JEE Chemistry Amines MCQs Set 30

Answer: (c) methylaniline

Explanation:
1. The compound has a methyl group on nitrogen of aniline.
2. Its correct names are N-methylaniline (common) and N-methylbenzenamine (IUPAC).
3. 'Methylaniline' without the locant N- is ambiguous, because it could mean a methyl group on the ring (a toluidine).

Teacher's Note:
1. Always write the locant N- when a substituent sits on the nitrogen atom of an amine.
2. Substituents on the ring get numbers; those on nitrogen get N.

Answer: (b) Gabriel phthalimide synthesis

Explanation:
1. Direct ammonolysis gives a mixture of 1°, 2°, 3° amines and a quaternary salt.
2. In Gabriel synthesis potassium phthalimide is alkylated once (the N has only one reactive N-H site) and then hydrolysed or treated with hydrazine.
3. Only the primary amine is obtained, free from 2° and 3° amines.

Teacher's Note:
1. Gabriel synthesis works well with primary alkyl halides.
2. It cannot be used for aryl amines because aryl halides do not undergo SN2.

Answer: (b) \( \ce{(CH3CH2)4N+OH^-} \)

Explanation:
1. Tetraethylammonium bromide is a salt of a strong base and strong acid, so it is neutral in water.
2. Tetraethylammonium hydroxide is a quaternary ammonium hydroxide, fully ionised to give \( \ce{OH^-} \).
3. So only the hydroxide is strongly basic.

Teacher's Note:
1. Quaternary ammonium hydroxides are strong bases like NaOH.
2. Their salts with halide ions are neutral.

Answer: (d) \( \ce{(CH3CH2)3N} \)

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

Teacher's Note:
1. In water the order changes (2° is often the strongest) because of solvation of the ammonium ion.
2. Gas-phase basicity follows the number of alkyl groups.

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

Explanation:
1. Boiling points: \( \ce{(CH3)3N} \) about 3 °C, \( \ce{CH3NHC2H5} \) about 37 °C, \( \ce{n-C3H7NH2} \) about 48 °C, methanol about 65 °C.
2. Tertiary amines cannot form hydrogen bonds with each other (no N-H), so they have the lowest boiling point among isomers.
3. Hence trimethylamine is the most volatile.

Teacher's Note:
1. Isomeric amines: boiling point 1° > 2° > 3°.
2. Methanol and amines with N-H are held together by hydrogen bonding.

Answer: (a) I > II > III

Explanation:
1. Oxygen is more electronegative than nitrogen, so the C-O and O-H bonds are more polar than C-N and N-H.
2. Ethanol has a dipole moment of about 1.7 D, ethylamine about 1.2 D.
3. Propane is almost non-polar (about 0.1 D).

Teacher's Note:
1. Dipole moment increases with the electronegativity difference in the polar bonds.
2. Alkanes are practically non-polar.

Answer: (a) \( \ce{FCH2CH2NH2} \)

Explanation:
1. Electron-withdrawing groups reduce the electron density on nitrogen and so reduce basicity.
2. Fluorine has a stronger -I effect than chlorine, and both exceed hydrogen.
3. So the order is \( \ce{CH3CH2NH2} \) > \( \ce{ClCH2CH2NH2} \) > \( \ce{FCH2CH2NH2} \).

Teacher's Note:
1. The stronger the -I group on the carbon chain, the weaker the amine.
2. -I effect: F > Cl > Br > I.

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

Explanation:
1. Methylamine has a free lone pair on nitrogen and is a base: it forms \( \ce{CH3NH3+} \) salts.
2. In acetamide the lone pair is delocalised into the carbonyl group, so it is only very weakly basic and does not form stable salts in aqueous acid.

Teacher's Note:
1. Amides are almost neutral because of resonance.
2. Alkylamines are moderately strong bases.

Answer: (b) ethyl chloroacetate

Explanation:
1. Potassium phthalimide is a base, so it would simply deprotonate the free \( \ce{ClCH2COOH} \) and no substitution would occur.
2. With ethyl chloroacetate the ester has no acidic proton, so SN2 gives N-phthalimidoacetic ester.
3. Hydrolysis then gives glycine.

Teacher's Note:
1. Avoid free acids in the alkylation step of Gabriel synthesis.
2. The ester group is hydrolysed in the last step.

Answer: (c) \( \ce{Me2C(OH)CH2CH3} \)

Explanation:
1. A primary amine with \( \ce{HONO} \) gives a diazonium ion that loses \( \ce{N2} \) to give a carbocation.
2. Neopentyl carbocation \( \ce{Me3CCH2+} \) is primary and unstable; a methyl shift gives the tertiary \( \ce{Me2C+CH2CH3} \).
3. Water attacks this to give \( \ce{Me2C(OH)CH2CH3} \) (2-methylbutan-2-ol).

Teacher's Note:
1. Aliphatic diazonium ions are very unstable and give rearranged carbocation products.
2. A 1,2-methyl shift converts the 1° cation into a 3° cation.

Answer: (c) \( \ce{CH3-N+#C^-} \)

Explanation:
1. This is the carbylamine (isocyanide) reaction, given only by primary amines.
2. Dichlorocarbene from chloroform and KOH reacts with the amine.
3. The product is methyl isocyanide, \( \ce{CH3-N+#C^-} \), with a foul smell.

Teacher's Note:
1. Carbylamine reaction is the test for primary amines (aliphatic and aromatic).
2. The isocyanide has N attached to the alkyl group.

Answer: (b) 1 > 3 > 2 > 4

Explanation:
1. Acetamidine on protonation gives a resonance-stabilised amidinium ion (two equivalent forms), so it is the strongest base (pKa of conjugate acid about 12).
2. Dimethylamine (about 10.7) is slightly stronger than ethylamine (about 10.6) in water.
3. Acetamide is a very weak base (the lone pair is delocalised).
4. Order: 1 > 3 > 2 > 4.

Teacher's Note:
1. Amidines and guanidine are strong bases due to resonance in their conjugate acids.
2. Amides are the weakest bases in the list.

Answer: (d) Dimethylamine

Explanation:
1. Aniline is weak because its lone pair is delocalised into the benzene ring.
2. Among methyl amines in water the order of pKb is \( \ce{(CH3)2NH} \) (3.27) < \( \ce{CH3NH2} \) (3.38) < \( \ce{(CH3)3N} \) (4.22).
3. So dimethylamine is the strongest base.

Teacher's Note:
1. In water, +I effect and solvation both matter; 2° amines come out strongest for methyl series.
2. Aryl amines are weaker than aliphatic amines.

Answer: (b) \( \ce{NH3 < CH3NH2 < (CH3)2NH} \)

Explanation:
1. Methyl groups donate electrons (+I) and increase electron density on N.
2. Ammonia has none, methylamine one, dimethylamine two.
3. So basicity increases: \( \ce{NH3 < CH3NH2 < (CH3)2NH} \).

Teacher's Note:
1. Replacing H on N by CH3 increases basicity in this series.
2. In water the trend reverses after two methyl groups (3° is weaker).

Answer: (d) ethylamine salt and methanoic acid

Explanation:
1. Isocyanides are hydrolysed by dilute acids, not by alkali.
2. The N-bonded alkyl group stays on nitrogen: \( \ce{C2H5NC + 2H2O -> C2H5NH2 + HCOOH} \).
3. In acid the amine is obtained as its salt.

Teacher's Note:
1. Isocyanide hydrolysis gives a primary amine plus formic acid.
2. Nitriles, in contrast, give carboxylic acids and ammonia.

Answer: (b) Wurtz reaction

Explanation:
1. Curtius (acyl azide) and Hofmann (amide) degradations give amines.
2. Hinsberg reagent separates 1°, 2° and 3° amines.
3. The Wurtz reaction couples alkyl halides with sodium to form alkanes; it has nothing to do with amines.

Teacher's Note:
1. Know the named reactions: Hofmann, Curtius, Schmidt, Lossen all give amines.
2. Hinsberg's reagent is benzenesulphonyl chloride.

Answer: (b) \( \ce{HN3} \)

Explanation:
1. This is the Schmidt reaction: a carboxylic acid reacts with hydrazoic acid in conc. \( \ce{H2SO4} \).
2. An acyl azide-type intermediate rearranges to an isocyanate, which hydrolyses.
3. The product is an amine with one carbon less, so X is \( \ce{HN3} \).

Teacher's Note:
1. Schmidt reaction: RCOOH + HN3 gives RNH2 + N2 + CO2.
2. Ammonia alone does not give the amine.

Answer: (b) cold dilute HCl

Explanation:
1. Benzylamine is a basic amine and dissolves in cold dilute HCl as \( \ce{C6H5CH2NH3+Cl^-} \).
2. Benzamide is a neutral amide and does not dissolve in cold dilute HCl.
3. Cold dilute NaOH reacts with neither.

Teacher's Note:
1. Amines dissolve in dilute acids, amides do not.
2. Heating with NaOH would hydrolyse benzamide and release ammonia, but not in the cold.

Answer: (d) an alkyl group with its electron pair to electron-deficient N atom

Explanation:
1. The N-bromoamide anion loses \( \ce{Br^-} \) to give an acyl nitrene (electron-deficient N).
2. The alkyl or aryl group migrates from carbon to nitrogen with its bonding electron pair, with retention of configuration.
3. This gives the isocyanate, which is hydrolysed to the amine.

Teacher's Note:
1. The migrating group keeps its electron pair and its stereochemistry.
2. Same rearrangement happens in Curtius, Schmidt and Lossen reactions.

Answer: (c) \( \ce{(CH3)3C-ND2} \)

Explanation:
1. Hofmann degradation converts the amide into \( \ce{(CH3)3C-NH2} \) (one carbon less).
2. In basic \( \ce{D2O} \) the N-H hydrogens exchange rapidly with D.
3. So the amine is obtained as \( \ce{(CH3)3C-ND2} \).

Teacher's Note:
1. Hofmann degradation retains the alkyl group structure, so tert-butyl stays unrearranged.
2. N-H protons exchange quickly with deuterated solvent.

Answer: (b) \( \ce{R-N=C=O} \)

Explanation:
1. In all four reactions an acyl nitrene (or equivalent) rearranges by migration of R from C to N.
2. The common product of this rearrangement is the isocyanate, \( \ce{R-N=C=O} \).
3. Hydrolysis of the isocyanate then gives the amine.

Teacher's Note:
1. Isocyanate hydrolysis gives carbamic acid, which loses \( \ce{CO2} \).
2. Hofmann: amide; Curtius: acyl azide; Schmidt: acid + \( \ce{HN3} \); Lossen: hydroxamic acid.

Answer: (b) Trimethylamine forms hydrogen bond neither with itself nor with water

Explanation:
1. Lower amines are soluble in water because they hydrogen bond with water, so (a) is true.
2. Trimethylamine has no N-H, so it cannot donate hydrogen bonds, but its lone pair accepts hydrogen bonds from water; (c) is true.
3. Statement (b) says there is no hydrogen bonding with water, which is false.

Teacher's Note:
1. A tertiary amine can accept but not donate a hydrogen bond.
2. Therefore it does not self-associate but dissolves in water.

Answer: (c) \( \ce{(CH3)3N + CH3OH} \)

Explanation:
1. Excess methyl iodide gives the quaternary salt [A] = \( \ce{(CH3)4N+I^-} \) (exhaustive methylation).
2. AgOH gives [B] = \( \ce{(CH3)4N+OH^-} \).
3. Heating gives Hofmann elimination if a beta-H is present; here no beta-hydrogen exists, so substitution occurs: \( \ce{(CH3)3N + CH3OH} \).

Teacher's Note:
1. Hofmann elimination needs a beta-hydrogen on the alkyl group.
2. With only methyl groups the hydroxide attacks the methyl carbon (SN2).

Answer: (b) \( \ce{H-CO-N(CH3)2} \)

Explanation:
1. N,N-dimethylformamide is hydrolysed to formic acid and dimethylamine, \( \ce{(CH3)2NH} \), a secondary amine.
2. Alkyl cyanides give acids (and ammonia).
3. A primary nitroparaffin gives a carboxylic acid and hydroxylamine; \( \ce{RCONH2} \) gives acid and ammonia.

Teacher's Note:
1. Hydrolysis of an N,N-disubstituted amide releases the secondary amine.
2. Check which nitrogen substituents are present in the amide.

Answer: (c) \( \ce{R-CN + H2O ->} \)

Explanation:
1. Ammonolysis of alkyl halide gives amines.
2. Reduction of an aldoxime with Na/ethanol gives a primary amine.
3. Hydrolysis of alkyl cyanide gives a carboxylic acid (via the amide) and not an amine.
4. Reduction of an amide gives an amine.

Teacher's Note:
1. Reduction converts C=N, C#N, C=O (in amides) to amines.
2. Hydrolysis of nitriles gives acids.