Question

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?

Short answer

A is dimethyl succinate

B is ethylene diacetate

Step-by-step solution

NMR Spectroscopy

NMR spectroscopy involves the study of transitions induced between the spin energy levels of the ¹H ¹³C and nuclei in a species in an applied magnetic field by absorption of electromagnetic radiation in the radiofrequency region.

This spectroscopy gives an idea about the nature of the immediate environment of each one of them, position within the molecule, etc.

Chemical shift

A chemical shift in terms of the magnetic field is given by:

$\delta =\frac{{\text{B}}_{\text{reference}}-{\text{B}}_{\text{sample}}}{{\text{B}}_{\text{reference}}}\times {10}^6$

Identification of compounds A and B

The condensed formula of dimethyl succinate

The ratio of absorbing signals is 4:6 or 2:3.

* Signal 1 corresponds to four protons which are obtained at 2.64 ppm.

* Signal 2 corresponds to six protons which are obtained at 3.69 ppm.

* Six hydrogen atoms are present with downfield absorption.

From all these, A is dimethyl succinate.

The condensed formula of ethylene diacetate

The ratio of absorbing signals is 6:4 or 3:2.

* Signal 1 corresponds to six protons which are obtained at 2.09 ppm.

* Signal 2 corresponds to four protons which are obtained at 4.27 ppm.

* Four hydrogen atoms are present with downfield absorption.

From all these, B is ethylene diacetate.