Question

The ¹H and ¹³C NMR spectra below belong to a compouns with formula C₆H₁₀O₂. Propose a structure for this compound.

Short answer

The structure for the compound which satisfies the spectral data has been found.

Step-by-step solution

Finding the peak for absorption of vinylic proton

The molecular formula for the compound is C₆H₁₀O₂. On ignoring hydrogen for the parent compound, the formula will be converted into C₆H₁₄. There is 4 hydrogen in the molecule and has two unsaturation units. The carboxyl carbonyl is one and the other is a double bond due to the peak for the vinylic proton absorption.

Finalizing the compound as an isolated methyl group

The three proton singlet in spectra at 1.8 δ represents an isolated methyl group. The singlet broad signal at 12.5 δ represents the carboxylic acid compound. The 3 proton triplet at 1.05 δ represents attachment of methyl group to methylene group and observed 2 proton multiplet at 2.25 δ. The observation of multiplet indicates the attachment of the methyl group.

The peak at 174.19 δ represents carboxyl carbon. The signal at 22.24 δ and 11.79 δ represents methylene and methyl group for double-bonded carbon. Two carbons in the double bond represent 126.62 δ and the other 146.71 δ represents the carbonyl group. The isolated methyl group indicates the signal at 12.89 δ.

Structure for the compound

The structure for the compound can be drawn as,