Question: Propose a structure consistent with each set of data.

a. Compound A:

Molecular formula: ${\mathit{C}}_{\mathbf{8}}{\mathit{H}}_{\mathbf{10}}\mathit{O}$

IR absorption at role="math" localid="1648804967273" $\mathbf{3150}\mathbf{-}\mathbf{2850}\mathbf{}\mathit{c}{\mathit{m}}^{\mathbf{-}\mathbf{1}}$

${}^{\mathbf{1}}\mathit{H}$NMR data: 1.4 (triplet, 3 H), 3.95 (quartet, 2 H), and 6.8–7.3 (multiplet, 5 H) ppm

b. Compound B:

Molecular formula: ${\mathit{C}}_{\mathbf{9}}{\mathit{H}}_{\mathbf{10}}{\mathit{O}}_{\mathbf{2}}$

IR absorption at role="math" localid="1648805107903" $\mathbf{1669}\mathbf{}\mathit{c}{\mathit{m}}^{\mathbf{-}\mathbf{1}}$

${}^{\mathbf{1}}\mathit{H}$NMR data: 2.5 (singlet, 3 H), 3.8 (singlet, 3 H), 6.9 (doublet, 2 H), and 7.9 (doublet, 2 H) ppm

The study of the interaction of matter with electromagnetic radiations (such as UV, IR, radio waves, etc) as a function of frequency (or wavelength) is known as spectroscopy.

Spectroscopy plays a great role in the characterization of various compounds.

The molecular formula of the compound is $C_8{H}_{10}O$.

IR absorptions at $3150-2850c{m}^{-1}$ corresponds to $s{p}^2$and $s{p}^3$C-H stretching.

Analyzing the ${}^{1}H$NMR data:

Triplet at 1.4 ppm (3H) corresponds to a methyl group adjacent to a $C{H}_2$group.

Quartet at 3.95 ppm (2H) corresponds to the $C{H}_2$ group adjacent to a $C{H}_3$group.

Multiplet at 6.8-7.3 ppm (5H) corresponds to the protons in a benzene ring.

On combining the points.

The obtained structure is:

Structure of compound A

The molecular formula of the compound is $C_9{H}_{10}{O}_2$.

IR absorption at $1669c{m}^{-1}$ corresponds to the CO $\left(C=O\right)$ stretching of an aldehyde or ketone.

Analyzing the ${}^{1}H$ NMR data:

• The singlet at 2.5 ppm (3H) corresponds to a deshielded methyl group and there is no hydrogen atom present adjacent to it.
• The singlet at 3.8 ppm (3H) corresponds to a deshielded$C{H}_3$group and the carbon atoms adjacent to the methyl group do not contain any hydrogen atoms.
• Doublet at 6.9 ppm (2H) corresponds to the aromatic protons.
• Doublet at 7.9 ppm (2H) corresponds to the aromatic protons.

The substituents on the benzene ring should be arranged in a manner that two of the hydrogen atoms on the benzene ring are more deshielded than the other two.

The proposed structure of the given compound is:

Structure of compound B