JEE Chemistry Amines MCQs Set 25

Answer: (c) Gattermann reaction

Explanation:
1. Diazonium salts react with copper powder and HBr (Gattermann reaction) or CuBr/HBr (Sandmeyer reaction) to replace the diazonium group by bromine.
2. Reimer-Tiemann gives salicylaldehyde, Friedel-Crafts forms C-C bonds on the ring and azo coupling gives azo dyes.

Teacher's Note:
1. Replacement of N2+ by halogen uses copper catalysts.
2. Gattermann uses copper powder, Sandmeyer uses cuprous halide.

Answer: (b) 5

Explanation:
1. The isomers are o-, m- and p-toluidine, benzylamine and N-methylaniline.
2. Together these make five.

Teacher's Note:
1. Count by type of amine: primary on ring (3), primary on side chain (1), secondary (1).
2. The formula has four degrees of unsaturation, all used by the ring.

Answer: (c) \( \ce{(CH3)3N} \)

Explanation:
1. Acylation needs an N-H bond that can be replaced by the acyl group.
2. Primary and secondary amines have N-H and are acylated.
3. Trimethylamine has no N-H and does not give an amide.

Teacher's Note:
1. Only 1 and 2 amines are acylated.
2. Tertiary amines have no replaceable hydrogen.

Answer: (a) Liebermann nitroso reaction

Explanation:
1. This is the reverse of the nitrosation of a secondary amine and is the test known as the Liebermann nitroso reaction.
2. The Etard, Fries and Perkin reactions are unrelated.

Teacher's Note:
1. Nitrosoamine with acid gives secondary amine back.
2. Phenol then gives a coloured indophenol-type product in the test.

Answer: (b) 1,3,5-tribromobenzene

Explanation:
1. Bromine water gives 2,4,6-tribromoaniline (A).
2. Diazotisation followed by \( \ce{H3PO2} \) replaces the diazonium group by hydrogen, giving 1,3,5-tribromobenzene.

Teacher's Note:
1. The amino group directs bromine to o- and p-positions and is then removed.
2. This is a route to meta-substituted benzenes.

Answer: (d) \( \ce{AgNO3} \) solution

Explanation:
1. Anilinium chloride is an ionic salt and gives a white precipitate of AgCl with \( \ce{AgNO3} \).
2. p-Chloroaniline has covalently bonded chlorine and gives no precipitate.
3. The other tests are given by both compounds.

Teacher's Note:
1. Ionic chloride precipitates with silver nitrate.
2. Both compounds are primary arylamines.

Answer: (d) \( \ce{CH3CN} \)

Explanation:
1. Nitrous acid converts a primary aliphatic amine into an alcohol, so B is ethylamine.
2. Reduction of \( \ce{CH3CN} \) gives \( \ce{CH3CH2NH2} \).
3. \( \ce{CH3CH2CN} \) would give propylamine and \( \ce{CH3NC} \) would give dimethylamine.

Teacher's Note:
1. Work backwards from the alcohol.
2. Nitrile reduction adds a CH2 to the amine.

Answer: (a) \( \ce{SnCl2/HCl/H2O} \), boil

Explanation:
1. Stephen's reaction uses \( \ce{SnCl2} \) and HCl to reduce a nitrile to an imine salt, which is hydrolysed to the aldehyde.
2. \( \ce{LiAlH4} \) would give the amine.

Teacher's Note:
1. Stephen reduction: nitrile to aldehyde.
2. Strong hydrides over-reduce to amines.

Answer: (d) \( \ce{CH3COOH} \)

Explanation:
1. Primary nitro compounds are hydrolysed by hot strong acid (Meyer reaction) to carboxylic acids and hydroxylamine.
2. Nitroethane gives acetic acid.

Teacher's Note:
1. Primary nitroalkane gives carboxylic acid with the same number of carbons.
2. Secondary nitroalkanes give ketones.

Answer: (a) p-toluidine: \( \ce{Br2/H2O} \), then \( \ce{HNO2} \), then \( \ce{H3PO2} \)

Explanation:
1. p-Toluidine is brominated at both positions ortho to the amino group, which are meta to the methyl group.
2. Diazotisation and reduction with \( \ce{H3PO2} \) removes the amino group and leaves 3,5-dibromotoluene.

Teacher's Note:
1. Use the amino group to place bromines, then remove it.
2. Direct bromination of toluene gives o- and p-products.

Answer: (c) N-cyclohexylcyclohex-1-en-1-amine (nitrogen attached to a ring carbon of a C=C)

Explanation:
1. An enamine has nitrogen attached directly to a carbon of a C=C double bond (N-C=C).
2. Only the cyclohexenylamine has this arrangement.
3. The imine has C=N and the other two have no C=C.

Teacher's Note:
1. Enamine = ene + amine.
2. Check for N bonded to a vinyl carbon.

Answer: (c) 1,3-dibromo-2-chlorobenzene

Explanation:
1. The first step replaces the amino group by chlorine (A: 1,3-dibromo-2-chloro-5-nitrobenzene).
2. Reduction converts the nitro group to amino (B).
3. Diazotisation and \( \ce{H3PO2} \) remove this amino group, leaving 1,3-dibromo-2-chlorobenzene.

Teacher's Note:
1. Track each substituent position.
2. The chlorine ends up between the two bromines.

Answer: (c) acetaldehyde

Explanation:
1. Partial reduction of the nitrile gives the imine \( \ce{CH3CH=NH} \) (X).
2. Hydrolysis of the imine with boiling water gives acetaldehyde and ammonia.

Teacher's Note:
1. Stephen reaction.
2. Imines hydrolyse to carbonyl compounds.

Answer: (d) phenyl isothiocyanate

Explanation:
1. Aniline reacts with carbon disulphide to give a dithiocarbamic acid.
2. \( \ce{HgCl2} \) removes \( \ce{H2S} \) (as HgS) and gives phenyl isothiocyanate, \( \ce{C6H5NCS} \).

Teacher's Note:
1. This is the mustard oil reaction.
2. The product has an N=C=S group.

Answer: (b) methyl alcohol

Explanation:
1. Aliphatic primary amines give unstable diazonium salts.
2. The salt loses nitrogen to give a carbocation and then mainly the alcohol, here methyl alcohol.

Teacher's Note:
1. Aliphatic diazonium salts decompose at once.
2. Alcohol is the main product.

Answer: (c) 3° > 2° > 1° > \( \ce{NH3} \)

Explanation:
1. In the gas phase there is no solvation, so only the inductive effect of alkyl groups matters.
2. More alkyl groups give greater electron density on nitrogen.
3. Hence 3° > 2° > 1° > \( \ce{NH3} \).

Teacher's Note:
1. Without solvent, basicity follows +I effect.
2. In water, solvation changes the order.

Answer: (b) benzylamine

Explanation:
1. Benzylamine is an aliphatic amine whose lone pair is localised on nitrogen.
2. The other three have the nitrogen lone pair delocalised into the ring or carbonyl group, so they are weaker bases.

Teacher's Note:
1. Lone pair availability decides basicity.
2. Aliphatic amines are stronger than aromatic amines.

Answer: (b) p-nitroaniline

Explanation:
1. The acetamido group is an ortho-para director and the bulky group hinders the ortho position, so p-nitroacetanilide is the major nitration product.
2. Alkaline hydrolysis then removes the acetyl group to give p-nitroaniline.

Teacher's Note:
1. Protect the amine as acetanilide before nitration.
2. Para product dominates.

Answer: (b) 2-bromo-4-methylaniline

Explanation:
1. Acetylation moderates the activating power of the amino group.
2. The acetamido group is a stronger director than methyl, so bromine enters ortho to NHCOCH3.
3. Hydrolysis removes the acetyl group to give 2-bromo-4-methylaniline.

Teacher's Note:
1. The stronger activating group controls the position.
2. Bromine goes ortho to the nitrogen group.

Answer: (b) nitration, chlorination, reduction

Explanation:
1. Nitration gives nitrobenzene, and the nitro group directs chlorine to the meta position.
2. Reduction of 3-chloronitrobenzene gives 3-chloroaniline.
3. The other orders put chlorine ortho or para to the nitrogen group.

Teacher's Note:
1. Meta director first, then reduce.
2. Order of steps decides the position.

Answer: (b) benzene

Explanation:
1. Nitration of benzene gives nitrobenzene (B).
2. Sn/HCl reduces it to aniline.

Teacher's Note:
1. Nitrobenzene is the precursor of aniline.
2. Toluene would give toluidine.