Chapter 23

Prominent peaks in the mass spectrum of a basic nitrogen compound have \(m / e\) values of \(87,72,57\), and 30 . The NMR spectrum shows only three proton resonances, having intensity ratios of \(9: 2: 2\) at \(0.9,1.3\), and \(2.3 \mathrm{ppm}\). Assign a structure to the compound and account for the fragment ions \(m / e 72,57\), and 30 .

Account for the following observations: a. The proton NMR spectrum of trimethylamine in nitromethane-D \(_{3}\left(\mathrm{CD}_{3} \mathrm{NO}_{2}\right)\) shows a single resonance near \(2.7 \mathrm{ppm} .\) On adding an equivalent of fluoroboric acid, \(\mathrm{HBF}_{4}\), the singlet at \(2.7 \mathrm{ppm}\) is replaced by a doublet at \(3.5 \mathrm{ppm}\) b. On adding trace amounts of trimethylamine to the solution described in Part a, the double at \(3.5\) ppm collapses to a singlet centered at \(3.5 \mathrm{ppm}\). As more trimethylamine is added, the singlet resonance moves progressively upfield.

Decide which member in each of the following pairs of compounds is the stronger base. Give your reasoning. a. \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{NH}_{2}\) or \(\left(\mathrm{CH}_{3} \mathrm{CH}_{2}\right)_{3} \mathrm{~N}\) b. \(\left(\mathrm{CH}_{2}\right)_{5} \mathrm{NH}\) or \(\left(\mathrm{CH}_{2}\right)_{2} \mathrm{NH}\) c. \(\mathrm{CF}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{NH}_{2}\) or \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{NH}_{2}\) d. \(\left(\mathrm{CH}_{3}\right)_{3} \stackrel{\oplus}{\mathrm{N}} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{NH}_{2}\) or \(\stackrel{\ominus}{\mathrm{O}}_{2} \mathrm{CCH}_{2} \mathrm{CH}_{2} \mathrm{NH}_{2}\)

Amidines, \(\mathrm{R}-\mathrm{C}\left(\mathrm{NH}_{2}\right)=\mathrm{NH}\), are stronger bases than saturated amines. Explain why this should be so, paying special attention to which nitrogen the proton adds to.

3-Nitrobenzenamine is less than \(1 / 100\) as strong a base as benzenamine, but is 23 times stronger than \(4-\) nitrobenzenamine. Remembering that the inductive effect falls of rapidly with the number of intervening bonds, why should 3nitrobenzenamine be a much weaker base than benzenamine itself, but substantially stronger than 4-nitrobenzenamine?

Offer plausible explanations of the following facts: a. Aza-2,4-cyclopentadiene (pyrrole) is unstable in acid solution and polymerizes. (Consider the effect of adding a proton to this molecule at the nitrogen and at carbon.) b. 1,3-Diaza-2,4-cyclopentadiene (imidazole) is a much stronger base than 1,3-diazabenzene (pyrimidine). c. The triaminomethyl cation, \(\left(\mathrm{NH}_{2}\right)_{3} \mathrm{C}\), is an exceptionally weak acid.

2-Amino-1,3-diazabenzene (2-aminopyrimidine) undergoes N-methylation with methyl iodide to give two isomeric products, A and B, of formula \(\mathrm{C}_{5} \mathrm{H}_{7} \mathrm{~N}_{3}\) (Section \(23-9 \mathrm{D}\) ). At high \(\mathrm{pH}\), the major methylation product is A, which is a weakly basic compound with \(\mathrm{p} K_{a}=3.82 . \mathrm{N}\) -Methylation in neutral conditions produces the more strongly basic compound \(\mathrm{B}\) with \(\mathrm{p} K_{a}=10.75 .\) Draw structures for the two isomers, \(\mathrm{A}\) and \(\mathrm{B}\), and explain why \(\mathrm{A}\) is a weak base and \(\mathrm{B}\) is a much stronger base. Why is A the predominant product under basic conditions? Give your reasoning.

