Formula: C₇H₁₆O₂

Answer: Propionaldehyde, diethyl acetal

Chemical Shift Assignments: ¹H NMR: δ 0.63 (3H, t, J = 8 Hz), 0.92 (6H, t, J = 7 Hz), 1.34 (2H, dq, J = 8 and 6 Hz), 3.20 (2H, dq, J = 7 and 9 Hz), 3.35 (2H, dq, J = 7 and 9 Hz), and 4.12 (1H, t, J = 6 Hz)

The degree of unsaturation is 0. Although this sample has some contamination, the impurities do not account for its complexity. This spectrum provides an example of diastereotopic protons. Let us first deal with the hydrogens on C₉, C₁₄, and C₂₆ (see structure). The C₂₆ CH₃- group at δ 0.63 (integral = 3H) is a triplet because of coupling (J = 8 Hz) to the vicinal CH₂ group. The same CH₂ group (C₁₄) causes the methine hydrogen H₁₃ at δ 4.12 to appear as a triplet. The C₁₄ methylene group is a doublet of quartets as a result of the unequal coupling to the C₂₆ methyl group and the C₉ methine hydrogen. The hydrogens H₇ and H₈ as well as H₁₉ and H₂₀ are diastereotopic. In the animation these diastereotopic hydrogens are in different shades of green. Hydrogens H₇ and H₁₉ are equivalent, as are H₈ and H₂₀. The diastereotopic hydrogens exist in different chemical and magnetic environments and therefore they have different chemical shifts. Thus, they couple to one another. They appear as a doublet of quartets, J = 7 and 9 Hz. Measure the coupling constants yourself. The two methyl groups at δ 0.92 are both triplets (J = 7 Hz), but the vicinal CH₂ groups appear as separate signals at δ 3.20 and 3.35. The ¹³C NMR spectrum is simpler. A plane of symmetry in the molecule reduces the number of singlets to five: 103.81 (C₉), 60.49 (C₂, C₁₅), 26.34 (C₁₄), 14.99 (C₁, C₁₈), and 8.64 (C₂₆) ppm. Note that the two oxygens attached to C₉ deshield it relative to C₂ and C₁₅. Return to Menu.