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Chapter 14: Q.21558-14-73P (page 569)

Question: The 1 H NMR spectrum of N,N-dimethylformamide shows three singlets at 2.9, 3.0, and 8.0 ppm. Explain why the two groups are not equivalent to each other, thus giving rise to two NMR signals.

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01

 1H NMR

The number of signals in a PMR spectrum is equal to the number of kinds of protons present in the molecule. The splitting of the NMR signal is caused by spin-spin coupling

02

Non-equivalent proton in NMR

The structure for N,N-dimethylformamide places the two CH3 groups in different environments and different chemical shifts. One CH3 group is cis to the O atom, and one is cis to the H atom.

This gives rise to two different absorptions for the CH3 groups.

Representation of cis

It should be noted that spin-spin splitting is observed only between non-equivalent neighboring protons.

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Most popular questions from this chapter

Identify products A and B from the given 1HNMR data.

a. Treatment of CH2=CHCOCH3 with one equivalent of HCl forms compound A. A exhibits the following absorptions in its 1HNMR spectrum: 2.2 (singlet, 3 H), 3.05 (triplet, 2 H), and 3.6 (triplet, 2H) ppm. What is the structure of A?

b.Treatment of acetone [(CH3)2C=O] with dilute aqueous base forms B. Compound B exhibits four singlets in its 1HNMR spectrum at 1.3 (6 H), 2.2 (3 H), 2.5 (2 H), and 3.8 (1 H) ppm. What is the structure of B?

Question: Describe the \({}^{\bf{1}}{\bf{H}}\)NMR spectrum of each compound. State how many NMR signals are present, the splitting pattern for each signal, and the approximate chemical shift.

a.

b.

c.

d.

The 1H NMR spectrum of CH3OH recorded on a 500 MHz NMR spectrometer consists of two signals, one due to the CH3 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 CH3 protons absorb upfield or downfield from the OH proton?

Question: 18-Annulene shows two signals in its 1 H NMR spectrum, one at 8.9 (12 H) and one at 鈥�1.8 (6 H) ppm. Using a similar argument to that offered for the chemical shift of benzene protons, explain why both shielded and deshielded values are observed for 18-annulene.

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

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

NMR (ppm):

1.87 (singlet, 6 H)

3.86 (singlet, 2 H)

b. C3H6Br2: IR peak at 3000鈥�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)
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