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Chapter 14: Q,21558-14-41P (page 562)

Question: How could you use chemical shift and integration data in 1H NMR spectroscopy to distinguish between CH3OCH2CH2OCH3and CH3OCH2OCH3? The 1H NMR spectrum of each compound contains only singlets.

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Answer

Step by step solution

01

Step 1:  1H NMR spectroscopy

The 1HNMR spectroscopy analyzes the chemical environment by coupling the protons.

02

Determination of compounds

  • If the proton is in the same chemical environment, there will be only one peak.
  • If the protons are in different chemical environments, there will be more than one peak.
03

Explanation

In the compound CH3OCH2CH2OCH3, CH2CH2 is in the same environment. Therefore, it has two types of protons.

Oxygen is an electronegative atom that gives a higher chemical shift value.

In the compound CH3OCH2OCH3, CH2 is in the same environment. Therefore, it has two types of protons.

Chemical shift and integration data

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Most popular questions from this chapter

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. Reaction of aldehyde D with amino alcohol E in the presence of NaH forms F (molecular formula C11H15NO2). F absorbs at 1730 cm-1in its IR spectrum. F also shows eight lines in its 13C-NMR spectrum, and gives the following -NMR spectrum: 2.32 (singlet, 6 H), 3.05 (triplet, 2 H), 4.20 (triplet, 2 H), 6.97 (doublet, 2 H), 7.82 (doublet, 2 H), and 9.97 (singlet, 1 H) ppm. Propose a structure for F. We will learn about this reaction in Chapter 18.

Question. When 2-bromo-3,3-dimethylbutane is treated with K+- OC(CH3)3, a single product T having molecular formula C6H12 is formed. When 3,3-dimethylbutan-2-ol is treated with H2SO4 , the major product U has the same molecular formula. Given the following -NMR data, what are the structures of T and U? Explain in detail the splitting patterns observed for the three split signals in T. 1 H NMR of T: 1.01 (singlet, 9 H), 4.82 (doublet of doublets, 1 H, J = 10, 1.7 Hz), 4.93 (doublet of doublets, 1 H, J = 18, 1.7 Hz), and 5.83 (doublet of doublets, 1 H, J = 18, 10 Hz) ppm 1 H NMR of U: 1.60 (singlet) ppm.

Question: (a) How many 1H NMR signals does each compound show? (b) Into how many peaks is each signal split?

Which of the highlighted carbon atoms in each molecule absorbs farther downfield?

a.

b.

c.

d.

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