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14.20 The DU equals 5. The peak at 190 δ results from a carbonyl carbon, which accounts for one DU. Since this carbon is bonded to one H, the unknown must be an aldehyde. The four signals between 140-120 δ are in the region for a benzene ring, which accounts for the other four degrees of unsaturation. There must be some O symmetry because only four C's appear in this region. To C-H account for six C's and five H's (C₇H₆O - CHO of aldehyde group), two of the CH signals must be due to two CH's each. The compound is benzaldehyde. 14.21 The DU is 4. The IR spectrum shows the presence of =CH (3100-3000 cm⁻¹), -CH (3000-2850 cm⁻¹), and possibly an aromatic ring (four bands in the 1600-1450 cm⁻¹ region). If an aromatic ring is present, it accounts for the DU of 4. The multiplet for 5 H's at 7.2 δ in the H-NMR spectrum confirms the presence of a monosubstituted aromatic ring. The two triplets at 3.6 δ and 3.2 δ, each representing two H's coupled to two H's, show the presence of a CH₂CH₂Br group. Assembly of these fragments, including the Br shown in the formula, shows that the unknown is 14.22 The DU for this compound is 3. The IR spectrum indicates the presence of -CH (3000-2850 cm⁻¹), C=N (2250 cm⁻¹), and C=O and C-O of an ester (1740 and 1200 cm⁻¹). The two H's appearing as a quartet at 4.3 δ in the spectrum are split by 3 H's. These must be the three H's responsible for the triplet at Therefore the unknown contains an ethyl group. Because the signal for the CH₂ of the ethyl group appears so far downfield (> 3.7 it must be bonded to the oxygen of the ester group. The singlet at 3.5 δ shows the presence of a CH₂ group with no nearby H's. Assembly of these fragments gives ethyl 2-cyanoethanoate. Note that the unsplit O CH₂ group at 3.45 δ is downfield because it is bonded to two electron withdrawing groups. 14.23 The DU for this compound is 1. The IR spectrum indicates the presence of -CH (3000-2850 cm⁻¹), and an ester group (C=O at 1733 cm⁻¹ and C-O at 1200 cm⁻¹). The actual number of hydrogens under each of the signals in the spectrum must be double that of the integral value in order to sum to the 10 H's in the formula. The signal at 4.2 δ is due to a CH₂ group that is bonded to the oxygen of the ester group. This CH₂ group is 221