Q.56. - Organic Chemistry

Question

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   $\mathbf{3150}\mathbf{-}\mathbf{2850}\mathbf{ }\mathit{c}{\mathit{m}}^{\mathbf{-}\mathbf{1}}$

${}^{\mathbf{1}}\mathit{H}$ 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   $\mathbf{1669}\mathbf{ }\mathit{c}{\mathit{m}}^{\mathbf{-}\mathbf{1}}$

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

Step-by-Step Solution

Answer

Answer

Step 1: Spectroscopy

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.

Step 2: The structure of compound A

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

IR absorptions at $3150-2850 c{m}^{-1}$  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

Step 3: The structure of compound B

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

IR absorption at $1669 c{m}^{-1}$  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