15597.ch10175-19810/30/0518:09Pag0189 EQA Solutions o Problems189 (D)=0.9 (doublet.6 H):Two identical CH3's,both split by a neighboring CH (1+1=2= doublet) =1.3(singlet,1 H):Most likely the OH(the broad signal gives it away) =1.45(quartet,2 H):CH2,split by three neighboring hydrogens(3+1=4=quartet) =1.7(multiplet,I H):CH,split by perhaps 7 neighboring hydrogens (eight lines are visible: there could always be more that are not) =triplet).and connected .The CH plit b C-on. CH neighbor- -must be the CH group.We can put the pieces together directly to give 1.7145 3713 (E)=0.8 (triplet.3 H):CHs,split by a neighboring CH2 8=13(amcss.4H:7? =1.5(quintet?.2 H):CHa,split by four (?)ncighboring hydrogens 8=3.0 (broad singlet.I H):OH again =3.5(triplet.2 H):CHa,split by a neighboring CHa and on the The pieces we can identify add up.Some guessing is nceded.Let's start by assuming that the CH groups at 15 an are atta 5 0 There's only on C way to put the pieces together to include everything.I-pentanol: CH, CH compounds (F)=0.9 (triplet.although a poor excuse for one.3 H):CHa.next to a CH =1.2 (doublet.3 H):CHs,next to a CH 14 (complex signal.4 H):?2? =1.6 (broad singlet.1 H):OH 3.(four.maybe five lines.1 H):CH.next toO.split by at least 3 and maybe 4 neighboring H's presumably also a 1.4.The pieces are(D) 0.9 (doublet, 6 H): Two identical CH3’s, both split by a neighboring CH (1  1 2 doublet) 1.3 (singlet, 1 H): Most likely the OH (the broad signal gives it away) 1.45 (quartet, 2 H): CH2, split by three neighboring hydrogens (3  1 4 quartet) 1.7 (multiplet, 1 H): CH, split by perhaps 7 neighboring hydrogens (eight lines are visible; there could always be more that are not) 3.7 (triplet, 2 H): CH2, split by a neighboring CH2 (2  1 3 triplet), and connected directly to O, according to the chemical shift These five groups add up to C5H12O, so we’ve found all the pieces in clearly separated signals. The CH group at 1.7 is reasonably assigned as the group responsible for splitting the 0.9 signal into a doublet. The CH2 group at 3.7 is presumably split by the other CH2 at 1.45; however, the latter is split by a third neighbor—must be the CH group. We can put the pieces together directly to give (E) 0.8 (triplet, 3 H): CH3, split by a neighboring CH2 1.3 (a mess, 4 H): ??? 1.5 (quintet?, 2 H): CH2, split by four (?) neighboring hydrogens 3.0 (broad singlet, 1 H): OH again 3.5 (triplet, 2 H): CH2, split by a neighboring CH2 and on the O The pieces we can identify add up to C3H8O. Some guessing is needed. Let’s start by assuming that the CH2 groups at 1.5 and 3.5 are attached to each other and see if it gets us anywhere. The CH3 then has to be attached to a CH2 that is buried in the signal at 1.3. Now our groups total up to C4H10O. There has to be one more CH2 group hiding at 1.3. So what do we have? There’s only one way to put the pieces together to include everything, 1-pentanol: CH3OCH2OCH2OCH2OCH2OOH. By the way, are you noticing how the chemical shifts of the OH groups of these compounds vary all over the place? That’s normal. (F) 0.9 (triplet, although a poor excuse for one, 3 H): CH3, next to a CH2 1.2 (doublet, 3 H): CH3, next to a CH 1.4 (complex signal, 4 H): ??? 1.6 (broad singlet, 1 H): OH 3.8 (four, maybe five lines, 1 H): CH, next to O, split by at least 3 and maybe 4 neighboring H’s Let’s look at these pieces. The CH3 at 1.2 could be attached to the CH at 3.8. The CH3 at 0.9 could be attached to a CH2 that is part of the 1.4 signal. That gives a total of C4H10O, so we need another CH2, presumably also at 1.4. The pieces are CH3 1.6 0.9 1.4 CH3 OH CH2 1.2 3.8 CH2 1.4 CH CH3 0.8 1.3 3.0 CH2 CH2 OH 1.5 3.5 CH2 1.3 CH2 CH2 CH2 CH3 CH OH . CH3 0.9 1.7 1.45 3.7 1.3 Solutions to Problems • 189 1559T_ch10_175-198 10/30/05 18:09 Page 189