A hydrocarbon, compound B, has molecular formula C6H6, and gave an NMR spectrum with two signals: delta 6.55 pm and delta 3.84 pm with peak ratio of 2:1. When warmed in pyridine for three hr, compound B quantitatively converts to benzene. Mild hydrogenation of B yielded another compound C with mass spectrum of m/z 82. Infrared spectrum showed no double bonds; NMR spectrum showed one broad peak at delta 2.34 ppm. With this information, address the following questions. a) How many rings are in compound C? b) How many rings are probably in B? How many double bonds are in B? c) Can you suggest a structure for compounds B and C? d) In the NMR spectrum of B, the up-field signal was a quintet, and the down field signal was a triplet. How must you account for these splitting patterns

Organic Chemistry
8th Edition
ISBN:9781305580350
Author:William H. Brown, Brent L. Iverson, Eric Anslyn, Christopher S. Foote
Publisher:William H. Brown, Brent L. Iverson, Eric Anslyn, Christopher S. Foote
Chapter13: Nuclear Magnetic Resonance Spectroscopy
Section: Chapter Questions
Problem 13.17P: Following are 1H-NMR spectra for compounds G, H, and I, each with the molecular formula C5H12O. Each...
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A hydrocarbon, compound B, has molecular formula C6H6, and gave an NMR spectrum with two signals: delta 6.55 pm and delta 3.84 pm with peak ratio of 2:1. When warmed in pyridine for three hr, compound B quantitatively converts to benzene. Mild hydrogenation of B yielded another compound C with mass spectrum of m/z 82. Infrared spectrum showed no double bonds; NMR spectrum showed one broad peak at delta 2.34 ppm. With this information, address the following questions.

  1. a) How many rings are in compound C?
  2. b) How many rings are probably in B? How many double bonds are in B?
  3. c) Can you suggest a structure for compounds B and C?
  4. d) In the NMR spectrum of B, the up-field signal was a quintet, and the down field signal was a triplet. How must you account for these splitting patterns?
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