NCERT Exemplar Solutions Class 12 Chemistry Ch 9 Amines PDF

Correct option: (ii) Triethylamine, (C2H5)3N.

Concept used. An amine's degree is decided by how many carbon substituents sit on the nitrogen, not by the carbon skeleton. 1^∘ has one C-N bond, 2^∘ has two, and 3^∘ has three. The classification of the carbon bearing -NH2 (e.g. tert-butyl) is irrelevant for amine ordering.

1. (i) 1-methylcyclohexylamine: -NH2 on a ring carbon ⇒ only one C-N bond ⇒ 1^∘.
2. (ii) Triethylamine N(C2H5)3: three ethyl groups on N ⇒ 3^∘.
3. (iii) tert-butylamine (CH3)3C-NH2: still only one C-N bond ⇒ 1^∘ (the carbon is 3^∘, not the amine).
4. (iv) N-methylaniline C6H5-NH-CH3: two C-N bonds ⇒ 2^∘.

Only triethylamine has three carbons on N; option (ii).

Correct option: (iv) N-methylprop-2-en-1-amine.

Concept used. For an amine, the IUPAC name is built from the longest carbon chain bearing -NH- (or -NH2), numbered so that the amine carbon takes the lowest locant. Any group on N is indicated by an italic ``N-'' prefix. Double bonds get their own locant (``-en-'').

1. Parent chain: the three-carbon CH2=CH-CH2- piece (the N is attached to C-1, the =CH- to C-2).
2. Suffix: ``-amine'' on C-1 ⇒ ``prop-2-en-1-amine''.
3. Substituent on N: -CH3 ⇒ prefix ``N-methyl''.
4. Full name: N-methylprop-2-en-1-amine. The common name ``allylmethylamine'' is non-IUPAC.

N-methylprop-2-en-1-amine; option (iv).

Correct option: (iii) (CH3)2NH, dimethylamine.

Concept used. In aqueous medium the strongest base combines electron-donating alkyl groups (raise lone-pair availability) with good hydration of the conjugate ammonium ion. 2^∘ aliphatic amines balance these factors best, giving the well-known order 2^∘ > 1^∘ ≈ 3^∘ > NH3 ≫ aryl amines.

1. (i) CH3NH2: one +I methyl ⇒ moderately basic.
2. (ii) NCCH2NH2: -C#N is strongly -I ⇒ withdraws density off N ⇒ weak base.
3. (iii) (CH3)2NH: two +I methyls + two N-H bonds still solvate the ammonium ion well ⇒ strongest.
4. (iv) C6H5NHCH3: lone pair on N delocalises into the ring ⇒ aryl amine, weakest of the four.

(CH3)2NH is the strongest aqueous base; option (iii).

Correct option: (i) Aniline.

Concept used. Brnsted basicity of an amine is set by how available the N lone pair is for protonation. Aromatic amines like aniline drain the lone pair into the ring by resonance ⇒ much weaker base than alkylamines (where the lone pair is fully localised on N and is reinforced by +I).

1. Aniline: lone pair on N overlaps with the benzene ring; four resonance structures put N⁺ on the ring ⇒ lone pair is partly delocalised, basicity low (pK_b ≈ 9.4).
2. Piperidine (pK_b ≈ 2.9): cyclic 2^∘ aliphatic amine, very strong base.
3. Cyclohexylamine (pK_b ≈ 3.4): strong +I from cyclohexyl, strong base.
4. Methylamine (pK_b ≈ 3.4): one +I methyl, moderately strong.

Aniline is the weakest Br"onsted base among the four; option (i).

Correct option: (iii) C6H5CH2Br (benzyl bromide).

Concept used. S_N1 proceeds through a carbocation. The reaction rate scales with the stability of that cation: resonance-stabilised cations (benzyl, allyl) and 3^∘ > 2^∘ alkyl cations win; aryl halides cannot do S_N1 because phenyl cations are extremely unstable.

1. (i) CH3-X and (iv) C2H5-X: primary alkyl halides ⇒ cation is CH3⁺ or C2H5⁺, both highly unstable ⇒ S_N1 does not run; S_N2 pathway only.
2. (ii) C6H5-Br: aryl halide; the C-Br bond is sp², S_N1 would require a phenyl cation ⇒ impossible.
3. (iii) C6H5CH2-Br: ionises to the resonance-stabilised benzylic cation C6H5-CH2⁺ ⇒ excellent S_N1 substrate.

Benzyl bromide; option (iii).

Correct option: (ii) LiAlH4 in ether.

Concept used. Aryl nitro compounds Ar-NO2 are reduced to aryl amines Ar-NH2 by dissolving-metal reductants in acidic medium (Sn/HCl, Fe/HCl), or by catalytic hydrogenation (H2/Pt, Pd, Ni). LiAlH4 is unsuitable: with aromatic nitro groups it tends to stop at azo (Ar-N=N-Ar) or hydrazo (Ar-NH-NH-Ar) compounds rather than going cleanly to Ar-NH2.

1. Acidic dissolving-metal: Ar-NO2 + 3 Sn + 7 HCl -> Ar-NH3⁺ Cl^- + 3 SnCl2 + 2 H2O.
2. Catalytic hydrogenation over Pt: Ar-NO2 + 3 H2 -> Ar-NH2 + 2 H2O.
3. LiAlH4 delivers H^- but does not cleanly cleave the second N-O bond on an aryl nitro group; product mixtures (azo/hydrazo) result.

LiAlH4 is the poor choice; option (ii).

Correct option: (iii) Potassium cyanide (KCN).

Concept used. To add one carbon while installing a 1^∘ amine, the standard route is R-X KCN R-CN LiAlH4 R-CH2-NH2. The nitrile carbon becomes the new -CH2- in the amine. The other reagents give amines but with no chain extension.

1. R-X + KCN -> R-CN + KX (S_N2).
2. R-CN + 4 [H] LiAlH4 R-CH2-NH2 (one C added as the new -CH2- next to N).
3. NaNH2 would just deprotonate or give nitrene; NaN3 gives R-N3 -> R-NH2 with no chain extension; phthalimide (Gabriel) gives R-NH2 also with no chain extension.

KCN; option (iii).

Correct option: (iv) Potassium phthalimide.

Concept used. Gabriel synthesis converts a 1^∘ alkyl halide into a pure 1^∘ amine without over-alkylation. The nitrogen source is potassium phthalimide, made by reacting phthalimide with KOH. Its anion Phth-N^- does S_N2 on R-X; subsequent acid or base hydrolysis releases R-NH2 plus phthalic acid.

1. Deprotonation of phthalimide with KOH: Phth-NH + KOH -> Phth-N^- K⁺ + H2O
2. Alkylation by S_N2: Phth-N^- K⁺ + R-X -> Phth-N-R + KX
3. Hydrolysis with aqueous acid: Phth-N-R + H3O⁺ -> R-NH2 + phthalic acid (hydrazine NH2NH2 gives the same amine plus phthalhydrazide as the by-product).

Nitrogen comes from potassium phthalimide; option (iv).

Correct option: (iii) reductive amination of a 1^∘ amine with an aldehyde.

Concept used. Reductive amination is the cleanest laboratory route to 2^∘ amines: a 1^∘ amine condenses with an aldehyde to give an imine, and the imine is reduced catalytically to the 2^∘ amine. No over-alkylation is possible because the carbonyl supplies exactly one carbon fragment.

1. R-NH2 + R2-CHO -> R-N=CH-R2 + H2O (imine; R2 is a generic second alkyl group).
2. R-N=CH-R2 + H2 Pt R-NH-CH2-R2 (2^∘ amine).
3. (i) gives a mixture: R2NH, R3N, R4N⁺ side products.
4. (ii) gives R-CH2-NH2, a 1^∘ amine (not 2^∘).
5. (iv) Gabriel gives only 1^∘ amines (single alkylation on phthalimide).

Reductive amination of a 1^∘ amine with an aldehyde; option (iii).

Correct option: (iv) LiAlH4 in ether.

Concept used. Reduction of an amide R-CONH2 with LiAlH4 converts the carbonyl C=O into CH2 and preserves the C-N bond without loss of carbon. So the carbon count is unchanged –- 2-phenylpropanamide (C6H5-CH(CH3)-CONH2, 9 C) → 2-phenylpropanamine (C6H5-CH(CH3)-CH2-NH2, 9 C).

1. R-CONH2 + 4 [H] LiAlH4 R-CH2-NH2 + H2O.
2. (ii) Br2/NaOH is Hoffmann bromamide ⇒ would lose one C, giving 1-phenylethanamine (asked in Q11).
3. (i) Plain H2 does not reduce amides; (iii) is a P/I combination for halide reduction, not amides.

LiAlH4/ether; option (iv).

Correct option: (ii) NaOH/Br2 (Hoffmann bromamide degradation).

Concept used. Hoffmann bromamide converts R-CONH2 into R-NH2, dropping the carbonyl carbon as Na2CO3. Apply to 2-phenylpropanamide C6H5-CH(CH3)-CONH2 (9 C) and the product is C6H5-CH(CH3)-NH2 (8 C) = 1-phenylethanamine.

1. R-CONH2 + Br2 + 4 NaOH -> R-NH2 + Na2CO3 + 2 NaBr + 2 H2O.
2. Mechanism: N-bromination → deprotonation → α-elimination to isocyanate R-N=C=O → hydrolysis to R-NH2.
3. (iv) LiAlH4 gives 2-phenylpropanamine (same C count), not 1-phenylethanamine.

