NMR: Novice Level, Spectrum 20

Formula: C₁₀H₁₈O₄

Answer: Diethyl adipate

Chemical Shift Assignments:

¹H NMR: δ 1.25 (3H, t, J = 7 Hz), 1.66 (2H, m), 2.31 (2H, m), and 4.12 (2H, quartet, 1H, J = 7 Hz)

¹³C NMR: 173.0, 60.2, 33.8, 24.3, and 14.1 ppm

The formula indicates two degree of unsaturation. The integration is half the number of hydrogens in the ¹H NMR integration. Moreover, the ¹³C spectrum reveals only six carbons. Symmetry is present. The signals at δ 1.25 and 4.12 in the ¹H NMR spectrum comprise an ethyl pattern with an attachment to a heteroatom-- namely, oxygen. The ¹³C NMR signal at 173 ppm indicates an ester. Thus, there are two -C(O)OC₂H₅ groups. These two groups constitute C₆H₁₀O₄, which leaves C₄H₈. These hydrogens are of two types, located at d 1.66 and 2.31. A chain of methylenes, -CH₂CH₂CH₂CH₂- , is in order. Clearly, the lower field signals are the two methylene groups adjacent to the carbonyl groups. A first approximation based upon chemical equivalence would argue that each of the two methylene groups should appear as a triplet. A careful inspection of these signals shows that they are broad and less well-defined than what one might expect. Eighteen of the preceding ¹H NMR spectra have involved only chemically equivalent protons. Because we have duplicate sets of chemically equivalent hydrogens, a new phenomenon can arise --- magnetic non-equivalence. Imagine that one pair of methylene hydrogens at δ 2.31 has spins ββ and the other pair at δ 2.31 has spins αβ. Now each of the methylene groups at δ 1.66 is in a slightly different magnetic environment. Consequently, they couple to one another to provide a complex pattern. The same argument can be made for the complexity of the signal at δ 2.31 caused by the hydrogens at δ 1.66. You may wish to re-evaluate the fine structure of ¹H NMR #5 in the Novice Level and explore example #18 in the Intermediate Level. Return to Menu.