Ressonância Magnética Nuclear

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212 Chapter 10 USING NUCLEAR MAGNETIC RESONANCE SPECTROSCOPY TO DEDUCE STRUCTURE (c) Cycloheptanone: Same reason as (b); note symmetry in molecule (d) The only example with alkene carbons (δ = 100-150 ppm) 58. (a) ¹³C: Seven different carbon signals show, six of which are CH₂ groups. Five signals, δ = 0.6 (broad singlet. 1 H), 0.8 (triplet, 3 H), 1.2 (broad, 8 H), 1.3 (multiplet, 2 H), 3.3 2 H). At this point. the signals at 1.2 and 1.3 are close to but the others can get us off to a good start. We can propose CH₂ (δ = 0.8) and CH₂ OH (δ = 3.3; the -OH proton is at δ = 0.6) as molecular fragments. These fragments add up to leaving C₃H₆ unaccounted for. There is no sign of any other signal for a CH₃ group in the 'H NMR spectrum; indeed. the ¹³C DEPT spectrum tells us that the remaining carbons are all CH₂ groups. So the answer has to be 1-heptanol. (Alternatives such as 3- or would show a methyl doublet near δ = 0.9 in the 'H NMR spectrum in addition to the triplet already present. and one more non-CH₂ peak in the ¹³C DEPT spectrum.) (b) ¹³C: Four signals for the seven carbons. so the molecule must have symmetry. ¹H: Four distin- guishable signals. δ = 0.9 (triplet. 6 H), 1.3 (quartet. 4 H), 2.8 (singlet. 2 H), 3.6 (singlet, 4 H). A couple of fragments may be discerned: (2X) CH, (2X) OH 1.3 0.9 3.6 2.8 These add up to C₆H₁₆O₂; adding in the quaternary carbon that shows up in the ¹³C (DEPT) spec- trum gives the complete molecular formula. What we have, therefore, are the four fragments shown above, plus a carbon atom. Connecting the four fragments to the four bonds of the seventh carbon gives us the solution: CH₂OH CH₂OH 59. An assortment of overlapping sharp singlets and doublets for the CH₃ groups is evident between δ = 0.6 and 1.1. Signals for the benzene H's are located between δ = 7.2 and 8.2. Three other signals may be interpreted as follows: CH₃| etc. C₆H₅COO δ = 4.8 H HH, H 2.4 5.4 The CH₂ at δ = 2.4 is split into a doublet by the neighboring CH. This CH is the cause of the δ = 4.85 signal, and its complex splitting results from the CH₂ neighbors on both sides. The expansion of this signal shows 9 lines. How can we explain its appearance? Recall Problem 36(h): The CH hydrogen is trans to two of its neighbors and cis to the other two. As in the earlier problem, the cis and trans coupling constants in the ring are different. Applying the type of analysis discussed in Section 10-8, we can attempt to analyze the