Question

Reaction of aldehyde D with amino alcohol E in the presence of NaH forms F (molecular formula \({{\bf{C}}_{{\bf{11}}}}{{\bf{H}}_{{\bf{15}}}}{\bf{N}}{{\bf{O}}_{\bf{2}}}\)). F absorbs at 1730 \({\bf{c}}{{\bf{m}}^{{\bf{ - 1}}}}\) in its IR spectrum. F also shows eight lines in its \({}^{{\bf{13}}}{\bf{C}}\)-NMR spectrum, and gives the following \({}^{\bf{1}}{\bf{H}}\)-NMR spectrum: 2.32 (singlet, 6 H), 3.05 (triplet, 2 H), 4.20 (triplet, 2 H), 6.97 (doublet, 2 H), 7.82 (doublet, 2 H), and 9.97 (singlet, 1 H) ppm. Propose a structure for F. We will learn about this reaction in Chapter 18.

Short answer

The structure of F is as follows:

Step-by-step solution

Reaction of tertiary amines

Due to lone pair of electrons, amines would act as nucleophiles and bond at an electrophilic center.

The benzaldehyde reacts with amino alcohol in the presence of the strong base sodium amide to form a tertiary ammonium salt.

Analyzing data

Given data

Chemical formula =\({{\rm{C}}_{11}}{{\rm{H}}_{{\rm{15}}}}{\rm{N}}{{\rm{O}}_{\rm{2}}}\)

IR absorption= 1730 \({\rm{c}}{{\rm{m}}^{{\rm{ - 1}}}}\)

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

Where,

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

On substituting values,

\(\begin{aligned}{c}{\rm{The degree of unsaturation }}\left( {{\rm{IHD}}} \right) = \frac{{\left( {\left( {{\rm{2}} \times 11{\rm{ + 2}}} \right){\rm{ - }}\left( {15 - 1} \right)} \right)}}{{\rm{2}}}\\ = 5\end{aligned}\)

The value of IHD suggests the presence of a benzene ring (3 pi-bonds and the ring) and another pi-bond. The eight signals in \({}^{{\rm{13}}}{\rm{C}}\)-NMR suggests the presence of eight types of carbon atoms.

From the NMR spectra,

• The singlet at 2.32 ppm attributes to the 6H atoms corresponding to the two methyl groups on the nitrogen atom.
• The triplet at 3.05 ppm attributes to the 2H atoms adjacent to the nitrogen atom
• The triplet at 4.20 ppm attributes to the 2H atoms adjacent to the hydroxyl group.
• The doublet at 6.97 ppm attributes to the 2H atoms on the benzene ring that are equivalent in nature.
• The doublet at 7.87 ppm attributes to the 2H atoms on the benzene ring that are equivalent in nature and are present next to the alpha-carbon atom.
• The singlet at 9.97 ppm attributes to the 1H atom of the 1H located para to the alpha- carbon atom.

The IR vibrational stretching at 1730 \({\rm{c}}{{\rm{m}}^{{\rm{ - 1}}}}\) indicates the presence of carbonyl stretching of a ketone group.

Proposing the structure