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Unformatted text preview: 649 AMINES 4-(N, N-dimethylamino)pyridine that is more basic. Note that protonation of the ring nitrogen permits delocalization of the dimethylamino lone pair and dispersal of the positive charge.
N(CH3)2 N(CH3)2 N N H H Most stable protonated form of
4-(N, N-dimethylamino)pyridine 22.47 The 1H NMR spectrum of each isomer shows peaks corresponding to ﬁve aromatic protons, so compounds A and B each contain a monosubstituted benzene ring. Only four compounds of molecular
formula C8H11N meet this requirement.
C6H5CH2NHCH3 C6H5NHCH2CH3 C6H5CHCH3 C6H5CH2CH2NH2 NH2
N-Methylbenzylamine N-Ethylaniline 1-Phenylethylamine 2-Phenylethylamine Neither 1H NMR spectrum is consistent with N-methylbenzylamine, which would have two
singlets due to the methyl and methylene groups. Likewise, the spectra are not consistent with
N-ethylaniline, which would exhibit the characteristic triplet–quartet pattern of an ethyl group. Although a quartet occurs in the spectrum of compound A, it corresponds to only one proton, not the
two that an ethyl group requires. The one-proton quartet in compound A arises from an H— C— CH3
unit. Compound A is 1-phenylethylamine.
Doublet ( 1.2 ppm) CH3
C6H5 C NH2
Singlet ( 1.3 ppm) H
Quartet ( 3.9 ppm) Compound B has an 1H NMR spectrum that ﬁts 2-phenylethylamine.
Singlet ( 1.1 ppm) C6H5CH2CH2NH2 at 22.48 Pair of triplets
2.75 ppm and 2.95 ppm Only the unshared electron pair on nitrogen that is not part of the electron cloud of the aromatic
system will be available for protonation. Treatment of 5-methyl- -carboline with acid will give the
N CH3 5-Methyl- -carboline Back Forward CH3 N N 22.49 N H H Write the structural formulas for the two possible compounds given in the problem and consider
how their 13C NMR spectra will differ from each other. Both will exhibit their CH3 carbons at high
ﬁeld signal, but they differ in the positions of their CH2 and quaternary carbons. A carbon bonded to Main Menu TOC Study Guide TOC Student OLC MHHE Website ...
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