Question

Question: Identify the structures of D and E, isomers of molecular formula ${\mathbf{C}}_{\mathbf{6}}{\mathbf{H}}_{\mathbf{12}}{\mathbf{NO}}_{\mathbf{2}}$, from their IR and${}^{\mathbf{1}}\mathbf{H}$NMR data. Signals at 1.35 and 1.60 ppm in the NMR spectrum of D and 1.90 ppm in the NMR spectrum of E are multiplets.

a. IR absorption for D at 1743cm^-1

b. IR absorption for E at 1746 cm^-1

Short answer

Answer

a.

b.

Step-by-step solution

Step-by-Step SolutionStep 1: IR spectroscopy

IR spectrometer is the device employed in IR spectroscopy, which generates the IR spectrum.

1H NMR spectrosopy

The chemical shift is one of the aspects that recognizes the molecular structure in an ¹H NMR spectrum. Protons found in the benzene have higher chemical shift values.

Identification of structures D and E

a. In molecule D, the IR absorption corresponding to 1743cm^-1 indicates the presence of the C=O group. In the¹H NMR data, the triplet at 0.9 ppm (3 H) indicates the presence of CH₃ adjacent to CH₂

The multiplet at 1.35 ppm (2 H) indicates the presence of CH₃ adjacent to CH₂ . The multiplet at 1.60 ppm (2 H) indicates the presence of CH₃ adjacent to CH₂ . The singlet at 2.1 ppm (3 H) arises from the connected to the C=O group.

The triplet at 4.1 ppm (2 H) indicates the adjacent to the electronegative oxygen atom and another .The structure of molecule D can be given as:

Structure of compound D