1559T_ch10_175-19810/30/0518:09Pa9e186 EQA 186.Chapter 10 USING NUCLEAR MAGNETIC RESONANCE SPECTROSCOPY TO DEDUCE STRUCTURE pors,giving the quintet at (c)The CH,Cl has a CH,as a neighbor and is the triplet at 8 =3.6 ppm:the CH,at C2 has three adithe quaret at Pm:and the CHCI has five neighbors and is the s 37.Use the N+1 rule:Number of lines=number of Neighbors The signal for the CH groups at0.9 will be split into a triplet by the two hydrogens on the ring CH2 groupo each one( 2+1 3).The signal for the .3 will between them because they are equivalent by symmetry and have the same chemical shift. Chemical-shift-cquivalent nucli do not give rise o (b)The signal for the CH3 gr s at =15 will be split into a doublet by the single hydro hboring CH goup =2).The signal for the CHgrp into a septet by the six hydrogens on the two neighboring CHa groups (6 +I =7). (e)All signals will be singlets.The CCl g is in between the CH,groups and the CH group. The hydroxy hydrogen does not split. (d)This spectrum will be messy.because three CHs groups have their resonances at about the sam it h latter in its neighborin sigr d th side (4+=5).Finally.the signal for the CH group at =1.5 will as a resu t of splitting by a total of eight eight similar chemical shift positions in the spectrum. (e)No splitting.All singlets. (f)The signal fo the CH roups atǒ 09 will b CH, and the ch a e signal t s),arising fron rogens on t (g)The signal for the CHa gr up at 8=1.0 will be split into a triplet by the neighboring CH group.The signal for the CH be by the combined effect e the CHy group at =3.4 will be a singlet. will be spi t into) The signal for the CH groups at=2.4 will be split into a triplet by each one's single neighboring CH2 eroup.Unfortunately,simplicity is not likely to b the case h ere:In ens that nto as many as nine (b) Each CH3 has a CH as a neighbor, giving the doublet at 1.5 ppm; and each CHCl has four neighbors, giving the quintet at 4.1 ppm. (c) The CH2Cl has a CH2 as a neighbor and is the triplet at 3.6 ppm; the CH2 at C2 has three neighbors and is the quartet at 2.1 ppm; and the CHCl has five neighbors and is the sextet at 4.2 ppm. 37. Use the N  1 rule: Number of lines number of Neighbors  1. (a) The signal for the CH3 groups at 0.9 will be split into a triplet by the two hydrogens on the neighboring CH2 group of each one (2  1 3). The signal for the CH2 groups at 1.3 will be split into a quartet by the three hydrogens on the neighboring CH3 group (3  1 4). Although each CH2 group has the other CH2 group as its neighbor also, splitting is not observed between them because they are equivalent by symmetry and have the same chemical shift. Chemical-shift-equivalent nuclei do not give rise to observable splitting with one another. (b) The signal for the CH3 groups at 1.5 will be split into a doublet by the single hydrogen on its neighboring CH group (1  1 2). The signal for the CH group at 3.8 will be split into a septet by the six hydrogens on the two neighboring CH3 groups (6  1 7). (c) All signals will be singlets. The CCl group is in between the CH3 groups and the CH2 group. The hydroxy hydrogen does not split its neighbor. (d) This spectrum will be messy, because three CH3 groups have their resonances at about the same chemical shift, 0.9, and they are not all equivalent. The signal for the two CH3 groups with the neighboring CH group will be split by the latter into a doublet. The signal for the remaining CH3 group at 0.9 will be split into a triplet by its neighboring CH2 group. The doublet and triplet signals will overlap in the spectrum. The signal for the CH2 group at 1.4 will be split into a quintet by the combined effect of its neighboring CH3 group and the CH group on its other side (4  1 5). Finally, the signal for the CH group at 1.5 will possess nine lines (a nonet) as a result of splitting by a total of eight neighboring hydrogens (two CH3 groups and one CH2 group). These latter two patterns, the quintet and the nonet, will also overlap, because of their similar chemical shift positions in the spectrum. (e) No splitting. All singlets. (f ) The signal for the CH3 groups at 0.9 will be split into a triplet by the two hydrogens on each one’s neighboring CH2 group. The signal for the CH2 groups at 1.3 will be split into a quintet by the combined effect of the neighboring CH3 group and the CH group on the other side (4  1 5). The signal for the CH group at 1.5 will be a septet (7 lines), arising from splitting by six hydrogens on the three neighboring CH2 groups. (g) The signal for the CH3 group at 1.0 will be split into a triplet by the neighboring CH2 group. The signal for the CH2 group at 1.7 will be split into a sextet by the combined effect of its neighboring CH3 group and the CH2 group on its other side (5  1 6). The signal for the CH2 group at 3.8 will be split into a triplet by its neighboring CH2 group. The signal for the CH3 group at 3.4 will be a singlet. (h) In the simplest possible scenario, the signal for the CH2 group at 1.5 will be split into a quintet by the combined effect of its two neighboring CH2 groups on either side (4  1 5). The signal for the CH2 groups at 2.4 will be split into a triplet by each one’s single neighboring CH2 group. Unfortunately, simplicity is not likely to be the case here: In cyclic compounds, coupling constants (J values) between nonequivalent hydrogens that are cis to each other are often different in magnitude from the J values between hydrogens that are trans. As discussed in Section 10-8, the usual consequence of this phenomenon is more lines. Taking this situation into account, the signal for the CH2 group at 1.5 may be split into as many as nine 186 • Chapter 10 USING NUCLEAR MAGNETIC RESONANCE SPECTROSCOPY TO DEDUCE STRUCTURE 1559T_ch10_175-198 10/30/05 18:09 Page 186