CBSE Class 12 Chemistry Amines MCQs Set 10

Answer: (b) Gabriel phthalimide synthesis

Explanation:
1. Gabriel phthalimide synthesis is used to get primary amines from alkyl halides without changing the number of carbon atoms.
2. Phthalimide \( \xrightarrow{\text{KOH (alc)}} \) Potassium phthalimide \( \xrightarrow{\text{R–X}} \) N-Alkylphthalimide \( \xrightarrow{\text{NaOH/H}_{2}\text{O}} \) Sodium phthalate + R–NH₂ (primary amine)

Teacher's Note:
1. Reaction with NH₃ also keeps the carbon count but gives a mixture of 1°, 2°, 3° amines and quaternary salt, so it is not the best method.

Answer: (d) N,N-Dimethylbutan-1-amine

Explanation:
1. ⁴CH₃–³CH₂–²CH₂–¹CH₂–N(CH₃)₂
2. IUPAC name: N,N-Dimethylbutan-1-amine

Teacher's Note:
1. Methyl groups on nitrogen are not part of the parent chain; they are shown with the locant N, once for each group (N,N-).

Answer: (d) N-methylprop-2-en-1-amine

Explanation:
1. CH₂=CHCH₂NHCH₃
2. IUPAC name: N-methylprop-2-en-1-amine

Teacher's Note:
1. Number the chain from the carbon bearing –NH– (C-1), so the double bond gets locant 2; “allylmethylamine” is only the common name.

Answer: (b) Triethylamine

Explanation:
1. Triethylamine [(C₂H₅)₃N] is a 3° or tertiary amine as nitrogen atom contains three ethyl groups.

Teacher's Note:
1. The degree of an amine depends on the number of carbon groups on nitrogen, not on the carbon bearing it; tert-butylamine is a primary amine.

Answer: (c) CH₃OH

Explanation:
1. Methylamine reacts with HNO₂ to form CH₃OH.
2. CH₃NH₂ + HNO₂ \( \xrightarrow{\text{NaNO}_{2}\text{ + HCl}} \) [CH₃–N₂⁺Cl⁻] \( \xrightarrow{\text{H}_{2}\text{O}} \) CH₃OH + N₂ + HCl

Teacher's Note:
1. Aliphatic diazonium salts are unstable and decompose at once, giving the alcohol and N₂ gas.

Answer: (b) N₂

Explanation:
1. Nitrogen gas is evolved.
2. R–NH₂ + HNO₂ \( \xrightarrow[\text{H}_{2}\text{O}]{\text{NaNO}_{2}\text{ + HCl}} \) R–OH + N₂↑ + HCl

Teacher's Note:
1. Brisk effervescence of N₂ with nitrous acid is the test for aliphatic primary amines.

Answer: (d) p-Nitroaniline < Aniline < p-Toluidine

Explanation:
1. The increasing order of basic strength is: p-Nitroaniline < Aniline < p-Toluidine
2. p-Nitroaniline: −I effect of −NO₂ group decreases the electron density on N-atom of −NH₂ group, so, it does not undergo protonation easily and hence, it is least basic amine.
3. Aniline: it does not contain −I or +I group.
4. p-Toluidine: +I effect of CH₃ group increases the electron density on the nitrogen atom of −NH₂ group, so, it undergoes protonation easily and hence, it is most basic amine.

Teacher's Note:
1. Use aniline as the reference point; then place the electron-withdrawing derivative below it and the electron-donating one above it.

Answer: (a) R–NC

Explanation:
1. In the carbylamine test, a primary amine reacts with chloroform and KOH to form alkyl isocyanide (i.e., R–NC) having unpleasant smell.

Teacher's Note:
1. Only primary amines (aliphatic or aromatic) give the carbylamine test; the foul smell of R–NC is the observation to write.

Answer: (a) Isobutyl amine

Explanation:
1. Gabriel phthalimide synthesis cannot be used for preparation of aromatic amines, as aromatic halides do not undergo nucleophilic substitution by salt formed by phthalimide.
2. It gives only primary aliphatic amines such as isobutyl amine, (CH₃)₂CHCH₂NH₂.

Teacher's Note:
1. Gabriel synthesis gives only 1° aliphatic amines: N-methylbenzylamine is a 2° amine and aniline is aromatic.

Answer: (b) (iv)

Explanation:
1. Primary and secondary amines form hydrogen bonds and hence are less volatile than corresponding alkanes.
2. Propane (iv) has the weakest intermolecular forces (no polarity, no H-bonding), so it is the most volatile.

Teacher's Note:
1. Most volatile means lowest boiling point; the compound with no N–H and no polar bond wins.

Answer: (c) Potassium cyanide, KCN

Explanation:
1. R–X + KCN → R–CN; reduction of the nitrile gives R–CH₂NH₂, a 1° amine with one extra carbon.