Hoffmann bromamide with NaOH/Br2; option (ii).

Correct option: (ii) ArCONH2, a primary aryl amide.

Concept used. The Hoffmann bromamide rearrangement requires a substrate with the -CONH2 functional group –- i.e. a primary amide. The two N–H bonds are essential (one is brominated, the other is deprotonated). Amines, nitro arenes and -CH2NH2 are not amides and do not undergo this rearrangement.

1. Generic Hoffmann: R-CONH2 + Br2 + 4 NaOH -> R-NH2 + Na2CO3 + 2 NaBr + 2 H2O.
2. Step 1: N-bromination of -CONH2 gives -CONHBr.
3. Step 2: deprotonation, α-elimination to acyl nitrene; alkyl/aryl group migrates from C to N.
4. Step 3: the isocyanate R-N=C=O hydrolyses to R-NH2 + CO2.
5. N-substituted amides (R-CONH-R2) lack the second N–H and do not undergo Hoffmann.

Only primary amide ArCONH2 undergoes Hoffmann; option (ii).

Correct option: (iv) II < I < III.

Concept used. Aromatic amine basicity is governed by how much electron density sits on N. Electron-donating para substituents (-CH3, +I, +H) push density onto N ⇒ stronger base. Electron-withdrawing groups (-NO2, -M, -I) pull density out of N via the ring ⇒ weaker base.

1. p-nitroaniline (II): -NO2 strongly withdraws by -M/-I ⇒ weakest.
2. Aniline (I): only the ring delocalisation drains N ⇒ intermediate.
3. p-toluidine (III): -CH3 donates by +I/hyperconjugation ⇒ strongest.

II < I < III; option (iv).

Correct option: (iii) CH3OH (methanol).

Concept used. 1^∘ aliphatic amines and HNO2 give an unstable aliphatic diazonium ion that fragments at once to N2 plus a carbocation; water traps the cation to give an alcohol. For methylamine the cation is CH3⁺ and the isolated organic product is methanol.

1. CH3NH2 + HNO2 -> CH3-N2⁺ + 2 H2O.
2. CH3-N2⁺ -> CH3⁺ + N2 (the gas).
3. CH3⁺ + H2O -> CH3OH + H⁺.

Methanol, CH3OH; option (iii).

Correct option: (ii) N2.

Concept used. Primary aliphatic amines react with HNO2 (generated in situ from NaNO2 + HCl) to give an unstable aliphatic diazonium salt R-N2⁺ X^-. Unlike aryl diazonium salts (which are stable at 0-5 ^∘C), the aliphatic one decomposes at once to release N2 and a carbocation that is trapped by water to give the corresponding alcohol.

1. CH3NH2 + HNO2 -> CH3-N2⁺ + 2 H2O (diazotisation).
2. Spontaneous: CH3-N2⁺ -> CH3⁺ + N2 (the gas evolved).
3. CH3⁺ + H2O -> CH3OH + H⁺ (final product methanol).

Brisk evolution of N2 gas confirms a 1^∘ aliphatic amine; option (ii).

Correct option: (iii) the nitronium ion NO2⁺.

Concept used. In the mixed-acid nitration of arenes, H2SO4 protonates HNO3 and water leaves to generate the nitronium ion NO2⁺. NO2⁺ is the actual electrophile that attacks the benzene ring.

1. HNO3 + H2SO4 -> H2NO3⁺ + HSO4^-.
2. H2NO3⁺ -> NO2⁺ + H2O.
3. C6H6 + NO2⁺ -> [arenium] -> C6H5-NO2 + H⁺.
4. NO, NO⁺ and NO2^- never enter the mechanism in this medium.

NO2⁺; option (iii).

Correct option: (iii) aromatic primary amine.

Concept used. Fe/HCl is a dissolving-metal acidic reductant that converts Ar-NO2 all the way to the 1^∘ aryl amine Ar-NH2. The same outcome as Sn/HCl or Zn/HCl, but cheaper ⇒ used industrially.

1. Overall: C6H5NO2 + 3 Fe + 7 HCl -> C6H5NH3⁺ Cl^- + 3 FeCl2 + 2 H2O, then base gives C6H5NH2.
2. FeCl2 hydrolyses (FeCl2 + 2 H2O -> Fe(OH)2 + 2 HCl), regenerating HCl ⇒ only catalytic acid is needed.

1^∘ aryl amine; option (iii).

Correct option: (ii) dimethylamine, (CH3)2NH.

Concept used. Reactivity towards dilute HCl tracks aqueous basicity. The well-known order is 2^∘ > 1^∘ ≈ 3^∘ > NH3 ≫ aryl amines. Dimethylamine (CH3)2NH sits at the top of that ladder in the methyl series.

1. (CH3)2NH: two +I methyls + one N-H for solvating H2N(CH3)2⁺ ⇒ strongest.
2. CH3NH2: one +I methyl + two N-H ⇒ moderate.
3. (CH3)3N: three +I methyls but zero N-H ⇒ poor cation solvation ⇒ weaker in water.
4. C6H5NH2: lone pair delocalised into ring ⇒ weakest.

(CH3)2NH is most reactive towards dilute HCl; option (ii).

Correct option: (i) an amide.

Concept used. Acylation of amines: the amine nitrogen attacks a carbonyl carbon of the acid anhydride, displacing carboxylate, to give an N-acyl amide. With a 1^∘ amine and acetic anhydride, the product is an N-substituted acetamide.

1. R-NH2 + (R2-CO)2O -> R-NH-CO-R2 + R2-COOH.
2. Example: C6H5-NH2 + (CH3CO)2O -> C6H5-NH-COCH3 + CH3COOH (acetanilide, the textbook example).
3. Pyridine is often added to mop up the carboxylic acid that is co-produced.

Amide (R-NH-CO-R2); option (i).

Correct option: (ii) Gatterman reaction.

Concept used. Two closely related transformations convert an aryl diazonium chloride into an aryl halide. Sandmeyer uses cuprous halide CuX in HX; Gatterman uses freshly precipitated copper powder in HX. The question shows plain Cu metal (not CuCl) above the arrow and emits CuCl as a product –- the textbook signature of the Gatterman variant.

1. Sandmeyer (with cuprous chloride in HCl): ArN2⁺ Cl^- + CuCl -> Ar-Cl + N2 + CuCl (Cu(I) shuttles an electron back and forth.)
2. Gatterman (with Cu metal in HCl; CuCl is generated in situ): ArN2⁺ Cl^- + Cu -> Ar-Cl + N2 + CuCl
3. The equation in the question matches the Gatterman pattern exactly.

Gatterman reaction; option (ii).

Correct option: (ii) Gabriel phthalimide synthesis.

Concept used. The question demands two things: (a) start from alkyl halide, (b) keep the same carbon count. Gabriel synthesis substitutes N from phthalimide onto R-X via S_N2 and then hydrolyses to R-NH2 –- the carbon skeleton from R-X is conserved exactly.

1. Reject (i) Hoffmann: starts from amide (R-CONH2), not R-X, and loses one C.
2. Reject (iii) Sandmeyer: starts from aryl diazonium, not alkyl halide.
3. (iv) NH3 + R-X works on alkyl halides but over-alkylates, giving a mix –- not the ``best'' method.
4. (ii) Gabriel: alkyl halide → pure 1^∘ amine, same C count, no over-alkylation.

Gabriel phthalimide synthesis; option (ii).

Correct option: (iv) Nitrobenzene.

Concept used. Azo coupling is an electrophilic aromatic substitution in which ArN2⁺ is a weak electrophile and so needs a strongly activated aromatic ring (e.g. -NH2, -OH, -OR). The ring in nitrobenzene is strongly deactivated by -NO2 (-M, -I) and refuses coupling.

1. Aniline (-NH2): strong activator ⇒ couples at p to give p-aminoazobenzene (yellow dye).
2. Phenol (-OH): strong activator ⇒ couples at p to give p-hydroxyazobenzene (orange dye).
3. Anisole (-OMe): activator (lone pair on O donates) ⇒ couples at p.
4. Nitrobenzene (-NO2): strongly deactivated ⇒ no coupling.

Nitrobenzene fails to couple; option (iv).

Correct option: (iii) Phenol.

Concept used. Basicity depends on (a) which heteroatom holds the lone pair (N less electronegative than O ⇒ N bases stronger), and (b) whether the lone pair is delocalised by an aromatic ring (-M via resonance ⇒ basicity drops sharply). Phenol combines both penalties –- oxygen and ring conjugation –- making it the weakest base in the set.

1. Cyclohexylamine: aliphatic amine, no resonance, lone pair on N ⇒ strongest (pK_b ≈ 3.4).
2. Aniline: aromatic amine, ring delocalisation drains lone pair ⇒ weaker (pK_b ≈ 9.4).
3. Cyclohexanol: aliphatic alcohol, O holds lone pair tightly ⇒ weaker base than amines.
4. Phenol: -OH on benzene ⇒ lone pair delocalised into ring; phenol is actually a weak acid (pK_a ≈ 10), not a base ⇒ weakest base.

Phenol is the weakest Br"onsted base; option (iii).

Correct option: (iv) Pyrrolidine.

Concept used. Pyrrolidine is a saturated 2^∘ aliphatic amine ⇒ lone pair on N is fully sp³ and available for protonation. Pyrrole is aromatic; its N lone pair sits in the aromatic π system and is not available for H⁺ acceptance. Aniline and NH3 are weaker than aliphatic amines.

1. Pyrrolidine: pK_b ≈ 2.9 –- a strong base (like piperidine).
2. NH3: pK_b ≈ 4.7.
3. Aniline: pK_b ≈ 9.4 (ring delocalisation drains lone pair).
4. Pyrrole: pK_b ≫ 14; using its N lone pair for H⁺ would destroy aromaticity ⇒ extremely weak base.

Pyrrolidine, option (iv), is by far the strongest base in the set.

Correct option: (i) NH2^- > OH^- > NH3 > H2O.

Concept used. Two trends operate together: (a) anionic forms (NH2^-, OH^-) are far stronger bases than the neutral parents (NH3, H2O), because the lone pair is unshared; (b) for the same charge type, nitrogen bases beat oxygen bases (N less electronegative ⇒ donates lone pair more readily).

1. NH2^-: anion on N ⇒ strongest base.
2. OH^-: anion on O ⇒ strong, but less so than NH2^- (O holds lone pair more tightly).
3. NH3: neutral N ⇒ moderate base (pK_b ≈ 4.7).
4. H2O: neutral O ⇒ weak base.

NH2^- > OH^- > NH3 > H2O; option (i).

Correct option: (ii) IV, propane.

Concept used. Volatility is the inverse of boiling point. Boiling point rises with the strength of intermolecular attractions –- the dominant one here is hydrogen bonding, which requires N-H or O-H. Propane has only weak London forces; trimethylamine has no N-H; methylethylamine has one N-H; n-propylamine has two N-H. So the BP ladder is propane < trimethylamine < methylethylamine < n-propylamine ⇒ propane is the most volatile.

1. Propane: BP -42 ^∘C (no H-bonding).
2. Trimethylamine (3^∘ amine): BP +3 ^∘C (no N-H).
3. N-methylethylamine (2^∘): BP ∼ 37 ^∘C (one N-H).
4. n-Propylamine (1^∘): BP 48 ^∘C (two N-H).

Propane (IV) is the most volatile; option (ii).

Correct option: (iii) Gabriel phthalimide synthesis.

Concept used. Audit each route for carbon-count change:

• (i) Nitrile reduction: R-C#N + 4 [H] -> R-CH2-NH2 –- the nitrile C becomes the new CH2 in the amine. Carbon count rises by 1 relative to the parent halide; relative to the nitrile, same. Depends on how you count ``reactant''.
• (ii) Amide reduction by LiAlH4: R-CONH2 -> R-CH2-NH2 –- same carbon count as the amide.
• (iii) Gabriel: Phth-NK + R-X -> R-NH2 –- same carbon count as the alkyl halide (the only new C comes from phthalimide, which is removed on hydrolysis).
• (iv) Hoffmann bromamide: R-CONH2 -> R-NH2 –- carbon count drops by 1.

1. NCERT key explicitly marks (iii) as the correct answer (Gabriel preserves the carbon count of the starting alkyl halide exactly).

Gabriel synthesis preserves the carbon count of the alkyl halide ⇒ option (iii).

Correct options: (iii) and (iv) –- iodobenzene and fluorobenzene.

Concept used. Sandmeyer's reaction needs the cuprous halide CuX (X = Cl, Br) to deliver X to the aryl radical generated from ArN2⁺. Cuprous iodide is not required: I^- from KI reacts directly with the diazonium salt –- so iodobenzene is made without Cu (not strictly ``Sandmeyer''). Cuprous fluoride doesn't work either; aryl fluorides use the Balz–Schiemann route via ArN2⁺ BF4^-.

1. Cl, Br: classic Sandmeyer with CuCl/HCl or CuBr/HBr.
2. I: needs only KI, no Cu; not a true Sandmeyer.
3. F: Sandmeyer fails; use HBF4 then heat (Balz–Schiemann).

Sandmeyer fails for Ar-I and Ar-F –- options (iii), (iv).

Correct options: (i), (ii) and (iii) –- Sn/HCl, Fe/HCl, and H2/Pd.

Concept used. Reducing C6H5-NO2 to C6H5-NH2 needs acidic dissolving-metal conditions (Sn/HCl, Fe/HCl, Zn/HCl) or catalytic hydrogenation (H2/Pt, Pd, Ni). Under basic/neutral conditions (Sn/NH4OH, LiAlH4), reduction halts at intermediate species (azoxybenzene, azobenzene, hydrazobenzene), not aniline.

1. C6H5NO2 + 3 Sn + 7 HCl -> C6H5NH3⁺ Cl^- + 3 SnCl2 + 2 H2O, then base gives C6H5NH2.
2. Fe/HCl is industrial: cheap, FeCl2 hydrolyses to regenerate HCl (catalytic acid).
3. Catalytic hydrogenation over Pd: C6H5NO2 + 3 H2 -> C6H5NH2 + 2 H2O.
4. Sn/NH4OH is basic ⇒ stops at azoxy/azo stage, not aniline.

(i), (ii), (iii) all give aniline; (iv) does not.

Correct options: (i) and (ii) –- R-NC and CHCl3.

Concept used. The carbylamine test for 1^∘ amines uses CHCl3 and alcoholic KOH. The dichlorocarbene CCl2 generated in situ reacts with R-NH2 to give the foul-smelling alkyl isocyanide (carbylamine) R-NC. Phosgene COCl2 and NaNO2/HCl are reagents of different reactions (Schotten-Baumann and diazotisation, respectively).

1. CHCl3 + KOH -> CCl2 + KCl + H2O (α-elimination).
2. R-NH2 + CCl2 + 2 KOH -> R-NC + 2 KCl + 2 H2O.
3. R-NC is the product whose obnoxious smell flags a 1^∘ amine.

Species involved: CHCl3 and R-NC; options (i), (ii).

Correct options: (ii) and (iii) –- ethanol and hypophosphorous acid.

Concept used. Replacing the -N2⁺ group of an aryl diazonium salt by an -H atom (i.e. deamination) is done with mild reducing agents that supply H^. or H^-. The two NCERT-listed reagents are ethanol (CH3CH2OH, which is oxidised to acetaldehyde) and hypophosphorous acid (H3PO2).

1. Ethanol: C6H5-N2⁺ Cl^- + CH3CH2OH -> C6H6 + N2 + HCl + CH3CHO.
2. Hypophosphorous acid: C6H5-N2⁺ Cl^- + H3PO2 + H2O -> C6H6 + N2 + H3PO3 + HCl.
3. SnCl2/HCl reduces Ar-N2⁺ to Ar-NH-NH2 (phenylhydrazine), not benzene. LiAlH4 is not the textbook reagent for this replacement.

Use CH3CH2OH or H3PO2; options (ii), (iii).

Correct options: (i) and (ii) –- p-bromoacetanilide (major) and o-bromoacetanilide (minor).

Concept used. Acetanilide C6H5-NHCOCH3 has the -NHCOCH3 group, a mild activator (lone pair on N delocalised into both the carbonyl and the ring), and an o/p-director with a strong steric bias for p. Br2/CH3COOH at room temperature gives mono-bromination at the activated o/p positions; over-bromination (as in aqueous Br2) does not occur in acetic acid solvent.

1. -NHCOCH3 activates the ring less strongly than -NH2 does (amide resonance into C=O steals some lone-pair density from the ring).
2. Bromination occurs only once ⇒ mono-substituted product.
3. Para is preferred over ortho on steric grounds (the bulky NHCOCH3 group blocks ortho approach).

Mainly p-bromoacetanilide with some o-bromoacetanilide; options (i), (ii).

Correct options: (i), (ii) and (iii) –- all three are valid arenium-ion (Wheland) intermediates for the bromination of aniline.

Concept used. An arenium ion (Wheland complex) forms when the electrophile Br⁺ adds to a ring carbon, generating a cyclohexadienyl cation. For an o/p-director like -NH2, the + charge can be delocalised onto the carbon bearing -NH2 via a resonance structure of type R2C=NH2⁺, giving an extra-stable iminium contributor at ortho and para positions only (not meta).

1. Ortho attack: Br goes to C-2; the + charge delocalises through three positions, one of which puts it on the N atom as =NH2⁺ at C-1 (structure (i)). Other resonance forms place + on C-3 or C-5 (structure (ii)).
2. Para attack: Br goes to C-4; the + charge delocalises and one resonance form again puts it on N as =NH2⁺ at C-1 (structure (iii)).
3. Meta attack (option (iv)) does not enjoy the iminium stabilisation ⇒ not a favoured arenium and not the answer.

Bromination via o- and p-arenium intermediates; options (i), (ii), (iii).

Correct options: (i) and (ii) –- isobutyl amine and 2-phenylethylamine.

Concept used. Gabriel synthesis works only when the alkyl halide can undergo S_N2 on the phthalimide anion. Two structural conditions: (a) the halide must be a 1^∘ (or 2^∘) alkyl halide, and (b) the product must be a primary amine (R-NH2). Aryl halides (C6H5-X) do not do S_N2 and 2^∘ amines (like N-methylbenzylamine) cannot come from a single alkylation of phthalimide.

1. (i) Isobutyl amine: from (CH3)2CH-CH2-Br (1^∘ alkyl halide) ⇒ Gabriel OK.
2. (ii) 2-Phenylethylamine: from C6H5CH2CH2-Br (1^∘ benzylic-adjacent) ⇒ Gabriel OK.
3. (iii) N-methylbenzylamine is 2^∘ (C6H5CH2-NH-CH3); Gabriel gives only 1^∘ amines ⇒ fails.
4. (iv) Aniline would need C6H5-Br + phthalimide anion ⇒ S_N2 on sp² carbon fails.

Only (i) and (ii) are accessible by Gabriel synthesis.

Correct options: (i) and (iii).

Concept used. Audit each equation against the standard reactivity rules:

• (i) Ammonolysis of 2^∘ alkyl halide with NH3 gives a 1^∘ amine ⇒ correct.
• (ii) Aqueous KOH on a 2^∘ alkyl halide gives the alcohol (substitution), not the alkene. Alkene formation needs alcoholic KOH (E2).
• (iii) Cyclohexyl chloride + alc. KOH → cyclohexene via E2 ⇒ correct.
• (iv) Isopropylamine + HNO2 gives isopropanol only at room temperature, not at 0 ^∘C as written for aryl diazotisation; the equation pattern is right (aliphatic diazonium → alcohol + N2), but the temperature label of 0 ^∘C is the giveaway for the aryl pattern. The NCERT key marks (iv) as incorrect because the equation includes the 0 ^∘C condition that belongs to aryl diazotisation.

1. (i): (CH3)2CHCl + 2 NH3 -> (CH3)2CHNH2 + NH4Cl –- one mole of NH3 is the nucleophile, the other neutralises HCl. Correct.
2. (ii): aqueous KOH supplies OH^-, the strong nucleophile in a polar medium. Substitution to give 2-propanol dominates; elimination is minor. The equation as written (forming the alkene) is wrong.
3. (iii): alcoholic KOH supplies OH^- in a less-solvating medium where the small base is forced to deprotonate β-H, giving cyclohexene by E2.
4. (iv): isopropylamine (1^∘ aliphatic) does give 2-propanol with HNO2, but it does so spontaneously at room temperature –- the 0 ^∘C label is reserved for aryl amines.

Correct equations: (i) and (iii).

Correct options: (i) and (ii) –- acylation (either reagent) first, then nitration gives p-nitroaniline as the major product.

Concept used. Direct nitration of aniline gives mostly the m-isomer because H2SO4 protonates -NH2 to the m-directing -NH3⁺. Acetylating the amine first (with CH3COCl or (CH3CO)2O in pyridine) converts -NH2 to the much less basic -NHCOCH3, which stays o/p-directing under the strong-acid nitration conditions and gives mainly p.

1. (i) Acetyl chloride/pyridine on aniline → acetanilide. Nitration of acetanilide → p-nitroacetanilide (major). Hydrolysis (acid) gives p-nitroaniline.
2. (ii) Acetic anhydride/pyridine → acetanilide; same outcome as (i).
3. (iii) Dil. HCl protonates aniline to anilinium (m-directing) before nitration ⇒ m-nitroaniline, not p.
4. (iv) Direct nitration: same anilinium-induced m-directing ⇒ m-nitroaniline mainly.

Acylate N first (any acyl chloride or anhydride) then nitrate; options (i), (ii).

Correct options: (i) and (ii) –- bromination of acetanilide and azo coupling.

Concept used. Electrophilic aromatic substitution (EAS) replaces a ring H with an electrophile while retaining aromaticity. The two-step arenium-ion mechanism applies. Reactions at the nitrogen of an amine (diazotisation, acylation) are not EAS even though aromatic amines are involved.

1. (i) Acetanilide + Br2/CH3COOH: Br⁺ attacks the ring at p ⇒ EAS.
2. (ii) Coupling: ArN2⁺ is the electrophile attacking another activated aromatic ring (phenol, aniline) ⇒ EAS.
3. (iii) Diazotisation: reaction at -NH2 (not on ring); NO⁺ attacks N ⇒ not EAS.
4. (iv) Acylation of aniline: (CH3CO)2O acylates N to give acetanilide ⇒ not EAS.

(i) and (ii) are EAS; (iii), (iv) react at N, not the ring.

Answer: HNO3 acts as a base in the nitrating mixture: it accepts H⁺ from H2SO4 and then loses water to generate the actual electrophile, the nitronium ion NO2⁺.

Concept used. Generating an electrophile strong enough to attack benzene requires the cooperation of two acids: H2SO4 is the stronger acid (proton donor) and HNO3 is the weaker acid; in their mutual presence, HNO3 behaves as a base.

1. Step 1 (acid-base): HNO3 + H2SO4 -> H2NO3⁺ + HSO4^-.
2. Step 2 (water loss): H2NO3⁺ -> NO2⁺ + H2O.
3. Step 3 (nitronium attack): C6H6 + NO2⁺ -> C6H5-NO2 + H⁺.
4. Overall: HNO3 provides the nitrogen by being protonated to give nitronium.

HNO3 acts as a base, supplying NO2⁺ (the nitronium electrophile) on protonation by H2SO4.

Answer: Acetylation converts the strongly activating but basic -NH2 into the mildly activating, non-basic -NHCOCH3. This prevents (a) protonation of -NH2 by H2SO4 to the m-directing -NH3⁺, and (b) oxidation of aniline by HNO3.

Concept used. Three problems arise if aniline is nitrated directly:

• H2SO4 protonates the basic -NH2 to -NH3⁺, which is strongly m-directing ⇒ mostly m-nitroaniline.
• HNO3 is a strong oxidant; aniline is electron-rich ⇒ partial oxidation gives tarry residues.
• Even if some unprotonated aniline reacts, over-nitration to 2,4,6- and beyond is possible.

1. Acetylation: C6H5-NH2 + (CH3CO)2O pyridine C6H5-NHCOCH3 + CH3COOH.
2. Acetanilide's N lone pair is partly delocalised into the carbonyl ⇒ N is much less basic (pK_b ≈ 13.6), so H2SO4 cannot protonate it.
3. -NHCOCH3 is still a mild activator and o/p-director; nitration gives mainly p-nitroacetanilide.
4. Acid hydrolysis: p-NO2-C6H4-NHCOCH3 + H3O⁺ -> p-NO2-C6H4-NH2 + CH3COOH.

Acetyl protection keeps N neutral, prevents oxidation, and steers nitration cleanly to the para position.

Product: Benzyl alcohol, C6H5CH2OH.

Concept used. Benzylamine is a 1^∘ aliphatic amine (the -NH2 sits on an sp³ benzylic carbon, not on the ring), so it follows the aliphatic-amine + HNO2 pattern: unstable diazonium ⇒ N2 escapes ⇒ H2O traps the carbocation ⇒ alcohol.

1. C6H5CH2NH2 + HNO2 -> C6H5CH2-N2⁺ + 2 H2O.
2. C6H5CH2-N2⁺ -> C6H5CH2⁺ + N2 (benzylic cation is resonance-stabilised, decomposition is fast).
3. C6H5CH2⁺ + H2O -> C6H5CH2OH + H⁺.

Benzyl alcohol, C6H5CH2OH, with brisk evolution of N2.

Answer: Lithium aluminium hydride, LiAlH4 in dry ether, or catalytic hydrogenation with sodium in alcohol (Mendius reduction).

Concept used. Nitrile reduction adds two H atoms across each of the two C-N π bonds, converting R-C#N into R-CH2-NH2 (a 1^∘ amine that has one more carbon than the parent halide).

1. R-C#N + 4 [H] LiAlH4 / ether R-CH2-NH2 (clean, single product).
2. Alternative: R-C#N + 4 [H] Na / C2H5OH R-CH2-NH2 (Mendius reduction).
3. Catalytic hydrogenation (H2/Ni, Pd, or Pt) also works but can over-reduce in some cases.

LiAlH4 in ether (or Na/C2H5OH) reduces nitriles cleanly to 1^∘ amines.

Answer: `A' is 3-nitrotoluene (m-nitrotoluene), CH3-C6H4-NO2 with NO2 at the meta position relative to CH3.

Concept used. Diazotisation at 0-5 ^∘C (273-278 K) converts Ar-NH2 to Ar-N2⁺; subsequent reduction with H3PO2/H2O replaces -N2⁺ by -H (deamination). The starting material has CH3 at C-4 and NO2 at C-2 relative to the NH2 group at C-1. After deamination at C-1, only CH3 and NO2 remain on the ring, CH3 at the original C-4 position and NO2 at the original C-2 –- but after dropping NH2, the ring is renumbered so that the substituents now end up meta to each other.

1. Diazotisation: Ar-NH2 + NaNO2 + 2 HCl -> Ar-N2⁺ Cl^- + NaCl + 2 H2O at 273-278 K.
2. Reduction: Ar-N2⁺ Cl^- + H3PO2 + H2O -> Ar-H + N2 + H3PO3 + HCl.
3. Ring map: C-1 was NH2, C-2 was NO2, C-4 was CH3. After removing NH2, the remaining substituents are NO2 and CH3 on positions that translate to 1,3 (meta).

`A' = 3-nitrotoluene (m-nitrotoluene), m-CH3-C6H4-NO2.

Answer: Hinsberg reagent is benzenesulphonyl chloride, C6H5-SO2Cl.

Concept used. The Hinsberg test uses C6H5SO2Cl to distinguish 1^∘, 2^∘ and 3^∘ amines. The reagent's S(=O)2 is a strong electrophile; an amine's lone pair displaces Cl^- to give a sulphonamide.

1. 1^∘ amine: R-NH2 + C6H5SO2Cl -> C6H5SO2NHR. The N-H is acidic (sulphonyl is strongly EW) ⇒ soluble in alkali.
2. 2^∘ amine: R2NH + C6H5SO2Cl -> C6H5SO2NR2. No N-H left ⇒ insoluble in alkali.
3. 3^∘ amine: no N-H to lose ⇒ does not react.

Hinsberg reagent = C6H5SO2Cl (benzenesulphonyl chloride); used to distinguish 1^∘/2^∘/3^∘ amines by alkali solubility.

Answer: Benzene diazonium chloride C6H5-N2⁺ Cl^- is thermally unstable above 5 ^∘C. On warming, even gently, it decomposes to give phenol, N2 and HCl (or chlorobenzene if dry); on prolonged storage even in solution it slowly decomposes by the same routes. So it must be prepared at 0-5 ^∘C and used immediately in the downstream reaction.

Concept used. The arenium-stabilised Ar-N#N⁺ cation is more stable than its aliphatic cousin (which dies on contact with water), but at higher temperatures the loss of N2 becomes spontaneous because N2 is an outstanding leaving group.

1. In water, > 5 ^∘C: C6H5-N2⁺ Cl^- + H2O -> C6H5-OH + N2 + HCl.
2. In dry solid form, on standing: C6H5-N2⁺ Cl^- -> C6H5-Cl + N2 (the dry salt can even detonate).

Thermal instability of Ar-N2⁺; prepared cold (0-5 ^∘C) and used at once to avoid loss of N2 to give phenol or chlorobenzene.

Answer: The N lone pair of aniline, originally free to push into the ring (strong +M activator), is now shared with the carbonyl O via the amide resonance R-NH-CO-R2 <-> R-NH⁺=C(O^-)-R2. The lone pair becomes less available to the ring ⇒ ring activation drops sharply.

Concept used. The amide carbonyl is a π-acceptor that competes with the benzene ring for the nitrogen lone pair. Resonance with C=O is energetically more favourable than resonance with the benzene ring (carbonyl is more polarisable), so most of the lone pair is locked into the N-C=O system.

1. In aniline: lone pair N -> ring (+M); ring activation ≈ 10⁶× benzene.
2. In acetanilide: lone pair is shared between N -> ring and N -> C=O; the carbonyl wins the larger share.
3. Net: NHCOCH3 is still an activator (mild) but much weaker than NH2 (strong).

Amide resonance with C=O pulls the lone pair off the ring, dropping the activating power of -NHCOCH3 relative to -NH2.

Concept used. Basicity (pK_b) is measured by how readily the heteroatom's lone pair grabs H⁺. The more available the lone pair (low electronegativity, good solvation of the resulting cation), the stronger the base.

1. Electronegativities: χ(N) = 3.04, χ(O) = 3.44.
2. Since O holds its lone pair more tightly than N, methylamine's N donates its lone pair to H⁺ more readily than methanol's O does.
3. Conjugate-acid view: CH3NH3⁺ (pK_a ≈ 10.6) is much weaker than CH3OH2⁺ (pK_a ≈ -2). Weaker conjugate acid ⇒ stronger base.

N is less electronegative than O, so its lone pair is more available for protonation; hence MeNH2 is the stronger base.

Answer: Pyridine acts as a base that neutralises the HCl (or CH3COOH) liberated during acylation, keeping the amine deprotonated (and therefore nucleophilic) and pushing the equilibrium towards the amide product. Pyridine also catalyses the reaction via a nucleophilic-catalysis pathway (N-acyl pyridinium intermediate).

Concept used. If the by-product acid (HCl from acyl chloride or CH3COOH from anhydride) is allowed to accumulate, it would protonate the amine to R-NH3⁺, which has no lone pair available for further attack and so the acylation stalls. Pyridine sops up that acid as C5H5N · H⁺ X^-.

1. Acylation: R-NH2 + R2-COCl -> R-NH-CO-R2 + HCl.
2. Pyridine traps HCl: HCl + C5H5N -> C5H5NH⁺ Cl^-.
3. Without pyridine: HCl + R-NH2 -> R-NH3⁺ Cl^-, the salt is unreactive.

Pyridine is the acid scavenger (and a mild nucleophilic catalyst); it keeps the amine free and pulls the acylation forward.

Answer: Mildly acidic (pH ≈ 4-5) –- or equivalently very weakly basic but not alkaline.

Concept used. Two species must coexist for the coupling to work: (a) the aryl diazonium cation ArN2⁺ (stable only below 5 ^∘C and in mildly acidic medium; strong base destroys it as diazoate Ar-N=N-O^-), and (b) the nucleophilic free amine ArNH2 (lost under strongly acidic conditions because -NH2 is protonated to the unreactive -NH3⁺).

1. In strong acid: ArNH2 + H⁺ -> ArNH3⁺, no lone pair, no coupling.
2. In strong base: ArN2⁺ + OH^- -> Ar-N=N-OH -> Ar-N=N-O^- (diazoate), no electrophile.
3. At pH ∼ 4-5 (mildly acidic / weakly basic), enough free aniline (lone pair) coexists with enough ArN2⁺ to react.

Mildly acidic conditions (pH 4–5); the medium keeps both partners alive for coupling.

Product: A mixture of 2-bromoaniline and 4-bromoaniline (mainly p-bromoaniline).

Concept used. Aniline is strongly activating; in aqueous bromine water it gives 2,4,6-tribromoaniline at once. In a non-polar solvent (CS2) the -NH2 is not protonated and the reaction is gentler –- bromination stops at the monosubstituted stage, giving mainly p-bromoaniline plus some o-bromoaniline.

1. Generate Br⁺: Br2 polarises on approach to the electron-rich ring.
2. EAS at o/p: -NH2 is an o/p-director by +M.
3. In CS2, only one Br installs ⇒ mono-bromination at o and p.

Mixture of 2-bromoaniline and 4-bromoaniline (p-product dominant).

Order: CH3CH2CH3 < CH3CH2NH2 < CH3CH2OH.

Concept used. Dipole moment μ scales with the electronegativity difference across the polar bond. χ(C) = 2.55, χ(N) = 3.04, χ(O) = 3.44. The C-O dipole (Δχ = 0.89) is much larger than the C-N dipole (Δχ = 0.49); the C-C bond in propane has Δχ = 0, so propane is essentially apolar.

1. Propane: only C-H and C-C bonds; μ ≈ 0.08 D.
2. Ethylamine: C-N and N-H polar bonds; μ ≈ 1.22 D.
3. Ethanol: C-O and O-H polar bonds; μ ≈ 1.69 D.

CH3CH2CH3 < CH3CH2NH2 < CH3CH2OH.

Answer: Structure –- CH2=CH-CH2-NH2. IUPAC name –- prop-2-en-1-amine.

Concept used. The common name ``allyl'' is the CH2=CH-CH2- group; adding -NH2 gives allylamine. To name it by IUPAC: longest chain that includes the amine carbon is propene; amine on C-1, double bond between C-2 and C-3 ⇒ prop-2-en-1-amine.

1. Structural drawing: H2C=CH-CH2-NH2 (3 carbons, terminal NH2, terminal C=C).
2. Functional priority: amine > alkene, so the chain is numbered from the amine end.
3. C-1: CH2-NH2; C-2 and C-3: CH=CH2.
4. Locants: ``2-en'' for the double bond, ``1-amine'' for -NH2 ⇒ prop-2-en-1-amine.

Allyl amine = CH2=CH-CH2-NH2 = prop-2-en-1-amine.

Answer: N,N-dimethylbenzenamine (also written N,N-dimethylaniline).

Concept used. For an aryl amine, the IUPAC name uses ``benzenamine'' (or the retained name ``aniline'') as the parent, with N-substituents prefixed by italic ``N-''. The compound C6H5-N(CH3)2 has two methyl groups on nitrogen ⇒ ``N,N-dimethylbenzenamine''.

1. Parent: benzene with -NH2 = benzenamine (or aniline).
2. N-substituents: two methyls on nitrogen ⇒ ``N,N-dimethyl''.
3. Full IUPAC name: N,N-dimethylbenzenamine; equivalent common name: N,N-dimethylaniline.

N,N-dimethylbenzenamine (= N,N-dimethylaniline).

Answer: Z is ethylmethylamine, CH3-NH-C2H5, a 2^∘ amine.

Concept used. Behaviour with the Hinsberg reagent (C6H5SO2Cl) is a fingerprint of amine class. (a) 1^∘ amine gives an alkali-soluble sulphonamide (acidic N-H). (b) 2^∘ amine gives an alkali-insoluble sulphonamide (no N-H). (c) 3^∘ amine does not react.

1. Molecular formula C3H9N has degree of unsaturation = 0, so Z is acyclic and saturated.
2. Possible structures: n-C3H7-NH2 (1^∘), (CH3)2CH-NH2 (1^∘), CH3-NH-C2H5 (2^∘), (CH3)3N (3^∘).
3. ``Reacts with Hinsberg'' ⇒ not 3<sup>∘</sup>. ``Solid insoluble in alkali'' ⇒ 2^∘ amine.
4. The only C3H9N 2^∘ amine is ethylmethylamine CH3-NH-C2H5.

Z is ethylmethylamine, CH3-NH-C2H5, which gives N-ethyl-N-methylbenzene sulphonamide (alkali-insoluble).

Method: Use the carbylamine reaction followed by catalytic hydrogenation. R-NH2 CHCl3 / KOH R-NC H2 / Pd R-NHCH3

Concept used. The carbylamine route gives the 2^∘ amine R-NH-CH3 in two clean steps: (a) formation of alkyl isocyanide R-NC from R-NH2 with CHCl3/alc. KOH (the two N-H are replaced by a single N=C bond); (b) catalytic hydrogenation of R-NC over Pd reduces the N=C bond to N-CH2- =N-CH3 (because the carbon already carries no other H), giving the 2^∘ amine without any further alkylation possible.

1. Step 1 (carbylamine): R-NH2 + CHCl3 + 3 KOH -> R-NC + 3 KCl + 3 H2O.
2. Step 2 (reduction): R-NC + 2 H2 Pd R-NH-CH3.
3. Net: R-NH2 -> R-NH-CH3, exactly one methyl added on N, no 3^∘ or 4^∘ side products.

R-NH2 CHCl3/KOH R-NC H2/Pd R-NHCH3 stops cleanly at the 2^∘ amine.

Product: p-Hydroxyazobenzene (when Ar = C6H5), the orange azo dye HO-C6H4-N=N-C6H5.

Concept used. Azo coupling: in mild alkaline medium, phenol is deprotonated to the highly activated phenoxide ion C6H5O^-, which attacks the weak electrophile ArN2⁺ at its para position to give an azo compound. The orange/yellow colour comes from the extended π-conjugated Ar-N=N-Ar2 chromophore.

1. Deprotonation: C6H5OH + OH^- -> C6H5O^- + H2O.
2. EAS coupling: ArN2⁺ + C6H5O^- -> p-HO-C6H4-N=N-Ar + H⁺.
3. For the generic aryl, the product is p-aryloxoazobenzene; for Ar = C6H5, p-hydroxyazobenzene.

Product = p-hydroxyazobenzene, an orange azo dye, formed by para coupling of phenoxide with aryl diazonium cation.

Concept used. Although aniline itself is a sparingly soluble oily liquid (the -NH2 lone pair is partly delocalised into the ring), its conjugate acid anilinium chloride C6H5NH3⁺ Cl^- is a fully ionic salt –- it dissolves in water just as ammonium chloride does.

1. C6H5NH2 + HCl -> C6H5NH3⁺ Cl^- (proton transfer).
2. The anilinium cation is hydrated by water (H-bonds to the three N-H bonds and ion–dipole attractions).
3. Cl^- is independently hydrated ⇒ the salt dissolves freely in aqueous HCl.

Aniline + HCl gives the water-soluble salt anilinium chloride, hence the apparent solubility in aqueous HCl.

Strategy. Hoffmann bromamide → carbylamine → catalytic hydrogenation.

1. [label=()]
2. c-C6H11-CONH2 Br2/KOH c-C6H11-NH2 (Hoffmann, -1 C).
3. c-C6H11-NH2 CHCl3/KOH c-C6H11-NC (carbylamine).
4. c-C6H11-NC H2/Pd c-C6H11-NH-CH3 (reduction).

Concept used. Two named reactions string together: (a) Hoffmann bromamide chops the carbonyl off the amide and gives the 1^∘ amine cyclohexylamine; (b) Carbylamine grafts a single methyl-equivalent (-NC) onto N which is then reduced to -NH-CH3. Cleaner than direct methylation, which over-alkylates.

1. Step 1: R-CONH2 + Br2 + 4 KOH -> R-NH2 + K2CO3 + 2 KBr + 2 H2O where R = cyclohexyl.
2. Step 2: R-NH2 + CHCl3 + 3 KOH -> R-NC + 3 KCl + 3 H2O.
3. Step 3: R-NC + 2 H2 Pd R-NH-CH3.

Cyclohexanecarboxamide Br2/KOH cyclohexylamine CHCl3/KOH cyclohexylisocyanide H2/Pd N-methylcyclohexylamine.

Answers.

• A: 2-(2-cyanoethyl)cyclohexan-1-one –- the -CH2-CH2-Cl tail of the substrate has been converted to -CH2-CH2-CN by S_N2 displacement of Cl^- by CN^-.
• B: 2-(3-aminopropyl)cyclohexan-1-one –- the nitrile of A has been reduced to a primary amine -CH2-CH2-CH2-NH2 (the chain has gained one CH2).

Concept used. Two single-step transformations: S_N2 with KCN on a primary alkyl chloride gives a nitrile (chain +1 C); H2/Pd reduces the nitrile to a 1^∘ amine (with no further change in carbon count).

1. Substitution: R-CH2-Cl + KCN -> R-CH2-CN + KCl (S_N2).
2. Reduction: R-CH2-CN + 2 H2 Pd R-CH2-CH2-NH2.
3. The cyclohexanone C=O does not reduce under H2/Pd at room temperature with normal pressure ⇒ ketone survives.

A = 2-(2-cyanoethyl)cyclohexan-1-one; B = 2-(3-aminopropyl)cyclohexan-1-one.

Part (i): Toluene → p-toluidine.

1. [label=()]
2. Nitration: C6H5CH3 HNO3 / H2SO4 p-O2N-C6H4-CH3 (with some o-isomer; separate by fractional crystallisation).
3. Reduction: p-O2N-C6H4-CH3 Fe / HCl p-H2N-C6H4-CH3 (p-toluidine).

Part (ii): p-toluidine diazonium chloride → p-toluic acid (p-CH3-C6H4-COOH).

1. [label=()]
2. Sandmeyer (with CuCN/KCN): p-CH3-C6H4-N2⁺ Cl^- + CuCN -> p-CH3-C6H4-CN + N2 + CuCl.
3. Acid hydrolysis: p-CH3-C6H4-CN + 2 H2O H⁺ / heat p-CH3-C6H4-COOH + NH3.

Concept used. Both halves are textbook: aromatic nitration then reduction (for NH2); Sandmeyer for N2⁺ -> CN then hydrolysis (for CN -> COOH).

(i) Toluene HNO3/H2SO4 p-nitrotoluene Fe/HCl p-toluidine. (ii) p-Toluidine diazonium chloride CuCN/KCN p-tolunitrile H2O/H⁺, Δ p-toluic acid.

Part (i): Nitrobenzene → acetanilide.

1. [label=()]
2. Reduce nitrobenzene to aniline: C6H5-NO2 Sn / HCl C6H5-NH2.
3. Acylate aniline with acetic anhydride in pyridine: C6H5-NH2 + (CH3CO)2O pyridine C6H5-NHCOCH3 + CH3COOH.

Part (ii): Acetanilide → p-nitroaniline.

1. [label=()]
2. Nitrate with mixed acid: C6H5-NHCOCH3 HNO3 / H2SO4 p-O2N-C6H4-NHCOCH3.
3. Acid hydrolysis: p-O2N-C6H4-NHCOCH3 + H2O H⁺ p-O2N-C6H4-NH2 + CH3COOH.

Concept used. Part (i) reduces nitrobenzene to aniline and acetylates. Part (ii) exploits acetanilide's o/p-directing -NHCOCH3 to control nitration to the para position, then hydrolyses off the acetyl group.

Sequences shown above. The protecting acetyl group is the trick that makes Part (ii) selective.

Major product: the azo dye coupling p-nitrophenyl diazonium with phenol, i.e. 4-nitro-4'-hydroxyazobenzene p-O2N-C6H4-N=N-C6H4-OH-p, not the p-toluene version.

Concept used. Azo coupling is electrophilic aromatic substitution; rate scales with the electrophilicity of the diazonium cation. -NO2 at the para position of the diazonium is strongly -M/-I and withdraws electron density, making the diazonium more electrophilic. -CH3 in the toluene-diazonium is mildly +I and weakly donates, lowering the electrophilicity. So the nitro-substituted diazonium wins the race.

1. Phenol in alkali → phenoxide (the nucleophile).
2. Two competing electrophiles in solution: p-O2N-C6H4-N2⁺ (more electrophilic) and p-CH3-C6H4-N2⁺ (less electrophilic).
3. Phenoxide couples preferentially with the stronger electrophile ⇒ p-nitrophenyldiazonium wins.
4. Product: p-O2N-C6H4-N=N-C6H4-OH, p-hydroxy-p'-nitroazobenzene (orange-red dye).

Phenol preferentially couples with p-nitrophenyldiazonium chloride because -NO2 raises the electrophilicity of the diazonium cation; product is 4-hydroxy-4'-nitroazobenzene.

Strategy. Use Br2/CH3COOH to brominate the ring of p-nitroaniline (the activating NH2 directs Br to its two ortho positions, here both meta to the para-NO2), then diazotise and remove the amine via H3PO2 to leave the three bromines on the ring.

1. [label=()]
2. Bromination at both ortho positions to NH2: p-O2N-C6H4-NH2 + 2 Br2 CH3COOH 2,6-dibromo-4-nitroaniline (the bromines sit at C-2 and C-6 of the aniline numbering, i.e. both ortho to NH2 and meta to NO2).
3. Diazotisation at 0-5 ^∘C: Ar-NH2 + NaNO2 + 2 HCl -> Ar-N2⁺ Cl^- + 2 H2O (Ar = 2,6-Br₂,4-NO₂-C₆H₂).
4. Sandmeyer with Cu2Br2/HBr to install the third Br at the position vacated by -N2⁺: Ar-N2⁺ Cl^- + Cu2Br2 -> Ar-Br + N2 + CuCl + CuBr giving 3,4,5-tribromonitrobenzene (positions 3,4,5 when numbered from the nitro carbon: Br, Br, Br on C-3, C-4, C-5, NO2 on C-1).

p-Nitroaniline Br2/CH3COOH 2,6-dibromo-4-nitroaniline NaNO2/HCl, 0 ^∘C diazonium Cu2Br2/HBr 3,4,5-tribromonitrobenzene.

Strategy. Direct nitration of aniline gives mainly the meta product (because -NH3⁺ formed in strong acid is m-directing). We therefore (a) install nitro first, (b) reduce to aniline, (c) protect -NH2 as acetanilide so that the free pair on N is calmed and o/p is selective, (d) nitrate to get the p-acetamido-nitro arene, (e) hydrolyse back to -NH2.

1. Nitration of benzene with conc. HNO3 and H2SO4: C6H6 + HNO3 -> C6H5-NO2 + H2O
2. Reduction of nitrobenzene with Sn and HCl: C6H5-NO2 + 6 [H] -> C6H5-NH2 + 2 H2O
3. Acetylation with acetic anhydride in pyridine (the -NHCOCH3 group is still activating and p-directing, but no longer protonated by H2SO4): C6H5-NH2 + (CH3CO)2O -> C6H5-NHCOCH3 + CH3COOH
4. Nitration of acetanilide with HNO3/H2SO4 gives clean para selectivity: C6H5-NHCOCH3 + HNO3 -> p-NO2-C6H4-NHCOCH3 + H2O
5. Acid hydrolysis of the acetyl group with H3O⁺: p-NO2-C6H4-NHCOCH3 + H2O -> p-NO2-C6H4-NH2 + CH3COOH

Benzene → nitrobenzene → aniline → acetanilide → p-nitroacetanilide → p-nitroaniline.

Strategy. Diazotise the amine to Ar-N2⁺, install -NO2 via Balz-Schiemann + radical pathway (but the standard NCERT key is even simpler): replace -N2⁺ by -NO2 using Sandmeyer-like conditions with NaNO2/Cu, then brominate at the now-activated meta position relative to -NO2.

1. [label=()]
2. Aniline + HNO2 at 273-278 K → benzenediazonium chloride C6H5-N2⁺ Cl^-.
3. C6H5-N2⁺ Cl^- + HBF4 → C6H5-N2⁺ BF4^- (diazonium tetrafluoroborate, more stable for isolation).
4. C6H5-N2⁺ BF4^- + NaNO2/Cu, heat → C6H5-NO2 + N2 (with copper by-products). NCERT route: -N2⁺ replaced by -NO2.
5. C6H5-NO2 + Br2/FeBr3 → m-Br-C6H4-NO2. The -NO2 is a strong m-director.

Concept used. Aniline cannot be directly nitrated at the meta position because -NH2 (or -NH3⁺) does not direct that way usefully. Going via the diazonium and converting -N2⁺ to -NO2 first removes the -NH2 ``label'' entirely; then bromination of nitrobenzene installs -Br at the meta position (since -NO2 is m-directing).

1. Step 1 & 2: convert -NH2 to -N2⁺ BF4^- for stability.
2. Step 3: replace -N2⁺ by -NO2 via NaNO2/Cu.
3. Step 4: bromination is directed by -NO2 to meta ⇒ m-bromonitrobenzene.

Aniline → C6H5N2⁺ Cl^- → C6H5N2⁺ BF4^- → C6H5NO2 → m-bromonitrobenzene (via Br2/FeBr3).

Part (i): Aniline → 3,5-dibromonitrobenzene.

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2. Aniline (CH3CO)2O / pyridine acetanilide.
3. Acetanilide HNO3 / H2SO4 p-nitroacetanilide.
4. Acid hydrolysis: p-nitroacetanilide H3O⁺ p-nitroaniline.
5. Bromination at the two ortho positions of NH2 (which are meta to NO2): p-nitroaniline Br2 / CH3COOH 2,6-dibromo-4-nitroaniline.
6. Diazotisation: 2,6-dibromo-4-nitroaniline NaNO2 / HCl, 273-278 K diazonium chloride.
7. Deamination with H3PO2: H3PO2 / H2O 3,5-dibromonitrobenzene (the amine is replaced by -H).

Part (ii): Aniline → 3,5-dibromo-4-iodonitrobenzene (one extra I relative to part (i)).

1. [label=()]
2. Follow Part (i) steps (a)–(e) to get the 2,6-dibromo-4-nitrobenzenediazonium chloride.
3. Replace -N2⁺ by -I using KI (no Cu required): diazonium KI 3,5-dibromo-4-iodonitrobenzene.

Concept used. The acetyl-protect-and-direct strategy installs NO2 at the para position. After deprotection, bromination installs two Br's at the ortho positions of NH2. The amine is then either deaminated (Part i) or replaced by iodide (Part ii) via the diazonium.

(i) Acylate → nitrate → hydrolyse → dibrominate → diazotise → H3PO2 deamination. (ii) Same first five steps; replace deamination with KI to install iodine.

Matches: (i) → (d), (ii) → (c), (iii) → (a), (iv) → (b).

Concept used. Each named reaction has a single defining description.

1. Ammonolysis: alkyl halide with ammonia, R-X + NH3 -> R-NH2 + HX (over-alkylates to give 1^∘/2^∘/3^∘/4^∘ mix) ⇒ matches (d).
2. Gabriel synthesis: phthalimide + KOH gives Phth-N^- K⁺, then + R-X and hydrolysis ⇒ matches (c).
3. Hoffmann bromamide: a primary amide reacts with bromine in aqueous NaOH to give a one-carbon-shorter primary amine, R-CONH2 + Br2 + 4 NaOH -> R-NH2 + Na2CO3 + 2 NaBr + 2 H2O ⇒ matches (a).
4. Carbylamine test: 1^∘ amine + CHCl3 + KOH → R-NC (foul-smelling isocyanide) –- positive only for 1^∘ amines ⇒ matches (b).

(i)→(d), (ii)→(c), (iii)→(a), (iv)→(b).

Matches: (i) → (b), (ii) → (a), (iii) → (d), (iv) → (c).

Concept used. Each compound has a single signature property in the list.

1. Benzenesulphonyl chloride (C6H5SO2Cl) is the literal Hinsberg reagent ⇒ matches (b).
2. Sulphanilic acid H2N-C6H4-SO3H exists in a zwitterionic tautomer H3N⁺-C6H4-SO3^- where the protonated amine balances the deprotonated sulphonate. This zwitterion is responsible for its high mp (∼ 288 ^∘C) and insolubility in non-polar solvents ⇒ matches (a).
3. Alkyl diazonium salts R-N2⁺ decompose at once (because the cation is not aromatic-stabilised), giving alcohols on water trapping ⇒ matches (d).
4. Aryl diazonium salts Ar-N2⁺ are stable at 0-5 ^∘C and undergo azo coupling with activated arenes (phenol, aniline) to give azo dyes (orange/red) ⇒ matches (c).

(i)→(b), (ii)→(a), (iii)→(d), (iv)→(c).

Correct option: (iii) A is correct, R is wrong.

Concept used. Acylation of an amine stops at the monosubstituted product because the freshly formed amide nitrogen R-NH-CO-R2 is much less nucleophilic than the starting amine (the carbonyl drains N via -M). Steric hindrance is a minor factor at best. The textbook reason is electronic, not steric.

1. Alkylation: R-NH2 + R2-X -> R-NH-R2, then R-NH-R2 is even more nucleophilic than R-NH2 (extra +I) ⇒ second alkylation runs faster ⇒ polysubstitution.
2. Acylation: R-NH2 + R2-COX -> R-NH-CO-R2, then R-NH-CO-R2 has its lone pair tied up in amide resonance ⇒ much less nucleophilic ⇒ second acylation is slow.
3. Reason as stated cites steric hindrance only –- this is not the dominant factor; the dominant factor is the loss of nucleophilicity.

Option (iii) –- assertion correct, reason wrong (the reason is electronic, not steric).

Correct option: (iii) –- A is correct, but R is wrong.

Concept used. Hoffmann bromamide converts a 1^∘ amide (R-CONH2) to a 1^∘ amine (R-NH2) by treatment with Br2/NaOH –- the substrate is an amide, not an amine. The assertion as worded is acceptable in the NCERT key (``given by primary amines'' here means the product is a 1^∘ amine; one-carbon shorter). The reason is factually wrong: in aqueous medium 2^∘ > 1^∘ in basicity.

1. Hoffmann: R-CONH2 + Br2 + 4 NaOH -> R-NH2 + Na2CO3 + 2 NaBr + 2 H2O.
2. Mechanism: N-bromination, deprotonation, α-elimination to isocyanate R-N=C=O, hydrolysis to amine. Only R-CONH2 (not the N-substituted amide R-CONHR) has the two acidic N–H needed.
3. R: aqueous basicity is 2^∘ > 1^∘ ≈ 3^∘ –- the reason is false.

Option (iii) –- assertion correct (per NCERT key), reason incorrect.

Correct option: (iv) both correct; R correctly explains A.

Concept used. N-ethylbenzenesulphonamide is C6H5-SO2-NH-C2H5 –- the Hinsberg product from a 1^∘ amine (EtNH2). The remaining N-H is rendered acidic by the two strongly electron-withdrawing S=O groups, so NaOH deprotonates it to a soluble sodium salt C6H5-SO2-N^- C2H5 Na⁺.

1. Resonance: R-SO2-N^- -R2 <-> R-S(=O)(-O^-)-N-R2 stabilises the conjugate base by delocalisation of the negative charge onto the sulphonyl oxygens.
2. pK_a of sulphonamide N-H is ∼ 10 (a weak acid, comparable to phenol), low enough for NaOH to deprotonate.
3. Sodium salt is ionic ⇒ water-soluble.

Option (iv): A correct, R correct, R explains A.

Correct option: (ii) both correct, but R is not the correct explanation of A.

Concept used. N,N-Diethylbenzenesulphonamide C6H5-SO2-N(Et)2 comes from a 2^∘ amine (Et2NH). The nitrogen has no N-H to lose, so NaOH has nothing to deprotonate and the sulphonamide stays insoluble in alkali. The fact that the sulphonyl group is electron-withdrawing is true, but that property would only matter if there were an N-H to acidify; since there isn't, that ``reason'' is real but irrelevant to the insolubility.

1. Structure: C6H5-SO2-N(C2H5)2 –- two ethyls on N, no N-H.
2. In NaOH: no acidic proton to remove ⇒ no salt formation ⇒ insoluble.
3. R is a true statement (sulphonyl group is indeed EW), but it does not explain why this particular sulphonamide is insoluble. The actual cause is the absence of N-H.

Option (ii): A correct, R correct (true statement), but R does not explain A.

Correct option: (iv) both correct; R correctly explains A.

Concept used. In the industrial Bechamp reduction (Ar-NO2 -> Ar-NH2) with Fe scrap, HCl and steam, HCl is effectively catalytic: the FeCl2 formed in the reduction hydrolyses with steam to give back HCl and Fe(OH)2. So only a small starting charge of HCl is needed –- it keeps getting regenerated.

1. Reduction: Ar-NO2 + 3 Fe + 6 HCl -> Ar-NH2 + 3 FeCl2 + 2 H2O.
2. Hydrolysis: FeCl2 + 2 H2O -> Fe(OH)2 + 2 HCl (steam-assisted).
3. Net effect: HCl from step 1 is regenerated in step 2 ⇒ HCl is catalytic, not stoichiometric.

Option (iv): A correct, R correct, R explains A. HCl is effectively catalytic via FeCl2 hydrolysis.

Correct option: (i) both A and R are wrong.

Concept used. Gabriel synthesis fails for aromatic amines because aryl halides (Ar-X) do not undergo S_N2 (the sp² carbon has no backside lobe; the ring π system blocks attack). So:

• Assertion is wrong –- aromatic 1^∘ amines (e.g. aniline) cannot be prepared by Gabriel.
• Reason is wrong –- aryl halides do not undergo S_N2 with phthalimide anion (or with most nucleophiles), they only undergo nucleophilic aromatic substitution under specific activating conditions (strong EW groups on the ring), which is not relevant here.

1. Gabriel substrate scope: 1^∘ and 2^∘ alkyl halides only.
2. Aryl halide C6H5-X + phthalimide anion → no reaction.
3. Aniline is therefore made by other routes (e.g. reduction of nitrobenzene).

Option (i) –- both A and R are wrong. Gabriel cannot make aromatic amines because aryl halides don't S_N2.

Correct option: (iv) –- both correct, and R correctly explains A.

Concept used. Basicity of an amine tracks electron density on N. Acetylation converts Ar-NH2 (aniline) to Ar-NH-COCH3 (acetanilide). The acetyl carbonyl is a strong π-acceptor: the N lone pair delocalises onto the carbonyl O (N-C=O ↔ N⁺=C-O^-). Density on N drops ⇒ basicity drops.

1. Aniline pK_b ≈ 9.4 (i.e. a weak base).
2. Acetanilide pK_b ≈ 13.6 –- about 4 orders of magnitude weaker.
3. Cause: the amide resonance -NH-CO-CH3 ↔ -NH⁺=C(O^-)-CH3 drains the lone pair from N onto O.

Option (iv) –- A correct, R correct, R explains A.

Concept used. Three diagnostic clues pin down each species: (1) ozonolysis of an alkene to two carbonyls fixes the position of the double bond; (2) Markovnikov addition of HCl converts an alkene into the more-substituted alkyl halide; (3) reaction with NaNO2/HCl followed by water on a 1^∘ aliphatic amine gives the corresponding alcohol via a short-lived diazonium ion.

1. Ozonolysis of A → 2 mol CH3CHO. Hence A is symmetrical with a C=C between two CH-CH3 units.
2. Therefore A = but-2-ene, CH3-CH=CH-CH3.
3. A + HCl: Markovnikov on a symmetrical alkene gives B = CH3-CHCl-CH2-CH3, 2-chlorobutane.
4. B + NH3 (1 mol): S_N2 at the 2^∘ centre gives C = CH3-CH(NH2)-CH2-CH3, butan-2-amine (a 1^∘ amine on a chiral sp³ carbon).
5. C + NaNO2/HCl + H2O: diazonium decomposes; water traps the cation with mostly retained configuration ⇒ D = CH3-CH(OH)-CH2-CH3, butan-2-ol –- optically active (chiral C).

A = but-2-ene; B = 2-chlorobutane; C = butan-2-amine; D = butan-2-ol (chiral, optically active).

Concept used. Six classical aniline tests –- (a) basicity (soluble in mineral acid), (b) carbylamine test (CHCl3/KOH on 1^∘ amine), (c) Hinsberg (1^∘ amine gives alkali-soluble sulphonamide), (d) diazotisation at 0-5 ^∘C, (e) coupling of Ar-N2⁺ with phenol in alkali to give an orange dye –- all fit aniline.

1. Formula C6H7N, sparingly water-soluble colourless liquid: aniline, C6H5-NH2. Hence A = C6H5NH2.
2. A + HCl -> C6H5NH3⁺ Cl^-, a salt soluble in water. B = anilinium chloride.
3. On heating, A + CHCl3 + 3 KOH -> C6H5-NC + 3 KCl + 3 H2O (the obnoxious smell). C = phenyl isocyanide C6H5NC.
4. A + C6H5SO2Cl -> C6H5-SO2-NH-C6H5 + HCl, where the remaining N-H is acidic and dissolves in NaOH. D = N-phenylbenzenesulphonamide.
5. A + NaNO2/HCl at 0-5 ^∘C → C6H5-N2⁺ Cl^-. E = benzene diazonium chloride.
6. E + C6H5OH/NaOH at 0 ^∘C → C6H5-N=N-C6H4-OH (para). F = p-hydroxyazobenzene (orange dye).

A = aniline, B = anilinium chloride, C = phenyl isocyanide, D = N-phenylbenzenesulphonamide, E = benzenediazonium chloride, F = p-hydroxyazobenzene.

Answers

• 1 (reagent): Sn / HCl (or Fe / HCl) –- reduction of Ar-NO2 to Ar-NH2.
• 2 (intermediate): 4-methyl-2-nitroacetanilide (a benzene ring with CH3 at C-4, NHCOCH3 at C-1, and NO2 at C-2; i.e. NO2 goes ortho to the NHCOCH3 since para is blocked by CH3).
• 3 (reagent): H3O⁺ (or H⁺ / H2O, Δ) –- acid hydrolysis to remove the acetyl group, giving 4-methyl-2-nitroaniline.
• 4 (intermediate): 4-methyl-2-nitrobenzenediazonium chloride 4-CH3-2-NO2-C6H3-N2⁺ Cl^-.
• 5 (reagent): H3PO2 / H2O (hypophosphorous acid) –- deamination: replaces -N2⁺ by -H, giving m-nitrotoluene.

Concept used. The route uses (a) Sn/HCl reduction to unlock the amine from p-nitrotoluene, (b) acetyl protection of the amine before nitration, (c) ortho-directed nitration of the acetamido-arene, (d) acid hydrolysis to free the amine, (e) diazotisation and (f) H3PO2 deamination to put a hydrogen where the amine was. The product is m-nitrotoluene (NO2 at C-3 of toluene), where NO2 and CH3 sit meta to each other –- a regiochemistry unreachable by direct nitration of toluene.

1. Step 1: p-nitrotoluene Sn / HCl p-toluidine.
2. Step 2: p-toluidine + (CH3CO)2O/pyridine → p-methylacetanilide (the amine is protected as acetamide).
3. Step 3 (intermediate 2): nitration with HNO3/H2SO4. Since para to the NHCOCH3 is blocked by CH3, NO2 enters ortho to the NHCOCH3 ⇒ 4-methyl-2-nitroacetanilide.
4. Step 4 (reagent 3): acid hydrolysis H3O⁺ removes acetyl → 4-methyl-2-nitroaniline.
5. Step 5 (intermediate 4): diazotisation with NaNO2/HCl at 0-5 ^∘C → diazonium chloride.
6. Step 6 (reagent 5): H3PO2/H2O deaminates -N2⁺ -> -H → m-nitrotoluene.

1 = Sn/HCl; 2 = 4-methyl-2-nitroacetanilide; 3 = H3O⁺ hydrolysis; 4 = 4-methyl-2-nitrobenzenediazonium chloride; 5 = H3PO2/H2O.