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Chapter 14: Q.21558-14-57P (page 564)

Question: Compound C has a molecular ion in its mass spectrum at 146 and a prominent absorption in its IR spectrum at 1762cm-1 . C shows the following 1H NMR spectral data: 1.47 (doublet, 3 H), 2.07 (singlet, 6 H), and 6.84 (quartet, 1 H) ppm. What is the structure of C?

Short Answer

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Answer

Step by step solution

01

Mass spectrometry

Mass spectrometry is used to determine the structure through the formation of the molecular ion.It also forms the stable fragments of the molecules and determines the m/z values.

02

1H NMR

The 1H NMR gives the location of the non-equivalent protons. The formula to calculate the number of peaks is .

For example, the methyl protons split the adjacent protons into a quartet.

03

Explanation

IR absorption at 1762cm-1 : C=O

NMR data: 1.47 (doublet, 3 H, group), 2.07 (singlet, 6 H, 2), and 6.84 (quartet, 1 H) ppm

Structure

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Most popular questions from this chapter

Rank each group of protons in order of increasing chemical shift.

a.

b.

Propose a structure consistent with each set of data.

a. Compound J: molecular ion at 72; IR peak at 1710cm-1 ; 1H -NMR data (ppm) at 1.0 (triplet, 3 H), 2.1 (singlet, 3 H), and 2.4 (quartet, 2 H)

b. Compound K: molecular ion at 88; IR peak at 3600–3200 ;1H-NMR data (ppm) at 0.9 (triplet, 3 H), 1.2 (singlet, 6 H), 1.5 (quartet, 2 H), and 1.6 (singlet, 1 H)

Question: Label the protons in each highlighted \({\bf{C}}{{\bf{H}}_{\bf{2}}}\) group as enantiotopic, diastereotopic, or homotopic.

a.

b.

c.

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)

The reaction of (CH3)3 CCHO with (C6H5)3 P=C(CH3)OCH3 , followed by treatment with aqueous acid, afford R . R has strong absorption in its IR spectrum at 1717 cm-1 and three singlets in its -NMR spectrum at 1.02 (9 H), 2.13 (3 H), and 2.33 (2 H) ppm. What is the structure of R? We will learn about this reaction in Chapter 21.
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