Question: Identify the structures of A and B, isomers of molecular formula${\mathbf{C}}_{\mathbf{10}}{\mathbf{H}}_{\mathbf{12}}{\mathbf{O}}_{\mathbf{2}}$ , from their IR data and¹H NMR spectra.

a. IR absorption for A at 1718cm^-1

b.IR absorption for B at 1740 cm^-1

IR spectroscopy can be employed to recognize unknown samples. Infrared radiation is utilized for analysis in this technique.

In ¹H NMR spectra, shielded protons comprise lower ppm values, while darkshielded protons comprise higher ppm values.

a In molecule A, the IR absorption at 1718cm^-1 corresponds to the C=O group. In the ¹H NMR data, the triplet at 1.4 ppm corresponds to adjacent to ${\mathrm{CH}}_3$the two hydrogens.

The singlet at 2.4 ppm corresponds to${\mathrm{CH}}_3$ a group. The quartet at 4.4 ppm corresponds to ${\mathrm{CH}}_2$a adjacent to the${\mathrm{CH}}_3$ . The doublet at 7.2 ppm indicates two hydrogens on the benzene ring. Hence the structure of the compound A can be given as:

Structure of compound A

b. In molecule B, the IR absorption at 1740 cm^-1 indicates the presence of the C=O group. In the ¹H NMR spectrum, the singlet at 2 ppm corresponds to the presence of group.

The triplet at 2.9 ppm corresponds to a situated next to the group. The triplet at 4.4 ppm corresponds to situated adjacent to the group. The multiplet at 7.3 ppm indicates five hydrogens and it is monosubstituted benzene.

Hence, the structure of compound B can be given as:

Structure of compound B