The degree of unsaturation is one. Because 11 of the 12 hydrogens are at δ 2.33 and higher field, an alkane chain of five carbons is expected (C₅H₁₁-). The remaining -CHO₂ can be either a formate [-O(C=O)H] or carboxylic acid group [-CO₂H]. The single degree of unsaturation is here.The formate is eliminated because the oxygen atom would be bound to a carbon bearing hydrogens absorbing in the region ~ δ 4.0. There are no signals there. It may not look like there is a signal at δ 9.56. Blow it up and it rises out of the baseline as a very broad singlet in the region for a carboxylic acid hydrogen. What about that C₅H₁₁- chain? There are 8 structural isomers (draw them) of C₅H₁₁CO₂H. Only one has a single methyl group with area = 3 and appearing as a triplet, that being the normal chain. Check your answer here. Thus, the answer is hexanoic acid. Note that the farther removed the carbon bound hydrogens are from the carbonyl group, the lower the δ-value is for their chemical shifts. The CH₂ groups at C₄ and C₅ are essentially indistinguishable at 300 MHz.

The ¹³C NMR spectrum displays a carbonyl carbon at 180.6 ppm. The remaining methylene carbons are upfield at 34.0, 31.4, 24.5, and 22.5 ppm. The methyl group is at the highest field, 13.8 ppm. Return to Menu.