Question 1

Write IUPAC names of the following compounds and classify them into primary, secondary and tertiary amines. (i) (CH 3) 2CHNH 2 (ii) CH 3(CH 2) 2NH 2 (iii) CH 3NHCH(CH 3) 2 (iv) (CH 3) 3CNH 2 (v) C 6H 5NHCH 3 (vi) (CH 3CH 2) 2NCH 3 (vii) m - BrC 6H 4NH 2

Accepted Answer

(i) 1-methylethylamine or propan-2-amine (iii) N-methyl-2-methylethylamine or N-methylpropan-2-amine (iv) 2-methylpropan-2-amine (v) N-methylbenzenamine or N-methylaniline (vii) 3-Bromoaniline or 3-Bromobenzenamine. The number of carbon groups tied directly to the nitrogen tells you primary, secondary, or tertiary — and the parent chain for naming is whichever carbon chain is bonded straight to that nitrogen. A primary amine (RNH 2) has one carbon group on N, a secondary amine (R 2NH) has two, and a tertiary amine (R 3N) has three. To name each one, find the longest carbon chain attached to the nitrogen, turn that chain into the "-amine" parent by dropping the "-e" of the alkane name and adding "-amine," and cite every other group on the nitrogen as an "N-" substituent prefix. (i) (CH 3) 2CHNH 2 — this is CH 3-CH(NH 2)-CH 3: a three-carbon chain with the amino group on the middle carbon, C-2. Numbering the chain to give the amine group the lowest locant, the parent is propan-2-amine. Only one carbon group (the isopropyl carbon skeleton) sits on the nitrogen, so this is a primary amine: propan-2-amine. (ii) CH 3(CH 2) 2NH 2 — this is CH 3-CH 2-CH 2-NH 2: a straight three-carbon chain with NH 2 on the terminal carbon, C-1. The parent is propan-1-amine, and since one carbon group is on nitrogen, this is a primary amine: propan-1-amine. (iii) CH 3NHCH(CH 3) 2 — this is CH 3-NH-CH(CH 3) 2: the nitrogen carries a methyl group on one side and an isopropyl group on the other. Between the two chains attached to N (methyl, one carbon, and propan-2-yl, three carbons), the longer one is chosen as the parent, so the base name is propan-2-amine, and the methyl group left over is cited as an N-substituent. Two carbon groups sit on the nitrogen, making it a secondary amine: N-methylpropan-2-amine. (iv) (CH 3) 3CNH 2 — this is (CH 3) 3C-NH 2: a tert-butyl carbon bonded to NH 2. Writing out the chain, the central carbon carries two methyl branches plus the chain, so the parent is propane with methyl substituents at C-2 and the amine also at C-2. That gives 2-methylpropan-2-amine. Only one carbon group is on the nitrogen, so this is a primary amine: 2-methylpropan-2-amine. (v) C 6H 5NHCH 3 — this is C 6H 5-NH-CH 3: a benzene ring bonded to nitrogen, and that same nitrogen also carries a methyl group. Because one of the two groups on nitrogen is the aryl (phenyl) ring, the parent name is the aromatic amine aniline (systematically benzenamine), and the methyl group is cited as an N-substituent. Two groups occupy the nitrogen, so this is a secondary amine: N-methylaniline (N-methylbenzenamine). (vi) (CH 3CH 2) 2NCH 3 — this is (CH 3-CH 2) 2N-CH 3: the nitrogen carries two ethyl groups and one methyl group, three carbon groups in total. The longest chain directly on nitrogen is ethyl (two carbons), so ethanamine is the parent; the remaining ethyl group and the methyl group are both cited as N-substituents, listed alphabetically. With three carbon groups on the nitrogen, this is a tertiary amine: N-ethyl-N-methylethanamine. (vii) m- BrC 6H 4NH 2 — this is a benzene ring bearing NH 2 and Br in a 1,3 (meta) relationship. Amino-substituted benzene is numbered from the amine carbon (C-1), so the bromine at the meta position falls at C-3. Since only the aryl ring sits on the nitrogen, this is a primary amine: 3-bromoaniline (3-bromobenzenamine). Answer: (i) propan-2-amine — primary; (ii) propan-1-amine — primary; (iii) N-methylpropan-2-amine — secondary; (iv) 2-methylpropan-2-amine — primary; (v) N-methylaniline — secondary; (vi) N-ethyl-N-methylethanamine — tertiary; (vii) 3-bromoaniline — primary.

Question 2

Give one chemical test to distinguish between the following pairs of compounds. (i) Methylamine and dimethylamine (ii) Secondary and tertiary amines (iii) Ethylamine and aniline (iv) Aniline and benzylamine (v) Aniline and N-methylaniline.

Accepted Answer

A primary amine still carries two hydrogens on nitrogen; a secondary amine has only one; and a benzene ring attached directly to - NH 2 is far more reactive toward electrophiles than one that has an NH 2 group parked one carbon away. Each pair below turns on exactly one of those structural facts, so a single reagent is enough to tell the two compounds apart in each case. (i) Methylamine, CH 3 NH 2, vs dimethylamine, ( CH 3) 2 NH — the carbylamine (isocyanide) test. Warm each compound with chloroform and alcoholic potassium hydroxide. Alcoholic KOH pulls a proton off CHCl 3 and the resulting trichloromethyl anion loses Cl^- to give dichlorocarbene,: CCl 2, a strong electrophile. This carbene can only complete the double insertion into an N–H bond when nitrogen carries two hydrogens to begin with — that is, only for a primary amine. CH 3 NH 2 + CHCl 3 + 3 KOH(alc.) Δ CH 3 N ≡ C + 3 KCl + 3 H 2 O Methylamine gives methyl isocyanide, CH 3 NC, an intensely disagreeable-smelling gas — a positive test. Dimethylamine's nitrogen already bears two carbon substituents and only one N–H; it cannot lose two hydrogens to a carbene, so it gives no reaction and no smell — a negative test. (ii) A secondary amine vs a tertiary amine — Hinsberg's test with benzenesulfonyl chloride, C 6 H 5 SO 2 Cl. A secondary amine, e.g. ( CH 3) 2 NH, still has one N–H to replace: ( CH 3) 2 NH + C 6 H 5 SO 2 Cl KOH C 6 H 5 SO 2 N ( CH 3) 2 + KCl + H 2 O The product, N,N -dimethylbenzenesulfonamide, has no hydrogen left on nitrogen at all, so it cannot be deprotonated by KOH — it stays as a water-insoluble solid even after excess alkali is added. A tertiary amine, e.g. ( CH 3) 3 N, has no N–H to begin with, so it does not react with benzenesulfonyl chloride at all; the amine is simply recovered unchanged, and — being basic — it dissolves cleanly in dilute HCl. So: insoluble product after KOH → secondary amine; unreacted layer that dissolves in dilute HCl → tertiary amine. (iii) Ethylamine, CH 3 CH 2 NH 2, vs aniline, C 6 H 5 NH 2 — the bromine water test. In aniline the lone pair on nitrogen delocalizes into the ring, making it strongly activated toward electrophilic substitution. Bromine water attacks it instantly, without any catalyst, substituting at both ortho positions and the para position: C 6 H 5 NH 2 + 3 Br 2( aq ) → C 6 H 2 Br 3 NH 2 ( white precipitate ) + 3 HBr The white precipitate is 2,4,6-tribromoaniline — an immediate positive test. Ethylamine has no aromatic ring at all, so there is nothing for bromine water to substitute onto; no precipitate forms — a negative test. (iv) Aniline, C 6 H 5 NH 2, vs benzylamine, C 6 H 5 CH 2 NH 2 — the same bromine water test, but now turning on where the - NH 2 sits. In benzylamine the amino group is bonded to a saturated CH 2 carbon, not to the ring carbon. The nitrogen lone pair therefore cannot conjugate with the ring — the - CH 2 NH 2 group is only a mild, uncoordinated activator, not the strong direct-conjugation activator that - NH 2 is in aniline. As a result benzylamine's ring does not react with bromine water at room temperature. Aniline again gives the instantaneous white precipitate of 2,4,6-tribromoaniline (as in part iii) — positive. Benzylamine gives no precipitate — negative. This one test cleanly separates the two, even though both are primary amines that would each answer "yes" to a carbylamine test. (v) Aniline, C 6 H 5 NH 2, vs N-methylaniline, C 6 H 5 NH ( CH 3) — back to the carbylamine test, since one is primary and the other secondary. Aniline is a primary amine (two hydrogens on nitrogen), so it undergoes the same dichlorocarbene double-insertion described in part (i): C 6 H 5 NH 2 + CHCl 3 + 3 KOH(alc.) Δ C 6 H 5 N ≡ C + 3 KCl + 3 H 2 O giving phenyl isocyanide, C 6 H 5 NC, with its characteristic offensive smell — positive. N-methylaniline's nitrogen is already substituted by one methyl group and carries only one N–H, so it cannot undergo the double insertion required to form an isocyanide — no foul smell, negative. Answer: (i) carbylamine test — methylamine gives foul-smelling CH 3 NC with CHCl 3 /alc. KOH, dimethylamine does not react; (ii) Hinsberg's test with C 6 H 5 SO 2 Cl — the secondary amine's sulfonamide stays insoluble in KOH, the tertiary amine is unreacted and dissolves in dilute HCl; (iii) bromine water — aniline gives an instant white precipitate of 2,4,6-tribromoaniline, ethylamine does not; (iv) bromine water again — aniline gives the white precipitate, benzylamine does not (its - NH 2 is not conjugated to the ring); (v) carbylamine test — aniline (primary) gives foul-smelling C 6 H 5 NC, N-methylaniline (secondary) does not react.

Question 3

Account for the following: (i) pK b of aniline is more than that of methylamine. (ii) Ethylamine is soluble in water whereas aniline is not. (iii) Methylamine in water reacts with ferric chloride to precipitate hydrated ferric oxide. (iv) Although amino group is o- and p- directing in aromatic electrophilic substitution reactions, aniline on nitration gives a substantial amount of m-nitroaniline. (v) Aniline does not undergo Friedel-Crafts reaction. (vi) Diazonium salts of aromatic amines are more stable than those of aliphatic amines. (vii) Gabriel phthalimide synthesis is preferred for synthesising primary amines.

