Question

A proton and Carbon-13 NMR for Catechol is described. First, the proton NMR spectrum. The solvent is CDCL3, which is probably the peak at 7.716. It has an integrated value of 0.904. There is a multiplet quartet at 6.819, 6.812, 6.807, 6.807, 6.8007 with an integrated value of 1.029. There is a quartet at 6.678, 6.671, 6.666, 6.659 with an integrated value of 1.000. The transmitter frequency is 500.132508 MHz. Second, the Carbon-13 NMR spectrum. There is a singlet at 206.323. There is a singlet at 145.919. There is a singlet at 120.752. There is a singlet at 116.153. There is a multiplet at 30.257, 30.103, 29.949, 29.795, 29.641, 29.487, 29.333. The transmitter frequency is 125.773542 MHz.

Describe the proton NMR and Carbon-13 NMR using the American Chemical Society (ACS) format for the experimental data.

Answer 1

American Chemical Society (ACS) format for the experimental data.

¹H NMR (500 MHz, CDCl₃, ppm) 6.810-6.80 (m, 1H), 6.678- 6.65 (q, 1H).

¹³C NMR (125 MHz, CDCl3, ppm) 206.3, 145.9, 120.7, 116.1, 30.2, 30.1, 29.9, 29.7, 29.6, 29.4, 29.3.

Correction:

In ¹H NMR the "OH" signals are missing. J values are missing (since the protons are from aromatic ring J value must be given.

Since its the ¹³C NMR spectra of Catechol. It should have 6 signals in ¹³C NMR. All signals are singlet, no multiplets are observed in ¹³C NMR.

In this case 11 signals in ¹³C NMR are observed the extra signals are due to solvent impurity.