5.11: Proton NMR problems
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- Solve unknown problems using 1H NMR spectra and molecular formula.
Helpful resources for solving these types of problems:
1. Degree of Unsaturation Equation
3. Coupling Constant Data Table
You may also want to read through some worked problems on how to solve unknown structure determination problems (Section 5.10).
Determine the structure for the unknown molecule with the molecular formula of C5H10O2.
1H NMR: The ratio of protons is 2:3:2:3. J = 7 Hz for all coupling.
- Answer
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Determine the structure for the unknown molecule with the molecular formula of C5H10O2.
1H NMR: The ratio of protons is 1:1:1:1:3:3. The peak at 9.5 ppm is a singlet. The peak at 2.41 ppm is sextet (J = 7 Hz). The peak at 1.72 ppm is a multiplet (J = 7Hz, 25 Hz). The peak at 1.53 ppm is a multiplet (J = 7 Hz, 25 Hz). The peak at 1.20ppm is a doublet (J = 7 Hz). The peak at 0.95 ppm is a triplet (J = 7 Hz).
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Note: The -CH2- protons are diastereotopic, so they show up differently in 1H NMR spectra.
Determine the structure for the unknown molecule with the molecular formula of C10H12O2.
1H NMR: The ratio of protons is 5:1:3:3. The peak at 7.5 ppm is a multiplet. The peak at 3.70 ppm is quartet (J = 7 Hz). The peak at 3.58 ppm is a singlet. The peak at 1.48 ppm is a doublet (J = 7 Hz).
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Determine the structure for the unknown molecule with the molecular formula of C10H15N.
1H NMR: The ratio of protons is 1:2:2:4:6. The peak at 7.3 ppm is a triplet (J = 2Hz) and the peak at 6.24 ppm is a doublet (J = 2 Hz).
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Determine the structure for the unknown molecule with the molecular formula of C7H14O.
1H NMR: The ratio of protons is 1:3:2:3:2:3. The peak at 4.36 ppm is a triplet of quartets (J = 17 Hz and 7Hz). The peak at 3.37 ppm is a singlet. The peak at 2.04 ppm is a doublet of triplets (J = 17 Hz and 7Hz). The peak at 1.56 ppm is a doublet (J = 7 Hz). The peak at 1.48 ppm is a sextet (J = 7 Hz). The peak at 0.85 ppm is a triplet (J = 7 Hz).
- Answer
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