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208 Chapter 10 USING NUCLEAR MAGNETIC RESONANCE SPECTROSCOPY TO DEDUCE STRUCTURE (c) The sketch should look like the spectrum of CH₃CH₂CH₂Br (text Figure 10-27). C2 gives a sextet (five neighbors), assuming similar coupling constants all around. CH₂ CH₂ CH₃ and and and CH₂ CH₂ CH₃ 10 9 8 7 6 5 4 3 2 1 0 ppm (δ) (d) As in (a), we again have a signal relatively downfield because of the group's location between two electronegative atoms. OCH₃ CH₃ CH OCH₃ 10 9 8 7 6 5 4 3 2 1 0 ppm (δ) 49. Two signals, δ = 0.9 (doublet) and δ = 1.4 (septet?) with an intensity of 12 H for the large signal and 2 H for the small one. Twelve equivalent hydrogens probably mean four equivalent CH₃ groups, which equals C₄H₁₂. That leaves C₂H₂ left to make up the molecule. The only way to do it that makes all the CH₃'s identical and split into a doublet is 2,3-dimethylbutane. Figure 10-22 shows the NMR of 2-iodopropane, another molecule containing the (CH₃)₂CH group. Again, the methyl signal is a doublet, but the CH signal is better resolved as a clean septet. The larger chemical shift difference between the two sets of signals in 2-iodopropane gives rise to a more nearly "first-order" appearance to its spectrum, compared with the spectrum of 2,3-dimethylbutane.