JEE Chemistry Amines MCQs Set 06

Answer: (c) \( \ce{C6H5N2Cl ->[Cu2Cl2/HCl] C6H5Cl} \)

Explanation:
1. Sandmeyer reaction uses cuprous halide (Cu₂Cl₂/HCl, Cu₂Br₂/HBr, Cu₂(CN)₂/KCN).
2. Cu powder with HCl is the Gattermann reaction; HBF₄ then heat is the Balz–Schiemann reaction.

Teacher's Note:
1. Sandmeyer means cuprous salt; Gattermann means copper powder.
2. Both replace the diazonium group by Cl, Br or CN.

Answer: (a) I > II > IV > III > V

Explanation:
1. Alcohols form the strongest hydrogen bonds (O–H is more polar than N–H), so I is highest.
2. Primary amines (two N–H) hydrogen bond more than secondary amines (one N–H); tertiary amines cannot hydrogen bond.
3. Order: I (97 °C) > II (48 °C) > IV (37 °C) > III (3 °C) > V (0 °C).

Teacher's Note:
1. Boiling point follows the extent of hydrogen bonding: alcohol > 1° amine > 2° amine > 3° amine > alkane of similar mass.
2. Compare molecular masses only when the hydrogen bonding is the same.

Answer: (a) \( \ce{C2H5NH2} \)

Explanation:
1. Hinsberg's reagent is benzenesulphonyl chloride.
2. A primary amine gives RNH–SO₂C₆H₅ which still has an acidic N–H and dissolves in NaOH.
3. Secondary amines give an insoluble sulphonamide; tertiary amines do not react.

Teacher's Note:
1. Soluble in alkali means the sulphonamide retains an N–H, which is true only for primary amines.
2. Remember: 1° soluble, 2° insoluble, 3° no reaction.

Answer: (a)

Explanation:
1. Gabriel synthesis needs an alkyl halide to react with potassium phthalimide by SN2.
2. Aryl halides do not undergo nucleophilic substitution, so aromatic primary amines such as p-toluidine cannot be made this way.

Teacher's Note:
1. Gabriel synthesis gives pure primary aliphatic amines only.
2. Benzylamine is possible because benzyl halide is an alkyl halide.

Answer: (c) \( \ce{CH3CN} \) and \( \ce{CH3CH2NH2} \)

Explanation:
1. CH₃Cl + KCN gives CH₃CN (A); partial hydrolysis gives CH₃CONH₂ (B).
2. PCl₅ dehydrates the amide back to CH₃CN (C).
3. LiAlH₄ reduces the amide to CH₃CH₂NH₂ (D).

Teacher's Note:
1. Primary amides lose water with P₂O₅ or PCl₅ to give nitriles.
2. LiAlH₄ converts the C=O of an amide into CH₂ without changing carbon count.

Answer: (b)

Explanation:
1. Secondary amines react with nitrous acid to give yellow oily N-nitrosoamines.
2. N-Methylaniline is a secondary amine and gives N-nitroso-N-methylaniline.
3. Primary amines evolve N₂ (or form diazonium salts) and the tertiary amine here gives only ring nitrosation or salt.

Teacher's Note:
1. Yellow oily nitrosoamine is the test for a secondary amine (Liebermann nitroso test).
2. Do not confuse it with primary amines, which give nitrogen gas.

Answer: (b) a, b

Explanation:
1. (a) is the Schmidt reaction: benzoic acid with hydrazoic acid gives aniline.
2. (b) is the Hofmann bromamide reaction: benzamide gives aniline.
3. (c) gives only benzamide, not aniline.
4. (d) Zn/NH₄Cl reduces nitrobenzene only as far as phenylhydroxylamine. Hence only (a) and (b) give aniline.

Teacher's Note:
1. Schmidt and Hofmann reactions both shorten the carbon chain by one carbon.
2. The product of nitro reduction depends on the medium: neutral gives hydroxylamine, strongly acidic gives aniline.

Answer: (b) Electrolysis in dil. \( \ce{H2SO4} \)

Explanation:
1. Zn/NaOH in methanol gives hydrazobenzene, sodium arsenite gives azoxybenzene and Zn/NaOH in water gives azobenzene.
2. Electrolytic reduction in dilute H₂SO₄ gives aniline.

Teacher's Note:
1. Neutral and alkaline reductants stop at bimolecular products (azoxy, azo, hydrazo).
2. Acidic reductants such as Sn/HCl or electrolysis in dilute acid give aniline.

Answer: (d)

Explanation:
1. In neutral medium (Zn/NH₄Cl) nitrobenzene is reduced through nitrosobenzene to N-phenylhydroxylamine.
2. So A is C₆H₅NHOH.

Teacher's Note:
1. Neutral reduction gives phenylhydroxylamine; it is used in the Tollens test confirmation of nitro compounds.
2. Memorise: acidic gives aniline, neutral gives hydroxylamine, alkaline gives azo products.

Answer: (c)

Explanation:
1. The NH₂ group strongly activates the ortho and para positions.
2. In bromine water all three positions react at once to give a white precipitate of 2,4,6-tribromoaniline.

