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Chapter 14: Q.21558-14-20P. (page 551)

Identify A and B, isomers of molecular formula C3H4Cl2 , from the given 1HNMR data: Compound A exhibits signals at 1.75 (doublet, 3H,J = 6.9 Hz) and 5.89 (quartet, 1H, J = 6.9 Hz) ppm. Compound B exhibits signals at 4.16 (singlet, 2 H), 5.42 (doublet, 1H, J = 1.9 Hz), and 5.59 (doublet, 1H, J = 1.9 Hz) ppm.

Short Answer

Expert verified

A is

B is

Step by step solution

01

Splitting of signals

Spin-spin coupling causes the splitting of many lines in the NMR spectrum into multiplets, that is, groups of two or more component peaks. This is known as spin-spin splitting, and this can be observed only when high-resolution spectrometers are used.

02

Rules for determining the multiplicity of signal

Some of the rules that help in determining the multiplicity of a signal are:

  • Equivalent protons do not give rise to observable splitting.
  • The midpoint of the multiplet represents the resonance position of a group and thus gives the chemical shift.
  • A signal in the proton NMR for a set of equivalent protons is split into (n + 1) component lines where n is the number of equivalent protons in the adjacent set.
03

Finding A and B

Representation of A

By observing this compound,

  • Ha gives a signal at 5.89 ppm due to 1H proton, and it will split into a quartet as the neighboring carbon atom has 3 hydrogen atoms in it. The splitting constant is also obtained at J = 6.9 Hz.
  • Hb gives a signal at 1. 75 ppm due to 3 H protons, and it will split into doublet as the neighboring carbon atom has 1 hydrogen atom in it. The splitting constant is also obtained at J = 6.9 Hz.

From these observations, this compound is confirmed as A.

Representation of B

By observing this compound,

  • The signal at 4.16 ppm is due to 2 H protons, and it will be a singlet as the neighboring carbon atom has no hydrogen atoms in it.
  • The signal at 5.42 ppm is due to 1 H protons, and it will split into doublet as the neighboring carbon atom has 1 hydrogen atom in it. The splitting constant is also obtained at J = 1.9 Hz.
  • The signal at 5.59 ppm is due to 1 H protons, and it will split into doublet as the neighboring carbon atom has 1 hydrogen atom in it. The splitting constant is also obtained at J = 1.9 Hz.

From these observations, this compound is confirmed as b.

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Most popular questions from this chapter

Question: How many 1H NMR signals does each natural product exhibit?

a.

b.

c.

d.

Question: How many \(^{\bf{1}}{\bf{H}}\) NMR signals would you expect for each compound?

a.

b.

c.

Identify the carbon atoms that give rise to the signals in the \)^{{\bf{13}}}{\bf{C}}\) NMR spectrum of each compound.

a. \){\bf{C}}{{\bf{H}}_{\bf{3}}}{\bf{C}}{{\bf{H}}_{\bf{2}}}{\bf{C}}{{\bf{H}}_{\bf{2}}}{\bf{C}}{{\bf{H}}_{\bf{2}}}{\bf{OH}}\); \)^{{\bf{13}}}{\bf{C}}\) NMR: 14, 19, 35, and 62 ppm

b. \){\left( {{\bf{C}}{{\bf{H}}_{\bf{3}}}} \right)_{\bf{2}}}{\bf{CHCHO}}\); \)^{{\bf{13}}}{\bf{C}}\) NMR: 16, 41, and 205 ppm

c. \){\bf{C}}{{\bf{H}}_{\bf{2}}}{\bf{ = CHCH}}\left( {{\bf{OH}}} \right){\bf{C}}{{\bf{H}}_{\bf{3}}}\); \)^{{\bf{13}}}{\bf{C}}\) NMR: 23, 69, 113, and 143 ppm

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.

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)

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