Question

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

A is

B is

Step-by-step solution

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.

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.

Finding A and B

Representation of A

By observing this compound,

• H_a 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.
• H_b 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.