The conjugate acid of \(\mathrm{N}, \mathrm{N}\) -dimethylbenzenamine has \(\mathrm{p} K_{a}=5.06\), whereas the conjugate acid of diphenyldiazene (azobenzene, \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{~N}=\mathrm{NC}_{6} \mathrm{H}_{5}\) ) has \(\mathrm{p} K_{a}=-2.5 .\) Yet for many years there was considerable controversy about where a proton adds to 4-(CH \(_{3}\) ) \(_{2} \mathrm{~N}-\mathrm{C}_{6} \mathrm{H}_{4} \mathrm{~N}=\mathrm{NC}_{6} \mathrm{H}_{5}\). Why is it not an open-and-shut case that a proton would add most favorably to the \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{~N}-\) nitrogen? Which of the two \(-\mathrm{N}=\mathrm{N}-\) nitrogens would you expect to be the more basic? Give your reasoning. (Consider the effect of the \(-\mathrm{N}=\mathrm{N}-\) group on the basicity of the \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{~N}-\) nitrogen and also the effect of the \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{~N}-\) group on the basicity of each of the \(-\mathrm{N}=\mathrm{N}-\) nitrogens.)

a. Explain why 1,3-diazacyclopentadiene (imidazole) is a much stronger acid than azacyclopentadiene (pyrrole). b. Would you expect benzenamine to be a stronger or weaker acid than cyclohexanamine? Give your reasoning.

Write a structural formula (one for each part) that fits the following descriptions. (These descriptions can apply to more than one structural formula.) a. A liquid basic nitrogen compound of formula \(\mathrm{C}_{3} \mathrm{H}_{7} \mathrm{~N}\) with \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{SO}_{2} \mathrm{Cl}\) and excess \(\mathrm{NaOH}\) solution gives a clear solution. This solution when acidified gives a solid product of formula \(\mathrm{C}_{9} \mathrm{H}_{11} \mathrm{O}_{2} \mathrm{NS}\). b. A liquid diamine of formula \(\mathrm{C}_{5} \mathrm{H}_{14} \mathrm{~N}_{2}\) with \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{SO}_{2} \mathrm{Cl}\) and \(\mathrm{NaOH}\) gives an insoluble solid. This solid dissolves when the mixture is acidified with dilute hydrochloric acid.

Show how the following compounds may be prepared from ammonia and the given starting materials: a. 1,2 -ethanediamine from ethene b. 2 -aminoethanol from ethene c. benzenamine from chlorobenzene

Assess the possibility of O-alkylation in the reaction of \(\mathrm{CH}_{2}=\mathrm{CH}-\mathrm{CH}_{2} \mathrm{Br}\) with \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{SO}_{2} \mathrm{NH}^{\ominus} \mathrm{Na}^{\oplus} .\) Give your reasoning.

Predict the products expected from the reactions of the following amines with nitrous acid (prepared from \(\mathrm{NaNO}_{2}+\mathrm{HCl}\) in aqueous solution): a. 2 -methylpropanamine b. azacyclopentane c. 2-butenamine d. 3-amino-2,3-dimethyl-2-butanol

Benzenediazonium chloride solvolyzes in water to give a mixture of benzenol and chlorobenzene. Some of the facts known about this and related reactions are 1\. The ratio \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{Cl} / \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OH}\) increases markedly with \(\mathrm{Cl}^{\ominus}\) concentration but the rate hardly changes at all. 2\. There is no rearrangement observed with 4-substituted benzenediazonium ions, and when the solvolysis is carried out in \(\mathrm{D}_{2} \mathrm{O}\), instead of \(\mathrm{H}_{2} \mathrm{O}\), no \(\mathrm{C}-\mathrm{D}\) bonds are formed to the benzene ring. 3\. 4-Methoxybenzenediazonium chloride solvolyzes about 30 times faster than 4-nitrobenzenediazonium chloride. 4\. Benzenediazonium salts solvolyze in \(98 \% \mathrm{H}_{2} \mathrm{SO}_{4}\) at almost the same rate as in \(80 \% \mathrm{H}_{2} \mathrm{SO}_{4}\) and, in these solutions, the effective \(\mathrm{H}_{2} \mathrm{O}\) concentration differs by a factor of 1000 . Show how these observations support an \(S_{\mathrm{N}} 1\) reaction of benzenediazonium chloride, and can be used to argue against a benzyne-type elimination-addition reaction with water acting as the \(E 2\) base (Section \(14-6 \mathrm{C}\) ) or an \(S_{\mathrm{N}} 2\) reaction with water as the nucleophile (Section 8-4, Mechanism B, and Section 14-6).

