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23.E: Organonitrogen Compounds I: Amines (Exercises)

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  • Exercise 23-1 Name the following substances by an accepted system (Section 7-8):

    a. \(\ce{(CH_3)_2N-CH=CH_2}\)


    c. \(\ce{H_2NCH_2CO_2H}\)

    d. \(\ce{H_2NCH_2CH_2CH_2CH_2NH_2}\)




    Exercise 23-2 How could one show with certainty that the peak at \(47 \: \text{Hz}\) with reference to TMS in the NMR spectrum of \(\ce{N}\)-ethylethanamine (Figure 23-5) is due to the \(\ce{N-H}\) resonance?

    Exercise 23-3 Show how structures can be deduced for the isomeric substances of molecular formula \(\ce{C_8H_{11}N}\), whose NMR and infrared spectra are shown in Figure 23-6.

    Figure 23-6: Infrared and NMR spectra of two isomeric compounds, (a) and (b), of formula \(\ce{C_8H_{11}N}\). The NMR spectra are at \(60 \: \text{MHz}\) relative to TMS. See Exercise 23-3.

    Exercise 23-4 The highest mass peak in the mass spectrum of a certain compound is \(m/e\) 73. The most abundant peak has \(m/e\) 58. Suggest a structure for the compound and explain how it could form an ion of \(m/e\) 58.

    Exercise 23-5 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 \(2.3 \: \text{ppm}\). Assign a structure to the compound and account for the fragment ions \(m/e\) 72, 57, and 30.

    Exercise 23-6* Explain why the configuration of the nitrogen in 1-ethylazacyclopropane, \(1\), is more stable than in triethylamine. Why is the configuration of oxazacyclopropanes, such as \(2\), exceptionally stable? (Consider the \(\pi\) molecular orbitals of an ethene bond, Figure 21-3, as a model for orbitals of the adjacent \(\ce{O}\) and \(\ce{N}\) atoms in the planar transition state for inversion in \(2\).)

    Figure 23-7 The proton NMR spectrum of 1,2,2-trimethylazacyclopropane, \(3\) at room temperature is shown in Figure 23-7. When the material is heated to \(110^\text{o}\), the two lines at \(63 \: \text{Hz}\) and \(70 \: \text{Hz}\) are found to have coalesced to a single line. At the same time, the lines at \(50 \: \text{Hz}\) and \(92 \: \text{Hz}\) coalesce to a single line. When the sample is cooled the spectrum changes back to that of Figure 23-7. Account for all the NMR lines of \(3\) and explain the effect of temperature on the spectrum. Review Section 9-10C.\(^3\)

    Figure 23-7: Proton NMR spectrum of 1,2,2-trimethylazacyclopropane at \(60 \: \text{MHz}\) relative to TMS at \(0.0 \: \text{ppm}\). See Exercise 23-7.

    Exercise 23-8* The \(\ce{^{19}F}\) spectrum of 4,4-difluoroazacyclohexane in acetone solution at \(25^\text{o}\) is a sharp, narrowly spaced 1:4:6:4:1 quintet; at \(-60^\text{o}\) it is a broad quartet with a chemical-shift difference of \(960 \: \text{Hz}\) and \(J\) of \(235 \: \text{Hz}\), and at \(-90^\text{o}\) it is a pair of overlapping quartets with chemical-shift differences and relative intensities of \(1050 \: \text{Hz}\) \(\left( 75\% \right)\) and \(700 \: \text{Hz}\) \(\left( 25\% \right)\), both with \(J\) of \(235 \: \text{Hz}\). Account for these changes in the \(\ce{^{19}F}\) spectra with temperature. Review Section 9-10C.\(^3\)

    Exercise 23-9 Account for the following observations:

    a. The proton NMR spectrum of trimethylamine in nitromethane-\(\ce{D_3}\) \(\left( \ce{CD_3NO_2} \right)\) shows a single resonance near \(2.7 \: \text{ppm}\). On adding an equivalent of fluoroboric acid, \(\ce{HBF_4}\), the singlet at \(2.7 \: \text{ppm}\) is replaced by a doublet at \(3.5 \: \text{ppm}\)

    b. On adding trace amounts of trimethylamine to the solution described in Part a, the double at \(3.5 \: \text{ppm}\) collapses to a singlet centered at \(3.5 \: \text{ppm}\). As more trimethylamine is added, the singlet resonance moves progressively upfield.

