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Chapter 14: Q.21558-14-55P (page 564)

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

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Answer

Step by step solution

01

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 .

02

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-H2-X2+N2+1

Here, X is the number of the halogen atoms.

03

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

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Most popular questions from this chapter

How many peaks are present in the NMR signal of each labelled proton?

a.

b.

c.

d.

Into how many peaks will the signal for each of the labeled protons be split?

a.

b.

c.

d.

e.

f.

g.

h.

i.

j.

Question:Propose a structure consistent with each set of spectral data:

a. C4H8Br2: IR peak at 3000–2850cm-1 ;

NMR (ppm):

1.87 (singlet, 6 H)

3.86 (singlet, 2 H)

b. C3H6Br2: IR peak at 3000–2850cm-1 ;

NMR (ppm):

2.4 (quintet)

3.5 (triplet)

c. C5H10O2: IR peak at 1740cm-1 ;

NMR (ppm):

1.15 (triplet, 3 H) 2.30 (quartet, 2 H)

1.25 (triplet, 3 H) 4.72 (quartet, 2 H)

d . C6H14O: IR peak at 3600-3200 cm-1 ;

NMR (ppm):

0.8 (triplet, 6 H) 1.5 (quartet, 4 H)

1.0 (singlet, 3 H) 1.6 (singlet, 1 H)

e. C6H14O: IR peak at 3000-2850cm-1 ;

NMR (ppm):

1.10 (doublet, relative area = 6)

3.60 (septet, relative area = 1)

f . C3H6O: IR peak at 1730cm-1 ;

NMR (ppm):

1.11 (triplet)

2.46 (multiplet)

9.79 (triplet)

Question: The \(^{\bf{1}}{\bf{H}}\) NMR spectrum of \({\bf{C}}{{\bf{H}}_{\bf{3}}}{\bf{OH}}\) recorded on a 500 MHz NMR spectrometer consists of two signals, one due to the \({\bf{C}}{{\bf{H}}_{\bf{3}}}\) protons at 1715 Hz and one due to the OH proton at 1830 Hz, both measured downfield from TMS. (a) Calculate the chemical shift of each absorption. (b) Do the \({\bf{C}}{{\bf{H}}_{\bf{3}}}\) protons absorb upfield or downfield from the OH proton?

Question: As we will learn in Chapter 20, reaction of (CH3)2CO withLiC≡CH followed by affords compound D, which has a molecular ion in its mass spectrum at 84 and prominent absorptions in its IR spectrum at 3600–3200, 3303, 2938, and 2120cm-1 . D shows the following 1 H NMR spectral data: 1.53 (singlet, 6 H), 2.37 (singlet, 1 H), and 2.43 (singlet, 1 H) ppm. What is the structure of D?
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