JEE Chemistry Amines MCQs Set 20

Answer: (d) Phenylmagnesium bromide treated with H₃O+

Explanation:
1. Soda lime decarboxylates salicylic acid to give phenol.
2. C₆H₅MgBr is a Grignard reagent; protonation by H₃O+ gives benzene, not phenol.
3. The Dow process (633 K, 300 atm, NaOH) converts chlorobenzene to sodium phenoxide and then phenol, and warming a diazonium salt with water gives phenol and nitrogen.

Teacher's Note:
1. Grignard reagents are strong bases and take up a proton from water or acid.
2. To make a phenol from an aryl Grignard reagent you would need an oxidant such as a borate and peroxide.

Answer: (c) nitration, chlorination, reduction

Explanation:
1. Nitration gives nitrobenzene; the nitro group is meta-directing.
2. Chlorination of nitrobenzene gives 3-chloronitrobenzene.
3. Reduction of the nitro group then gives 3-chloroaniline. Reducing first would make the group ortho/para-directing.

Teacher's Note:
1. Introduce the meta-directing group before the halogen when a 1,3 relationship is wanted.
2. Reduce last because the amino group directs ortho and para.

Answer: (b)

Explanation:
1. Hofmann bromamide degradation removes one carbon: benzamide gives aniline (A).
2. Aniline reacts with ethanoyl chloride at nitrogen to give an amide.
3. B is acetanilide, N-phenylethanamide.

Teacher's Note:
1. The Hofmann reaction shortens the carbon chain by one carbon.
2. An amine is acylated on nitrogen, not on the ring.

Answer: (c) acetanilide

Explanation:
1. Aniline is a nucleophile through the nitrogen lone pair.
2. It attacks the acyl chloride carbonyl and chloride is lost; base removes the HCl.
3. The product is acetanilide (N-phenylethanamide).

Teacher's Note:
1. This is acylation (protection) of an aromatic amine.
2. Acetophenone would need Friedel-Crafts acylation of benzene.

Answer: (d)

Explanation:
1. Acetylation gives N-(4-methylphenyl)ethanamide; the amide group is only moderately activating.
2. Bromine enters ortho to the NHCOCH3 group because the para position is blocked by the methyl group.
3. Acid hydrolysis removes the acetyl group and gives 2-bromo-4-methylaniline.

Teacher's Note:
1. Acetylation prevents polybromination and oxidation of the amino group.
2. The acetyl group is removed by hydrolysis as the last step.

Answer: (a) CH₃COCH₃

Explanation:
1. C₆H₅NC is formed in the carbylamine reaction with aniline (X) and chloroform (Y).
2. Chloroform is made from acetone, ethanol or ethanal by the haloform reaction with bleaching powder (chlorine and slaked lime).
3. Among the options only acetone gives chloroform.

Teacher's Note:
1. The haloform reaction needs a methyl ketone or a group that is oxidised to one.
2. Methanol has no methyl ketone-type structure.

Answer: (d) addition of AgNO₃

Explanation:
1. Anilinium chloride is an ionic salt and gives chloride ions in solution, so AgNO₃ gives a white precipitate of AgCl.
2. 4-Chloroaniline contains covalently bound chlorine and gives no precipitate.
3. The other tests apply to primary aromatic amines and are positive for both.

Teacher's Note:
1. Look for a test that depends on free chloride ions.
2. The other three reactions target the amine, which both compounds contain.

Answer: (d) ArCONH₂

Explanation:
1. The reaction converts a primary amide to a primary amine with one carbon less.
2. Only ArCONH₂ is a primary amide.

Teacher's Note:
1. The reagent is bromine with aqueous alkali.
2. The carbonyl carbon is lost as carbonate.

Answer: (c) NaOH/Br₂

Explanation:
1. C₆H₅CH(CH₃)CONH₂ has three side-chain carbons and the product C₆H₅CH(CH₃)NH₂ has two.
2. One carbon is lost, so the Hofmann bromamide reaction (NaOH/Br₂) is needed.

Teacher's Note:
1. Count the carbons to decide between reduction and degradation.
2. The migrating group keeps its configuration.

Answer: (c) Gabriel phthalimide synthesis

Explanation:
1. Gabriel synthesis gives pure primary amines from alkyl halides.
2. Reaction with NH₃ gives a mixture of primary, secondary, tertiary amines and a quaternary salt.
3. Hofmann degradation shortens the chain and Sandmeyer reaction is for diazonium salts.

Teacher's Note:
1. Phthalimide anion can be alkylated only once.
2. Hydrolysis or hydrazinolysis then frees the primary amine.

Answer: (b) CH₃NH₂

Explanation:
1. This is the Hofmann bromamide reaction.
2. Acetamide loses its carbonyl carbon as carbonate and gives methylamine.

Teacher's Note:
1. The amine has one carbon less than the amide.
2. The reaction goes through an isocyanate.

Answer: (b) H₃PO₂

Explanation:
1. Hypophosphorous acid reduces the diazonium group and replaces it by hydrogen (deamination).
2. The other acids are not reducing agents of this type.

Teacher's Note:
1. Hypophosphorous acid and ethanol are the two common reducing agents.
2. Deamination is used to remove an amino group after it has served as a directing group.

Answer: (d) phenol

Explanation:
1. X is benzenediazonium chloride.
2. Warming a diazonium salt with water replaces the diazonium group by OH, and nitrogen is evolved.
3. Y is phenol.

Teacher's Note:
1. Diazonium salts are unstable above 5 degrees C.
2. Hydrolysis of the diazonium ion gives a phenol.

Answer: (b) Gattermann reaction

Explanation:
1. When copper powder with HCl is used the reaction is the Gattermann reaction.
2. The Sandmeyer reaction uses cuprous chloride dissolved in HCl.

Teacher's Note:
1. Remember: Sandmeyer uses Cu(I) salt, Gattermann uses copper powder.
2. Both give aryl chlorides or bromides.

Answer: (c)

Explanation:
1. The nitro group is meta-directing, so bromination gives 3-bromonitrobenzene.
2. Hydrogenation gives 3-bromoaniline (A).
3. Diazotisation and the Sandmeyer reaction with Cu₂Cl₂ give 1-bromo-3-chlorobenzene.

Teacher's Note:
1. The meta relationship set in the first step is retained.
2. Sandmeyer replacement keeps the position of the original amino group.

Answer: (b)

Explanation:
1. Both OH and NH2 are para to each other; in weakly acidic solution the phenol is un-ionised but the amine is free and is the stronger activating group.
2. Coupling therefore takes place ortho to the NH2 group (para is occupied).
3. The azo group ends up ortho to the amino group.

Teacher's Note:
1. Near pH 4-6 amines couple, while phenols couple best at pH 9-10 as the phenoxide ion.
2. Para position of the activating group is blocked, so coupling goes ortho.

Answer: (a)

Explanation:
1. In alkaline solution the phenol is converted into the phenoxide ion, a much stronger activating group than NH2.
2. The para position is occupied, so coupling occurs ortho to the O- group.
3. The azo group is ortho to OH.

Teacher's Note:
1. pH controls which group in a molecule directs coupling.
2. Phenoxide ion is the most reactive coupling partner.

Answer: (b) A-s, B-r, C-p, D-q

Explanation:
1. (A) Ammonolysis is the reaction of alkyl halides with ammonia (s).
2. (B) Gabriel synthesis uses phthalimide, KOH and then R-X (r).
3. (C) The Hofmann reaction gives an amine with one carbon less (p).
4. (D) The carbylamine reaction is a test for primary amines (q).

Teacher's Note:
1. Match each reaction with its defining reagent or feature, then check the whole code.
2. Eliminate options with one wrong pair.

Answer: (a) benzamide

Explanation:
1. 4-Bromoaniline is made from aniline by acetylation (R = acetanilide), para-bromination and hydrolysis.
2. R comes from aniline (Q) and Q is formed from P by the Hofmann reaction, so P is benzamide.

Teacher's Note:
1. Work backwards from the product.
2. Hofmann degradation turns benzamide into aniline.

Answer: (d) p-cresol

Explanation:
1. Diazotisation gives p-toluenediazonium chloride.
2. Boiling with water replaces the diazonium group by OH.
3. The product is 4-methylphenol (p-cresol).

Teacher's Note:
1. The ring methyl group remains unchanged.
2. Hydrolysis of a diazonium salt gives a phenol.