Compounds A and B have molecular formula C₉H₁₀O . Identify their structures from the ¹HNMR and IR spectra given.

Spectroscopy is an important characterization technique used in analytical chemistry.

A combination of various spectrums can be used to identify the structure of an unknown compound.

The molecular formula of the given compound is C₉H₁₀O .

IR absorption at 1700 cm^-1 indicates the presence of a carbonyl (C=O) group.

IR absorption at 2700 cm^-1 is the characteristic of aldehydic CH stretching.

Triplet at 1.2 ppm indicates that the carbon adjacent to the one to which the concerned proton contains two hydrogen atoms. (According to the n+1 rule, where n indicates the number of hydrogen atoms on the adjacent carbon).

Quartet at 2.7 (number of adjacent hydrogen atoms is 3).

Doublet at 7.3 (2 hydrogen atoms on a benzene ring).

Doublet at 7.7 (2 hydrogen atoms on benzene ring present adjacent to an electron-withdrawing group).

(Two signals in the aromatic region indicate that the compound is para-substituted).

The singlet is at 9.9 (characteristic of the aldehydic proton).

Combining all the points, the predicted structure of compound A is:

Compound A

IR absorption at 1700 cm^-1 indicates the presence of a carbonyl (C=O) group.

IR absorption at 2700 cm^-1 is characteristic of aldehydic CH stretching.

Two triplets at 2.85 and 2.95 represents -CH₂-CH₂ group.

The multiplet at 7.2 corresponds to aromatic protons. (the benzene ring is mono-substituted).

The singlet is at 9.8 (characteristic of an aldehydic proton).

Combining all the points, the predicted structure of compound B is:

Compound B