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b) The DU is 2. The IR spectrum shows the presence of a C=O group and sp³ C-H bonds. Because there is no indication of a C=C, the other DU is probably due to a ring. The spectrum is not very helpful. The presence of only four signals in the spectrum indicates that the compound has symmetry. The signal at ~214 δ is due to a carbonyl carbon that is attached to no hydrogens, so the compound is a ketone. Since there is only one carbonyl group in the compound, there are probably two of each of the three CH₂ groups shown in the spectrum. This gives the proper formula of C₇H₁₂O. To assemble these fragments into a compound, start with the C=O group. Bond a chain of three CH₂ groups to each side and then connect the last C's to form a ring. The unknown is cycloheptanone. c) This compound has a DU of 5. The two bands in the region of 2830- 2700 cm⁻¹ and the carbonyl absorption near 1730 cm⁻¹ suggest that it is an aldehyde (not conjugated). The presence of sp²-hybridized C-H bonds (3100-3000 cm⁻¹) and the four bands in the 1600-1450 cm⁻¹ region, along with the high DU, suggest the presence of an aromatic ring. In the NMR spectrum, the signal for 5 H's in the aromatic region (7.2 confirms the presence of a monosubstituted aromatic ring. The hydrogen of the aldehyde, at 9.7 δ, is a doublet, so the aldehyde group must be attached to a CH group. The H that appears as a quartet at 3.6 δ must be coupled to the three H's that appear as a doublet at 1.5 δ. This H must also be coupled to the aldehyde H, but the coupling is too small to be seen without expanding the peak. The shows the aldehyde carbon at 200 δ. The three types of aromatic CH's around 130- 125 δ are part of the monosubstituted benzene ring. The C of the aromatic ring that is bonded to the CH₃ O substituent appears at 147 δ. We know the CH (at CH-CH ~53 δ) is bonded to the aldehyde carbonyl and the methyl group (at ~ 14 δ) from the spectrum. Its final bond must be to the aromatic ring. 218