Accepted Answer

(i) Aniline's nitrogen lone pair is tied up in conjugation with the benzene ring, so it is far less available to accept a proton than the lone pair on methylamine's nitrogen. In aniline, C6H5-NH2, nitrogen sits directly on an sp2 ring carbon, and its lone pair overlaps with the ring's π system — resonance delocalizes that electron density out to the ortho and para carbons. Spreading the lone pair over the ring like this means it is much less free to grab an incoming proton, so aniline behaves as a weak base. In methylamine, CH3-NH2, there is no ring to delocalize into; instead the methyl group is a weak electron donor (+I effect) that pushes extra electron density onto nitrogen, leaving the lone pair fully available and making methylamine the stronger base. Since pK b = - log K b, a stronger base has the smaller pK b — this is the step people get backwards. So methylamine (stronger base) has the smaller pK b, and aniline (weaker base) has the larger pK b. (ii) Ethylamine's hydrocarbon part is small enough for water's hydrogen-bonded structure to absorb it; aniline's benzene ring is not. Both ethylamine (CH3-CH2-NH2) and aniline (C6H5-NH2) can hydrogen-bond to water through their N-H bonds and nitrogen lone pair. Solubility is a balance between that favourable hydrogen bonding and the cost of squeezing a hydrocarbon group into water's hydrogen-bonded network. Ethylamine's ethyl group is small, so the cost is low and it mixes with water in all proportions. Aniline's phenyl ring is a large, flat, hydrophobic surface — fitting it into water breaks far more of water's own hydrogen bonds than aniline's N-H and lone pair can repay, so aniline is only sparingly soluble in water. (iii) Methylamine is a strong enough base to build up hydroxide ion in water on its own, and that hydroxide precipitates iron(III) as hydrated ferric oxide. Methylamine hydrolyzes water: CH3-NH2 + H 2O ⇌ CH3-NH3+ + OH-. Being a stronger base than ammonia, this equilibrium sits far enough to the right to generate a significant OH- concentration. That hydroxide then attacks the Fe 3 + supplied by ferric chloride solution: FeCl 3 + 3 OH- → Fe(OH) 3 + 3 Cl-. Freshly precipitated Fe(OH) 3 loses water on standing to give the reddish-brown hydrated ferric oxide, Fe2O3·xH2O, which is the precipitate observed — the same result ammonium hydroxide gives with FeCl 3, produced here because methylamine is at least as strong a base. (iv) Under the strongly acidic nitrating mixture, part of the aniline is protonated to the anilinium ion, and it is that ion — not free aniline — that is meta-directing. The nitrating mixture, concentrated HNO 3 with concentrated H 2SO 4, is strongly acidic, and in it a substantial fraction of aniline is protonated on nitrogen: C6H5-NH2 + H^+ → C6H5-NH3+ (anilinium ion). Once the lone pair is holding that extra proton, it can no longer donate into the ring by resonance, so -NH3+ behaves like any other positively charged, electron-withdrawing group: it deactivates the ring and sends the incoming NO 2 + to the meta position. The aniline that remains unprotonated still nitrates at ortho and para as expected, but because so much of the substrate exists as the anilinium ion under these conditions, a substantial share of the overall product comes out as m-nitroaniline alongside the o- and p-nitroaniline. (v) The AlCl 3 catalyst attaches to aniline's basic nitrogen before the reaction can touch the ring, turning the amino group into a deactivating, meta-directing one and stalling the reaction. Friedel-Crafts alkylation and acylation both need AlCl 3 as a Lewis acid to generate the electrophile. But AlCl 3 is itself a Lewis acid hunting for a lone pair, and aniline's nitrogen lone pair is far more available than the ring's π electrons, so AlCl 3 binds there instead: C6H5-NH2 + AlCl 3 → C6H5-NH2·AlCl3, a salt-like adduct in which nitrogen carries a formal positive charge. With the lone pair now tied up on aluminium, the group left on the ring is strongly electron-withdrawing and meta-directing rather than the usual activating, o/p-directing -NH2. The ring is deactivated toward electrophilic attack, so Friedel-Crafts alkylation or acylation fails to proceed to any useful extent. (vi) An aromatic diazonium ion's positive charge is spread into the benzene ring by resonance; an aliphatic diazonium ion has no π system to spread it into, so it collapses the moment it forms. An aromatic diazonium salt such as benzenediazonium chloride, C6H5-N2+ Cl-, delocalizes the positive charge on the terminal nitrogen through the ring: the C-N bond gains partial double-bond character, and resonance structures push positive charge onto the ortho and para ring carbons as well as onto nitrogen. This lowers the ion's energy enough that the salt can be kept and used near 0-5 °C. An aliphatic diazonium ion, R-N2+, has no adjacent π system to accept that charge — the only support is the alkyl group's weak +I effect, which is not enough. Aliphatic diazonium salts therefore decompose the instant they are generated, even below 0 °C, releasing N 2 gas and leaving a carbocation that reacts further by elimination, rearrangement, or substitution. (vii) Gabriel synthesis lets the alkyl halide react with nitrogen only once, because after the first substitution nitrogen has no N-H left to attack with again; direct amination with ammonia has no such limit and keeps alkylating. Treating an alkyl halide (R-X) directly with ammonia gives primary amine R-NH2, but that amine is itself as good a nucleophile as ammonia, so it goes on reacting with more R-X to give secondary amine, then tertiary amine, then the quaternary ammonium salt R4N+X- — a mixture that is difficult to separate cleanly. Gabriel phthalimide synthesis sidesteps this. Phthalimide's N-H is flanked by two carbonyl groups, which makes it acidic; alcoholic KOH removes that proton to give potassium phthalimide, where nitrogen is negatively charged and strongly nucleophilic. Potassium phthalimide displaces the halide from R-X in an SN 2 step to give N-alkylphthalimide, in which nitrogen already carries the alkyl group plus the two ring carbonyls and has no N-H left — so a second R-X cannot react there, and over-alkylation is structurally impossible. Hydrolysis of the N-alkylphthalimide (aqueous acid or base, or hydrazinolysis) then cleaves both C-N bonds to the carbonyls, releasing the pure primary amine R-NH2 together with phthalic acid (or its hydrazide). Because the method is restricted to a single substitution event, it is preferred whenever a primary amine free of secondary/tertiary contamination is needed — though it works only with alkyl halides, since aryl halides do not undergo this substitution and so aromatic primary amines cannot be made this way. Answer: (i) Aniline's lone pair is delocalized into the ring, making it a weaker base with the larger pK b; methylamine's +I methyl group makes it the stronger base with the smaller pK b. (ii) Ethylamine's small alkyl group fits water's hydrogen-bonded structure, but aniline's bulky hydrophobic ring disrupts it, so aniline is only sparingly soluble. (iii) Methylamine's basic hydrolysis generates enough OH- to precipitate Fe 3 + as hydrated ferric oxide, Fe2O3·xH2O. (iv) In the strongly acidic nitrating mixture much of the aniline exists as the meta-directing anilinium ion, C6H5-NH3+, giving a substantial share of m-nitroaniline alongside the o/p products. (v) AlCl 3 coordinates to aniline's nitrogen lone pair, converting -NH2 into a deactivating, meta-directing group and blocking Friedel-Crafts alkylation/acylation. (vi) Aromatic diazonium ions are resonance-stabilized by the ring and so are isolable near 0-5 °C, while aliphatic diazonium ions lack such stabilization and decompose instantly with loss of N2. (vii) Gabriel phthalimide synthesis lets the alkyl halide react with nitrogen only once (via potassium phthalimide, then hydrolysis), giving pure primary amine R-NH2 without secondary/tertiary/quaternary by-products, though it is limited to alkyl halides and cannot make aromatic primary amines.

Question 4

Arrange the following: (i) In decreasing order of the pK b values: C 2H 5NH 2, C 6H 5NHCH 3, (C 2H 5) 2NH and C 6H 5NH 2 (ii) In increasing order of basic strength: C 6H 5NH 2, C 6H 5N(CH 3) 2, (C 2H 5) 2NH and CH 3NH 2 (iii) In increasing order of basic strength: (a) Aniline, p-nitroaniline and p-toluidine (b) C 6H 5NH 2, C 6H 5NHCH 3, C 6 H 5 CH 2 NH 2. (iv) In decreasing order of basic strength in gas phase: C 2H 5NH 2, (C 2H 5) 2NH, (C 2H 5) 3N and NH 3 (v) In increasing order of boiling point: C 2H 5OH, (CH 3) 2NH, C 2H 5NH 2 (vi) In increasing order of solubility in water: C 6H 5NH 2, (C 2H 5) 2NH, C 2 H 5 NH 2.

Accepted Answer

(i) C 6 H 5 NH 2 < C 6 H 5 NHCH 3 < C 2 H 5 NH 2 < ( C 2 H 5 )2NH (ii) C 6 H 5 NH 2 < C 6 H 5 N( CH 3 )2 < CH 3 NH 2 < ( C 2 H 5 )2NH (iii) (a) p-nitroaniline < aniline < p-toluidine (b) C 6 H 5 NH 2 < C 6 H 5 NHCH 3 < C 6 H 5 CH 2 NH 2 (iv) ( C 2 H 5 )3N ( C 2 H 5 )2NH C 2 H 5 NH 2 NH 3 (vi) C 6 H 5 NH 2 < ( C 2 H 5 )2NH < C 2 H 5 NH 2 (ii) Propan-1-amine (vi) N-Ethyl-N-methylethanamine (v) ( CH 3 )2NH < C 2 H 5 NH 2 < C 2 H 5 OH Notes. The base strength of an amine is a tug‑of‑war between three effects: the +I (electron‑releasing) push of alkyl groups on nitrogen, the loss of that electron density when the lone pair conjugates into a benzene ring, and — in water only — how well the protonated nitrogen can hydrogen‑bond with solvent. p K b runs opposite to basicity: the weaker the base, the larger its p K b. Working through each part with that in mind: (i) Ethylamine C 2 H 5 NH 2, N‑methylaniline C 6 H 5 NHCH 3, diethylamine ( C 2 H 5) 2 NH, aniline C 6 H 5 NH 2. In both C 6 H 5 NH 2 and C 6 H 5 NHCH 3 the nitrogen lone pair is conjugated directly into the ring — it delocalises over the ortho/para carbons instead of sitting on N ready to grab a proton — so both are weak bases; the extra - CH 3 on the second one pushes a little electron density back onto N by its +I effect, making C 6 H 5 NHCH 3 slightly the stronger of the two. Neither aliphatic amine has a ring to conjugate into, so both are far more basic; between them, the ammonium ion from ( C 2 H 5) 2 NH still carries one N–H bond to hydrogen‑bond with water even though it has two ethyl groups pushing electron density onto N, so in aqueous solution diethylamine outranks ethylamine. Basicity rises as C 6 H 5 NH 2 < C 6 H 5 NHCH 3 < C 2 H 5 NH 2 < ( C 2 H 5) 2 NH, so p K b falls the other way: C 6 H 5 NH 2 C 6 H 5 NHCH 3 C 2 H 5 NH 2 ( C 2 H 5) 2 NH (ii) Aniline C 6 H 5 NH 2, N,N‑dimethylaniline C 6 H 5 N(CH 3) 2, diethylamine ( C 2 H 5) 2 NH, methylamine CH 3 NH 2. Aniline is again the weakest base — the lone pair is tied up in ring conjugation with nothing to offset it. C 6 H 5 N(CH 3) 2 has the same conjugation problem, but two methyl groups feed +I electron density onto N, lifting it just above plain aniline — still, it is an aromatic amine and stays well below any aliphatic amine. CH 3 NH 2, with no ring at all, is far more basic than either aromatic amine, and ( C 2 H 5) 2 NH tops the list for the same reason as in (i) — two +I ethyl groups plus one N–H left for solvation. So, increasing basic strength: C 6 H 5 NH 2 < C 6 H 5 N(CH 3) 2 < CH 3 NH 2 < ( C 2 H 5) 2 NH (iii)(a) Aniline, p‑nitroaniline, p‑toluidine. These three differ only in the para substituent on the ring, so the whole ranking comes down to whether that substituent adds or removes electron density from the already-weakened nitrogen. The - NO 2 group in p‑nitroaniline (4‑nitroaniline) is strongly electron‑withdrawing by both resonance and induction — it pulls density out of the ring and, through the ring, further off nitrogen — so this is the weakest base of the three. The - CH 3 group in p‑toluidine (4‑methylaniline) is electron‑donating, feeding density back into the ring and partially compensating for the conjugation loss at N, so it is the strongest base of the three, with plain aniline in between: p -nitroaniline < aniline < p -toluidine (iii)(b) C 6 H 5 NH 2, C 6 H 5 NHCH 3, C 6 H 5 CH 2 NH 2 (benzylamine). The key difference is where the nitrogen sits relative to the ring. In aniline and N‑methylaniline, N is bonded directly to the ring carbon, so its lone pair is part of the conjugated system and is not fully available. In benzylamine, the - CH 2- group sits between the ring and the nitrogen, breaking that conjugation — the lone pair on N has no path into the ring and behaves like an ordinary aliphatic amine's lone pair, making C 6 H 5 CH 2 NH 2 much the strongest base of the three; between the two aromatic amines, the methyl group again makes C 6 H 5 NHCH 3 slightly stronger than C 6 H 5 NH 2: C 6 H 5 NH 2 < C 6 H 5 NHCH 3 < C 6 H 5 CH 2 NH 2 (iv) Gas‑phase basicity of C 2 H 5 NH 2, ( C 2 H 5) 2 NH, ( C 2 H 5) 3 N, NH 3. Take away the solvent and the whole "N–H bonds needed for hydration" argument from (i)–(ii) disappears — there is no water to hydrogen‑bond with the cation. What is left is pure inductive/polarisability stabilisation of the positive charge that appears on nitrogen once it is protonated: every alkyl group replacing an N–H adds +I electron density and adds polarisable C–H bonds that spread the positive charge out, stabilising the ammonium cation. So basicity now increases monotonically with the number of ethyl groups, with no reversal at the tertiary amine: ( C 2 H 5) 3 N ( C 2 H 5) 2 NH C 2 H 5 NH 2 NH 3 (v) Boiling points of C 2 H 5 OH, ( CH 3) 2 NH, C 2 H 5 NH 2. All three have almost the same molar mass (46, 45 and 45 g/mol), so the ranking is fixed entirely by hydrogen‑bond strength and the number of H‑bond donors available per molecule. Oxygen is more electronegative than nitrogen, so the O–H···O hydrogen bond in ethanol is stronger than any N–H···N bond, putting C 2 H 5 OH at the top. Between the two amines, C 2 H 5 NH 2 is a primary amine with two N–H bonds available to hydrogen‑bond to neighbouring molecules, while ( CH 3) 2 NH is a secondary amine with only one N–H bond — fewer hydrogen bonds per molecule means a lower boiling point, even though the formula weight is almost identical: ( CH 3) 2 NH < C 2 H 5 NH 2 < C 2 H 5 OH (vi) Solubility in water of C 6 H 5 NH 2, ( C 2 H 5) 2 NH, C 2 H 5 NH 2. Solubility in water depends on how much of the molecule can hydrogen‑bond with water versus how much of it is a hydrophobic hydrocarbon block that water has to make room for. Aniline carries a bulky, non‑polar phenyl ring attached directly to the only polar group it has, so most of the molecule resists solvation — it is the least soluble of the three. Between the aliphatic amines, both dissolve freely, but diethylamine carries two hydrophobic ethyl groups against ethylamine's one, so ethylamine — the smaller, less hydrocarbon‑heavy molecule — is the more soluble of the two: C 6 H 5 NH 2 < ( C 2 H 5) 2 NH < C 2 H 5 NH 2 Answer: (i) C 6 H 5 NH 2 C 6 H 5 NHCH 3 C 2 H 5 NH 2 ( C 2 H 5) 2 NH (decreasing p K b ); (ii) C 6 H 5 NH 2 < C 6 H 5 N(CH 3) 2 < CH 3 NH 2 < ( C 2 H 5) 2 NH; (iii‑a) p‑nitroaniline < aniline < p‑toluidine; (iii‑b) C 6 H 5 NH 2 < C 6 H 5 NHCH 3 < C 6 H 5 CH 2 NH 2; (iv) ( C 2 H 5) 3 N ( C 2 H 5) 2 NH C 2 H 5 NH 2 NH 3 (gas phase); (v) ( CH 3) 2 NH < C 2 H 5 NH 2 < C 2 H 5 OH (boiling point); (vi) C 6 H 5 NH 2 < ( C 2 H 5) 2 NH < C 2 H 5 NH 2 (solubility).