Teacher's Note:
1. Nitrile reduction is the ascent route (one C more); Hoffmann bromamide is the descent route (one C less).

Answer: (b) NaOH/Br₂

Explanation:
1. The product has one carbon less than the amide, so Hoffmann bromamide degradation (Br₂/NaOH) is used.
2. C₆H₅CH(CH₃)CONH₂ + Br₂ + 4NaOH → C₆H₅CH(CH₃)NH₂ + Na₂CO₃ + 2NaBr + 2H₂O

Teacher's Note:
1. Compare this with MCQ 3: same starting amide, but a different product decides the reagent.

Answer: (d) CH₃—CH₂—COOH

Explanation:
1. CH₃CH₂COOH + NH₃ → CH₃CH₂COO⁻NH₄⁺ (B) →(heat) CH₃CH₂CONH₂ (C)
2. CH₃CH₂CONH₂ + Br₂ + KOH → CH₃CH₂NH₂ (ethylamine). So A is propanoic acid.

Teacher's Note:
1. Work backwards: Hoffmann degradation loses one carbon, so the acid must have one carbon more than ethylamine.

Answer: (d) Potassium phthalimide, C₆H₄(CO)₂N⁻K⁺

Explanation:
1. Phthalimide with ethanolic KOH forms potassium phthalimide; this is alkylated by R–X and the product is hydrolysed to give RNH₂.

Teacher's Note:
1. Remember the three steps of Gabriel synthesis: KOH, then R–X, then alkaline hydrolysis.

Answer: (d) Arylamines react with nitrous acid to produce phenols.

Explanation:
1. Primary arylamines react with nitrous acid at 273–278 K to give arenediazonium salts, not phenols. Statements (a), (b) and (c) are true.

Teacher's Note:
1. Only aliphatic diazonium salts decompose at once (giving alcohols); arenediazonium salts are stable in the cold.

Answer: (c) (CH₃)₂NH

Explanation:
1. In aqueous medium, dimethylamine is the strongest base among these due to +I effect of two methyl groups and good solvation of its cation.
2. The –CN group (−I) and the phenyl ring (resonance) lower the basicity of (b) and (d).

Teacher's Note:
1. For methyl amines in water the order is 2° > 1° > 3° > NH₃.

Answer: (a) Aniline, C₆H₅NH₂

Explanation:
1. In aniline, the lone pair on nitrogen is delocalised over the benzene ring, so it is least available for accepting a proton. The others are aliphatic amines.

Teacher's Note:
1. A cyclohexyl ring is not aromatic – cyclohexylamine behaves like an ordinary alkylamine.

Answer: (d) II < I < III

Explanation:
1. –NO₂ (electron-withdrawing) decreases and –CH₃ (electron-donating) increases the electron density on N.
2. Hence the order is p-nitroaniline < aniline < p-toluidine, i.e., II < I < III.

Teacher's Note:
1. Use aniline as the reference point and move substituted anilines up (EDG) or down (EWG).

Answer: (b) (CH₃)₂NH

Explanation:
1. Reactivity towards an acid depends on basic strength. In aqueous medium (CH₃)₂NH is the strongest base, so it reacts most readily.

Teacher's Note:
1. Dilute HCl means aqueous medium, so use the aqueous basicity order, not the gas-phase order.

Answer: (b) RI > RBr > RCl

Explanation:
1. The C–X bond strength decreases from C–Cl to C–I, so I⁻ is the best leaving group. Order: RI > RBr > RCl.

Teacher's Note:
1. The same order holds for most nucleophilic substitution reactions of alkyl halides.

Answer: (c) C₆H₅CH₂Br

Explanation:
1. S_N1 depends on the stability of the carbocation. C₆H₅CH₂Br gives the benzyl carbocation, which is stabilised by resonance.

Teacher's Note:
1. Aryl halides like C₆H₅Br do not undergo S_N1 – the phenyl cation is very unstable.

Answer: (a) primary amines

Explanation:
1. Only primary amines (aliphatic and aromatic) on heating with chloroform and ethanolic KOH give foul-smelling isocyanides.

Teacher's Note:
1. Secondary and tertiary amines give no carbylamine reaction, so the test identifies 1° amines.

Answer: (b) CH₃NH₂

Explanation:
1. A primary amine gives N-alkylbenzenesulphonamide, C₆H₅SO₂NHCH₃, whose N–H is acidic, so it dissolves in alkali.
2. 2° amines give sulphonamides with no H on N (insoluble in alkali); 3° amines do not react.

Teacher's Note:
1. Hinsberg test summary: 1° – soluble in alkali, 2° – insoluble solid, 3° – no reaction.

Answer: (c) NO₂⁺

Explanation:
1. HNO₃ + 2H₂SO₄ → NO₂⁺ + H₃O⁺ + 2HSO₄⁻; the nitronium ion is the electrophile.

Teacher's Note:
1. In the nitrating mixture HNO₃ behaves as a base, accepting a proton from H₂SO₄.

Answer: (d) All of the above

Explanation:
1. Nitrobenzene is reduced to aniline by Sn/HCl, by Fe/HCl and by catalytic hydrogenation with H₂/Pd. Hence all three reagents give aniline.

Teacher's Note:
1. Fe/HCl is preferred in industry because only a small amount of acid is needed.

Answer: (d) Fluorobenzene

Explanation:
1. In Sandmeyer reaction, –N₂⁺ is replaced by Cl, Br or CN using CuCl/HCl, CuBr/HBr or CuCN/KCN.
2. Fluorobenzene is prepared by the Balz–Schiemann reaction (via ArN₂⁺BF₄⁻) and iodobenzene by warming with KI.

Teacher's Note:
1. Remember: Cl, Br, CN – Sandmeyer (Cu(I) salts); I – KI; F – HBF₄, then heat.

Answer: (b) ArCONH₂

Explanation:
1. Hoffmann Bromamide degradation reaction is shown by ArCONH₂ by which amide is converted into amine via undergoing intramolecular migration of phenyl group.
2. C₆H₅CONH₂ (Benzamide) + Br₂ + 4KOH → C₆H₅NH₂ (Aniline) + K₂CO₃ + 2KBr + 2H₂O

Teacher's Note:
1. Only primary amides (–CONH₂) give this reaction; the aryl or alkyl group migrates from carbon to nitrogen.

Answer: (c) A is true, but R is false.

Explanation:
1. Lower aliphatic amines are soluble in water because they can form hydrogen bonds with water molecules. However, solubility decreases with increase in molar mass of amines due to increase in size of the hydrophobic alkyl part.

Teacher's Note:
1. The –NH₂ group forms similar hydrogen bonds in all amines; it is the growing hydrophobic hydrocarbon part that lowers solubility.

Answer: (c) A is true, but R is false.

Explanation:
1. Amides on treatment with Br₂ and NaOH give primary amines with one carbon atom less, so A is correct.
2. In aqueous solution secondary amines are generally more basic than primary amines, so R is incorrect.

Teacher's Note:
1. Hoffmann bromamide degradation is a reaction of amides; the primary amine is the product, not the reactant.

Answer: (c) A is true, but R is false.

Explanation:
1. Gabriel phthalimide synthesis gives pure primary amines only; it cannot be used for 2° or 3° amines. So A is correct and R is incorrect.

Teacher's Note:
1. The phthalimide nitrogen has only one replaceable H, so only one alkyl group can be introduced.

Answer: (c) A is true, but R is false.

Explanation:
1. Oxygen is more electronegative than nitrogen, so alcohols are more polar than amines and form stronger hydrogen bonds with water. A is correct; R is incorrect.

Teacher's Note:
1. Hydrogen bonding of O–H is stronger than that of N–H – the same reason alcohols boil higher than amines of similar mass.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. FeCl₂ formed in the reaction is hydrolysed by steam and releases HCl, which is used again in the reduction. Therefore only a small amount of HCl is needed to start the reaction. R explains A.

Teacher's Note:
1. This is why Fe/HCl is preferred over Sn/HCl for reducing nitro compounds on a large scale.

Answer: (c) A is true, but R is false.

Explanation:
1. Ammonium sulphide is a mild reducing agent that selectively reduces only one nitro group, giving m-nitroaniline. The reason given is incorrect.

Teacher's Note:
1. Selective reduction of one –NO₂ group is done with (NH₄)₂S or NaHS; Sn/HCl would reduce both groups.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. C₆H₅SO₂NHC₂H₅ has an H on nitrogen which is made strongly acidic by the electron-withdrawing sulphonyl group. It reacts with alkali to form a soluble salt. R explains A.

Teacher's Note:
1. This solubility in alkali is the basis of identifying primary amines in the Hinsberg test.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. n-Propylamine (a primary amine) has N–H bonds and forms intermolecular hydrogen bonds.
2. Trimethylamine has no H on nitrogen, so it cannot form hydrogen bonds; hence its boiling point is lower. The reason explains the assertion.

Teacher's Note:
1. For isomeric amines the boiling point order is 1° > 2° > 3°, because of the number of N–H bonds available for hydrogen bonding.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. Both statements are true. Aniline does not undergo Friedel–Crafts reaction because, being a base, it forms a salt with the Lewis acid AlCl₃; the positive nitrogen then deactivates the ring.
2. The reason only states the type of reaction and does not explain why aniline fails to react.

Teacher's Note:
1. The real cause is the acid–base reaction between –NH₂ and AlCl₃ – mention salt formation in written answers.