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200 Chapter 10 USING NUCLEAR MAGNETIC RESONANCE SPECTROSCOPY TO DEDUCE STRUCTURE 38. (a) The spectrum shows two signals, at δ = 1.1 and 3.3. The δ = 1.1 signal is for 9 equivalent hydrogens. and the δ = 3.3 signal is for 2 equivalent hydrogens. A good way to get nine equivalent hydrogens is with a (CH₃)₃C group. The other two hydrogens must be on a separate, single carbon, because the (CH₃)₃C group contains all but one of the five carbons in the formula. The downfield location of these two hydrogens suggests their carbon is attached to the So, + CH₂ as a plausible structure. 1.1 3.3 (b) Somewhat similar: two signals, δ = 1.9 and 3.8. The signal for six equivalent hydrogens is probably due to two methyl groups on the same carbon (CH₃-C-CH₃). The two-hydrogen signal can only be a because there are only four carbons in the molecule. So, Br CH₂ 3.8 Br 1.9 39. (a) The spectrum has two signals in a 3: intensity ratio. The molecule has eight hydrogens, so there must be one group of six equivalent H's and another of two equivalent H's (6:2 = 3:1). Two equivalent CH₃'s and a CH₂ account for all but the two oxygen atoms in the formula. The larger signal at δ = 3.3 is just right for hydrogens on carbons attached to an oxygen. The downfield location for the small signal (δ = 4.4) is consistent with attachment of carbon to more than one oxygen. Putting it all together we get 4.4 Equivalent. at 3.3 (b) Again two signals, but now in a 9:1 intensity ratio. Reasoning as in (a), there are three equivalent CH₃'s, each attached to an and a CH attached to more than one oxygen, as indicated by its downfield chemical shift (δ = 4.9). The only consistent structure is then (CH₃O)₃CH 3.3 4.9 (c) Two signals that are equal in intensity imply two different groups. each with six equivalent H's. The signal at δ = 3.1 could imply two equivalent groups, and the signal at δ = 1.2 suggests two equivalent CH₃ groups not attached to oxygen. These all add up to C₄H₁₂O₂, leaving one carbon unaccounted for in the formula of the molecule The fifth carbon can be used to connect the other four groups: (CH₃O)₂C(CH₃)₂, which is the answer. By comparison. 1,2-dimethoxyethane has two signals in a 3:2 (= 6:4) ratio, and they are both in a region consistent with H's on a carbon attached to a single as the structure requires.