    Exercise 23-10 Decide which member in each of the following pairs of compounds is the stronger base. Give your reasoning.

    a. \(\ce{CH_3CH_2NH_2}\) or \(\ce{(CH_3CH_2)_3N}\)
    b. \(\ce{(CH_2)_5NH}\) or \(\ce{(CH_2)_2NH}\)
    c. \(\ce{CF_3CH_2CH_2NH_2}\) or \(\ce{CH_3CH_2CH_2NH_2}\)
    d. \(\ce{(CH_3)_3} \overset{\oplus}{\ce{N}} \ce{CH_2CH_2NH_2}\) or \(\overset{\ominus}{\ce{O_2}} \ce{CCH_2CH_2NH_2}\)

    Exercise 23-11 Draw atomic-orbital models for benzenamine and its conjugate acid and describe the features of these models that account for the low base strength of benzenamine relative to saturated amines.

    Exercise 23-12 Amidines, \(\ce{R-C(NH_2)=NH}\), are stronger bases than saturated amines. Explain why this should be so, paying special attention to which nitrogen the proton adds to.

    Exercise 23-13 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 3-nitrobenzenamine be a much weaker base than benzenamine itself, but substantially stronger than 4-nitrobenzenamine?

    Exercise 23-14 Indicate whether the following equilibria would have \(K\) greater, or less, than unity. This is equivalent to asking which amine is the stronger base. Give a reason for your answer.






    Exercise 23-15 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, \(\ce{(NH_2)_3} \overset{\oplus}{\ce{C}}\), is an exceptionally weak acid.

    Exercise 23-16 The p\(K_a\) of the conjugate acid of caffeine (Figure 23-1) is 10.61. Calculate the \(K_b\) of caffeine. Write structures for the possible conjugate acids of caffeine in which the added proton is attached to one or the other of the nitrogens of the five-membered ring. Use the resonance method to determine which of these two nitrogens will be the preferred protonation site of caffeine. Give your reasoning.

    Exercise 23-17* 2-Amino-1,3-diazabenzene (2-aminopyrimidine) undergoes \(\ce{N}\)-methylation with methyl iodide to give two isomeric products, A and B, of formula \(\ce{C_5H_7N_3}\) (Section 23-9D). At high pH, the major methylation product is A, which is a weakly basic compound with p\(K_a =\) 3.82. \(\ce{N}\)-Methylation in neutral conditions produces the more strongly basic compound B with p\(K_a =\) 10.75. Draw structures for the two isomers, A and B, and explain why A is a weak base and B is a much stronger base. Why is A the predominant product under basic conditions? Give your reasoning.

    Exercise 23-18* The conjugate acid of \(\ce{N}\),\(\ce{N}\)-dimethylbenzenamine has p\(K_a =\) 5.06, whereas the conjugate acid of diphenyldiazene (azobenzene, \(\ce{C_6H_5N=NC_6H_5}\)) has p\(K_a =\) -2.5. Yet for many years there was considerable controversy about where a proton adds to 4-\(\ce{(CH_3)_2N-C_6H_4N=NC_6H_5}\). Why is it not an open-and-shut case that a proton would add most favorably to the \(\ce{(CH_3)_2N}-\) nitrogen? Which of the two \(\ce{-N=N}-\) nitrogens would you expect to be the more basic? Give your reasoning. (Consider the effect of the \(\ce{-N=N}-\) group on the basicity of the \(\ce{(CH_3)_2N}-\) nitrogen and also the effect of the \(\ce{(CH_3)_2N}-\) group on the basicity of each of the \(\ce{-N=N}-\) nitrogens.)

    Exercise 23-19

    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.

    Exercise 23-20 Show the products you would expect to be obtained in each of the following reactions:


    b. \(\ce{H_2N(CH_2)_4NH_2}\) (1 mole) \(+ \ce{C_6H_5SO_2Cl}\) (2 moles) \(\overset{\ce{NaOH}}{\longrightarrow}\)

    c. \(\ce{CH_3CH_2CHO} + \ce{CH_3CH_2NH_2} \overset{\ce{H}^\oplus}{\longrightarrow}\)



    Exercise 23-21 2,4-Pentanedione reacts with methanamine to give a product of composition \(\ce{C_6H_{11}NO}\) that is an equilibrium mixture of three isomers. The NMR spectrum of the mixture indicates that all three have strong hydrogen bonding. Draw the structures of the three isomers and indicate the nature of the hydrogen bonding.