Question 5

How will you convert: (i) Ethanoic acid into methanamine (ii) Hexanenitrile into 1-aminopentane (iii) Methanol to ethanoic acid (iv) Ethanamine into methanamine (v) Ethanoic acid into propanoic acid (vi) Methanamine into ethanamine (vii) Nitromethane into dimethylamine (viii) Propanoic acid into ethanoic acid?

Accepted Answer

Each of these eight conversions is really the same two-part puzzle: does the carbon count need to change, and does the functional group need to change — pick the one reagent that does that single job, then chain the minimum number of steps. (i) Ethanoic acid into methanamine A one-carbon acid cannot be pushed down to a shorter amine by substitution — the carbon has to be removed deliberately, and Hofmann's bromamide degradation is the reaction built for exactly that. Ethanoic acid, CH 3 COOH, is first treated with ammonia to form its ammonium salt, ammonium ethanoate, CH 3 COONH 4. Heating this salt drives off a molecule of water, closing the C–N bond and giving ethanamide, CH 3 CONH 2. Ethanamide is then treated with bromine in aqueous sodium hydroxide (Hofmann bromamide degradation): the amide nitrogen is brominated to CH 3 CONHBr, base removes the remaining N–H proton, and as bromide leaves nitrogen the methyl group migrates from carbon to nitrogen in the same step, giving an isocyanate, CH 3 N=C=O. The isocyanate is hydrolysed by the alkaline medium to a carbamate, which loses CO 2 (trapped as carbonate) to release the amine. Because the carbonyl carbon leaves as carbonate, the product has one carbon fewer than the amide: methanamine, CH 3 NH 2. CH 3 COOH NH 3 CH 3 COONH 4 Δ, - H 2 O CH 3 CONH 2 Br 2/ NaOH CH 3 NH 2 (ii) Hexanenitrile into 1-aminopentane Reducing the nitrile directly would keep all six carbons; to lose one, hydrolyse it to the acid first and then apply the same bromamide degradation as in (i). Hexanenitrile, CH 3( CH 2) 4 CN (the nitrile carbon is counted as C 1 of the six-carbon chain), is hydrolysed by boiling with aqueous acid: water adds across the C ≡ N in two steps through an amide intermediate to give the carboxylic acid, hexanoic acid, CH 3( CH 2) 4 COOH, releasing ammonium ion. Hexanoic acid is converted to ammonium hexanoate with ammonia, and heating this salt gives hexanamide, CH 3( CH 2) 4 CONH 2, by loss of water. Treating hexanamide with bromine and sodium hydroxide runs the Hofmann degradation exactly as above, stripping off the carbonyl carbon as carbonate and leaving the five-carbon amine, pentan-1-amine (1-aminopentane), CH 3( CH 2) 4 NH 2. CH 3( CH 2) 4 CN H 3 O^+, Δ CH 3( CH 2) 4 COOH NH 3 salt Δ, - H 2 O CH 3( CH 2) 4 CONH 2 Br 2/ NaOH CH 3( CH 2) 4 NH 2 (iii) Methanol to ethanoic acid Going from one carbon to two needs a carbon-adding step — a cyanide substitution is the only reaction here that actually builds a new C–C bond. Methanol, CH 3 OH, is converted to bromomethane, CH 3 Br, using red phosphorus and bromine (or PBr 3 ), which replaces the –OH with –Br. Treating bromomethane with alcoholic potassium cyanide runs an S N 2 substitution: the cyanide carbon (nucleophile, attacking through carbon because carbon is more electronegative-shy but nitrogen's lone pair is less available for bonding through N in KCN with primary alkyl halides) displaces bromide, giving ethanenitrile (methyl cyanide), CH 3 CN — the new C–C bond is exactly the carbon added. Hydrolysing this nitrile by boiling with aqueous acid converts the nitrile to the carboxylic acid via the amide, giving ethanoic acid, CH 3 COOH. CH 3 OH Red P / Br 2 CH 3 Br KCN (alc.) CH 3 CN H 3 O^+, Δ CH 3 COOH (iv) Ethanamine into methanamine A primary aliphatic amine cannot be shortened directly; deamination through nitrous acid first turns it into an alcohol, which is then rebuilt into the same one-carbon amine target as in (i). Ethanamine, CH 3 CH 2 NH 2, is treated with nitrous acid ( NaNO 2 + HCl, 273–278 K). The diazonium ion formed from a primary aliphatic amine is unstable even in the cold and decomposes at once, releasing nitrogen gas and leaving ethanol, CH 3 CH 2 OH. Oxidising ethanol with acidified potassium dichromate (or alkaline KMnO 4 ) takes the primary alcohol through the aldehyde to the carboxylic acid, ethanoic acid, CH 3 COOH. From here the route is identical to part (i): ammonia gives ammonium ethanoate, heating dehydrates it to ethanamide, CH 3 CONH 2, and bromine/sodium hydroxide runs the Hofmann degradation, removing the carbonyl carbon to leave methanamine, CH 3 NH 2. CH 3 CH 2 NH 2 HNO 2 CH 3 CH 2 OH [ O ] CH 3 COOH NH 3, Δ CH 3 CONH 2 Br 2/ NaOH CH 3 NH 2 (v) Ethanoic acid into propanoic acid Adding one carbon to an acid means reducing it all the way down to the alkyl halide first, so cyanide substitution has a leaving group to displace. Ethanoic acid, CH 3 COOH, is reduced by lithium aluminium hydride to ethanol, CH 3 CH 2 OH (the carbonyl is reduced past the aldehyde stage to the primary alcohol). Treating ethanol with PBr 3 (or red P/ Br 2 ) replaces –OH with –Br, giving bromoethane, CH 3 CH 2 Br. Alcoholic potassium cyanide displaces the bromide in an S N 2 step, adding the cyanide carbon to give propanenitrile, CH 3 CH 2 CN. Acid hydrolysis of this nitrile (boiling with aqueous H 3 O^+ ) converts it through the amide to the three-carbon acid, propanoic acid, CH 3 CH 2 COOH. CH 3 COOH LiAlH 4 CH 3 CH 2 OH PBr 3 CH 3 CH 2 Br KCN (alc.) CH 3 CH 2 CN H 3 O^+, Δ CH 3 CH 2 COOH (vi) Methanamine into ethanamine Reducing a carbylamine off methanamine only ever gives a secondary amine (dimethylamine) — to reach a primary two-carbon amine, deaminate to the alcohol and rebuild through a nitrile instead. Methanamine, CH 3 NH 2, is diazotised with nitrous acid ( NaNO 2 + HCl, 273–278 K); the resulting diazonium ion decomposes immediately, expelling nitrogen and giving methanol, CH 3 OH. Methanol is converted to bromomethane, CH 3 Br, with red phosphorus and bromine. Alcoholic potassium cyanide displaces the bromide ( S N 2 ) to add one carbon, giving ethanenitrile, CH 3 CN. Reducing this nitrile with H 2/ Ni (or LiAlH 4 ) adds four hydrogens across the C ≡ N triple bond, converting the nitrile carbon into a - CH 2 NH 2 group and giving the two-carbon primary amine, ethanamine, CH 3 CH 2 NH 2. CH 3 NH 2 HNO 2 CH 3 OH Red P / Br 2 CH 3 Br KCN (alc.) CH 3 CN H 2/ Ni CH 3 CH 2 NH 2 (vii) Nitromethane into dimethylamine Here the isocyanide-reduction route that failed in (vi) is exactly the right tool, because the target is a secondary amine, not a primary one. Nitromethane, CH 3 NO 2, is reduced with hydrogen over nickel (or Fe/HCl), which converts the nitro group to an amino group, giving methanamine, CH 3 NH 2. Methanamine is heated with chloroform and alcoholic potassium hydroxide (the carbylamine reaction): the base generates dichlorocarbene from chloroform, which inserts into the N–H bond and eliminates two equivalents of chloride to give methyl isocyanide, CH 3 NC — this is the diagnostic, foul-smelling carbylamine test for a primary amine. Reducing methyl isocyanide with sodium and ethanol (or H 2/ Ni ) adds four hydrogens across the isocyanide carbon, converting - N ≡ C into - NH - CH 3 and giving the secondary amine, dimethylamine (N-methylmethanamine), CH 3 NHCH 3. CH 3 NO 2 H 2/ Ni CH 3 NH 2 CHCl 3/ KOH (alc.) CH 3 NC Na/EtOH CH 3 NHCH 3 (viii) Propanoic acid into ethanoic acid Shortening a three-carbon acid to two carbons runs the same Hofmann degradation as (i), but the degradation lands on an amine, so a diazotisation-and-oxidation tail is needed to bring it back to an acid. Propanoic acid, CH 3 CH 2 COOH, is converted to ammonium propanoate with ammonia, and heating dehydrates the salt to propanamide, CH 3 CH 2 CONH 2. Treating propanamide with bromine in sodium hydroxide runs the Hofmann bromamide degradation (bromination of the amide nitrogen, base-promoted loss of bromide with migration of the ethyl group to nitrogen, hydrolysis of the resulting isocyanate, and loss of CO 2 as carbonate), removing the carbonyl carbon and giving ethanamine, CH 3 CH 2 NH 2. Ethanamine is diazotised with nitrous acid ( NaNO 2 + HCl, 273–278 K); the diazonium salt decomposes at once to release nitrogen and give ethanol, CH 3 CH 2 OH. Oxidising ethanol with acidified potassium dichromate (or alkaline KMnO 4 ) carries the primary alcohol through the aldehyde to the carboxylic acid, ethanoic acid, CH 3 COOH — two carbons, as required. CH 3 CH 2 COOH NH 3, Δ CH 3 CH 2 CONH 2 Br 2/ NaOH CH 3 CH 2 NH 2 HNO 2 CH 3 CH 2 OH [ O ] CH 3 COOH Answer: (i) CH 3 COOH → CH 3 COONH 4 → CH 3 CONH 2 Br 2/ NaOH CH 3 NH 2 (methanamine). (ii) CH 3( CH 2) 4 CN hydrolysis CH 3( CH 2) 4 COOH → amide Br 2/ NaOH CH 3( CH 2) 4 NH 2 (1-aminopentane). (iii) CH 3 OH Red P / Br 2 CH 3 Br KCN CH 3 CN hydrolysis CH 3 COOH (ethanoic acid). (iv) CH 3 CH 2 NH 2 HNO 2 CH 3 CH 2 OH [ O ] CH 3 COOH → amide Br 2/ NaOH CH 3 NH 2 (methanamine). (v) CH 3 COOH LiAlH 4 CH 3 CH 2 OH PBr 3 CH 3 CH 2 Br KCN CH 3 CH 2 CN hydrolysis CH 3 CH 2 COOH (propanoic acid). (vi) CH 3 NH 2 HNO 2 CH 3 OH Red P / Br 2 CH 3 Br KCN CH 3 CN H 2/ Ni CH 3 CH 2 NH 2 (ethanamine). (vii) CH 3 NO 2 H 2/ Ni CH 3 NH 2 CHCl 3/ KOH CH 3 NC Na/EtOH CH 3 NHCH 3 (dimethylamine). (viii) CH 3 CH 2 COOH → amide Br 2/ NaOH CH 3 CH 2 NH 2 HNO 2 CH 3 CH 2 OH [ O ] CH 3 COOH (ethanoic acid).

