Question

The structure of L is as follows:

In the presence of a small amount of acid, a solution of acetaldehyde \(\left( {{\bf{C}}{{\bf{H}}_{\bf{3}}}{\bf{CHO}}} \right)\)in methanol \(\left( {{\bf{C}}{{\bf{H}}_{\bf{3}}}{\bf{OH}}} \right)\) was allowed to stand and a new compound L was formed. L has a molecular ion in its mass spectrum at 90 and IR absorptions at 2992 and 2941 \({\bf{c}}{{\bf{m}}^{{\bf{ - 1}}}}\). L shows three signals in its\({}^{{\bf{13}}}{\bf{C}}\)-NMR at 19, 52, and 101 ppm. \({}^{\bf{1}}{\bf{H}}\)-NMR spectrum of L is given below. What is the structure of L?

Short answer

The structure of L is as follows:

Step-by-step solution

Acetal formation

One mole of acetaldehyde reacts with two moles of methanol in an acidic medium to form 1,1-Dimethoxyethane. Acetals are important groups found in sugar.

Analyzing the data

Given Data

• Mass/charge ratio = 90
• IR absorption= 2992 \({\rm{c}}{{\rm{m}}^{{\rm{ - 1}}}}\) and 2941 \({\rm{c}}{{\rm{m}}^{{\rm{ - 1}}}}\)

The molecular formula of the compound is evaluated from the value of molecular ion\(\)

\(\begin{aligned}{c}{\rm{Number of carbon atoms = }}\frac{{{\rm{90}}}}{{{\rm{12}}}}\\{\rm{ = 7}}\;{\rm{remainder}}\;{\rm{6}}\end{aligned}\)

This indicates that the compound may have 7 C atoms and 6 hydrogen atoms,

Replace one carbon with 7 hydrogen atoms,

Replace one CH with an O atom and again replace one CH by an O atom.

The molecular formula was found to be \({{\rm{C}}_{\rm{4}}}{{\rm{H}}_{10}}{{\rm{O}}_2}\).

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

Where,

n= Number of carbon atoms

x= (Number of hydrogen atoms) + (Number of halogen atoms) - (Number of nitrogen atoms)

On substituting values,

\(\begin{aligned}{c}{\rm{The degree of unsaturation }}\left( {{\rm{IHD}}} \right){\rm{ = }}\frac{{\left( {\left( {{\rm{2 \times 4 + 2}}} \right){\rm{ - 10}}} \right)}}{{\rm{2}}}\\{\rm{ = 0}}\end{aligned}\)

The value of unsaturation suggests the absence of pi-bond.

From the\({}^{\rm{1}}{\rm{H}}\)-NMR Spectra,

• The doublet at 1.2 ppm attributes to 3H atoms due to the split of 1H atom,
• The singlet at 3.3 ppm attributes to 6H atoms due to two methyl groups,
• The quartet at 4.8 ppm attributes to 1H atoms due to the split of three adjacent H atoms,

From the\({}^{{\rm{13}}}{\rm{C}}\)-NMR Spectra, the three signals suggest the presence of three types of carbon atoms,

• The signal at 19 ppm attributes to C of C-H bond,
• The signal at 52 ppm attributes to C of C-C bond,
• The signal at 101 ppm attributes to C of C-O bond,

The IR vibrational stretching at 2992 \({\rm{c}}{{\rm{m}}^{{\rm{ - 1}}}}\) and 2941 \({\rm{c}}{{\rm{m}}^{{\rm{ - 1}}}}\)indicates the presence of \({{\rm{C}}_{{\rm{s}}{{\rm{p}}^{\rm{3}}}}}{\rm{ - H}}\) bond.

Proposing structure