Question 1

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 2

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 3

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 4

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 9.11–9.14 (part 2 of 2)

Exercise 9.11

Exercise 9.12

Exercise 9.13

Exercise 9.14