Question 6

Describe a method for the identification of primary, secondary and tertiary amines. Also write chemical equations of the reactions involved.

Accepted Answer

Hinsberg's test tells the three classes apart by how many hydrogens are left on nitrogen after one substitution — a primary amine still has an N-H that is acidic, a secondary amine has none, and a tertiary amine never had one to begin with. The reagent used is benzenesulfonyl chloride, C6H5-SO2-Cl (Hinsberg's reagent). An amine is shaken with this reagent in the presence of aqueous KOH (or NaOH), and the product's solubility in excess alkali is what actually distinguishes the three types. Step 1 — reaction with a primary amine, R-NH2 The nitrogen's lone pair attacks the sulfur of C6H5-SO2-Cl, chloride leaves, and one N-H is replaced by the benzenesulfonyl group: R-NH2 + C6H5-SO2-Cl → C6H5-SO2-NH-R + HCl The product is N-alkylbenzenesulfonamide, C6H5-SO2-NH-R. It still carries one hydrogen on nitrogen, and that hydrogen is made strongly acidic by the adjacent electron-withdrawing -SO2- group (it pulls electron density away from nitrogen, the same way it does in sulfonic acids). Because this N-H is acidic, the amide dissolves in the KOH already present, by simple acid-base proton transfer: C6H5-SO2-NH-R + KOH → C6H5-SO2-N(K)-R + H 2O This gives a clear, homogeneous solution — the visible signature of a primary amine. Step 2 — reaction with a secondary amine, R2NH The same substitution happens once, since a secondary amine has exactly one N-H to lose: R2NH + C6H5-SO2-Cl → C6H5-SO2-NR2 + HCl The product, N,N-dialkylbenzenesulfonamide (C6H5-SO2-NR2), now has zero hydrogens on nitrogen — both positions are occupied by alkyl groups. With no N-H left to ionize, this compound has no acidic proton to hand to KOH, so it does not dissolve; it separates out as an insoluble oily layer or solid. Step 3 — reaction with a tertiary amine, R3N A tertiary amine has no hydrogen on nitrogen at all, so there is nothing for benzenesulfonyl chloride to substitute: R3N + C6H5-SO2-Cl → no reaction The two reagents simply do not combine. On shaking with KOH, the layers stay separate; the amine itself is still a base, so if dilute HCl is added instead it dissolves by simple protonation — R3N + HCl → R3NH+Cl− — while unreacted C6H5-SO2-Cl stays behind, unaffected. This lack of reaction, plus solubility in acid rather than in alkali, is what marks a tertiary amine. Reading the result - Alkaline mixture turns into one clear solution → primary amine (the sulfonamide's N-H was acidic enough to dissolve as its potassium salt). - Alkaline mixture leaves an insoluble oily layer or precipitate → secondary amine (the disubstituted sulfonamide has no acidic N-H). - No visible reaction with the alkaline reagent; the amine layer dissolves only when dilute HCl is added → tertiary amine (there was no N-H for benzenesulfonyl chloride to attack in the first place). Answer: A primary, secondary, or tertiary amine is identified by treating it with benzenesulfonyl chloride (Hinsberg's reagent) and aqueous KOH. A primary amine (R-NH2) gives C6H5-SO2-NH-R, which has an acidic N-H and dissolves in KOH as its potassium salt (clear solution). A secondary amine (R2NH) gives C6H5-SO2-NR2, which has no N-H and stays insoluble in KOH (oily layer/precipitate). A tertiary amine (R3N) has no N-H to react at all, so it does not react with the reagent, but instead dissolves in dilute HCl to give R3NH+Cl−.

Question 7

Write short notes on the following: (i) Carbylamine reaction (ii) Diazotisation (iii) Hofmann’s bromamide reaction (iv) Coupling reaction (v) Ammonolysis (vi) Acetylation (vii) Gabriel phthalimide synthesis.

Accepted Answer

(i) Carbylamine reaction — the diagnostic smell test that only a primary amine gives. A primary amine, aliphatic or aromatic, warmed with chloroform and ethanolic potassium hydroxide produces an isocyanide (carbylamine) carrying a foul, fish-like smell — this is used as a chemical test to distinguish primary amines from secondary and tertiary ones, since only a primary amine has the two N–H hydrogens the mechanism needs. Mechanism, step by step: 1. KOH deprotonates chloroform, CHCl 3, giving the trichloromethyl carbanion CCl 3^-. 2. This carbanion expels a chloride ion ( α -elimination), generating the electron-deficient dichlorocarbene,: CCl 2. 3. The lone pair on the amine nitrogen attacks the electrophilic carbene carbon, forming a new C–N bond and displacing one chloride, giving an imidoyl chloride intermediate. 4. A second molecule of KOH removes the remaining N–H proton and the last chloride leaves, forming the carbon–nitrogen triple bond of the isocyanide. For ethanamine: CH 3 CH 2 NH 2 + CHCl 3 + 3 KOH Δ CH 3 CH 2 NC + 3 KCl + 3 H 2 O Product: ethyl isocyanide (ethyl carbylamine), CH 3 CH 2 NC. Aniline gives phenyl isocyanide, C 6 H 5 NC, by the same sequence. (ii) Diazotisation — converting a primary aromatic amine into a diazonium salt that is stable only in the cold. An aromatic primary amine treated with nitrous acid — generated in situ from sodium nitrite and a mineral acid such as HCl, kept at 273–278 K (0–5 °C) — is converted into a diazonium salt. The cold is the step people skip: above roughly 283 K the diazonium salt decomposes (to a phenol, with loss of nitrogen gas), so diazotisation is always run in an ice bath. Mechanism, step by step: 1. Sodium nitrite and hydrochloric acid react to form nitrous acid, HNO 2, which is protonated further to the true electrophile, the nitrosonium ion, NO^+. 2. The lone pair on the amine nitrogen of aniline attacks NO^+, giving an N-nitrosoamine intermediate, C 6 H 5- NH - N = O. 3. A proton shifts from nitrogen to oxygen (tautomerisation), giving C 6 H 5- N = N - OH. 4. Protonation of the OH group, then loss of a water molecule, gives the resonance-stabilised diazonium cation, C 6 H 5- N ≡ N^+. C 6 H 5 NH 2 + NaNO 2 + 2 HCl 273 – 278 K C 6 H 5 N 2^+ Cl^- + NaCl + 2 H 2 O Product: benzenediazonium chloride, C 6 H 5 N 2^+ Cl^-. (iii) Hofmann's bromamide degradation — the reaction that shortens a chain by one carbon while turning an amide into an amine. An amide, treated with bromine in aqueous or ethanolic sodium hydroxide, degrades to a primary amine having one carbon fewer than the amide it came from — the original carbonyl carbon is lost as carbon dioxide (trapped as carbonate in the basic medium), which is exactly why the amine is "one carbon short." Mechanism, step by step: 1. Bromine reacts with hydroxide to give the hypobromite ion, OBr^-. 2. Base deprotonates the amide N–H, and the resulting nitrogen anion attacks OBr^-, giving an N-bromoamide, R-CO-NHBr. 3. Base removes the remaining N–H proton of the N-bromoamide, giving the anion R-CO-N^-- Br. 4. This anion rearranges: the group R migrates from carbon to nitrogen in the same step that bromide leaves — a concerted 1,2-shift with no free nitrene or carbocation — giving an isocyanate, R-N=C=O. 5. Hydroxide hydrolyses the isocyanate, first to an unstable carbamate ion, R-NH-COO^-, which loses CO 2 (as carbonate) to give the primary amine. For ethanamide (acetamide): CH 3 CONH 2 + Br 2 + 4 NaOH → CH 3 NH 2 + 2 NaBr + Na 2 CO 3 + 2 H 2 O Product: methanamine (methylamine), CH 3 NH 2 — one carbon fewer than the acetamide it came from. (iv) Coupling reaction — using a diazonium salt to build a coloured azo dye. A diazonium salt reacts with a second, strongly activated aromatic ring — typically a phenol (mildly alkaline medium) or an aromatic amine (mildly acidic medium) — at 273–278 K, to give a brightly coloured azo compound, Ar-N=N-Ar '. The extended conjugation through the - N=N - (azo) linkage joining the two rings is what produces the colour, which is why azo compounds are used as dyes. Mechanism: electrophilic aromatic substitution, with the diazonium ion as a weak electrophile. 1. In phenol under mildly alkaline conditions, the phenoxide ion is a strongly activated ring — the negative oxygen pushes electron density into the ring, most strongly to the para position. 2. The terminal nitrogen of the diazonium ion, Ar-N ≡ N^+, attacks the para carbon of the phenoxide ring. 3. Loss of the para-hydrogen as a proton restores aromaticity and completes the substitution. C 6 H 5 N 2^+ Cl^- + C 6 H 5 OH NaOH, cold p -HO-C 6 H 4- N=N-C 6 H 5 + HCl Product: p-hydroxyazobenzene (4-(phenyldiazenyl)phenol), an orange azo dye. (v) Ammonolysis — replacing a halogen on an alkyl halide with an amino group, using ammonia as the nucleophile. An alkyl halide, heated with excess ammonia in a sealed tube (ethanolic ammonia keeps both reagents in one liquid phase at the temperature needed), undergoes nucleophilic substitution in which the halide is displaced by - NH 2. It is called ammonolysis because the C–X bond is broken by ammonia the way hydrolysis breaks a bond using water. Mechanism: S N2 at the halogen-bearing carbon. 1. The lone pair on ammonia's nitrogen attacks the electrophilic carbon from the side opposite the halide (backside attack). 2. As the new C–N bond forms, the C–X bond breaks simultaneously in one concerted step and halide ion leaves. 3. The immediate product is an alkylammonium halide salt, freed to the amine by a further molecule of ammonia (which removes the proton). CH 3 Br + NH 3 excess, sealed tube CH 3 NH 3^+ Br^- NH 3 CH 3 NH 2 + NH 4 Br The real complication: the primary amine formed is itself a nucleophile and can attack a second molecule of alkyl halide, giving successively a secondary amine, a tertiary amine, and finally a quaternary ammonium salt. A large excess of ammonia is used specifically to suppress this and favour the primary amine, methanamine (methylamine), CH 3 NH 2, as the major product. (vi) Acetylation — acylating the N–H of an amine to protect it or moderate its reactivity. Acetylation introduces an acetyl group, CH 3 CO -, onto the nitrogen of an amine, most often using acetic anhydride or acetyl chloride with a trace of pyridine — pyridine neutralises the acid released (HCl or CH 3 COOH ) and stops the amine being tied up as its unreactive protonated salt. Mechanism: nucleophilic acyl substitution at the carbonyl carbon. 1. The lone pair on the amine nitrogen attacks the electrophilic carbonyl carbon of acetic anhydride, ( CH 3 CO ) 2 O. 2. A tetrahedral intermediate forms and collapses by ejecting acetate ion, the better leaving group, restoring the C=O double bond. 3. A proton is removed from nitrogen (by acetate or by pyridine) to give the neutral amide. For aniline: C 6 H 5 NH 2 + ( CH 3 CO ) 2 O pyridine C 6 H 5 NHCOCH 3 + CH 3 COOH Product: N-phenylacetamide (acetanilide ), C 6 H 5 NHCOCH 3. Aniline is acetylated deliberately before nitration because the less strongly activating, harder-to-oxidise acetamido group directs cleanly to the para position without the side reactions free aniline undergoes with the strongly acidic, oxidising nitrating mixture. (vii) Gabriel phthalimide synthesis — a route to a pure primary amine with no over-alkylation, but only from an alkyl halide. Phthalimide is first converted to its potassium salt with ethanolic KOH; the salt is heated with an alkyl halide, and the resulting N-alkylphthalimide is hydrolysed (acid, alkali, or — better — hydrazinolysis with NH 2 NH 2 ) to release a pure primary amine, free of any secondary or tertiary amine — the advantage that makes this method better than direct ammonolysis for making primary amines. Mechanism, step by step: 1. KOH deprotonates the acidic N–H of phthalimide (acidic because the anion is stabilised by delocalisation into both flanking carbonyls), giving potassium phthalimide. 2. The phthalimide nitrogen anion, a good nucleophile, displaces halide from the alkyl halide by S N2 (backside attack, as in part v), giving N-alkylphthalimide. 3. Hydrolysis breaks both C–N bonds of the imide, releasing the primary amine and phthalic acid (or, with hydrazine, phthalhydrazide). For bromoethane: Phthalimide KOH Potassium phthalimide CH 3 CH 2 Br N -ethylphthalimide H 3 O^+/Δ CH 3 CH 2 NH 2 + phthalic acid Product: ethanamine (ethylamine), CH 3 CH 2 NH 2. This method fails for making an aromatic primary amine directly, because an aryl halide cannot undergo the S N2 displacement step 2 needs. Answer: (i) Carbylamine reaction: 1° amine + CHCl 3 + alcoholic KOH → isocyanide (foul smell); test for 1° amines. (ii) Diazotisation: aromatic 1° amine + NaNO 2/ HCl, 273–278 K → diazonium salt. (iii) Hofmann bromamide reaction: amide + Br 2/ NaOH → 1° amine with one carbon less (via an isocyanate intermediate). (iv) Coupling reaction: diazonium salt + phenol/aromatic amine → coloured azo compound, by electrophilic aromatic substitution. (v) Ammonolysis: alkyl halide + excess NH 3 (sealed tube) → 1° amine, by S N2; excess ammonia suppresses over-alkylation. (vi) Acetylation: amine + acetic anhydride/pyridine → N-acetyl amide, by nucleophilic acyl substitution; used to protect – NH 2 before nitration. (vii) Gabriel phthalimide synthesis: potassium phthalimide + alkyl halide, then hydrolysis → pure 1° amine (works only for alkyl, not aryl, halides).

