JEE Chemistry Amines MCQs Set 04

Answer: (c) basic

Explanation:
1. Aniline accepts a proton from the acid on its lone pair and forms the anilinium salt.
2. Accepting a proton is basic behaviour.

Teacher's Note:
1. Aniline is a weak base.
2. It forms salts with strong acids such as HCl and \( \ce{H2SO4} \).

Answer: (d) benzylamine

Explanation:
1. Diazotisation needs a primary aromatic amine in which \( \ce{-NH2} \) is attached directly to the ring.
2. Aniline, p-toluidine and p-nitroaniline all qualify.
3. Benzylamine has \( \ce{-CH2NH2} \), so it gives an alcohol instead.

Teacher's Note:
1. The amino group must sit on the benzene ring itself.
2. Side-chain amines do not give stable diazonium salts.

Answer: (c) N-methylaniline

Explanation:
1. Secondary amines react with nitrous acid to give N-nitrosoamines, \( \ce{R2N-N=O} \).
2. N-Methylaniline is a secondary amine.
3. Primary amines give diazonium salts or alcohols instead.

Teacher's Note:
1. N-nitrosoamines are yellow oils and are carcinogenic.
2. Liebermann's nitroso test identifies secondary amines.

Answer: (b) Hofmann bromamide reaction

Explanation:
1. In the Hofmann bromamide reaction the N-bromoamide anion loses bromide to form an acyl nitrene-type species that rearranges to isocyanate.
2. Carbylamine and Reimer-Tiemann reactions involve dichlorocarbene, and Friedel-Crafts involves a carbocation.

Teacher's Note:
1. Nitrene is the nitrogen analogue of a carbene.
2. Carbene: carbylamine and Reimer-Tiemann.

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

Explanation:
1. Aldehydes, formic acid and phenylhydroxylamine all reduce Tollens' reagent to silver.
2. Aniline is not a reducing agent for it.

Teacher's Note:
1. Phenylhydroxylamine gives the Mullikan-Barker test.
2. Aniline is oxidised only by strong oxidants.

Answer: (b) p-nitroaniline

Explanation:
1. Acetanilide is ortho and para directing, and the bulky acetamido group favours para.
2. Hydrolysis removes the acetyl group and gives p-nitroaniline as the main product.

Teacher's Note:
1. Steric hindrance favours the para isomer.
2. This is the standard route to pure p-nitroaniline.

Answer: (a) \( \ce{C6H5-N=N-Cl} \) (covalent azo chloride)

Explanation:
1. Benzenediazonium chloride is an ionic salt in which the diazonium ion \( \ce{C6H5-N+#N} \) is stabilised by delocalisation into the ring.
2. Forms b, c and d are different ways of writing this ionic structure.
3. The covalent form with N=N-Cl is wrong.

Teacher's Note:
1. Diazonium salts are ionic.
2. The positive charge is delocalised over the ring, which makes the aromatic salt stable at 0-5 C.

Answer: (a) only (a)

Explanation:
1. Chloroform with alcoholic KOH is the carbylamine reagent.
2. Hofmann degradation uses Br2/KOH, and Reimer-Tiemann uses chloroform with aqueous NaOH on phenol.
3. The mustard oil reaction uses \( \ce{CS2} \) and \( \ce{HgCl2} \). So only (a) applies.

Teacher's Note:
1. Alcoholic KOH points to carbylamine.
2. Aqueous NaOH with chloroform points to Reimer-Tiemann.

Answer: (c) p-bromoacetanilide

Explanation:
1. The acetamido group is an ortho and para director.
2. The para product is favoured because the group is bulky, so p-bromoacetanilide is the major product.

Teacher's Note:
1. Acetylation also limits the reaction to monobromination.
2. Hydrolysis then gives p-bromoaniline.

Answer: (a) \( \ce{ArX + KCN} \)

Explanation:
1. Aryl halides do not undergo nucleophilic substitution by cyanide under normal conditions because the C-X bond has partial double bond character.
2. Diazonium salts with CuCN (Sandmeyer) and amide dehydration with \( \ce{P2O5} \) or \( \ce{SOCl2} \) give aromatic nitriles.

Teacher's Note:
1. Aryl halides are very unreactive to simple SN reactions.
2. Use the diazonium route to make aryl cyanides.

Answer: (b) o-, p- and o-, p-directing

Explanation:
1. Diazotisation needs a primary amino group, so X is \( \ce{NH2} \), an ortho and para director.
2. Warming the diazonium salt with water gives phenol, so Y is OH, also an ortho and para director.

Teacher's Note:
1. Both \( \ce{NH2} \) and OH activate the ring.
2. Both direct ortho and para.

Answer: (a) methyl isocyanate

Explanation:
1. The gas was methyl isocyanate, \( \ce{CH3NCO} \).
2. It is a highly toxic liquid that boils near 39 C and evaporates easily.

Teacher's Note:
1. Methyl isocyanate is an intermediate in making carbamate pesticides.
2. Water reacts with it exothermically.

