Question

Compound O has a molecular formula \({{\bf{C}}_{{\bf{10}}}}{{\bf{H}}_{{\bf{12}}}}{\bf{O}}\) and shows an IR absorption at 1687 \({\bf{c}}{{\bf{m}}^{{\bf{ - 1}}}}\). The \({}^{\bf{1}}{\bf{H}}\)-NMR spectrum of O is given below. What is the structure of O?

Short answer

The arrangement of atoms or elements of a compound to satisfy the energy barriers and geometry of the molecule is known as the structural formula of the compound.

Step-by-step solution

Structural formula

The arrangement of atoms or elements of a compound to satisfy the energy barriers and geometry of the molecule is known as the structural formula of the compound.

Analysing data

1. Given data:

Chemical formula =\({{\rm{C}}_{{\rm{10}}}}{{\rm{H}}_{{\rm{12}}}}{\rm{O}}\)

IR absorption= 1687 \({\rm{c}}{{\rm{m}}^{{\rm{ - 1}}}}\)

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

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) \(\begin{aligned}{c} = \frac{{\left( {\left( {{\rm{2}} \times 10{\rm{ + 2}}} \right){\rm{ - 12}}} \right)}}{{\rm{2}}}\\ = 5\end{aligned}\).

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

From the NMR spectra,

1. The multiplet at 7.4-8 ppm attributes to the 5H atoms on the benzene ring that has a standard chemical shift from 6.5-8 ppm. This leaves behind three more carbon atoms.
2. The triplet at 1.0 ppm attributes to the 3H atoms due to the split of 2H adjacent protons.
3. The sextet at 1.7 ppm attributes to the 2H atoms due to the split of protons of ethyl group.
4. The triplet at 2.9 ppm attributes to the 2H atoms due to the split of 2H adjacent protons.
5. The IR vibrational stretching at 1687 \({\rm{c}}{{\rm{m}}^{{\rm{ - 1}}}}\) indicates the presence of a ketone group.

Proposing structure

Molecular structure of O