    Exercise 23-22 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 \(\ce{C_3H_7N}\) with \(\ce{C_6H_5SO_2Cl}\) and excess \(\ce{NaOH}\) solution gives a clear solution. This solution when acidified gives a solid product of formula \(\ce{C_9H_{11}O_2NS}\).

    b. A liquid diamine of formula \(\ce{C_5H_{14}N_2}\) with \(\ce{C_6H_5SO_2Cl}\) and \(\ce{NaOH}\) gives an insoluble solid. This solid dissolves when the mixture is acidified with dilute hydrochloric acid.

    Exercise 23-23 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

    Exercise 23-24 Show how a mixture of amines prepared from 1-bromobutane and an excess of butanamine may be resolved into its components by reaction with the anhydride of 1,4-butanedioic acid, \(\ce{(CH_2)_2(CO)_2O}\), separation of the products through advantage of their solubility properties in acid or base, and regeneration of the corresponding amines (Section 18-10C). Write equations for the reactions involved.

    Exercise 23-25* Assess the possibility of \(\ce{O}\)-alkylation in the reaction of \(\ce{CH_2=CH-CH_2Br}\) with \(\ce{C_6H_5SO_2NH}^\ominus \ce{Na}^\oplus\). Give your reasoning.

    Exercise 23-26 The tertiary amine, \(\ce{C_6H_5CH_2N(CH_3)_2}\), reacts with nitrous acid to give benzenecarbaldehyde and \(\ce{N}\)-nitroso-\(\ce{N}\)-methylmethanamine (\(\ce{N}\)-nitrosodimethylamine):

    \[2 \ce{C_6H_5CH_2N(CH_3)_2} + 4 \ce{HONO} \rightarrow 2 \ce{C_6H_5CHO} + 2 \ce{(CH_3)_2N-NO} + 3 \ce{H_2O} + \ce{N_2O}\]

    A possible reaction sequence that explains the formation of benzenecarbaldehyde involves nitrosation, \(E2\) elimination, hydrolysis, and finally nitrosation. Write each of the steps involved in this sequence. Formation of \(\ce{N_2O}\) appears to take place by dimerization of the hypothetical substance \(\ce{HNO}\) \(\left( 2 \ce{HNO} \rightarrow \ce{N_2O} + \ce{H_2O} \right)\).

    Exercise 23-27

    a. Write two valence-bond structures for \(\ce{N}\)-nitroso-\(\ce{N}\)-methylmethanamine and show how these structures explain the fact that the \(\ce{N}\)-nitrosamine is a much weaker base than \(\ce{N}\)-methylmethanamine.

    b. \(\ce{N}\)-Nitroso-\(\ce{N}\)-methylmethanamine shows two separate methyl resonances in its proton NMR spectrum. These collapse to a single resonance when the material is heated to \(190^\text{o}\) and reappear on cooling. Studies of the changes in the line shapes with temperature show that the process involved has an energy barrier of about \(23 \: \text{kcal mol}^{-1}\). Explain why there should be separate methyl peaks in the NMR and why they should coalesce on heating. (Review Section 9-10C. You also may wish to read Section 27-2.)

    Exercise 23-28

    a. When ethyl aminoethanoate \(\left( \ce{H_2NCH_2CO_2C_2H_5} \right)\) is treated with nitrous acid in the presence of a layer of diethyl ether, a yellow compound known as ethyl diazoethanoate \(\left( \ce{N_2CHCO_2C_2H_5} \right)\) is extracted into the ethyl layer. What is the probable structure of this compound and what is the mechanism by which it is formed?

    b. Would you expect the same type of reaction sequence to occur with ethyl 3-aminopropanoate? Explain.

    Exercise 23-29 Predict the products expected from the reactions of the following amines with nitrous acid (prepared from \(\ce{NaNO_2} + \ce{HCl}\) in aqueous solution):

    a. 2-methylpropanamine
    b. azacyclopentane
    c. 2-butenamine
    d. 3-amino-2,3-dimethyl-2-butanol

    Exercise 23-30 The following sequence is very useful for expanding the ring size of a cyclic ketone:

    List reagents, conditions, and the important intermediates for the sequence, noting that several individual synthetic steps may be required. (Refer to Table 23-6 for amine synthesis.)

    Exercise 23-31* The reaction of nitrous acid with 3-butenamine, \(\ce{CH_2=CHCH_2CH_2NH_2}\), has been found to give the following mixture of alcohols: 3-buten-1-ol \(\left( 45\% \right)\), 3-buten-2-ol \(\left( 21\% \right)\), 2-buten-1-ol \(\left( 7\% \right)\), cyclobutanol \(\left( 12\% \right)\), and cyclopropylmethanol \(\left( 15\% \right)\). Show how each of these products may be formed from the 3-butenyl cation.

