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
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
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