Propose a structure consistent with each set of data.

a. C₉H₁₀O₂: IR absorption at 1718cm^-1

b. C₉H₁₂ : IR absorption at 2850–3150cm^-1

The consecutive arrangement of different atoms in a compound in accordance with the geometry of the species gives the molecular structure of the compound.

Chemical formula =C₉H₁₀O₂

IR absorption= 1718cm^-1

The degree of unsaturation (IHD) $=\frac{\left(\right(2n+2)-x)}{2}$

where

n= number of carbon atoms

x= (Number of hydrogen atoms) + (Number of halogen atoms) - (Number of nitrogen atoms)

On substituting the values,

The degree of unsaturation (IHD) $=\frac{\left(\right(2\times 9+2)-10}{2}$

= 5

The value of IHD suggests the presence of a benzene ring (3 pi-bonds and the ring) and another pi-bond.

From the NMR spectra,

This leaves behind three more carbon atoms;

b) the quartet at 4.4 ppm attributes to the protons having 3H on the neighboring C. This downfield shift indicates the presence of an electronegative element attached to the carbon;

c) the triplet at 1.4 ppm attributes to the protons having 2H on the neighboring C; and

d)the IR vibrational stretching at 1718 indicates the presence of carbonyl stretching of a ketone group.

Chemical formula =C₉H₁₂

IR absorption= 3150-2850 cm^-1

The degree of unsaturation (IHD) ,$=\frac{\left(\right(2n+2)-x}{2}$

where

n= number of carbon atoms

x= (Number of hydrogen atoms) + (Number of halogen atoms) - (Number of nitrogen atoms)

On substituting the values,

the degree of unsaturation (IHD) $=\frac{\left(\right(2\times 9+2)-12}{2}$

=4

The value of IHD suggests the presence of a benzene ring (3 pi-bonds and the ring).

From the NMR spectra,

This leaves behind three more carbon atoms;

The required molecular structure is as follows:

a.

b.