The degree of unsaturation is 5. Five of the hydrogen atoms are in the aromatic region. They comprise three different chemical shifts in a ratio of 2:2:1. This information accounts for C₆H₅-, a phenyl group. The remainder is CHO, which can be only an aldehyde (δ 10.00). The electron-withdrawing effect of the carbonyl group causes the hydrogens of the aromatic ring to have different chemical shifts [ortho (2), meta (2), and para (1)]. Clearly, the para hydrogen is at δ 7.62 because there is only one of them. If you blow-up the spectrum in this area, you will see that this signal is a triplet of triplets (J = 7 and ~1 Hz). These coupling constants are typical of ortho (red-green) at meta (violet-green) coupling, respectively. The triplets are formed by coupling to the two ortho and two meta hydrogens. Of the remaining 2H signals, which one is the pair of ortho hydrogens? Observe that the signal at δ 7.86 is a doublet of triplets, where J = 7 Hz for the doublet and J= ~1 Hz for the triplet coupling. Only the ortho hydrogens are positioned to have "ortho" and "meta" partners. The meta hydrogens at δ 7.52 appear as a triplet (J = 7 Hz).

The ¹³C spectrum of this sample is contaminated. The signals of benzaldehyde at: 192.71, 136.62, 134.73, 129.95, and 129.27. Visible signals of the inmpurity are at 133.79, 130.37, and 128.65. The weakest signal of benzaldehyde at 136.62 ppm (ipso) is as intense as the peaks assigned to the impurity. The single para carbon of benzaldehyde is at 134.73 ppm, which is of lower intensity than the two ortho carbons at 129.95 ppm and the two meta carbons at 129.27 ppm. The ortho carbons are slightly deshielded by the electron-withdrawing CHO group. The carbon of the aldehyde is at 192.71 ppm and strong owing to the aldehyde hydrogens direct attachment to the carbon. What is the impurity? Benzaldehyde is easily air oxidized to benzoic acid: ¹H NMR; ¹³C NMR.

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