1H NMR Interpretation Tricks are useful for organic synthesis structure determination. This article discusses the approach to NMR problems.

Problem

An organic compound (C7H12O2) exhibited the following data in the 1H NMR spectrum.
7.10 (1H, d t, J 16 and 7.2Hz),
5.90 (1H, d t, J 16 and 2 Hz),
4.1 (2H, q, J 7.2Hz,)
2.10 (2H, m),
1.25(3H, t, J 7.2Hz ),
0.90 (3H, t, J = 7.2 Hz) ppm.
The compound, among the choices given below, is:

solution 1H NMR Interpretation Tricks

Four steps involved in solving this problem

  1. Identification of chemical nonequivalent protons
  2. prediction of the multiplicity of protons
  3. allotting chemical shift values
  4. interpreting coupling constants of protons

First step: Identification of chemical nonequivalent protons

Number of signals is equal to number of non equivalent sets of protons

Ha = 3H, Hb = 2H, Hc = 1H, Hd = 1H, He = 2H, Hf = 3H

Ha = 3H, Hb = 2H, Hc = 1H, Hd = 1H, He = 2H, Hf = 3H

Ha = 3H, Hb = 2H, Hc = 1H, Hd = 1H, He = 2H, Hf = 3H

Ha = 3H, Hb = 2H, Hc = 1H, Hd = 1H, He = 2H, Hf = 3H

Analysis: NMR data provided in the problem has a similar number of chemical non-equivalent protons as shown in all options

7.10 (1H, d t, J 16 and 7.2Hz),
5.90 (1H, d t, J 16 and 2 Hz),
4.1 (2H, q, J 7.2Hz,)
2.10 (2H, m),
1.25(3H, t, J 7.2Hz ),
0.90 (3H, t, J = 7.2 Hz) ppm.

Conclusion: all options have a similar number of chemical non-equivalent protons. Hence cannot interpret the final structure just based on the identification of different proton signals.

Second step: prediction of the multiplicity of protons

Multiplicity of protons = n+1 (n = number of neighborhood protons)

Chemical non equivalent protons: Ha = 3H , Hb = 2H, Hc = 1H, Hd = 1H, He = 2H, Hf = 3H

Multiplicity :

  • Ha multiplicity = n+1
    here n = Hb ie 2H(neighborhood protons)
    Ha = 2 + 1 = 3 triplet (t)
  • Hb multiplicity = n+1
    here n = Ha + Hc ie 3H + 1H = 4H
    Hb = 4 + 1 = 5 multiplet (m)
  • Hc multiplicity = n+1
    here n = Hb + Hd ie 2H + 1H = 3H
    Hc = 3 + 1 = 4 quartet (q) (d t) coupling based interactions
  • Hd multiplicity = n+1
    here n = Hc ie 1H
    Hd = 1 + 1 = 2 doublet (d) (d t) coupling based interactions
  • He multiplicity = n+1
    here n = Hf ie 3H
    He = 3 + 1 = 4 quartet (q)
  • Hf multiplicity = n+1
    here n = He ie 2H
    Hf = 2 + 1 = 3 triplet (t)

Chemical non equivalent protons and multiplicity: Ha = 3H(t) , Hb = 2H(m), Hc = 1H(q) (d t), Hd = 1H(d) (d t), He = 2H(q), Hf = 3H(t)

Analysis of all four options show similar multiplicity as per the data comparison

7.10 (1H, d t, J 16 and 7.2Hz),
5.90 (1H, d t, J 16 and 2 Hz),
4.1 (2H, q, J 7.2Hz,)
2.10 (2H, m),
1.25(3H, t, J 7.2Hz ),
0.90 (3H, t, J = 7.2 Hz) ppm.

Conclusion: all options have a similar multiplicity patterns. Hence cannot interpret the final structure just based on the multiplicity.

Third step: allotting chemical shift values

Analysis Chemical shift values of options A and B matches with given NMR data. while option C and D do not match as

  • He here is next to carbonyl carbon that does not match the value 4.1.
  • Also, chemical shift values of unsaturated protons Hc and Hd vary depending upon the position of the oxygen atom.

7.10 (1H, d t, J 16 and 7.2Hz),
5.90 (1H, d t, J 16 and 2 Hz),
4.1 (2H, q, J 7.2Hz,)
2.10 (2H, m),
1.25(3H, t, J 7.2Hz ),
0.90 (3H, t, J = 7.2 Hz) ppm.

Conclusion Options C and D can be eliminated based on chemical shift value anlysis.

Fourth step: interpreting coupling constants of protons

The coupling constant: the separation between two adjacent lines in the NMR signal. Possible couplings include

  • CIS/ TRANS coupling: three bond coupling cis (5 to 13 Hz) Trans (13 to 18), trans coupling constant greater than cis.
  • allylic coupling: four bond coupling (0 to 2 Hz)
  • vicinal coupling: three bond coupling (0 to 13 Hz)

Hd and Hc interacts to give doublet (16 Hz), Hd and Hb interacts to give doublet of triplet (2 Hz)

Analysis Option A with trans coupling matches with the NMR data given in the problem as shown below

7.10 (1H, d t, J 16 and 7.2Hz),
5.90 (1H, d t, J 16 and 2 Hz),
4.1 (2H, q, J 7.2Hz,)
2.10 (2H, m),
1.25(3H, t, J 7.2Hz ),
0.90 (3H, t, J = 7.2 Hz) ppm.

Conclusion option A is the correct answer for the given problem.

1H NMR Interpretation Tricks: Here we explained in detail four steps chronologically to solve the NMR problem is to clarify concepts to students. while solving in the exam to reach the final answer depending upon time constraints and demand of question you can start directly from chemical shift values or coupling constant analysis to eliminate options.

Also read: Mass spectral fragmentation

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