Question

Question: Identify the structures of isomers A and B (molecular formula C₉H₁₀O ).

Compound A: IR peak at 1742 cm^-1 ; 1 H NMR data (ppm) at 2.15 (singlet, 3 H), 3.70 (singlet, 2 H), and 7.20 cm^-1(broad singlet, 5 H). Compound B: IR peak at 1688 ; 1 H NMR data (ppm) at 1.22 (triplet, 3 H), 2.98 (quartet, 2 H), and 7.28–7.95 (multiplet, 5 H).

Short answer

Answer

a.

b.

Step-by-step solution

IR spectroscopy

The IR spectroscopy measures the stretching frequency in cm^-1 .

The value of the stretching frequency decreases when the corresponding group is adjacent to the electron donor species, and it increases when any electron-withdrawing species is adjacent to the corresponding atom.

Degree of unsaturation

The degree of unsaturation gives the number of bonds and rings present in the given molecule. It is calculated by the following formula:

$Degreeofunsaturation=C_n-\frac{H}{2}-\frac{X}{2}+\frac{N}{2}+1$

Here, X is the number of halogen atoms.

Explanation

The degree of unsaturation for is shown as:

a. Compound A:

IR absorption at 1742cm^-1 : C=O

NMR data: Absorptions: singlet at 2.15 (3 H) ( CH₃group) singlet at 3.70 (2 H) (CH₂ group) broad singlet at 7.20 (5 H) (likely a monosubstituted benzene ring)

Isomer of C₉H₁₀O

1. Compound B:

IR absorption at 1688cm^-1 : C=O

NMR data: Absorptions: triplet at 1.22 (3 H) (CH₃ group split by 2 H's) quartet at 2.98 (2 H) ( CH₂group split by 3 H's) multiplet at 7.28–7.95 (5 H) (likely a monosubstituted benzene ring)

Isomer of C₉H₁₀O