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Chapter 14: Q.21558-14-21P. (page 552)

How many signals are present in the 1HNMR spectrum for each molecule? What splitting is observed in each signal?

a.

b.

c.

Short Answer

Expert verified

a.

three signals (splitting observed: a - singlet; b-singlet; c-singlet).

b.

four signals (splitting observed: a - triplet; b-multiplet; c-triplet).

c.

three signals (splitting observed: a - doublet; b-multiplet; c-singlet).

Step by step solution

01

 1HNMR Spectroscopy

The number of signals in a proton NMR ( 1HNMR) spectroscopy depends on the number of protons and their chemical environment.

Chemically equivalent protons resonate at the same frequency and exhibit a single signal.

The hydrogen atoms attached to heteroatoms give only a single signal.

02

Splitting of signals in  1HNMR spectroscopy

The splitting of the 1HNMR signal of a particular proton/hydrogen depends on the number of hydrogens present on the adjacent carbon atoms.

The splitting and multiplicity of a hydrogen atom are predicted using the n+1 rule, where n represents the number of hydrogens present on the carbon atoms adjacent to the hydrogen atom under consideration.

03

Signals in  1HNMR spectrum of the given compounds

a. Since the compound consists of three chemically distinct groups, it exhibits three signals.

Non-equivalent protons in compound a

The a, b, and c hydrogen atoms exhibit a singlet each, as, in the case of a and b, there are no hydrogen atoms on the carbon adjacent to it. In c, the hydrogen is attached to a heteroatom.

b. The given compound exhibits four NMR signals.

Non-equivalent protons in compound b

The multiplicity of a is 2 + 1 = 3 (triplet).

The multiplicity of b is 5 + 1 = 6 (multiplet)

The multiplicity of c is 2 + 1 =3 (triplet).

The multiplicity of d is a singlet.

c. The given compound exhibits three NMR signals.

Non-equivalent protons in compound c

The multiplicity of a is 1 + 1 =2 (doublet).

The multiplicity of b is 6 + 1 = 7 (multiplet)

The multiplicity of c is a singlet.

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Most popular questions from this chapter

Compound A exhibits two signals in its 1H NMR spectrum at 2.64 and 3.69 ppm and the ratio of the absorbing signals is 2:3. Compound B exhibits two signals in its 1H NMR spectrum at 2.09 and 4.27 ppm and the ratio of the absorbing signals is 3:2. Which compound corresponds to dimethyl succinate and which compound corresponds to ethylene diacetate?

Question: The \(^{\bf{1}}{\bf{H}}\) NMR spectrum 1,2-dimethoxyethane \(\left( {{\bf{C}}{{\bf{H}}_{\bf{3}}}{\bf{OC}}{{\bf{H}}_{\bf{2}}}{\bf{C}}{{\bf{H}}_{\bf{2}}}{\bf{OC}}{{\bf{H}}_{\bf{3}}}} \right)\) recorded on a 300 MHz NMR spectrometer consists of signals at 1017 Hz and 1065 Hz downfield from TMS. (a) Calculate the chemical shift of each absorption. (b) At what frequency would each absorption occur if the spectrum were recorded on a 500 MHz NMR spectrometer?

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: How can you use 1H NMR spectroscopy to distinguish between CH2=C(Br)CO2CH3 and methyl (E)-3-bromopropenoate, BrCH=CHCO2CH3?

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