1559r.ch10.175-19810/30/0518:09Page183 EQA Solutions to Problems183 32.(a)The spectrum shows two signals,at=1.1 and 3.3.The=1.1 signal is for 9 equivalent single carbon.because the (H The downfield location of these two hydrogens suggests their carbon is attached to the CI.So. (CH3);C-+-CH2-+-Cl(CH3);C-CH2-Cl 33 as a plausible structure. (b)Somewhat similar:two signals.=19 and 3.8.The signal for six signal can only be a-CH2-because there are only four carbons in the molecule.So. ，←3.8 CH-C-CH +-CH-+(2x)-Br CH- equivalent CH for all but the two oxygen atoms in the formula.The large signal at3.3 is just right for hydrogens on carbons attached to an oxygen.The downfield xygen. CH3-O-CH2-O-CH3 4.4 (b)Again two signals,but now in a 9:I intensity ratio.Reasoning as in (a).there are three equivalent (CHOCH 334.9 二 @that are intimply differento.cach wit油出 imply two used to connect the thr four groups:(CH,O)C(CH nal whic answe in a region consist CH.OCH2CH2OCHsrequires.32. (a) The spectrum shows two signals, at 1.1 and 3.3. The 1.1 signal is for 9 equivalent hydrogens, and the 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 C1. So, (b) Somewhat similar: two signals, 1.9 and 3.8. The signal for six equivalent hydrogens is probably due to two methyl groups on the same carbon (CH3OCOCH3). The two-hydrogen signal can only be a OCH2O, because there are only four carbons in the molecule. So, 33. (a) The spectrum has two signals in a 3
1 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 3.3 is just right for hydrogens on carbons attached to an oxygen. The downfield location for the small signal ( 4.4) is consistent with attachment of carbon to more than one oxygen. Putting it all together we get (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 oxygen, and a CH attached to more than one oxygen, as indicated by its downfield chemical shift ( 4.9). The only consistent structure is then (c) Two signals that are equal in intensity imply two different groups, each with six equivalent H’s. The signal at 3.1 could imply two equivalent CH3OO groups, and the signal at 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 (C5H12O2). 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 oxygen, as the structure CH3OCH2CH2OCH3 requires. 4.9 (CH3O)3CH 3.3 CH3 O CH2 O CH3 4.4 Equivalent, at 3.3 (2) Br Br Br CH3 C C CH3   CH2 CH3 CH3 CH2 1.9 3.8 3.3 as a plausible structure. (CH3)3C CH  2 Cl Cl (CH3)3C 1.1  CH2 Solutions to Problems • 183 1559T_ch10_175-198 10/30/05 18:09 Page 183