Question 8

Accomplish the following conversions: (i) Nitrobenzene to benzoic acid (ii) Benzene to m-bromophenol (iii) Benzoic acid to aniline (iv) Aniline to 2,4,6-tribromofluorobenzene (v) Benzyl chloride to 2-phenylethanamine (vi) Chlorobenzene to p-chloroaniline (vii) Aniline to p-bromoaniline (viii) Benzamide to toluene (ix) Aniline to benzyl alcohol.

Accepted Answer

Each of these nine conversions is a synthesis problem: read the target functional group, find the one reagent that installs exactly that group, and never let a step disturb a substituent that is already correctly placed. (i) Nitrobenzene to benzoic acid A nitro group cannot be oxidised straight to –COOH — there is no C–H bond on a ring carbon for an oxidant to attack, and the –COOH carbon does not even exist yet. The nitrogen has to be turned into a leaving group and a new carbon attached in its place. Step 1 — reduction: C 6H 5NO 2 with Sn/HCl (then NaOH to free the base) adds six [H] to the nitro group, giving C 6H 5NH 2 (aniline ). Step 2 — diazotisation: C 6H 5NH 2 with NaNO 2 /HCl at 273–278 K. The nitrosonium ion NO^+ (generated in situ from NaNO 2 + HCl) attacks the lone pair on the amine nitrogen; loss of water gives the diazonium ion, C6H5N2+Cl− (benzenediazonium chloride). Step 3 — Sandmeyer reaction: C6H5N2+Cl− with CuCN/KCN. Cyanide ion, activated by Cu(I), displaces N 2 from the diazonium carbon, giving C 6H 5CN (benzonitrile ). Step 4 — nitrile hydrolysis: C 6H 5CN with H 3O^+ (or NaOH then acidify), heat. Water attacks the nitrile carbon, passing through the amide C 6H 5CONH 2, and hydrolyses further to the acid with loss of NH3. Product: benzoic acid, C6H5COOH. (ii) Benzene to m-bromophenol Br and OH must end up meta to each other, but –OH itself is an ortho/para director — so the ring has to be built with a meta-directing group first, and only swapped for –OH at the very last step. Step 1 — nitration: C 6H 6 with conc. HNO 3 / H 2SO 4 (~320 K). The electrophile NO 2 + substitutes a ring H, giving C 6H 5NO 2 (nitrobenzene ). Step 2 — bromination: C 6H 5NO 2 with Br 2 /Fe, heat. Because –NO2 is deactivating and meta-directing, Br^+ enters the position meta to it, giving 1-bromo-3-nitrobenzene (m-bromonitrobenzene). Step 3 — reduction: m-bromonitrobenzene with Sn/HCl reduces –NO2 to –NH2 without touching the C–Br bond, giving 3-bromoaniline (m-bromoaniline). Step 4 — diazotisation: m-bromoaniline with NaNO 2 /HCl at 273–278 K gives the diazonium salt, 3-bromobenzenediazonium chloride. Step 5 — hydrolysis: warming this diazonium salt with water — the oxygen of water attacks the diazonium carbon, N 2 leaves as gas — gives the phenol directly. Product: m-bromophenol (3-bromophenol), with –Br and –OH in a 1,3 relationship on the ring. (iii) Benzoic acid to aniline Losing exactly one carbon while keeping the ring intact is the signature of the Hofmann bromamide degradation — this route goes through the amide, never through a reduction (reduction of –COOH would keep the extra carbon, e.g. as –CH2OH or –CH3). Step 1 — amide formation: C 6H 5COOH with NH 3 gives ammonium benzoate, which on heating loses H 2O to give C 6H 5CONH 2 (benzamide ). Step 2 — Hofmann degradation: C 6H 5CONH 2 with Br 2 /NaOH (aq). Br 2 / OH^- first converts the amide N–H to N–Br; base then removes the remaining N–H, and the resulting nitrogen anion drives a 1,2-shift of the phenyl group from carbon to nitrogen with loss of Br^-, generating an isocyanate, C6H5–N=C=O. Aqueous base hydrolyses this isocyanate to a carbamate that decomposes (losing CO 2 as carbonate) to the free amine. Product: aniline, C 6H 5NH 2 — one carbon fewer than the acid, exactly as the Hofmann degradation always delivers. (iv) Aniline to 2,4,6-tribromofluorobenzene The –NH2 group activates the ring so strongly that bromination needs no catalyst and no direction — it happens at all three open ortho/para positions in one step. Removing the amine afterwards without disturbing the three new C–Br bonds is exactly the job the Balz–Schiemann reaction does. Step 1 — bromination: C 6H 5NH 2 with excess Br 2 (aqueous), room temperature. The amine lone pair delocalises heavily into the ring, making the two ortho and the one para position electron-rich enough to brominate spontaneously, giving 2,4,6-tribromoaniline. Step 2 — diazotisation: 2,4,6-tribromoaniline with NaNO 2 /HCl at 273–278 K gives 2,4,6-tribromobenzenediazonium chloride. Step 3 — Balz–Schiemann reaction: this diazonium salt is treated with HBF 4, precipitating the diazonium tetrafluoroborate, which is then heated. On heating it decomposes: fluoride from the BF 4 − ion takes the place of N 2 at the diazonium carbon (through an aryl cation), releasing N 2 and BF3. Product: 2,4,6-tribromofluorobenzene. (v) Benzyl chloride to 2-phenylethanamine One extra carbon is needed between the ring and the amine nitrogen, so ammonia acting directly on the chloride would give the wrong compound (benzylamine, one carbon short). Cyanide is the nucleophile that both substitutes and extends the chain by a carbon. Step 1 — nucleophilic substitution: C 6H 5CH 2Cl with KCN (alcoholic). The carbon lone pair of CN^- attacks the benzylic carbon from the back side, displacing Cl^- in an SN 2 step (fast here because the developing negative charge in the transition state is stabilised by the adjacent ring), giving C 6H 5CH 2CN (phenylacetonitrile). Step 2 — reduction: C 6H 5CH 2CN with H 2 /Ni (or LiAlH 4, then H 3O^+ workup). Hydrogen adds across the C≡N triple bond; the nitrile carbon becomes a –CH2–NH2 group. Product: 2-phenylethanamine, C6H5–CH2–CH2–NH2 (phenethylamine) — the new CH 2 from the nitrile plus the original benzylic CH 2 give the two-carbon bridge to nitrogen. (vi) Chlorobenzene to p-chloroaniline Chlorine is a deactivating substituent but still an ortho/para director (its lone pair conjugates into the ring even as its electronegativity withdraws density inductively), so nitration lands mostly at the positions needed here. Step 1 — nitration: C 6H 5Cl with conc. HNO 3 / H 2SO 4 (~300 K) gives a mixture of o-chloronitrobenzene and p-chloronitrobenzene; the para isomer is separated (it is higher-melting and less soluble, so it is isolated by fractional crystallisation). Step 2 — reduction: p-chloronitrobenzene with Sn/HCl (or Fe/HCl) reduces only the –NO2 group to –NH2; the C–Cl bond is not involved in this step and survives unchanged. Product: p-chloroaniline (4-chloroaniline). (vii) Aniline to p-bromoaniline Direct bromination of aniline over-reacts: the ring is so electron-rich that all three open positions brominate at once (as in part iv), giving the tribromo compound, not the mono-para one wanted here. Protecting the nitrogen as an amide first tones the ring down enough for a clean, single substitution. Step 1 — acetylation: C 6H 5NH 2 with (CH 3CO) 2O (acetic anhydride), in pyridine. The amine's lone pair attacks the anhydride's carbonyl carbon, displacing acetate, giving the amide C 6H 5NHCOCH 3 (acetanilide ); the nitrogen lone pair is now partly delocalised onto the carbonyl oxygen, so the ring is only mildly activated instead of strongly activated. Step 2 — bromination: C 6H 5NHCOCH 3 with Br 2 in glacial CH 3COOH, room temperature. The bulky –NHCOCH3 group sterically disfavours attack next to itself, so bromination goes predominantly to the para position, giving p-bromoacetanilide. Step 3 — hydrolysis: p-bromoacetanilide with H 3O^+ (or aqueous NaOH), heat. Water attacks the amide carbonyl carbon and the acetyl group is removed as acetic acid, regenerating the free amine. Product: p-bromoaniline (4-bromoaniline). (viii) Benzamide to toluene There is no single reagent that turns –CONH2 directly into –CH3. The nitrogen has to be stripped off completely as an amine, and then a fresh methyl group has to be built onto the ring by carbon chemistry — two unrelated operations chained together. Step 1 — Hofmann degradation: C 6H 5CONH 2 with Br 2 /NaOH (aq), by the same rearrangement described in part (iii), gives C 6H 5NH 2 (aniline ), with the carbon of the original amide lost as carbonate. Step 2 — diazotisation: C 6H 5NH 2 with NaNO 2 /HCl at 273–278 K gives C6H5N2+Cl−. Step 3 — deamination: this diazonium salt is treated with H 3PO 2 / H 2O (warm). H 3PO 2 supplies a hydrogen in place of the N 2 + group (H3PO2 is oxidised to H 3PO 3 as it does so), replacing –N2+ with –H and giving back C 6H 6 (benzene) — the amino group is now gone entirely, with nothing else on the ring to disturb. Step 4 — Friedel–Crafts alkylation: C 6H 6 with CH 3Cl and anhydrous AlCl3. AlCl 3 polarises the C–Cl bond of CH 3Cl to generate an electrophilic methyl species; the ring's π electrons attack this carbon, forming an arenium (Wheland) intermediate, which then loses H^+ to restore aromaticity. Product: toluene, C6H5CH3. (ix) Aniline to benzyl alcohol Benzyl alcohol carries its oxygen one carbon away from the ring, not directly on it, so this route must go through a nitrile and stop cleanly at the aldehyde oxidation level — reducing all the way to the carboxylic acid or overshooting to the amine would both give the wrong compound. Step 1 — diazotisation: C 6H 5NH 2 with NaNO 2 /HCl at 273–278 K gives C6H5N2+Cl−. Step 2 — Sandmeyer reaction: C6H5N2+Cl− with CuCN/KCN gives C 6H 5CN (benzonitrile ), N 2 being displaced by cyanide as in part (i). Step 3 — controlled (partial) reduction: C 6H 5CN with DIBAL-H at low temperature, followed by aqueous acid workup. DIBAL-H delivers a single hydride to the nitrile carbon and forms a stable imine–aluminium complex that does not reduce further at low temperature; acidic hydrolysis of this imine on workup gives the aldehyde, C 6H 5CHO (benzaldehyde ), rather than over-reducing to the amine. Step 4 — reduction of the aldehyde: C 6H 5CHO with NaBH 4 (or LiAlH4), then H 3O^+ workup. Hydride from BH 4 − attacks the carbonyl carbon; protonation of the resulting alkoxide on workup gives the alcohol. Product: benzyl alcohol, C6H5CH2OH. Answer: (i) C 6H 5NO 2 →(Sn/HCl) C 6H 5NH 2 →(NaNO2/HCl, 273–278 K) C6H5N2+Cl− →(CuCN/KCN) C 6H 5CN →(H3O+, heat) C 6H 5COOH (benzoic acid). (ii) C 6H 6 →(HNO3/H2SO4) C 6H 5NO 2 →(Br2/Fe) m-BrC 6H 4NO 2 →(Sn/HCl) m-BrC 6H 4NH 2 →(NaNO2/HCl, 273–278 K) diazonium salt →(H2O, warm) m-bromophenol. (iii) C 6H 5COOH →(NH3, then heat) C 6H 5CONH 2 →(Br2/NaOH, Hofmann degradation) aniline (C6H5NH2). (iv) C 6H 5NH 2 →(Br2 excess) 2,4,6-tribromoaniline →(NaNO2/HCl, 273–278 K) diazonium salt →(HBF4, then heat, Balz–Schiemann) 2,4,6-tribromofluorobenzene. (v) C 6H 5CH 2Cl →(KCN, alc.) C 6H 5CH 2CN →(H2/Ni) C 6H 5CH 2CH 2NH 2 (2-phenylethanamine). (vi) C 6H 5Cl →(HNO3/H2SO4) p-chloronitrobenzene (separated from ortho) →(Sn/HCl) p-chloroaniline. (vii) C 6H 5NH 2 → ((CH 3CO) 2O) acetanilide →(Br2/CH3COOH) p-bromoacetanilide →(H3O+, heat) p-bromoaniline. (viii) C 6H 5CONH 2 →(Br2/NaOH) aniline →(NaNO2/HCl, 273–278 K) diazonium salt →(H3PO2/H2O) benzene →(CH3Cl/anhyd. AlCl3) toluene. (ix) C 6H 5NH 2 →(NaNO2/HCl, 273–278 K) diazonium salt →(CuCN/KCN) benzonitrile →(DIBAL-H, then H3O+) benzaldehyde → (NaBH 4) benzyl alcohol.

Question 9

Give the structures of A, B and C in the following reactions: ————→ NaCN ———————→ OH − —————→ NaOH Br + (i) CH CH I A B 2 C 3 2 Partial hydrolysis ————→ CuCN —————→ H O/H + ———→ NH (ii) C H N Cl A 2 B 3 C 6 5 2 Δ (iii) CH CH Br ———→ KCN A ————→ LiAlH B ————→ HNO C 4 2 3 2 0 C ° ————→ Fe/HCl ——————→ NaNO + HCl —————→ H O/H + (iv) C H NO A 2 B 2 C 6 5 2 273 K Δ (v) CH COOH ———→ NH A ————→ NaOBr B ——————→ NaNO /HCl C 3 2 3 Δ C H NO ————→ Fe/HCl A ————→ HNO B ————→ C H OH C (vi) 2 6 5 6 5 2 273K

Accepted Answer

Each arrow here is a name reaction — identify which one, and the structure of A, B or C follows from what that reaction is known to do to a nitrile, an amide, an amine or a diazonium salt. Work each of the six chains link by link. (i) CH 3 CH 2 Br NaCN A Partial hydrolysis B Br 2/ NaOH C Bromoethane has an electrophilic carbon bonded to bromine, a good leaving group. Cyanide ion, CN^-, attacks that carbon from the back side ( S N2 ) and displaces Br^-. The carbon of CN^- bonds to what was the alkyl carbon, so the chain gains one carbon: A = CH 3 CH 2 CN ( propanenitrile, common name propionitrile ) Partial hydrolysis is the step people rush past — controlled hydrolysis (dilute acid, or MnO 2 /mild conditions) stops the nitrile at the amide stage instead of running it all the way to the acid, because the amide's carbonyl is less electrophilic than the nitrile's carbon once the first water has added: B = CH 3 CH 2 CONH 2 ( propanamide ) Br 2/ NaOH on a primary amide is the Hofmann bromamide degradation. Bromine brominates the amide nitrogen, base removes the remaining N–H, and the resulting nitrogen anion pushes the alkyl group across to nitrogen with loss of bromide (a nitrene/isocyanate pathway), and the isocyanate is then hydrolysed and decarboxylated. The net effect is that the carbonyl carbon is stripped out as carbonate — the product amine has one carbon FEWER than the amide it came from, not the same count: C = CH 3 CH 2 NH 2 ( ethanamine, i.e. ethylamine ) (ii) C 6 H 5 N 2^+ Cl^- CuCN A H 3 O^+/ H 2 O B NH 3 C Benzenediazonium chloride with cuprous cyanide is the Sandmeyer reaction: the - N 2^+ group is replaced by - CN, the copper(I) ion mediating a radical substitution at the ring carbon: A = C 6 H 5 CN ( benzonitrile ) Acidic aqueous hydrolysis ( H 3 O^+/ H 2 O, with heat) takes the nitrile all the way through the amide to the carboxylic acid — this is complete hydrolysis, unlike step (i): B = C 6 H 5 COOH ( benzoic acid ) Ammonia is a base, not a dehydrating agent; without a heating step to drive off water it only deprotonates the acid to its ammonium salt (it does not by itself become the amide — that needs the Δ that appears explicitly in part (v) but not here): C = C 6 H 5 COONH 4 ( ammonium benzoate ) (iii) CH 3 CH 2 Br KCN A LiAlH 4 B HNO 2, 0^° C C Exactly the same S N2 displacement as (i), just with K^+ as the counter-ion: A = CH 3 CH 2 CN ( propanenitrile ) LiAlH 4 delivers hydride twice to the nitrile carbon and, after the imine intermediate is reduced again, converts - C ≡ N fully to - CH 2 NH 2 — a net addition of four hydrogens, and the nitrile carbon becomes a CH 2 attached to NH 2, so the carbon count is unchanged from the nitrile: B = CH 3 CH 2 CH 2 NH 2 ( propan-1-amine, n -propylamine ) A primary aliphatic amine plus nitrous acid never survives as a diazonium salt. HNO 2 (generated in the flask) converts the - NH 2 to - N 2^+, but an alkyl- N 2^+ is far too unstable to isolate even at 0^° C: it loses N 2 gas immediately, and water displaces the nitrogen, giving the alcohol: C = CH 3 CH 2 CH 2 OH ( propan-1-ol ) (iv) C 6 H 5 NO 2 Fe/HCl A NaNO 2 + HCl, 273 K B H 3 O^+/ H 2 O, Δ C Fe/HCl is a standard nitro-to-amine reduction (six electrons and six protons delivered to the nitro group, iron being oxidised to Fe^2+/ Fe^3+ ): A = C 6 H 5 NH 2 ( aniline ) NaNO 2 + HCl at 273 K generates HNO 2 in situ, which diazotises the aromatic primary amine. Here the diazonium survives, because the ring's π system delocalises the positive charge onto nitrogen far better than an alkyl chain can, and the low temperature suppresses decomposition: B = C 6 H 5 N 2^+ Cl^- ( benzenediazonium chloride ) Warming that diazonium salt with water ( H 3 O^+/ H 2 O, Δ ) lets water's oxygen attack the terminal nitrogen; N 2 leaves as gas and the ring is left bonded to - OH: C = C 6 H 5 OH ( phenol ) (v) CH 3 COOH NH 3, Δ A NaOBr B NaNO 2/ HCl C Here the Δ is written on the arrow, so ammonia first forms the ammonium salt and heat then drives off water, dehydrating the salt to the amide: A = CH 3 CONH 2 ( ethanamide, acetamide ) NaOBr (bromine dissolved in NaOH) is the same Hofmann bromamide degradation as in part (i). Acetamide's R-group is just CH 3-, and the carbonyl carbon is removed: B = CH 3 NH 2 ( methanamine, methylamine ) NaNO 2/ HCl again diazotises a primary aliphatic amine, and again the alkyl diazonium salt cannot survive — it decomposes with loss of N 2 and water takes its place: C = CH 3 OH ( methanol ) (vi) C 6 H 5 NO 2 Fe/HCl A HNO 2, 273 K B C 6 H 5 OH C Same reduction as (iv): A = C 6 H 5 NH 2 ( aniline ) Same cold diazotisation as (iv): B = C 6 H 5 N 2^+ Cl^- ( benzenediazonium chloride ) The last step is a diazonium coupling reaction, not a hydrolysis — phenol is used as the nucleophile instead of water. The diazonium ion is a weak electrophile, so it only attacks a strongly activated ring: phenol's - OH is a powerful ortho/para director, and the para position (less hindered than ortho) is where the diazonium nitrogen bonds to the ring carbon, keeping the - N=N - linkage (an azo group) intact rather than losing nitrogen: C = C 6 H 5- N=N - C 6 H 4- OH (para) (p -hydroxyazobenzene, 4-(phenyldiazenyl)phenol ) Answer: (i) A= CH 3 CH 2 CN (propanenitrile), B= CH 3 CH 2 CONH 2 (propanamide), C= CH 3 CH 2 NH 2 (ethanamine). (ii) A= C 6 H 5 CN (benzonitrile), B= C 6 H 5 COOH (benzoic acid), C= C 6 H 5 COONH 4 (ammonium benzoate). (iii) A= CH 3 CH 2 CN (propanenitrile), B= CH 3 CH 2 CH 2 NH 2 (propan-1-amine), C= CH 3 CH 2 CH 2 OH (propan-1-ol). (iv) A= C 6 H 5 NH 2 (aniline), B= C 6 H 5 N 2^+ Cl^- (benzenediazonium chloride), C= C 6 H 5 OH (phenol). (v) A= CH 3 CONH 2 (ethanamide), B= CH 3 NH 2 (methanamine), C= CH 3 OH (methanol). (vi) A= C 6 H 5 NH 2 (aniline), B= C 6 H 5 N 2^+ Cl^- (benzenediazonium chloride), C= C 6 H 5- N=N - C 6 H 4- OH, para (p-hydroxyazobenzene).

Question 10

An aromatic compound ‘A’ on treatment with aqueous ammonia and heating forms compound ‘B’ which on heating with Br 2 and KOH forms a compound ‘C’ of molecular formula C 6 H 7 N. Write the structures and IUPAC names of compounds A, B and C.

Accepted Answer

A carbon atom disappears only when an amide is broken down by bromine and alkali — the Hofmann bromamide degradation — so the fastest way into this problem is to work backward from that one fact: C is a primary amine, B is the amide that is one carbon larger, and the "aqueous ammonia + heat" step that makes B is just ammonia doing a nucleophilic acyl substitution on an acid chloride. Step 1 — identify C from its molecular formula. For C 6H 7N, the degree of unsaturation is DoU = (2(6)+2+1-7)/2 = 8/2 = 4. Four degrees of unsaturation with six carbons is exactly a benzene ring and nothing else. With one nitrogen left over, the only structure that fits is a benzene ring carrying one -NH 2 group directly on the ring: C 6H 5-NH 2. Here C 6H 5- is the phenyl group (a benzene ring missing one hydrogen) and -NH 2 is the primary amino group. This is aniline; its systematic IUPAC name is benzenamine. So C = C 6H 5NH 2, aniline (benzenamine). Step 2 — identify B from the Br 2 /KOH step (Hofmann bromamide degradation). This named reaction takes a primary amide R-CO-NH 2 and hands back the amine R-NH 2 with the carbonyl carbon removed as CO 2 — the product amine always has one carbon fewer than the amide. Since C is C 6H 5-NH 2, the amide B must have been C 6H 5-CO-NH 2: benzamide. The mechanism, step by step: 1. Bromine brominates the acidic N–H of the amide (KOH mops up the HBr formed), giving N-bromobenzamide, C 6H 5-CO-NHBr. 2. KOH removes the remaining N–H proton, giving the anion C 6H 5-CO-N^-Br. 3. This anion rearranges intramolecularly: the phenyl group migrates from the carbonyl carbon onto the electron-poor nitrogen at the same instant that bromide ion leaves — a concerted 1,2-shift, not a free radical or free carbocation step. The product of this migration is phenyl isocyanate, C 6H 5-N=C=O. 4. Hydroxide ion attacks the isocyanate carbon of C 6H 5-N=C=O, adding across the N=C to give the carbamate ion C 6H 5-NH-COO^-. 5. This carbamate loses carbon dioxide (as carbonate, with the excess KOH present) — the decarboxylation step that actually removes the extra carbon — leaving the nitrogen attached straight to the ring: C 6H 5-NH 2, aniline, plus potassium carbonate and potassium bromide. Net equation: C 6H 5CONH 2 + Br 2 + 4KOH → C 6H 5NH 2 + K 2CO 3 + 2KBr + 2H 2O. So B = C 6H 5CONH 2, benzamide (systematic name: benzenecarboxamide). Step 3 — identify A from the ammonia step. B (benzamide) is an amide, and the cleanest way to build an amide from "an aromatic compound treated with aqueous ammonia, with heating," is nucleophilic acyl substitution of ammonia on the corresponding acid chloride. So A must be benzoyl chloride, C 6H 5-CO-Cl. Mechanism: the lone pair on the nitrogen of NH 3 attacks the electrophilic carbonyl carbon of C 6H 5COCl, pushing electron density onto the carbonyl oxygen and forming a tetrahedral intermediate. Chloride ion then leaves from that intermediate, the C=O is re-formed, and a second molecule of ammonia removes the proton from the resulting -NH 2^+- center. The heating is used simply to drive the reaction to completion and expel the ammonium chloride byproduct. C 6H 5COCl + 2NH 3 Δ C 6H 5CONH 2 + NH 4Cl. So A = C 6H 5COCl, benzoyl chloride (systematic name: benzenecarbonyl chloride). Putting the three together, the whole sequence reads: C 6H 5COCl [Δ] NH 3(aq) C 6H 5CONH 2 [KOH] Br 2 C 6H 5NH 2. Answer: A is benzoyl chloride, C 6H 5COCl (IUPAC: benzenecarbonyl chloride); B is benzamide, C 6H 5CONH 2 (IUPAC: benzenecarboxamide); C is aniline, C 6H 5NH 2 (IUPAC: benzenamine) — formed from B by the Hofmann bromamide degradation, which strips the carbonyl carbon out of the amide as CO 2.

Question 11

Complete the following reactions: (i) C H NH + CHCl + alc.KOH → 6 5 2 3 (ii) C H N Cl + H PO + H O → 6 5 2 3 2 2 (iii) C H NH + H SO ( ) → conc. 6 5 2 2 4 (iv) C H N Cl + C H OH → 6 5 2 2 5 + ( ) → (v) C H NH Br aq 6 5 2 2 ( ) (vi) C H NH + CH CO O → 6 5 2 3 2 ———————→ () i HBF (vii) C H N Cl 4 6 5 2 () ii NaNO /Cu, Δ 2

Accepted Answer

Every one of these seven reactions turns on the same fact: the lone pair on the amine nitrogen (or the leaving - N 2^+ group of a diazonium salt) decides what happens — so track that lone pair, or that leaving group, through each step. (i) C 6 H 5 NH 2 + CHCl 3 + alc. KOH → — the carbylamine reaction. Alcoholic KOH first pulls two protons off chloroform in succession (α-elimination), kicking out two chloride ions and leaving a highly reactive, electron-deficient carbon species called dichlorocarbene,: CCl 2. Being a primary aromatic amine, aniline still has a nucleophilic lone pair on nitrogen; that lone pair attacks the electron-deficient carbon of the carbene. The addition intermediate then loses two more HCl molecules (mopped up by the remaining KOH) to convert the single C–N bond into a C≡N triple bond. The net change is that the two remaining N–H bonds of aniline are replaced by one N=C(terminal) linkage. Product: phenyl isocyanide (phenyl carbylamine), C 6 H 5 –N ≡ C — this foul-smelling isocyanide is exactly what the "carbylamine test" for primary amines detects; secondary and tertiary amines have no N–H lone pair positioned this way and give no reaction. C 6 H 5 NH 2 + CHCl 3 + 3 KOH → C 6 H 5 NC + 3 KCl + 3 H 2 O (ii) C 6 H 5 N 2 Cl + H 3 PO 2 + H 2 O → — reductive deamination. Hypophosphorous acid, H 3 PO 2, is a mild reducing agent. It supplies a hydride-like hydrogen to the terminal nitrogen of the diazonium ion, which collapses by expelling N 2 gas — the classic, very stable leaving group of any diazonium chemistry — and leaving a hydrogen in place of the group that used to be - N 2^+ Cl^-. H 3 PO 2 itself is oxidised to phosphorous acid, H 3 PO 3, and the chloride is released as HCl. This reaction is how chemists remove an - NH 2 group after having used it only to direct an earlier substitution (deamination). Product: benzene, C 6 H 6. C 6 H 5 N 2 Cl + H 3 PO 2 + H 2 O → C 6 H 6 + H 3 PO 3 + N 2 + HCl (iii) C 6 H 5 NH 2 + H 2 SO 4( conc. ) → — sulphonation through a salt intermediate. At room temperature the nitrogen lone pair of aniline simply grabs a proton from the strong acid — this is an acid-base reaction, not a substitution — giving the anilinium hydrogensulphate salt, C 6 H 5 NH 3^+ HSO 4^-. Because the ring nitrogen is now positively charged and can no longer donate its lone pair into the ring, the ring is temporarily deactivated. Only on strong heating (453–473 K, the industrial "baking process") does the electrophile SO 3 / HSO 3^+ attack the ring, and it does so at the position para to nitrogen (the position left most accessible and still favoured once the salt re-equilibrates with free aniline in solution). Losing a proton restores aromaticity, and the product settles as the internal salt (zwitterion) in which the sulphonic acid proton has migrated onto the amino nitrogen. Product: sulphanilic acid (4-aminobenzenesulfonic acid), written as the zwitterion ^+ H 3 N–C 6 H 4 –SO 3^- (para-substituted). C 6 H 5 NH 2 + H 2 SO 4 → C 6 H 5 + N H 3 HSO 4^- 453 – 473 K p -H 2 N–C 6 H 4 –SO 3 H + H 2 O (iv) C 6 H 5 N 2 Cl + C 2 H 5 OH → — ethanol as the reducing agent. This is the same reductive-deamination idea as part (ii), but now ethanol is the hydrogen source. Ethanol's α C–H hydrogen is transferred to the diazonium nitrogen, expelling N 2 gas and leaving a hydrogen in the ring position that used to carry - N 2^+. In donating that hydrogen, ethanol itself is oxidised: the C–H and O–H bonds around that carbon rearrange into a C=O, so CH 3 CH 2 OH becomes CH 3 CHO. Product: benzene, C 6 H 6 (with acetaldehyde as the by-product that carries away the "missing" hydrogen). C 6 H 5 N 2 Cl + C 2 H 5 OH → C 6 H 6 + N 2 + HCl + CH 3 CHO (v) C 6 H 5 NH 2 + Br 2( aq ) → — bromination needs no catalyst here. Unlike benzene, which needs a Lewis-acid catalyst ( FeBr 3 ) to polarise Br 2 before the ring can attack it, aniline's ring is already so electron-rich — the nitrogen lone pair is delocalised into the ring, most heavily onto the ortho and para carbons — that bromine water reacts directly and instantly. Each of the two ortho positions and the one para position is activated, so all three positions are substituted in one go (three successive electrophilic aromatic substitutions, each following the arenium-ion mechanism: ring attacks Br 2, a C–Br bond forms, a proton is lost to restore aromaticity). Product: 2,4,6-tribromoaniline, seen as an immediate white precipitate — this is the standard qualitative test distinguishing aniline from benzene. C 6 H 5 NH 2 + 3 Br 2( aq ) → 2,4,6 -Br 3 C 6 H 2 NH 2 + 3 HBr (vi) C 6 H 5 NH 2 + ( CH 3 CO ) 2 O → — acetylation of the amine. The nitrogen lone pair of aniline attacks the electrophilic carbonyl carbon of one C=O group in acetic anhydride. This gives a tetrahedral intermediate at that carbon, which collapses by kicking out the acetate ion, CH 3 COO^-, as the leaving group (breaking the anhydride's central C–O bond) while re-forming the C=O. The expelled acetate then removes a proton from the now positively charged nitrogen, restoring a neutral amide. Net effect: one H on nitrogen is replaced by an acetyl group, - COCH 3. Product: N-phenylacetamide, commonly called acetanilide, C 6 H 5 –NH–CO–CH 3. C 6 H 5 NH 2 + ( CH 3 CO ) 2 O → C 6 H 5 NHCOCH 3 + CH 3 COOH (vii) C 6 H 5 N 2 Cl treated (i) with HBF 4, then (ii) with NaNO 2/ Cu, Δ — two different fates for the same diazonium salt. (i) HBF 4, then heat — the Balz–Schiemann reaction. Fluoroboric acid first swaps the counter-ion: chloride is exchanged for tetrafluoroborate, precipitating the far more thermally stable salt benzenediazonium tetrafluoroborate, C 6 H 5 N 2^+ BF 4^-. When this dry solid is heated, it decomposes: the C^- N 2^+ bond breaks, releasing N 2 gas and BF 3, and a fluoride ion (transferred from the departing BF 4^- ) bonds to the carbon left behind. This is the standard route to an aryl fluoride, since fluoride is too poor a nucleophile to be introduced by any direct substitution. Product: fluorobenzene, C 6 H 5 F. C 6 H 5 N 2 Cl + HBF 4 → C 6 H 5 N 2^+ BF 4^- + HCl; C 6 H 5 N 2^+ BF 4^- Δ C 6 H 5 F + N 2 + BF 3 (ii) NaNO 2/ Cu, Δ — a Sandmeyer-type substitution. Copper powder catalyses the exchange of the diazonium group for the nitrite group supplied by sodium nitrite: the C^- N 2^+ bond breaks, N 2 is expelled, and the nitrogen of NO 2^- bonds to the ring carbon in its place, with NaCl as the by-product. Product: nitrobenzene, C 6 H 5 NO 2. C 6 H 5 N 2 Cl + NaNO 2 [Δ] Cu C 6 H 5 NO 2 + N 2 + NaCl Answer: (i) phenyl isocyanide, C 6 H 5 NC; (ii) benzene, C 6 H 6; (iii) sulphanilic acid, p -H 2 N–C 6 H 4 –SO 3 H; (iv) benzene, C 6 H 6 (+ CH 3 CHO ); (v) 2,4,6-tribromoaniline; (vi) acetanilide, C 6 H 5 NHCOCH 3; (vii) fluorobenzene C 6 H 5 F via HBF 4/Δ, and nitrobenzene C 6 H 5 NO 2 via NaNO 2/ Cu, Δ.

Question 12

Why cannot aromatic primary amines be prepared by Gabriel phthalimide synthesis?

Accepted Answer

A nucleophile can only attack a carbon it can reach from behind, and in an aryl halide the halogen-bearing carbon is locked inside the ring where there is no "behind" to reach. Step 1 — What the Gabriel synthesis actually needs The Gabriel phthalimide synthesis makes a primary amine in two stages: 1. Phthalimide (a cyclic imide, C 6 H 4( CO ) 2 NH ) is treated with alcoholic KOH. The N–H is acidic (flanked by two carbonyls), so it is deprotonated to the phthalimide anion, C 6 H 4( CO ) 2 N^- K^+. This anion is the nucleophile. 2. The potassium phthalimide is then heated with an alkyl halide, R–X (R = alkyl, X = Cl, Br, I). The nitrogen lone pair attacks the carbon bearing the halogen from the side opposite to X, the C–X bond breaks, and the C–N bond forms, giving N-alkylphthalimide. This is a plain S N2 displacement. 3. N-alkylphthalimide is hydrolysed (acid, base, or hydrazine/Ising method) to release the primary amine R–NH 2 and phthalic acid (or its hydrazide). The entire method stands or falls on step 2 being a working S N2 reaction, because that is the only step that actually installs the nitrogen on the target carbon. Step 2 — Why an aryl halide fails step 2 To make an aromatic primary amine, say aniline ( C 6 H 5 NH 2 ), by this route, the alkyl halide would have to be replaced by an aryl halide such as chlorobenzene, C 6 H 5 Cl. Two independent features of C 6 H 5 Cl shut down the S N2 step: - Geometry. The carbon carrying Cl is one of the six sp2 carbons of the benzene ring. For the phthalimide nitrogen to displace Cl^- by backside attack, it would have to approach that carbon from the side directly opposite the C–Cl bond — but that side is occupied by the plane and the π -electron cloud of the aromatic ring itself. There is no accessible backside, so the nucleophile physically cannot line up for inversion. - Bond strength. The lone pair on chlorine conjugates with the ring's π system (resonance donation into the ring), which gives the C–Cl bond partial double-bond character. A bond with double-bond character is shorter and stronger than an ordinary single C–X bond, so even where geometry were not a problem, the bond is far harder to break heterolytically. Because of these two effects, aryl and vinyl halides are, as a rule, inert to nucleophilic substitution under the conditions the Gabriel synthesis uses. Potassium phthalimide simply does not react with chlorobenzene (or bromobenzene, iodobenzene) to give N-phenylphthalimide, so the sequence never gets past step 2, and no aniline-type product is ever formed. Step 3 — The product actually named Since the alkylation step fails outright, there is no N-arylphthalimide intermediate to hydrolyse, and hence no way to obtain an aromatic primary amine such as aniline ( C 6 H 5 –NH 2 ) or p -toluidine ( CH 3 –C 6 H 4 –NH 2 ) from this method. The Gabriel synthesis is therefore restricted to primary aliphatic amines (from alkyl halides, where the halogen-bearing carbon is sp3 and open to backside attack), and cannot be used for aromatic primary amines. Answer: Aromatic primary amines cannot be made by the Gabriel phthalimide synthesis because the method's key step is an S N2 displacement of halide by the phthalimide anion on the halogen-bearing carbon. In an aryl halide (e.g., C 6 H 5 Cl ) that carbon is sp2 and part of the ring, so backside attack is blocked, and the C–X bond has extra strength from resonance with the ring; aryl halides therefore do not undergo nucleophilic substitution with potassium phthalimide, so the required N-arylphthalimide intermediate — and hence the aromatic amine — is never formed.

Question 13

Write the reactions of (i) aromatic and (ii) aliphatic primary amines with nitrous acid.

Accepted Answer

A primary amine's fate with nitrous acid turns on one thing: whether the diazonium ion it forms can spread its positive charge into a ring. Aromatic ones can, and survive in the cold; aliphatic ones can't, and fall apart the instant they form. Nitrous acid, HNO 2, is too unstable to keep in a bottle, so it is always generated in situ by treating sodium nitrite with a mineral acid in an ice bath: NaNO 2 + HCl 273 - 278 K HNO 2 + NaCl Protonation of HNO 2 followed by loss of water gives the true electrophile, the nitrosonium ion NO^+, and it is this species that the amine nitrogen's lone pair attacks in both cases below. (i) Aromatic primary amine — aniline, C 6 H 5 -NH 2 (a benzene ring with one -NH 2 group): The lone pair on the -NH 2 nitrogen attacks the nitrogen of NO^+, giving an N-nitrosamine, C 6 H 5 -NH-N=O. A proton then shifts from nitrogen to the nitroso oxygen and that oxygen leaves as water, generating the diazonium ion. Carried out below 278 K (the salt starts hydrolysing to phenol once the solution warms past about 283 K), the overall reaction — called diazotization — is: C 6 H 5 -NH 2 + HNO 2 + HCl 273 - 278 K C 6 H 5 -N 2^+ Cl^- + 2 H 2 O The product is benzenediazonium chloride, C 6 H 5 -N 2^+ Cl^-. It can be kept (in the cold) because the positive charge on the terminal nitrogen is delocalised back into the benzene ring through the attached sp ^2 carbon — the -N 2^+ group is conjugated with the ring π system, so the ion is resonance-stabilised rather than a bare, isolated cation. (ii) Aliphatic primary amine — ethylamine, CH 3 CH 2 -NH 2 (an ethyl group, CH 3 CH 2 -, bonded to -NH 2 ): The same first step occurs — the amine nitrogen attacks NO^+^- and an aliphatic diazonium ion, CH 3 CH 2 -N 2^+, forms briefly. But here the C-N bond is a plain σ bond to a saturated alkyl carbon with no π system to delocalise the charge into, so the ion is far too unstable to isolate even in the cold: it collapses immediately, expelling nitrogen gas and leaving an ethyl carbocation, CH 3 CH 2^+, which water then captures: CH 3 CH 2 -NH 2 + HNO 2 → CH 3 CH 2 -OH + N 2 + H 2 O The product is ethanol, CH 3 CH 2 -OH (propan-... no, ethan-1-ol by IUPAC name), evolved together with bubbles of nitrogen gas. Because the actual intermediate is a free carbocation, this is a substitution at a cation, not a clean one-product reaction: the same carbocation can also lose a proton to give a minor amount of ethylene ( CH 2 =CH 2 ) or capture chloride ion to give a trace of ethyl chloride ( CH 3 CH 2 -Cl ), so the alcohol is the major product among a mixture. The step people miss: it is the absence of a ring to delocalise into — not any difference in how NO^+ attacks — that decides everything here. Both amines form a diazonium ion by the identical first step; only the aromatic one has somewhere to park the extra positive charge, so only the aromatic one survives to be isolated as a salt. This contrast is also a working test: an aliphatic 1° amine gives immediate, visible effervescence of N 2 gas with nitrous acid at ordinary temperature, while an aromatic 1° amine gives a clear, gas-free solution of the diazonium salt in the cold. Answer: (i) Aromatic 1° amine + HNO 2 /HCl at 273–278 K undergoes diazotization to a stable diazonium salt — aniline, C 6 H 5 -NH 2, gives benzenediazonium chloride, C 6 H 5 -N 2^+ Cl^- (stabilised by resonance with the ring). (ii) Aliphatic 1° amine + HNO 2 forms an unstable alkyl diazonium ion that decomposes at once, releasing N 2 gas and giving mainly the corresponding alcohol — ethylamine, CH 3 CH 2 -NH 2, gives ethanol, CH 3 CH 2 -OH, along with N 2 and H 2 O (plus minor alkene/alkyl halide by-products).

Question 14

Give plausible explanation for each of the following: (i) Why are amines less acidic than alcohols of comparable molecular masses? (ii) Why do primary amines have higher boiling point than tertiary amines? (iii) Why are aliphatic amines stronger bases than aromatic amines?

Accepted Answer

(i) Acidity is decided by how well the atom left behind can hold the negative charge after the proton leaves — and nitrogen is worse at that job than oxygen. When an alcohol R - OH loses its O - H proton, the charge lands on oxygen, giving the alkoxide ion R - O^-. When an amine R - NH 2 loses one of its N - H protons, the charge lands on nitrogen, giving the amide ion R - NH^-. Electronegativity on the Pauling scale is O = 3.44 and N = 3.04. Oxygen pulls electron density toward itself more strongly than nitrogen does, so it can accommodate an extra lone pair (the negative charge of R - O^- ) far more comfortably than nitrogen can accommodate the same extra lone pair in R - NH^-. The amide ion is therefore much higher in energy — much less stable — than the alkoxide ion. A weak, unwilling-to-form conjugate base means the parent compound holds on to its proton tightly, i.e. is a weak acid. Because R - NH^- is so unstable, amines resist giving up their N - H proton ( p K a of a typical amine is around 35–38) far more than alcohols resist giving up their O - H proton ( p K a around 16–18). That is why amines are markedly less acidic than alcohols of comparable molecular mass. (ii) Boiling point among amines tracks how many N - H bonds are left to hydrogen-bond with, and a tertiary amine has none. A primary amine R - NH 2 carries two N - H bonds per molecule, a secondary amine R 2NH carries one, and a tertiary amine R 3N carries zero — all three bonds from nitrogen are used up joining alkyl groups, so there is no hydrogen left on nitrogen at all. Hydrogen bonding requires a hydrogen attached to a small, highly electronegative atom (here, nitrogen) reaching across to a lone pair on a neighbouring molecule. A primary amine molecule can donate hydrogen bonds through both of its N - H hydrogens, so primary amine molecules link up into an extended hydrogen-bonded network in the liquid. Breaking that network to let a molecule escape into the vapour phase costs extra energy, which shows up as a higher boiling point. A tertiary amine molecule has no N - H hydrogen to donate, so it cannot hydrogen-bond to another molecule of itself at all — only the much weaker dipole-dipole and van der Waals forces act between tertiary amine molecules. With less energy needed to separate the molecules, the tertiary amine boils at a lower temperature. This is why, for amines of comparable molecular mass, the boiling-point order is primary secondary tertiary — for example propan-1-amine, CH 3CH 2CH 2NH 2 (boiling point about 49^° C ), boils well above trimethylamine, (CH 3) 3N (boiling point about 3^° C ), even though both share the molecular formula C 3H 9N. (iii) Basicity depends on how available the nitrogen lone pair is to grab an incoming proton, and in an aromatic amine that lone pair is partly given away to the ring. In an aliphatic amine such as ethanamine, CH 3CH 2NH 2, the attached alkyl group is electron-releasing (the +I inductive effect). It pushes electron density onto nitrogen, so the lone pair sitting on nitrogen is more concentrated and more eager to accept a proton. The alkyl group also stabilizes the resulting alkylammonium ion, CH 3CH 2NH 3^+, through the same +I effect, which makes protonation even more favourable. That is why aliphatic amines are stronger bases than ammonia itself, and much stronger bases than aromatic amines. In an aromatic amine such as aniline, C 6H 5NH 2, nitrogen is bonded directly to an sp^2 ring carbon, and its lone pair is delocalized by resonance into the benzene π system — the electron density is spread out over the ring (concentrated at the ortho and para carbons) rather than sitting fully on nitrogen. A delocalized lone pair is a poor proton acceptor, so aniline is reluctant to protonate. Worse still, once aniline is protonated to the anilinium ion, C 6H 5NH 3^+, that resonance delocalization is destroyed completely — nitrogen no longer has a lone pair to share with the ring — so the conjugate acid is left relatively high in energy (destabilized) compared with the neutral amine. Both effects push in the same direction and make aromatic amines much weaker bases than aliphatic ones: aniline has p K b ≈ 9.4 (its conjugate acid p K a ≈ 4.6 ), while ethylamine has p K b ≈ 3.3 (its conjugate acid p K a ≈ 10.7 ). Answer: (i) Amines are less acidic than alcohols because losing N - H gives the amide ion R - NH^-, which nitrogen (less electronegative than oxygen) stabilizes far worse than oxygen stabilizes the alkoxide ion R - O^- from an alcohol. (ii) Primary amines ( R - NH 2, two N - H bonds) hydrogen-bond extensively with each other, while tertiary amines ( R 3N, no N - H bond) cannot hydrogen-bond at all and rely only on weaker dipole-dipole/van der Waals forces, so primary amines boil higher than tertiary amines of comparable molecular mass. (iii) Aliphatic amines are stronger bases than aromatic amines because the aromatic ring in compounds like aniline delocalizes the nitrogen lone pair by resonance (lost entirely on protonation), making the lone pair less available and the conjugate acid less stable, whereas the alkyl group's +I effect in aliphatic amines pushes electron density onto nitrogen and stabilizes the conjugate acid, favouring protonation.

NCERT Solutions · Class 12 Chemistry Amines

Exercises 9.1–9.10 (part 1 of 2)

Exercise 9.1

Exercise 9.2

Exercise 9.3

Exercise 9.4

Exercise 9.5

Exercise 9.6

Exercise 9.7

Exercise 9.8

Exercise 9.9

Exercise 9.10

Exercises 9.11–9.14 (part 2 of 2)

Exercise 9.11

Exercise 9.12

Exercise 9.13

Exercise 9.14