CH140 13C NMR: 1Η ΝMR: 29.7, t 29.8, q (3) 46.4, s 60.0, t 0.91, s, 9H 1.53, t, J = 7.3 Hz, 2H 2.13, bs, lH (exchanges) 3.70, t, J = 7.3 Hz, 2H CH100 13C NMR: 'H NMR: 23.5, t 26.7, t (2) 68.9, t (2) 1.7, m, 6H 3.72, t, J = 7.4 Hz, 4H lim PPha A C12H1402 13C NMR: 'H NMR: 2.27, s, 3H 27.2, q 68.8, t 72.9, t 4.20, d, J = 4.5 Hz, 2H 4.57, s, 2H 127.7, d (2) 127.9, d 6.35, d, J = 16.1 Hz, IH 6.80, dt, J= 16.1, 4.5 Hz, IH 7.3, m, 5H 128.5, d (2) 130.4, d 137.6, s 142.9, d 198.0, s
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