Question

Question: A compound of molecular formula C₄H₈0₂ shows no IR peaks at 3600–3200 or 1700 cm^-1 . It exhibits one singlet in its ¹H NMR spectrum at 3.69 ppm, and one line in its ¹³C NMR spectrum at 67 ppm. What is the structure of this unknown?

Short answer

Answer

Step-by-step solution

1H  NMR

The NMR spectra is obtained by employing continuous wave method in ¹H NMR. The chemical shift of ¹H NMR ranges from 0 to 14 ppm.

 13C NMR

The various types of carbon atoms can be identified with the help of signals produced by the ¹³C NMR spectrum.

Predicting the structure of the compound with molecular formula  C4H802

The given molecular formula is C₄H₈0₂ . The double bond equivalence (DBE) of the molecule can be found as:

DBE=C+1 -$\frac{H}{2}-\frac{X}{2}+\frac{N}{2}$

Where,

• C denotes the number of carbon atoms,
• H denotes the number of hydrogen atoms,
• X denotes the number of halogen atoms and
• N denotes the number of nitrogen atoms.

The values can be substituted as:

DBE=4+1-$\frac{8}{2}-0+0$

=5-4

=1

The double bond equivalence is one which indicates that the compound contains one double bond or one ring in the compound. It does not exhibit IR peaks at 3600–3200 or which indicates the absence of carbonyl and alcohol group. Hence, the structure of the compound is given as:

Structure of the compound with molecular formula C₄ H₈ 0₂

All the protons and carbon are equivalent for this molecule and it exhibits one singlet at 3.69 ppm for NMR spectrum and one line at 67 ppm is observed for the ¹³C spectrum.