Q.6. - Organic Chemistry

Question

Question: What is the structure of a compound of molecular formula ${\mathit{C}}_{\mathbf{10}}{\mathit{H}}_{\mathbf{14}}{\mathit{O}}_{\mathbf{2}}$  that shows a strong IR absorption at 3150–2850 $\mathit{c}{\mathit{m}}^{\mathbf{-}\mathbf{1}}$ and gives the following 1 H NMR absorptions: 1.4 (triplet, 6 H), 4.0 (quartet, 4 H), and 6.8 (singlet, 4 H) ppm

Step-by-Step Solution

Answer

Answer 

Structure of compound 

Step 1: Significance of IR and NMR data

IR and NMR data can provide chemical information about the compound that can help in determining the structure of an organic molecule

 

The presence of a specific functional group, chemical environment of an atom can be deduced using IR and NMR spectra data.

Step 2: Elucidating the structure of organic compound

Molecular formula:  $C_{10}{H}_{14}{O}_2$

 $$\begin{gathered} Degree of unsaturation = \frac{2\times 10+2+0-14-0}{2} \\ =\frac{22-14}{2} \\ =4 \end{gathered}$$

 

 Degree of unsaturation 4 suggests presence of benzene ring in the molecule.

IR absorption at $3150-2850 c{m}^{-1}$  is for $s{p}^2$ C-H stretching and $s{p}^3$ C-H stretching

 

NMR data suggests there are three sets of protons.

 

Therefore, based on the data, the structure can be:

1,4-diethoxybenzene