Answer: (b) \( \ce{(CH3)3C-NH-CH3} \)

Explanation:
1. Reduction of an isocyanide \( \ce{R-NC} \) gives \( \ce{R-NH-CH3} \).
2. The isocyanide carbon becomes the methyl group on nitrogen.
3. So the product is N-methyl-tert-butylamine.

Teacher's Note:
1. Isocyanide reduction gives a secondary amine with one methyl group.
2. Nitrile reduction gives a primary amine instead.

Answer: (c) both of the above

Explanation:
1. A is aniline and B is benzenediazonium chloride.
2. Both ethanol and hypophosphorous acid reduce the diazonium group and replace it by hydrogen, giving benzene.

Teacher's Note:
1. Ethanol is oxidised to acetaldehyde in this reaction.
2. \( \ce{H3PO2} \) gives better yields.

Answer: (a) A-a, B-c, C-b, D-d

Explanation:
1. Aniline is the starting material for azo dyes (A-a).
2. Nitrobenzene is a solvent for Friedel-Crafts reactions (B-c).
3. Sulphanilamide is a sulpha drug (C-b).
4. TNT is an explosive (D-d).

Teacher's Note:
1. Learn the uses of each aromatic nitrogen compound.
2. Sulphanilamide comes from sulphanilic acid chemistry.

Answer: (a) a halide group

Explanation:
1. Sandmeyer reaction uses cuprous halides or cuprous cyanide.
2. The diazonium group is replaced by Cl, Br or CN.
3. So the halide group is the answer from the options.

Teacher's Note:
1. Sandmeyer = Cu(I) salt.
2. Gattermann uses copper powder with HX.

Answer: (a) methyl isocyanide

Explanation:
1. Methyl isocyanide gives \( \ce{CH3NHCH3} \), dimethylamine, a secondary amine.
2. Acetamide, methyl cyanide and nitroethane all give primary amines.

Teacher's Note:
1. Only isocyanides give secondary amines on reduction.
2. Remember this against nitrile reduction.

Answer: (b) p-(dimethylamino)azobenzene, \( \ce{C6H5-N=N-C6H4-N(CH3)2} \)

Explanation:
1. Benzenediazonium chloride is a weak electrophile and attacks the para position of the activated N,N-dimethylaniline ring.
2. This coupling gives an azo compound with the -N=N- link, a coloured dye (yellow).

Teacher's Note:
1. Coupling happens at the para position of the activated ring.
2. Azo compounds have the N=N chromophore.

Answer: (b) N-nitroso-N-methylaniline, \( \ce{C6H5N(CH3)-N=O} \)

Explanation:
1. A secondary amine reacts with nitrous acid to give an N-nitrosoamine.
2. N-Methylaniline is a secondary aromatic amine, so the N-nitroso compound forms.

Teacher's Note:
1. Secondary amine plus \( \ce{HNO2} \) gives a yellow N-nitrosoamine.
2. This is the Liebermann nitroso reaction.

Answer: (a) \( \ce{CH3CH2CH2NH2} \)

Explanation:
1. Primary amines give \( \ce{RNHSO2C6H5} \), which still has an acidic N-H and dissolves in KOH.
2. Secondary amines give an insoluble sulphonamide, tertiary amines do not react, and alcohols are not sulphonylated this way.

Teacher's Note:
1. This is Hinsberg's test.
2. Primary: soluble in alkali, secondary: insoluble, tertiary: no reaction.

Answer: (a) N-phenylhydroxylamine

Explanation:
1. Neutral reduction with Zn/\( \ce{NH4Cl} \) stops at the hydroxylamine stage.
2. The product is \( \ce{C6H5NHOH} \).
3. It is used in the Mullikan-Barker test.

Teacher's Note:
1. Acidic medium gives aniline, neutral medium gives hydroxylamine.
2. Alkaline medium gives bimolecular products.

Answer: (d) \( \ce{H3PO2} \) and \( \ce{H2O} \)

Explanation:
1. The diazonium group is to be replaced by hydrogen.
2. \( \ce{H3PO2} \) with water does this.
3. Warming with water would give a phenol, and \( \ce{Cu2Cl2} \) would introduce chlorine.

Teacher's Note:
1. Use \( \ce{H3PO2} \) to remove an amino group via diazonium salt.
2. This helps place groups in positions not reachable directly.

Answer: (c) 1,3-dinitrobenzene

Explanation:
1. The nitro group is a meta director.
2. The second nitro group enters the 3-position, giving 1,3-dinitrobenzene.

Teacher's Note:
1. Meta directing groups are all electron withdrawing.
2. Only about 6 percent ortho/para products form.

Answer: (b) \( \ce{C6H5OH} \)

Explanation:
1. Aniline gives benzenediazonium chloride with \( \ce{HNO2} \) at low temperature.
2. Warming with dilute acid in water hydrolyses it to phenol with loss of \( \ce{N2} \).

Teacher's Note:
1. Water or dilute acid with warming gives phenol.
2. \( \ce{H3PO2} \) gives benzene and \( \ce{SnCl2/HCl} \) gives phenylhydrazine.