Question

Question: Reaction of C₆H₅CH₂CH₂OH with CH₃COCl affords compound W, which has molecular formula C₁₀H₁₂O₂. W shows prominent IR absorptions at 3088–2897, 1740, and 1606cm^-1 . W exhibits the following signals in its 1 H NMR spectrum: 2.02 (singlet), 2.91 (triplet), 4.25 (triplet), and 7.20–7.35 (multiplet) ppm. What is the structure of W? We will learn about this reaction in Chapter 22.

Short answer

Answer

Step-by-step solution

IR spectroscopy

The IR spectroscopy gives the signals of ketone as 1740cm^-1 . The double bond stretching frequency comes in the range of 3300 to 3000cm^-1 .

Degree of unsaturation

The degreeof unsaturation helps determine the number of bonds and rings, if present in the given molecule. It is calculated by the following formula:

$Degreeofunsaturatio=Cn-\frac{H}{2}-\frac{X}{2}+\frac{N}{2}+1$

Here, X is the number of the halogen atoms.

Explanation

IR absorption at 1740 : C=O, 3088-2897cm^-1 : benzene double bond, 1606cm^-1 :ester group

NMR data: Absorptions: singlet at 2.02 (singlet, CH₃ group), 2.91 (triplet, CH₂ group), 4.25 (triplet, ), and 7.20–7.35 (multiplet, benzene ring) ppm

Structure formation