Treatment of a mixture of 2,2'-dimethylhydrazobenzene and hydrazobenzene with acid gives only 4,4'- diaminobiphenyl and 4,4 '-diamino-2,2'-dimethylbiphenyl. What does this tell you about the mechanism of this type of rearrangement? Write a mechanism for the rearrangement of hydrazobenzene that is in accord with the acid catalysis (the rate depends on the square of the \(\mathrm{H}^{\oplus}\) concentration) and the lack of mixing of groups as described above.

Show how the following transformations may be achieved. List reagents and approximate reaction conditions. a. 3-bromopropene to 3-butenamine b. cyclohexanone to cyclohexamine c. benzenecarboxylic acid to phenylmethanamine (not N-phenylmethanamine) d. benzenecarbaldehyde to N-methylphenylmethanamine \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{NHCH}_{3}\right)\)

The point of this exercise is to show that reactions of known stereospecificity can be used to establish configuration at chiral centers. A carboxylic acid of \((+)\) optical rotation was converted to an amide by way of the acyl chloride. The amide in turn was converted to a primary amine of one less carbon atom than the starting carboxylic acid. The primary amine was identified as 2\(S\) -aminobutane. What was the structure and configuration of the \((+)\) -carboxylic acid? Indicate the reagents you would need to carry out each step in the overall sequence \(\mathrm{RCO}_{2} \mathrm{H} \rightarrow \mathrm{RCOCl} \rightarrow \mathrm{RCONH}_{2} \rightarrow \mathrm{RNH}_{2}\).

Cleavage of \(\mathrm{C}-\mathrm{N}\) bonds by catalytic hydrogenation is achieved much more readily with diphenylmethanamine or triphenylmethanamine than with alkanamines. Explain why this should be so on the basis that the cleavage is a homolytic reaction.

Explain why the nitration of benzenamine to give 2- and 4-nitrobenzenamines is unsatisfactory with nitric acid-sulfuric acid mixtures. Show how this synthesis could be achieved by suitably modifying the amine function.

Compound A is chiral and is a liquid with the formula \(\mathrm{C}_{5} \mathrm{H}_{11} \mathrm{O}_{2} \mathrm{~N}\). A is insoluble in water and dilute acid but dissolves in sodium hydroxide solution. Acidification of a sodium hydroxide solution of chiral A gives racemic A. Reduction of chiral A with hydrogen over nickel produces chiral compound B of formula \(\mathrm{C}_{5} \mathrm{H}_{13} \mathrm{~N}\). Treatment of chiral B with nitrous acid gives a mixture containing some chiral alcohol \(\mathrm{C}\) and some 2 -methyl-2-butanol. Write structures for compounds \(\mathrm{A}, \mathrm{B}\), and C that agree with all the given facts. Write balanced equations for all the reactions involved. Show your reasoning. In this type of problem, one should work backward from the structures of the final products, analyzing each reaction for the structural information it gives. The key questions to be inferred in the preceding problem are (a) What kind of chiral compound or compounds could give 2-methyl-2-butanol and a chiral alcohol with nitrous acid? (b) What kinds of compounds could give \(\mathrm{B}\) on reduction? (c) What does the solubility behavior of A indicate about the type of compound that it is? (d) Why does chiral A racemize when dissolved in alkali?

What reagents and conditions would you use to prepare 2 -methylpropanamine by the following reactions: a. Hofmann rearrangement b. Schmidt rearrangement c. Curtius rearrangement d. Gabriel synthesis e. lithium aluminum hydride reaction

Write structural formulas for substances (one for each part) that fit the following descriptions: a. an aromatic amine that is a stronger base than benzenamine b. a substituted phenol that would not be expected to couple with benzenediazonium chloride in acid, alkaline, or neutral solution c. a substituted benzenediazonium chloride that would be a more active coupling agent than benzenediazonium chloride itself d. methyl \(Z\) -benzenediazoate e. the important resonance structures of the ammonium salt of N-nitroso-N- phenylhydroxylamine (Cupferron)

Diazotization of 4-chlorobenzenamine with sodium nitrite and hydrobromic acid yields a diazonium salt solution that couples with N,N-dimethylbenzenamine to give substantial amounts of 4 -dimethylamino-4'-bromoazobenzene. Explain.

Explain why triphenylamine is a much weaker base than benzenamine and why its electronic absorption spectrum is shifted to longer wavelengths compared with the spectrum of benzenamine. Would you expect \(\mathrm{N}\) -phenylcarbazole to be a stronger, or weaker, base than triphenylamine? Explain.