Teacher's Note:
1. Bromine water with aniline gives an immediate white precipitate, a test for aniline.
2. To get a mono bromo product, protect the amine by acetylation first.

Answer: (b) \( \ce{C2H5NCS} \)

Explanation:
1. A primary amine adds to CS₂ to give a dithiocarbamic acid.
2. HgCl₂ removes H₂S and gives the alkyl isothiocyanate C₂H₅NCS, a mustard oil with a pungent smell.

Teacher's Note:
1. This is the mustard oil reaction, a test for primary amines.
2. The product has N attached to the alkyl group, so it is an isothiocyanate.

Answer: (a) \( \ce{CH3NCO} \)

Explanation:
1. The gas released was methyl isocyanate, CH₃–N=C=O (MIC), an intermediate in the manufacture of carbamate pesticides.

Teacher's Note:
1. Methyl isocyanate has N attached to the methyl group and the formula CH₃NCO.
2. It is highly toxic and reacts violently with water.

Answer: (c) A Schiff base

Explanation:
1. An aldehyde condenses with a primary amine with loss of water to give an imine, C₆H₅CH=NC₆H₅.
2. Such imines are called Schiff bases (benzylideneaniline here).

Teacher's Note:
1. Aldehyde (or ketone) plus primary amine gives a Schiff base.
2. Schiff reagent is a different thing: it is a dye solution used to test aldehydes.

Answer: (b) Schotten–Baumann reaction

Explanation:
1. Treating an amine (or alcohol) with benzoyl chloride in the presence of aqueous NaOH or pyridine is the Schotten–Baumann reaction.
2. The product from a primary amine is an N-substituted benzamide.

Teacher's Note:
1. Schotten–Baumann means benzoylation in alkaline medium.
2. Base removes the HCl formed and drives the reaction forward.

Answer: (c) \( \ce{HBF4/\Delta} \)

Explanation:
1. Fluorobenzene is made by the Balz–Schiemann reaction: the diazonium chloride is converted to the insoluble fluoroborate with HBF₄, then heated.
2. CuF and Cu/HF do not work, and HBF₄ with NaNO₂/Cu gives the nitro compound.

Teacher's Note:
1. Fluorine is introduced via diazonium tetrafluoroborate on heating.
2. Iodine goes in with KI; Cl, Br, CN need copper salts.

Answer: (b) An alkyl isocyanide

Explanation:
1. This is the carbylamine reaction; the primary amine reacts with dichlorocarbene from CHCl₃/KOH.
2. The product is an alkyl isocyanide (carbylamine), RNC, with a very offensive smell.

Teacher's Note:
1. The carbylamine reaction is the test for primary amines, aliphatic or aromatic.
2. Secondary and tertiary amines do not respond.

Answer: (b) A primary amine

Explanation:
1. PCl₅ dehydrates acetamide to acetonitrile (A).
2. LiAlH₄ reduces the nitrile to ethylamine, CH₃CH₂NH₂, a primary amine.

Teacher's Note:
1. Nitrile reduction always gives a primary amine with the same carbon count as the nitrile.
2. Hydrolysis of nitrile, in contrast, gives acid.

Answer: (b) \( \ce{CH3NH2} \)

Explanation:
1. The foul smell is from an isocyanide formed in the carbylamine reaction, which only primary amines give.
2. Of the compounds listed, only CH₃NH₂ is a primary amine (the amide CH₃CONH₂ is not an amine).

Teacher's Note:
1. Carbylamine reaction needs a free –NH₂ on an alkyl or aryl carbon.
2. Amides do not give this test.

Answer: (d) Aniline

Explanation:
1. Fe/HCl (like Sn/HCl) is a strongly acidic reducing medium that reduces NO₂ completely to NH₂.
2. The product is aniline, formed as anilinium chloride and liberated with alkali.

Teacher's Note:
1. Metal with acid reduces nitro to amine completely; this is the industrial route to aniline.
2. Compare with neutral and alkaline media, which stop early.

Answer: (b) Stephen's reaction

Explanation:
1. Reduction of a nitrile with SnCl₂/HCl gives an imine hydrochloride, which hydrolyses to the aldehyde. This is Stephen's reduction.
2. Mendius reduction gives an amine; Rosenmund reduces acid chlorides; Perkin gives cinnamic acids.

Teacher's Note:
1. Stephen reaction: nitrile to aldehyde with SnCl₂/HCl then water.
2. DIBAL-H also stops at the aldehyde stage.

Answer: (d) Acetone

Explanation:
1. All four have nearly the same molar mass (58–60).
2. Acetone (56 °C) and propanal (49 °C) are polar carbonyl compounds with dipole attraction; methoxyethane (7 °C) is much less polar and n-butane (0 °C) is non-polar.
3. Acetone has the highest dipole moment among them, so the highest boiling point.

Teacher's Note:
1. For similar masses, boiling point rises with polarity: ketone > aldehyde > ether > alkane.
2. None of these forms hydrogen bonds with itself.