    Exercise 23-32* How could one determine experimentally how much of the propene formed in the reaction of propanamine with nitrous acid arises from the propyl cation and how much from the isopropyl cation?

    Exercise 23-33 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 \(\ce{C_6H_5Cl}\)/\(\ce{C_6H_5OH}\) increases markedly with \(\ce{Cl}^\ominus\) 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 \(\ce{D_2O}\), instead of \(\ce{H_2O}\), no \(\ce{C-D}\) 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 \(98\%\) \(\ce{H_2SO_4}\) at almost the same rate as in \(80\%\) \(\ce{H_2SO_4}\) and, in these solutions, the effective \(\ce{H_2O}\) concentration differs by a factor of 1000.

    Show how these observations support an \(S_\text{N}1\) reaction of benzenediazonium chloride, and can be used to argue against a benzyne-type elimination-addition reaction with water acting as the \(E2\) base (Section 14-6C) or an \(S_\text{N}2\) reaction with water as the nucleophile (Section 8-4, Mechanism B, and Section 14-6).

    Exercise 23-34 Indicate how you could prepare each of the following compounds, starting with benzene. One of the steps in each synthesis should involve formation of a diazonium salt. (Review Sections 22-4 and 22-5 if necessary.)

    a. monodeuteriobenzene
    b. 2-cyano-1-isopropylbenzene
    c. 4-tert-butylbenzenol
    d. 3-hydroxyphenylethanone
    e. 3-chloronitrobenzene
    f. 3-iodobenzenecarboxylic acid
    g. 4-chlorophenyl azide
    h. 4-methylphenylhydrazine
    i. 2-amino-4'-methylazobenzene
    j. 2-chloro-1-phenylpropane

    Exercise 23-35* Some of the rearrangements of arenamines, \(\ce{ArNHY}\), to \(\ce{Y-Ar-NH_2}\) shown in Section 23-10D proceed by an intermolecular mechanism involving acid-catalyzed cleavage of the \(\ce{N-Y}\) bond followed by a normal electrophilic substitution of the aromatic ring. Show the steps in this mechanism for \(\ce{Y} = \ce{NO}\) and \(\ce{Cl}\).

    Exercise 23-36* 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 \(\ce{H}^\oplus\) concentration) and the lack of mixing of groups as described above.

    Exercise 23-37* Show how the following substances could be prepared by a suitable \(\ce{ArNRY}\)-type rearrangement.


    b. 4-amino-3-methylbenzenol


    d. \(\ce{N}\)-methyl-1,4-benzenediamine

    Exercise 23-38 What is the oxidation state of each nitrogen in each of the following substances?


    b. \(\ce{F_2N-NF_2}\)

    c. \(\ce{(CH_3)_2C=NOH}\)

    d. \(\ce{CH_3N_2^+Cl^-}\)

    e. \(\ce{CH_3N=C} \colon\)


    g. \(\ce{CH_3-N=} \overset{\oplus}{\ce{N}} = \overset{\ominus}{\ce{N}}\)

    h. \(\ce{CH_3NCl_2}\)



    Exercise 23-39 Show how one could synthesize and resolve the oxide from \(\ce{N}\)-ethyl-\(\ce{N}\)-methyl-2-propenamine with the knowledge that amine oxides are somewhat basic substances having \(K_b\) values of about \(10^{-11}\) \(\left( K_a \sim 10^{-3} \right)\).

    Exercise 23-40 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 \(\ce{N}\)-phenylmethanamine)
    d. benzenecarbaldehyde to \(\ce{N}\)-methylphenylmethanamine \(\left( \ce{C_6H_5CH_2NHCH_3} \right)\)

    Exercise 23-41 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 \(\left( + \right)\) 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 \(\left( + \right)\)-carboxylic acid? Indicate the reagents you would need to carry out each step in the overall sequence \(\ce{RCO_2H} \rightarrow \ce{RCOCl} \rightarrow \ce{RCONH_2} \rightarrow \ce{RNH_2}\).

    Exercise 23-42 Draw the structures of the products expected to be formed in the following reactions:

    a. \(\ce{(CH_3)_2CHCH_2CONH_2} \underset{\ce{NaOH}}{\overset{\ce{Br_2}}{\longrightarrow}}\)




    Exercise 23-43 Cleavage of \(\ce{C-N}\) 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.

    Exercise 23-44 Write the steps involved in (a) the formation of triphenylmethanamine from triphenylmethyl chloride in aqueous ammonia containing sodium hydroxide and (b) the hydrolysis of triphenylmethanamine in aqueous ethanoic acid. (This is an unusually facile heterolytic cleavage of a saturated \(\ce{C-N}\) bonds.)

    Exercise 23-45 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.

    Exercise 23-46 Suggest protecting groups and reaction sequences whereby the following transformations could be achieved:

    a. 3-amino-1-propanol to 3-aminopropanoic acid
    b. 4-(2-aminoethyl)benzenamine to 2-(4-nitrophenyl)ethanamine

    Exercise 23-47 Write equations for a practical laboratory synthesis of each of the following compounds from the indicated starting materials. Give reagents and conditions.

    a. \(\ce{(CH_3)_3CCH_2NH}\) from \(\ce{(CH_3)_3CCO_2H}\)
    b. 1,6-hexanediamine from butadiene

    Exercise 23-48 Write a structure of at least one substance that fits each of the following descriptions. (Different structures may be written for each part.)

    a. a water-insoluble, acid-soluble nitrogen compound that gives no nitrogen gas with nitrous acid
    b. a compound that gives off water on heating to \(200^\text{o}\)
    c. a chiral ester that hydrolyzes to give only achiral compounds

    Exercise 23-49 Compound A is chiral and is a liquid with the formula \(\ce{C_5H_{11}O_2N}\). 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 \(\ce{C_5H_{13}N}\). Treatment of chiral B with nitrous acid gives a mixture containing some chiral alcohol C and some 2-methyl-2-butanol. Write structures for compounds A, 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 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?

    Exercise 23-50 Arrange the following pairs of substances in order of expected base strengths. Show your reasoning.

    a. \(\ce{N}\),\(\ce{N}\)-dimethylmethanamine and trifluoro-\(\ce{N}\),\(\ce{N}\)-bis(trifluoromethyl)methanamine
    b. phenylmethanamine and 4-methylbenzenamine
    c. ethanenitrile and azabenzene
    d. methamidine \(\ce{[HC(=NH)NH_2]}\) and methanamide (review Exercise 23-12)
    e. \(\ce{N}\)-methylazacyclopropane and \(\ce{N}\)-methylazacyclopentane (review Section 11-8B)

    Exercise 23-51 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

    Exercise 23-52 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 \(\ce{N}\)-nitroso-\(\ce{N}\)-phenylhydroxylamine (Cupferron)

    Exercise 23-53 Diazotization of 4-chlorobenzenamine with sodium nitrite and hydrobromic acid yields a diazonium salt solution that couples with \(\ce{N}\),\(\ce{N}\)-dimethylbenzenamine to give substantial amounts of 4-dimethylamino-4'-bromoazobenzene. Explain.

    Exercise 23-54* Hypophophorous acid, \(\ce{H_3PO_2}\), in the presence of copper reduces aryldiazonium salts to arenes (Table 23-4) by a radical-chain mechanism with formation of \(\ce{H_3PO_3}\). Cupric copper, \(\ce{Cu}\)(II), initiates the chain by reducing \(\ce{H_3PO_2}\) to \(\ce{H_2} \overset{\cdot}{\ce{P}} \ce{O_2}\). Write a chain mechanism for reduction of \(\ce{ArN_2^+}\) to \(\ce{ArH}\) that involves \(\ce{H_2} \overset{\cdot}{\ce{P}} \ce{O_2}\) in the chain-propagating steps.

    Exercise 23-55 Give the principal product(s) to be expected from the following reactions:

    a. \(\ce{C_6H_5N_2^+Cl^-} + \ce{C_6H_5NHCH_3} \underset{0^\text{o}}{\overset{\text{pH 7}}{\longrightarrow}}\)

    b. p-\(\ce{CH_3C_6H_4NHNHC_6H_5} \overset{\ce{H_2SO_4}}{\longrightarrow}\)



    Exercise 23-56 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 \(\ce{N}\)-phenylcarbazole to be a stronger, or weaker, base than triphenylamine? Explain.

    \(^3\)You also may wish to read ahead in Section 27-2.


    • John D. Robert and Marjorie C. Caserio (1977) Basic Principles of Organic Chemistry, second edition. W. A. Benjamin, Inc. , Menlo Park, CA. ISBN 0-8053-8329-8. This content is copyrighted under the following conditions, "You are granted permission for individual, educational, research and non-commercial reproduction, distribution, display and performance of this work in any format."