Question

Chem 08LC Lab Manual Fall 2019 3) Critical analysis (5 points) (a) Your TA will give you a 'H NMR spectrum of the 3-nitroaniline product. Using your NMR knowledge and the special NMR section in the lab book (especially pages 48 O2N. and 49), assign the 4 peaks in the spectrum to each of the protons in the product you will need to use coupling pattern and chemical shift to complete the assignment. (b) Explain your assignment, specifically how you differentiated between HA and He НА NH2 На Но Не

Answer 1

Answer to part a:

Assignment of H_C: The peak at 7.26 ppm is a triplet signal and has a J value of 7.9 Hz. This is typical value for the vicinal coupling of the aromatic protons. The proton H_C has two hydrogen atoms in the adjacent carbon atoms. Therefore, H_C shows shows a triplet with coupling constant value for vicinal coupling for the aromatic protons.

Assignment of H_A: The proton H_A should show a triplet with a coupling constant value for the 4 bond coupling (J ~ 2.1 Hz). The triplet pattern is for the fact that H_A has 4 bond coupling with two protons H_D and H_B. Such type of signal; a triplet with J value of 2.1 Hz is observed at 7.38 ppm. Therefore, the signal at 7.38 ppm is assigned to H_A proton.

Assignments of H_B and H_D: The H_B and H_D protons would exhibit similar pattern NMR signals. The coupling for H_B proton can be explained in the following way.

H_B proton at first couples with H_C proton to give rise a doublet. The doublet further spits to doublet of doublet due to the coupling with H_D proton. The doublet of doublet further splits to doublet of doublet of doublet due to the coupling with H_A proton. The coupling for H_D can also explained in the same way.

Note that the coupling of H_B to H_A and H_D protons in each case is 4 bond coupling. However, the 4 bond coupling first happens with the H_D protons due to the fact that the H_D proton is attached to a more electronegative carbon atom than that of the carbon atom for the H_A proton. The H_D attached carbon atom is in ortho position to the NO₂ group (electron withdrawing group), and in para position to the NH₂ group. whereas the H_A proton is in ortho positions to the NO₂ and NH₂ groups. Since, the +R or -R effect is prominent in ortho position than the para position. Therefore, the H_D proton attached carbon atom is more electronegative than the H_A attached proton. Due to same reason for the +R and -R effect, the doublet of doublet of doublet for H_D proton would be in downfield compared to the doublet of doublet of doublet for H_B proton. Two doublet of doublet of doublets are found at 6.95 and 7.31 ppm. The signal at 7.31 ppm is assigned to the H_D proton whereas the signal at 6.95 ppm is assigned to the H_B proton.

Answer to part b:

As explained above, the signal for the H_A proton is different from the signal of H_D proton by the coupling constant value.