Answer: (c) Propanenitrile

Explanation:
1. LiAlH₄ reduces R–CN to R–CH₂NH₂. Propanenitrile (CH₃CH₂CN) gives CH₃CH₂CH₂NH₂.
2. Propyl cyanide (butanenitrile) gives butanamine, and N-methylpropanamide gives N-methylpropanamine.

Teacher's Note:
1. Name nitriles by the total carbon chain: propyl cyanide has four carbons.
2. The reduced amine has the same carbon count as the nitrile.

Answer: (c) Benzenamine

Explanation:
1. A larger pK_b means a weaker base.
2. Aryl amines are much weaker bases than aliphatic ones because the lone pair is delocalised into the ring.
3. Aniline (pK_b about 9.4) is weaker than N-methylaniline (about 9.2), while triethylamine (3.3) and benzylamine (4.7) are far stronger.

Teacher's Note:
1. Maximum pK_b means minimum basicity.
2. Aniline is the weakest base among these choices.

Answer: (a) Primary amine

Explanation:
1. The test uses CHCl₃ and alcoholic KOH and the foul-smelling isocyanide is formed only by primary amines.

Teacher's Note:
1. Primary amines (aliphatic and aromatic) give the carbylamine test.
2. The test is very sensitive and also detects traces of chloroform.

Answer: (b) Aniline

Explanation:
1. Coupling with aniline in weakly acidic medium gives p-aminoazobenzene, which is yellow.
2. Phenol couples in alkaline medium to give an orange dye.

Teacher's Note:
1. Aniline coupling gives a yellow dye; phenol gives orange; beta-naphthol gives red.
2. Coupling occurs at the para position.

Answer: (c) X = \( \ce{(CH3)2CHNH2} \); Y = \( \ce{(CH3)2CHOH} \)

Explanation:
1. Hofmann bromamide reaction of 2-methylpropanamide gives isopropylamine (one carbon fewer), so X = (CH₃)₂CHNH₂.
2. A primary aliphatic amine with HNO₂ gives the alcohol; Y = (CH₃)₂CHOH (propan-2-ol) with N₂ released.

Teacher's Note:
1. Hofmann reaction retains the carbon skeleton except for the carbonyl carbon, which is lost as carbonate.
2. Aliphatic primary amine plus HNO₂ gives alcohol (plus N₂ gas).

Answer: (c) \( \ce{C6H5COOH} \)

Explanation:
1. X is benzenediazonium chloride; CuCN (Sandmeyer) gives benzonitrile Y.
2. Boiling with aqueous acid hydrolyses the nitrile through the amide to benzoic acid (Z).

Teacher's Note:
1. Sandmeyer cyanation followed by hydrolysis converts an aryl amine into an aryl carboxylic acid.
2. Mild hydrolysis would stop at the amide, but boiling gives the acid.

Answer: (b)

Explanation:
1. Diazotisation gives C₆H₅N₂⁺Cl⁻.
2. Coupling at the para position of N,N-dimethylaniline gives the azo dye p-(N,N-dimethylamino)azobenzene, (CH₃)₂N–C₆H₄–N=N–C₆H₅.

Teacher's Note:
1. Azo coupling forms an –N=N– bridge between the two rings.
2. The first ring (from aniline) has no extra group after diazotisation.

Answer: (c) \( \ce{CH3-NH-CH2CH3} \)

Explanation:
1. A sulphonamide insoluble in alkali comes from a secondary amine (no N–H left).
2. Among isomers of C₃H₉N, the secondary amine is CH₃NHCH₂CH₃.
3. Option (4) is C₉H₁₃N and does not fit the formula; (1) gives a soluble product; (2) does not react.

Teacher's Note:
1. Check the molecular formula first, then the Hinsberg behaviour.
2. Secondary amine gives an alkali-insoluble solid.

Answer: (d) \( \ce{CH3CH2CONH2} \)

Explanation:
1. The Hofmann bromamide reaction needs an unsubstituted (primary) amide, RCONH₂.
2. Only propanamide fits; N-methylbenzamide is N-substituted and the others are not amides.

Teacher's Note:
1. Hofmann reaction: RCONH₂ with Br₂/KOH gives RNH₂.
2. N-substituted amides do not undergo this reaction.

Answer: (d) \( \ce{C6H5NH2} \)

Explanation:
1. This is the Gabriel phthalimide synthesis, which needs R–Cl to be an alkyl halide.
2. Primary, benzylic and secondary alkyl halides can be used, but an aryl halide cannot, so aniline cannot be formed.

Teacher's Note:
1. Gabriel synthesis is for primary aliphatic and benzylic amines.
2. Aromatic amines need a different route, such as nitro reduction.

Answer: (b)

Explanation:
1. Ammonia opens the anhydride ring to form the amic acid, which on heating loses water and closes to the imide.
2. The final product is phthalimide.

Teacher's Note:
1. Phthalimide is made by heating phthalic anhydride with ammonia.
2. It is the starting compound of Gabriel synthesis.