JEE Chemistry Amines MCQs Set 07

Answer: (a) \(-\)CONH₂ and \(-\)NH₂

Explanation:
1. Acetamide is dehydrated by \(\ce{PCl5}\) to methyl cyanide (A), \(\ce{CH3CN}\).
2. Partial hydrolysis of the nitrile with water gives back the amide (B), \(\ce{CH3CONH2}\), with the group \(-\)CONH₂.
3. Reduction of the amide gives ethylamine (C), \(\ce{CH3CH2NH2}\), with the group \(-\)NH₂.

Teacher's Note:
1. Amide to nitrile is dehydration, nitrile to amide is partial hydrolysis.
2. Amides and nitriles are both reduced to primary amines.

Answer: (b) Phthalic acid and methylamine

Explanation:
1. KOH removes the acidic N\(-\)H proton to give potassium phthalimide (A).
2. This reacts with methyl bromide to give N-methylphthalimide (B).
3. Acid hydrolysis of B gives phthalic acid and methylamine (Gabriel synthesis of a primary amine).

Teacher's Note:
1. The Gabriel synthesis gives only primary amines, and the alkyl group of the halide ends up on nitrogen.
2. Aromatic halides cannot be used because they do not undergo nucleophilic substitution easily.

Answer: (d) \(\ce{C6H5NO2}\)/KOH

Explanation:
1. Nitration or sulphonation of nitrobenzene, nitration of phenol and Friedel-Crafts alkylation of benzene all involve attack by an electrophile.
2. Nitrobenzene is electron-poor because of the \(-\)NO₂ group, so the hydroxide ion (a nucleophile) can attack the ring and give o-nitrophenol.
3. Hence nitrobenzene with KOH is nucleophilic aromatic substitution.

Teacher's Note:
1. An electron-withdrawing group such as \(-\)NO₂ activates the ring towards nucleophiles.
2. Electron-donating groups do the reverse.

Answer: (a) \(-\)NO₂, \(-\)NH₂, \(-\)N=N\(-\)

Explanation:
1. Sn/HCl reduces nitrobenzene (A, \(-\)NO₂) to aniline (B, \(-\)NH₂).
2. Aniline with nitrous acid at 0 \(^\circ\)C gives benzenediazonium chloride (C, \(-\)N=N\(-\) group, written as \(-\)N₂⁺).
3. Ethanol replaces the diazonium group by hydrogen and forms benzene.

Teacher's Note:
1. Reduction, diazotisation and then replacement by hydrogen is a standard way to remove an amino group.
2. Diazonium salts must be prepared at 0 to 5 \(^\circ\)C.

Answer: (d) Dimethylamine is less basic than methylamine

Explanation:
1. Alkyl groups push electrons towards nitrogen, so methylamine is a stronger base than ammonia.
2. Primary and secondary amines form N\(-\)H\(\cdots\)N hydrogen bonds, so ethylamine boils higher than propane of nearly the same molar mass.
3. In water the secondary amine dimethylamine (pK_b = 3.27) is more basic than methylamine (pK_b = 3.38), so statement (d) is wrong.

Teacher's Note:
1. In aqueous solution the order of basic strength of methyl amines is \((\ce{CH3})_2\ce{NH} > \ce{CH3NH2} > (\ce{CH3})_3\ce{N} > \ce{NH3}\).
2. This is a balance of the +I effect and solvation of the ammonium ion.

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

Explanation:
1. The carbylamine test (chloroform and alcoholic KOH, heat) is given only by primary amines, which form foul-smelling isocyanides.
2. Ethylamine is a primary aliphatic amine.
3. Urea and acetamide are amides, which do not give the test.

Teacher's Note:
1. Secondary and tertiary amines do not give the carbylamine test.
2. The foul smell of the isocyanide is the sign of a positive test.

Answer: (a) Electrophilic substitution

Explanation:
1. The diazonium cation is a weak electrophile.
2. It attacks the electron-rich ring of the naphtholate ion at the carbon next to the \(-\)OH group and replaces hydrogen.
3. This coupling is therefore an electrophilic aromatic substitution.

Teacher's Note:
1. Coupling with phenols is done in mildly alkaline medium and with amines in mildly acidic medium.
2. The product contains the \(-\)N=N\(-\) azo group, which makes it coloured.

Answer: (a) Benzene

Explanation:
1. The Schotten-Baumann reaction is the benzoylation of compounds containing \(-\)OH or \(-\)NH₂ groups with benzoyl chloride in the presence of aqueous NaOH.
2. Phenol, aniline and benzyl alcohol all fit this description.
3. Benzene has no such group; its benzoylation is a Friedel-Crafts acylation using \(\ce{AlCl3}\).

Teacher's Note:
1. Schotten-Baumann conditions need a base to take up the HCl formed.
2. Friedel-Crafts acylation needs a Lewis acid instead.

Answer: (c) Bimolecular nucleophilic substitution reaction

Explanation:
1. Ammonia, with its lone pair, is the nucleophile.
2. It attacks the carbon of the primary alkyl halide from the back and bromide leaves in a single step.
3. The rate depends on both reactants, so this is an S_N2 (bimolecular nucleophilic substitution) reaction.

Teacher's Note:
1. A methyl halide is too unhindered for S_N1 and has no tendency to form a carbocation.
2. Excess ammonia stops further alkylation of the amine.

Answer: (b) A: primary amine; B: methyl ketone; C: primary amine; D: alcohol

Explanation:
1. Reduction of a nitrile gives the primary amine \(\ce{RCH2NH2}\) (A).
2. A Grignard reagent adds to the nitrile and hydrolysis of the imine formed gives the methyl ketone \(\ce{RCOCH3}\) (B).
3. Hydrolysis of an isocyanide gives a primary amine \(\ce{RNH2}\) (C) and formic acid, and a primary aliphatic amine with nitrous acid gives an alcohol (D) with release of \(\ce{N2}\).

Teacher's Note:
1. Nitriles and isocyanides both lead to primary amines.
2. Nitrous acid converts primary aliphatic amines into alcohols.

Answer: (c) Cyclohexylamine

Explanation:
1. A Schiff base (imine) has the group \(>\)C=N\(-\)R and needs a primary amine, \(\ce{RNH2}\), which loses water with the carbonyl group.
2. Cyclohexylamine is the only primary amine here.
3. Pyrrolidine and N-methylcyclohexanamine are secondary amines, which give enamines, and the tertiary amine does not react.

Teacher's Note:
1. Primary amine gives an imine, secondary amine gives an enamine.
2. Tertiary amines have no N\(-\)H and cannot form a stable condensation product.

Answer: (a) Cyclohex-2-en-1-amine

Explanation:
1. Release of \(\ce{N2}\) with nitrous acid shows a primary aliphatic amine, so the N-methyl and N-ethyl compounds (secondary amines) are ruled out.
2. Decolourising bromine water in \(\ce{CCl4}\) shows a C=C bond.
3. Cyclohexa-2,5-dien-1-amine has a plane of symmetry through C-1 and C-4, so it is optically inactive; cyclohex-2-en-1-amine has a stereocentre at C-1 and is optically active.

Teacher's Note:
1. A compound with a mirror plane is optically inactive.
2. Check for the primary amine first, then for chirality.