Question

Question: Compound X (molecular formula ${\mathbf{\text{C}}}_{\mathbf{\text{10}}}{\mathbf{\text{H}}}_{\mathbf{\text{12}}}\mathbf{\text{O}}$) was treated with, ${\mathbf{\text{NH}}}_{\mathbf{\text{2}}}{\mathbf{\text{NH}}}_{\mathbf{\text{2}}}$, ${}^{\mathbf{-}}\mathbf{\text{OH}}$to yield compound Y (molecular formula ${\mathbf{\text{C}}}_{\mathbf{\text{10}}}{\mathbf{\text{H}}}_{\mathbf{\text{14}}}$). Based on the${}^{\mathbf{\text{1}}}\mathbf{\text{H}}$NMR spectra of X and Y were given below, what are the structures of X and Y?

Short answer

Answer

Compound X structure

Compound Y structure

Step-by-step solution

NMR Spectra

The representation of the plot of the applied radiofrequency against the absorption is known as NMR spectra. The signals obtained are referred to as resonance, and the chemical shift is the frequency.

Structure determination

Compound X

$\begin{array}{c}\text{1.\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Double\hspace{0.17em}\hspace{0.17em}bond\hspace{0.17em}\hspace{0.17em}equivalence\hspace{0.17em}}=\text{C}+1-\frac{\text{H}+\text{X}-\text{N}}{2}\\ =10+1-\frac{12+0-0}{2}\\ =11-6\\ =5\end{array}$

The double bond equivalence is 5. Therefore the molecule contains a benzene ring.

2.

One doublet at 1.2 ppm - 6 hydrogens - 6 H’s adjacent to 1 H

One septet at 3.5 ppm - 1 hydrogen - 1 H adjacent to 6 H’s

Multiplet at 7.4 - 8.1 - 5 hydrogens - monosubstituted benzene ring.

Based on the date, the structure of compound X is:

Compound Y

1. When the compound X is treated with , then the carbonyl group of X is reduced to a group.

2.

One doublet at 0.9 ppm - 6 hydrogens - 6 H’s adjacent to 1 H

One multiplet at 1.8 ppm - 1 hydrogen - 1 H adjacent to many H’s.

One doublet at 2.5 ppm - 2 hydrogens - 2 H’s adjacent to 1 H

Multiplet at 7.1 - 7.3 - 5 hydrogens - monosubstituted benzene ring.

Based on the date, the structure of compound Y is: