આ MCQ મોડ્યુલ આના પર આધારિત છે: NCERT Exercises and Solutions: Amines
આ મૂલ્યાંકન આના પર આધારિત હશે: NCERT Exercises and Solutions: Amines
મૂલ્યાંકન બનાવવામાં તેમની સામગ્રી સામેલ કરવા ચિત્રો, PDF અથવા Word દસ્તાવેજ અપલોડ કરો.
Amines can be considered as derivatives of ammonia obtained by replacement of hydrogen atoms with alkyl or aryl groups. Replacement of one hydrogen atom gives R–NH₂, a primary amine; secondary amines are R₂NH or R–NHR′, and tertiary amines R₃N. Secondary and tertiary amines are simple if the groups are the same and mixed if they differ. Like ammonia, all three types carry one unshared electron pair on nitrogen, due to which they behave as Lewis bases.
Amines are usually formed from nitro compounds, halides, amides and imides. They exhibit hydrogen bonding, which influences their physical properties. In alkylamines a combination of electron-releasing, steric and hydrogen-bonding factors influences the stability of the substituted ammonium cations in protic polar solvents and thus affects basic nature. Alkylamines are stronger bases than ammonia. In aromatic amines, electron-releasing and electron-withdrawing groups respectively increase and decrease basic character; aniline is a weaker base than ammonia.
Reactions of amines are governed by the availability of the unshared pair of electrons on nitrogen. The influence of the number of hydrogen atoms at the nitrogen atom on the type of reaction and nature of products is responsible for the identification and distinction between primary, secondary and tertiary amines. p-Toluenesulphonyl chloride is used for this identification. The presence of an amino group in an aromatic ring enhances reactivity of aromatic amines, and this reactivity can be controlled by acylation — treating with acetyl chloride or acetic anhydride. Tertiary amines like trimethylamine are used as insect attractants.
Aryldiazonium salts, usually obtained from arylamines, undergo replacement of the diazonium group with a variety of nucleophiles to provide advantageous methods for producing aryl halides, cyanides, phenols and arenes by reductive removal of the diazo group. Coupling reactions of aryldiazonium salts with phenols or arylamines give rise to azo dyes.
(i) (CH₃)₂CHNH₂ → propan-2-amine, 1°. (ii) CH₃(CH₂)₂NH₂ → propan-1-amine, 1°.
(iii) CH₃NHCH(CH₃)₂ → N-methylpropan-2-amine, 2°. (iv) (CH₃)₃CNH₂ → 2-methylpropan-2-amine, 1° (nitrogen carries only one carbon).
(v) C₆H₅NHCH₃ → N-methylaniline (N-methylbenzenamine), 2°. (vi) (CH₃CH₂)₂NCH₃ → N-ethyl-N-methylethanamine, 3°.
(vii) m-BrC₆H₄NH₂ → 3-bromoaniline (3-bromobenzenamine), 1°.
(i) Methylamine and dimethylamine. Carbylamine test. Methylamine (1°) with CHCl₃ and ethanolic KOH gives the foul-smelling methyl isocyanide; dimethylamine (2°) gives no such smell.
(ii) Secondary and tertiary amines. Hinsberg test. The 2° amine reacts with benzenesulphonyl chloride to give a sulphonamide insoluble in alkali; the 3° amine does not react at all.
(iii) Ethylamine and aniline. Azo dye test — diazotise each at 273–278 K and couple with alkaline 2-naphthol. Aniline gives a brilliant orange-red azo dye; ethylamine, being aliphatic, simply evolves nitrogen and gives ethanol, with no dye. (Alternatively, aniline gives a white precipitate with bromine water; ethylamine does not.)
(iv) Aniline and benzylamine. Bromine water. Aniline, having the –NH₂ attached directly to the highly activated ring, gives an immediate white precipitate of 2,4,6-tribromoaniline. Benzylamine's ring is not activated, so no precipitate forms.
(v) Aniline and N-methylaniline. Carbylamine test. Aniline is a primary amine and gives the offensive phenyl isocyanide; N-methylaniline is secondary and does not respond.
(i) pKb of aniline is more than that of methylamine. In aniline the lone pair on nitrogen is in conjugation with the benzene ring and is delocalised over it, so it is much less available for protonation. In methylamine the +I effect of the methyl group actually increases electron density on nitrogen. Hence aniline is the weaker base and its pKb is higher (9.38 against 3.38).
(ii) Ethylamine is soluble in water whereas aniline is not. Ethylamine forms hydrogen bonds with water through its small –NH₂ group and short ethyl chain. In aniline the large hydrophobic benzene ring dominates the molecule, so hydrogen bonding with water cannot compensate and solubility is very low.
(iii) Methylamine in water reacts with ferric chloride to precipitate hydrated ferric oxide. Methylamine is a base and in water it produces hydroxide ions: CH₃NH₂ + H₂O → CH₃NH₃⁺ + OH⁻. These hydroxide ions then react with Fe³⁺: Fe³⁺ + 3OH⁻ → Fe(OH)₃, precipitated as hydrated ferric oxide, Fe₂O₃·xH₂O.
(iv) Aniline on nitration gives a substantial amount of m-nitroaniline. Nitration is carried out in a strongly acidic medium, in which aniline is protonated to the anilinium ion. The –N⁺H₃ group is meta-directing, so alongside the ortho and para products from the unprotonated amine, a significant amount of the meta isomer is formed.
(v) Aniline does not undergo Friedel–Crafts reaction. The Lewis acid catalyst AlCl₃ forms a salt with the lone pair on nitrogen. The nitrogen thereby acquires a positive charge and becomes a strong deactivating group, so the ring is no longer susceptible to electrophilic attack.
(vi) Diazonium salts of aromatic amines are more stable than those of aliphatic amines. In an arenediazonium ion the positive charge is delocalised into the benzene ring by resonance, which stabilises it enough to survive in cold solution. An alkyldiazonium ion has no such delocalisation and decomposes immediately with loss of nitrogen.
(vii) Gabriel phthalimide synthesis is preferred for synthesising primary amines. It gives an exclusively primary amine. The nitrogen is held between the two carbonyl groups of the phthalimide ring and can be alkylated only once, so the secondary, tertiary and quaternary products that contaminate ammonolysis cannot form.
(i) Decreasing order of pKb: C₆H₅NH₂ > C₆H₅NHCH₃ > C₂H₅NH₂ > (C₂H₅)₂NH. (Higher pKb = weaker base, so aniline first.)
(ii) Increasing order of basic strength: C₆H₅NH₂ < C₆H₅N(CH₃)₂ < CH₃NH₂ < (C₂H₅)₂NH.
(iii)(a) p-Nitroaniline < aniline < p-toluidine. (iii)(b) C₆H₅NH₂ < C₆H₅NHCH₃ < C₆H₅CH₂NH₂.
(iv) Decreasing order of basic strength in the gas phase: (C₂H₅)₃N > (C₂H₅)₂NH > C₂H₅NH₂ > NH₃ — the pure inductive order, with no solvation to disturb it.
(v) Increasing order of boiling point: (CH₃)₂NH < C₂H₅NH₂ < C₂H₅OH. (No N–H in the tertiary sense is not the issue here — the 2° amine has one N–H, the 1° amine has two, and the alcohol's O–H bonding is strongest of all.)
(vi) Increasing order of solubility in water: C₆H₅NH₂ < (C₂H₅)₂NH < C₂H₅NH₂.
(i) Ethanoic acid into methanamine (2 C → 1 C, descent). CH₃COOH —[NH₃, Δ]→ CH₃CONH₂ —[Br₂ + 4NaOH]→ CH₃NH₂.
(ii) Hexanenitrile into 1-aminopentane (6 C → 5 C). CH₃(CH₂)₄CN —[H₃O⁺]→ CH₃(CH₂)₄COOH —[NH₃, Δ]→ CH₃(CH₂)₄CONH₂ —[Br₂/NaOH]→ CH₃(CH₂)₄NH₂.
(iii) Methanol to ethanoic acid (1 C → 2 C, ascent). CH₃OH —[HI or PCl₅]→ CH₃I —[KCN]→ CH₃CN —[H₃O⁺, hydrolysis]→ CH₃COOH.
(iv) Ethanamine into methanamine (2 C → 1 C). C₂H₅NH₂ —[HNO₂]→ C₂H₅OH —[oxidation, KMnO₄]→ CH₃COOH —[NH₃, Δ]→ CH₃CONH₂ —[Br₂/NaOH]→ CH₃NH₂.
(v) Ethanoic acid into propanoic acid (2 C → 3 C). CH₃COOH —[LiAlH₄]→ CH₃CH₂OH —[PCl₅ or HI]→ CH₃CH₂Cl —[KCN]→ CH₃CH₂CN —[H₃O⁺]→ CH₃CH₂COOH.
(vi) Methanamine into ethanamine (1 C → 2 C). CH₃NH₂ —[HNO₂]→ CH₃OH —[PCl₅]→ CH₃Cl —[KCN]→ CH₃CN —[LiAlH₄]→ CH₃CH₂NH₂.
(vii) Nitromethane into dimethylamine. CH₃NO₂ —[Sn/HCl reduction]→ CH₃NH₂; then alkylate with methyl iodide, CH₃NH₂ + CH₃I → (CH₃)₂NH (separated from over-alkylated products).
(viii) Propanoic acid into ethanoic acid (3 C → 2 C). CH₃CH₂COOH —[NH₃, Δ]→ CH₃CH₂CONH₂ —[Br₂/NaOH]→ CH₃CH₂NH₂ —[HNO₂]→ CH₃CH₂OH —[oxidation]→ CH₃COOH.
Method: the Hinsberg test. Treat the amine with benzenesulphonyl chloride (Hinsberg's reagent) and then add aqueous KOH.
Primary: C₆H₅SO₂Cl + H₂N–R → C₆H₅SO₂NHR + HCl. The N–H is made strongly acidic by the electron-withdrawing sulphonyl group, so the product dissolves in alkali.
Secondary: C₆H₅SO₂Cl + HNR₂ → C₆H₅SO₂NR₂ + HCl. With no N–H remaining the product is not acidic and is insoluble in alkali.
Tertiary: no reaction, since there is no hydrogen on nitrogen to be substituted.
Confirmatory test for 1°: the carbylamine reaction, R–NH₂ + CHCl₃ + 3KOH → R–NC + 3KCl + 3H₂O, which only primary amines give.
(i) Carbylamine reaction. Primary amines heated with chloroform and ethanolic KOH give foul-smelling isocyanides; 2° and 3° amines do not respond, so it is a test for 1° amines.
(ii) Diazotisation. Conversion of a primary aromatic amine into a diazonium salt with NaNO₂ and HCl at 273–278 K.
(iii) Hofmann's bromamide reaction. An amide with Br₂ in aqueous or ethanolic NaOH gives a primary amine with one carbon fewer, the alkyl group migrating from the carbonyl carbon to nitrogen.
(iv) Coupling reaction. A diazonium salt reacts with phenol or aniline at the para position to give a coloured azo compound containing the –N=N– link; an electrophilic substitution used to make dyes.
(v) Ammonolysis. Cleavage of the C–X bond of an alkyl halide by ammonia to give an amine; carried out in a sealed tube at 373 K and giving a mixture unless a large excess of ammonia is used.
(vi) Acetylation. Replacement of a hydrogen of –NH₂ or >N–H by an acetyl group using acetic anhydride or acetyl chloride, usually in the presence of pyridine; used to protect the amino group of aniline.
(vii) Gabriel phthalimide synthesis. Potassium phthalimide alkylated with R–X and then hydrolysed with alkali gives an exclusively primary amine; not applicable to aromatic primary amines.
(i) Nitrobenzene to benzoic acid. C₆H₅NO₂ —[Fe/HCl]→ C₆H₅NH₂ —[NaNO₂/HCl, 273–278 K]→ C₆H₅N₂⁺Cl⁻ —[CuCN]→ C₆H₅CN —[H₃O⁺]→ C₆H₅COOH.
(ii) Benzene to m-bromophenol. C₆H₆ —[HNO₃/H₂SO₄]→ C₆H₅NO₂ —[Br₂/FeBr₃]→ m-bromonitrobenzene (–NO₂ is meta-directing) —[Fe/HCl]→ m-bromoaniline —[NaNO₂/HCl, 273–278 K]→ diazonium salt —[warm H₂O, 283 K]→ m-bromophenol.
(iii) Benzoic acid to aniline. C₆H₅COOH —[NH₃, Δ]→ C₆H₅CONH₂ —[Br₂ + 4NaOH]→ C₆H₅NH₂.
(iv) Aniline to 2,4,6-tribromofluorobenzene. C₆H₅NH₂ —[Br₂(aq)]→ 2,4,6-tribromoaniline —[NaNO₂/HCl, 273–278 K]→ diazonium salt —[HBF₄]→ fluoroborate —[heat]→ 2,4,6-tribromofluorobenzene.
(v) Benzyl chloride to 2-phenylethanamine. C₆H₅CH₂Cl —[KCN]→ C₆H₅CH₂CN —[LiAlH₄]→ C₆H₅CH₂CH₂NH₂.
(vi) Chlorobenzene to p-chloroaniline. C₆H₅Cl —[conc. HNO₃/H₂SO₄]→ p-chloronitrobenzene —[Fe/HCl]→ p-chloroaniline.
(vii) Aniline to p-bromoaniline. Protect first: C₆H₅NH₂ —[(CH₃CO)₂O]→ acetanilide —[Br₂/CH₃COOH]→ p-bromoacetanilide —[H₃O⁺ hydrolysis]→ p-bromoaniline. (Direct bromination would give the 2,4,6-tribromo compound.)
(viii) Benzamide to toluene. C₆H₅CONH₂ —[Br₂/NaOH]→ C₆H₅NH₂ —[NaNO₂/HCl]→ C₆H₅N₂⁺Cl⁻ —[CuCN]→ C₆H₅CN —[H₃O⁺]→ C₆H₅COOH —[LiAlH₄]→ C₆H₅CH₂OH —[HI/red P or PCl₅ then Zn–Hg/HCl]→ C₆H₅CH₃.
(ix) Aniline to benzyl alcohol. C₆H₅NH₂ —[NaNO₂/HCl, 273–278 K]→ C₆H₅N₂⁺Cl⁻ —[CuCN]→ C₆H₅CN —[H₃O⁺]→ C₆H₅COOH —[LiAlH₄]→ C₆H₅CH₂OH.
(i) CH₃CH₂I —[NaCN]→ A = CH₃CH₂CN; —[partial hydrolysis, OH⁻]→ B = CH₃CH₂CONH₂; —[Br₂/NaOH]→ C = CH₃CH₂NH₂ (ethanamine).
(ii) C₆H₅N₂⁺Cl⁻ —[CuCN]→ A = C₆H₅CN; —[H₂O/H⁺]→ B = C₆H₅COOH; —[NH₃, Δ]→ C = C₆H₅CONH₂ (benzamide).
(iii) CH₃CH₂Br —[KCN]→ A = CH₃CH₂CN; —[LiAlH₄]→ B = CH₃CH₂CH₂NH₂; —[HNO₂, 0 °C]→ C = CH₃CH₂CH₂OH (propan-1-ol) + N₂.
(iv) C₆H₅NO₂ —[Fe/HCl]→ A = C₆H₅NH₂; —[NaNO₂ + HCl, 273 K]→ B = C₆H₅N₂⁺Cl⁻; —[H₂O/H⁺, warm]→ C = C₆H₅OH (phenol).
(v) CH₃COOH —[NH₃, Δ]→ A = CH₃CONH₂; —[NaOBr]→ B = CH₃NH₂; —[NaNO₂/HCl]→ C = CH₃OH (methanol) + N₂.
(vi) C₆H₅NO₂ —[Fe/HCl]→ A = C₆H₅NH₂; —[HNO₂, 273 K]→ B = C₆H₅N₂⁺Cl⁻; —[C₂H₅OH]→ C = C₆H₆ (benzene) + N₂ + CH₃CHO.
Given: an aromatic compound A with aqueous ammonia and heating forms B, which on heating with Br₂ and KOH forms C of molecular formula C₆H₇N.
Work backwards. C₆H₇N is aniline, C₆H₅NH₂ — so C = aniline. Br₂/KOH is the Hoffmann bromamide degradation, which needs an amide with one carbon more, so B = benzamide, C₆H₅CONH₂. An aromatic compound giving an amide with aqueous ammonia on heating is the acid, so A = benzoic acid, C₆H₅COOH.
IUPAC names: A = benzoic acid; B = benzamide; C = aniline (benzenamine).
(i) C₆H₅NH₂ + CHCl₃ + alc. KOH → C₆H₅NC (phenyl isocyanide) + 3KCl + 3H₂O.
(ii) C₆H₅N₂⁺Cl⁻ + H₃PO₂ + H₂O → C₆H₆ + N₂ + H₃PO₃ + HCl.
(iii) C₆H₅NH₂ + H₂SO₄ (conc.) → anilinium hydrogensulphate, which on heating at 453–473 K gives sulphanilic acid (p-aminobenzene sulphonic acid).
(iv) C₆H₅N₂⁺Cl⁻ + C₂H₅OH → C₆H₆ + N₂ + CH₃CHO + HCl.
(v) C₆H₅NH₂ + Br₂(aq) → 2,4,6-tribromoaniline (white precipitate) + 3HBr.
(vi) C₆H₅NH₂ + (CH₃CO)₂O → C₆H₅NHCOCH₃ (acetanilide) + CH₃COOH.
(vii) C₆H₅N₂⁺Cl⁻ —[(i) HBF₄]→ C₆H₅N₂⁺BF₄⁻ —[(ii) NaNO₂/Cu, Δ]→ C₆H₅NO₂ + N₂ + NaBF₄.
Gabriel phthalimide synthesis requires the phthalimide anion to displace a halide by nucleophilic substitution on the alkyl halide. Aryl halides do not undergo nucleophilic substitution with the anion formed by phthalimide: the C–X bond in an aryl halide has partial double-bond character because of resonance with the ring, making it shorter and stronger, and the flat ring carbon is not open to backside attack. Hence aniline and other aromatic primary amines cannot be prepared by this route.
(i) Aromatic primary amines. React at 273–278 K to give relatively stable arenediazonium salts: C₆H₅NH₂ + NaNO₂ + 2HCl → C₆H₅N₂⁺Cl⁻ + NaCl + 2H₂O. No nitrogen is evolved at this temperature; the salt is used immediately for substitution or coupling.
(ii) Aliphatic primary amines. Form highly unstable alkyldiazonium salts that decompose at once, liberating nitrogen quantitatively and giving alcohols: R–NH₂ + HNO₂ → R–OH + N₂↑ + H₂O. This quantitative evolution of nitrogen is used in the estimation of amino acids and proteins.
(i) Why are amines less acidic than alcohols of comparable molecular mass? Acidity depends on how readily the O–H or N–H bond releases a proton and how stable the resulting anion is. Oxygen (electronegativity 3.5) is more electronegative than nitrogen (3.0), so the O–H bond is more polar and breaks more readily, and the resulting alkoxide ion (RO⁻) accommodates the negative charge far better than the amide ion (RNH⁻). Hence alcohols are more acidic than amines.
(ii) Why do primary amines have higher boiling points than tertiary amines? A primary amine has two hydrogen atoms on nitrogen and undergoes extensive intermolecular hydrogen bonding. A tertiary amine has no hydrogen on nitrogen and cannot self-associate at all. More energy is therefore needed to separate the molecules of a primary amine, giving it the higher boiling point.
(iii) Why are aliphatic amines stronger bases than aromatic amines? In aliphatic amines the alkyl groups are electron releasing (+I) and push electron density towards nitrogen, making the lone pair more available for protonation and stabilising the resulting cation. In aromatic amines the lone pair is delocalised into the benzene ring by resonance, so it is much less available, and protonation would destroy that stabilising delocalisation. Hence aliphatic amines are the stronger bases.
1. Outline the route from benzene to p-chloroaniline and explain the order of the two substitutions. L4 Analyse
2. Explain why the amino group must be introduced and then removed in the synthesis of 1,3,5-tribromobenzene. L5 Evaluate
3. Two reagents are offered for converting benzenediazonium chloride to benzene: hypophosphorous acid and ethanol. Write both equations and name the oxidation product in each case. L2 Understand
4. Fill in the blanks: Aniline cannot be nitrated directly because it is ______ by the nitrating mixture and because it is ______ in the acidic medium to an ion that is ______ directing. L1 Remember
5. A batch of azo dye comes out pale and weakly coloured. Suggest two chemical reasons connected to the diazonium step and how each would be corrected. L5 Evaluate
For each pair choose: (A) Both A and R are true and R is the correct explanation of A. (B) Both A and R are true but R is not the correct explanation of A. (C) A is true but R is false. (D) A is false but R is true.
Assertion (A): Methylamine in water precipitates hydrated ferric oxide from a ferric chloride solution.
Reason (R): Methylamine is a base and generates hydroxide ions in aqueous solution.
Assertion (A): Alcohols are more acidic than amines of comparable molecular mass.
Reason (R): Oxygen is more electronegative than nitrogen, so the O–H bond is more polar and the resulting alkoxide ion is better stabilised.
Assertion (A): p-Chloroaniline is best prepared by chlorinating benzene first and nitrating afterwards.
Reason (R): The nitro group is a meta director, so nitrating first would place the chlorine meta rather than para.
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