Question

Draw the structure of the compound identified by the following simulated 1H and 13C NMR spectra. The molecular formula of the compound is C10H12O. (Blue numbers next to the lines in the 1H NMR spectra indicate the integration values.)

Answer 1

Concepts and reason

The concept used in this problem is Spectroscopy. Spectroscopy is a technique used to elucidate the molecular structure of unknown sample. It is non-destructive technique, and a very small amount of sample is required. The interaction of matter with electromagnetic radiation will result into the spectra recorded will further help in elucidation of structure.

Fundamentals

Various techniques are followed before elucidation of the final molecular structure. In which NMR technique is most important. NMR records the induced transition in spin energy levels of the nucleus by the absorption of radio waves in presence of high magnetic field. Mostly it records about the proton, carbon, and other NMR active nuclei.

But proton (¹H) and carbon (¹³C) are main key factor which helps in structural elucidation.

NMR doesn’t detect the presence of protons; perhaps it distinguishes the proton according to their different chemical environment.

Through 1HNMR, number, nature and position of hydrogen can be detected. Whereas, through ¹³C NMR number nature of carbon can be detected.

Now, firstly the Double Bond Equivalent (DBE) is to be calculated, which determines the degree of unsaturation.

$D.B.E = \left[ {{{\rm{C}}_n} - \frac{{{M_v}}}{2} + \frac{{{T_v}}}{2} + 1} \right]$

Here, ${{\rm{C}}_n}$ is the number of carbon atoms, ${M_v}$ is the monovalent species, and ${T_v}$ is the trivalent species.

The given molecular formula is ${{\rm{C}}_{{\rm{10}}}}{{\rm{H}}_{{\rm{12}}}}{\rm{O}}$ .

That is D.B.E for this molecule will be as follows:

$\begin{array}{c}\\D.B.E = \left[ {10 - \frac{{12}}{2} + \frac{0}{2} + 1} \right]\\\\ = 5\\\end{array}$

This shows that the molecule may have one ring and four double bonds. This confirms the possibility of the presence of an aromatic ring.

The ¹HNMR data is given as follows:

The number of peaks in ¹HNMR spectrum observed is 6. This shows that 6 chemically different protons present in the compound.

Characteristic peaks which help the most in elucidation are as follows:

1. Two peaks for 2H each at the range of 6-8 ppm indicate them as the aromatic ring proton. Doublet in a peak show that both non-equivalent protons couple each other. Confirms 1, 4 disubstituted ring.

2. The peak at the range between 9 to 10 ppm, indicates the presence of aldehydic group (-CHO).

3. The 3 peaks near the range of 2 to 4, indicates the presence of -CH₃, -CH₂, group. The doublet and triplet shows that these methyl and methylene group have coupling with each other.

4. One peak at near to 4 ppm, single peak indicates that it is attached with the aldehydic group. Hence downfield recorded in the chemical shift value.

The ¹³CNMRdata is given as follows:

This spectrum data indicates that there are 6 different peaks showing the presence of 6 different carbons types in the structure of the compound.

So, the structure elucidated by interpreting all data of spectrum is of 2-(4-ethylphenyl) acetaldehyde.

Ans:

The structure of the given organic compound elucidated by the interpretation of spectral data is 2-(4-ethylphenyl) acetaldehyde.