Chapter 14: Q55P (page 527)

Reaction of \({{\bf{C}}_{\bf{6}}}{{\bf{H}}_{\bf{5}}}{\bf{C}}{{\bf{H}}_{\bf{2}}}{\bf{C}}{{\bf{H}}_{\bf{2}}}{\bf{OH}}\) with \({\bf{C}}{{\bf{H}}_{\bf{3}}}{\bf{COCl}}\) affords compound W, which has molecular formula \({{\bf{C}}_{{\bf{10}}}}{{\bf{H}}_{{\bf{12}}}}{{\bf{O}}_{\bf{2}}}\). W shows prominent IR absorptions at 3088–2897, 1740, and 1606 \({\bf{c}}{{\bf{m}}^{{\bf{ - 1}}}}\) . 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

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Step by step solution

IR spectroscopy

The IR spectroscopy gives the signals of ketone as 1740 \({\rm{c}}{{\rm{m}}^{{\rm{ - 1}}}}\). The double bond stretching frequency comes in the range of 3300 to 3000 \({\rm{c}}{{\rm{m}}^{{\rm{ - 1}}}}\) .

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:

\({\rm{Degree}}\,{\rm{of}}\,{\rm{unsaturation}} = {{\rm{C}}_{\rm{n}}}{\rm{ - }}\frac{{\rm{H}}}{{\rm{2}}}{\rm{ - }}\frac{{\rm{X}}}{{\rm{2}}}{\rm{ + }}\frac{{\rm{N}}}{{\rm{2}}}{\rm{ + 1}}\)

Here, X is the number of the halogen atoms.

Explanation

IR absorption at 1740\({\rm{c}}{{\rm{m}}^{{\rm{ - 1}}}}\): C=O, 3088-2897\({\rm{c}}{{\rm{m}}^{{\rm{ - 1}}}}\): benzene double bond, 1606\({\rm{c}}{{\rm{m}}^{{\rm{ - 1}}}}\):ester group

NMR data: Absorptions: singlet at 2.02 (singlet, \({\rm{C}}{{\rm{H}}_{\rm{3}}}\) group), 2.91 (triplet, \({\rm{C}}{{\rm{H}}_2}\) group), 4.25 (triplet, \({\rm{C}}{{\rm{H}}_2}\)), and 7.20–7.35 (multiplet, benzene ring) ppm