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(a) The organic compound can be identified by studying magnetic behavior. This technique is nuclear magnetic resonance spectroscopy or NMR spectroscopy.

This method helps us to find the structure and molecular formula of any organic compound.

The given compound A has the chemical shift values 11.8 ppm, 7.3 ppm, 3.8 ppm, and 1.5 ppm. The given compound bears 5 elements of unsaturation. The hydrogen sets are 1H, 5H, 1H, and 3H.

The chemical shift value of 11.8 ppm with 1H corresponds to the COOH group. The chemical shift value of 7.3 ppm with 5H corresponds to the mono-substituted benzene ring.

The chemical shift value 3.8 ppm with 1H with quartet corresponds to the CHCH₃group. The chemical shift value 1.5 ppm with 3H with doublet corresponds to the CHCH₃ group. Thus, the chemical structure of compound A is:

(b) The given compound B has the chemical shift values 12.1 ppm, 6.2 ppm, 5.7 ppm, and 1.9 ppm. The given compound bears 2 elements of unsaturation. The hydrogen sets are 1H, 1H, 1H, and 3H.

The chemical shift value of 12.1 ppm with 1H corresponds to the COOH group. The chemical shift value 6.2 ppm with 1H corresponds to the H-C=C and it is a singlet. The chemical shift value 5.7 ppm with 1H with singlet corresponds to the H-C=C.

The chemical shift value 1.9 ppm with 3H with singlet corresponds to the vinyl CH₃ group. Thus, the chemical structure of compound B is:

(c) The given compound C has the chemical shift values 12.15 ppm, 7.1 ppm, 5.8 ppm, and 2.2-0.9 ppm. The given compound bears 2 elements of unsaturation. The hydrogen sets are 1H, 1H, and 1H.

The chemical shift value of 12.15 ppm with 1H corresponds to the COOH group. The chemical shift value 7.1 ppm with 1H corresponds to the H-C=C-COOH and it is a multiplet.

The chemical shift value 5.8 ppm with 1H with doublet corresponds to the C=C(H)COOH. The chemical shift value 2.2-0.9 ppm corresponds to the vinyl CH₂CH₂CH₃ group. Thus, the chemical structure of compound C is: