1559T_ch22_389-40811/9/0515:29Pa9e40 EQA Solutions to Problems.401 cordigoegpotcatoa1compounds:h call the McLaff t results in cleav -C,H4m=110 OH HO which is the enol isomer of the methyl ketone structure shown above The IR data is t l,th NMR contansofnominhl evryni ohyro are (b .78(s)4.33(d 220 3.85(sCH30 >0.93( can give a d from which loss of a methyl group (m/z 15) gives the m/z 122 peak. Meanwhile the “tricky” peak at m/z 195 is related to the m/z 194 fragment identified above; in fact, it is that piece plus a hydrogen. That is, Where did that come from? Recall the McLafferty fragmentation of carbonyl compounds: It results in cleav￾age between the and  carbons according to the following process: which is the enol isomer of the methyl ketone structure shown above. The IR data is fairly straightforward to interpret. The band at 972 cm1 derives from a bending motion of the trans alkene double bond, that at 1660 cm1 from the amide CPO stretch, 3016 cm1 from the alkene and arene COH stretches, and 3445 and 3541 cm1 from the amide NOH and phenol OOH stretches. Finally, the NMR contains a wealth of information with virtually every individual group of hydrogens clearly resolved. The assignments are made easier if you pay attention to the integrations and the splittings, which con￾form to the N  1 rule. Here they are (br broad signal; qn quintet): The OH and NH hydrogens are indistinguishable to the NMR in this molecule and combine to give a broad signal at  5.82 ppm. Unusually, the NH hydrogen splits the neighboring benzylic CH2 group, whose signal at  4.33 ppm is a doublet (this assignment must be made by process of elimination—no other CH2 group in the molecule can give a doublet in the NMR). O HO CH3O N H 5.82 (br) 6.73 and 6.85 (d) 5.33 (br) 0.93 (d) 3.85 (s) 6.78 (s) 5.82 (br) 2.18 (t) 1.62 (qn) 4.33 (d) 1.35 (qn) 2.20 (m) 1.97 (q) HO CH3O N H OH CH2  H HO CH3O N H C8H14 (m/z 110)  O O HO CH3O N CH3 H  Solutions to Problems • 401 1559T_ch22_389-408 11/9/05 15:29 Page 401