JEE Chemistry Amines MCQs Set 19

Answer: (b) benzylamine

Explanation:
1. In benzylamine the nitrogen is not attached to the ring and its lone pair is localised, so it is a typical aliphatic amine.
2. In aniline the lone pair is delocalised, in p-nitroaniline even more, and in acetanilide it is drawn towards the carbonyl group.

Teacher's Note:
1. Benzylamine (pKb about 4.7) is about 100,000 times more basic than aniline (pKb 9.4).

Answer: (a) (i), (ii) and (iv) only

Explanation:
1. Statements (i), (ii) and (iv) are correct.
2. (iii) is wrong: in the cold, aryl amines give diazonium salts; phenols form only when these salts are warmed with water.

Teacher's Note:
1. Aryl amines give diazonium salts at 273-278 K, which are the starting point for many substitutions.

Answer: (b) base

Explanation:
1. Sulphuric acid is the stronger acid and protonates nitric acid: \(\ce{HNO3 + H2SO4 -> H2NO3+ + HSO4^-}\).
2. Nitric acid therefore acts as a base, and loses water to give the nitronium ion \(\ce{NO2+}\).

Teacher's Note:
1. Sulphuric acid is the catalyst; nitric acid is the base in the nitrating mixture.

Answer: (d) \(\ce{C6H5CH2OH}\)

Explanation:
1. Diazotisation gives benzenediazonium chloride (A).
2. CuCN gives benzonitrile (B).
3. Hydrogenation gives benzylamine (C).
4. A primary aliphatic amine with nitrous acid gives the alcohol, so D = benzyl alcohol.

Teacher's Note:
1. Benzylamine behaves as an aliphatic amine towards nitrous acid, giving the alcohol and nitrogen.

Answer: (c) 4-(dimethylamino)azobenzene

Explanation:
1. The diazonium ion couples at the para position of N,N-dimethylaniline (which is strongly activated by the dimethylamino group).
2. The product is \(\ce{C6H5-N=N-C6H4-N(CH3)2}\), 4-(dimethylamino)azobenzene, a coloured azo dye.

Teacher's Note:
1. Azo dyes have the characteristic -N=N- chromophore joining two aromatic rings.

Answer: (d) an alkyl isocyanide

Explanation:
1. This is the carbylamine reaction.
2. The primary amine reacts with dichlorocarbene to form an alkyl isocyanide (foul-smelling).

Teacher's Note:
1. The carbylamine test is a test for primary amines, aliphatic and aromatic.

Answer: (a) \(\ce{(CH3)2NH}\)

Explanation:
1. A smaller pKb means a stronger base.
2. In aqueous solution dimethylamine (pKb 3.27) is the strongest among these; methylamine is 3.38, trimethylamine 4.22 and aniline 9.38.

Teacher's Note:
1. pKb for methylamines in water: secondary < primary < tertiary.

Answer: (c) 5

Explanation:
1. Each acetylation replaces an N-H hydrogen with \(\ce{COCH3}\), adding 42 amu.
2. Number of amino groups = (390 - 180)/42 = 5.

Teacher's Note:
1. Mass gain per acetyl group = 43 - 1 = 42.

Answer: (c) \(\ce{C2H5NC}\) and 3KCl

Explanation:
1. This is the carbylamine reaction: \(\ce{RNH2 + CHCl3 + 3KOH -> RNC + 3KCl + 3H2O}\).
2. The products are ethyl isocyanide and potassium chloride.

Teacher's Note:
1. Balance the equation: three Cl atoms combine with three K atoms.

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

Explanation:
1. In the gas phase there is no solvation, so only the +I effect matters.
2. Each methyl group increases the electron density on nitrogen, so basicity rises: NH3 < CH3NH2 < (CH3)2NH (< (CH3)3N).

Teacher's Note:
1. In the gas phase the order is tertiary > secondary > primary > NH3; in water, solvation changes the order.

Answer: (a) \(\ce{RCONHBr}\) and \(\ce{RNCO}\)

Explanation:
1. The amide is first brominated at nitrogen to give N-bromoamide RCONHBr.
2. Base removes a proton, and the alkyl group migrates from carbon to nitrogen with loss of bromide, giving the isocyanate RNCO.
3. Hydrolysis of RNCO gives the amine and carbonate.

Teacher's Note:
1. The key step of the Hofmann rearrangement is migration of R from C to N.

Answer: (d) 2,4,6-tribromofluorobenzene

Explanation:
1. Aniline with bromine water gives 2,4,6-tribromoaniline.
2. Diazotisation gives the diazonium salt and \(\ce{HBF4}\) gives the tetrafluoroborate.
3. Heating the dry salt (Balz-Schiemann reaction) replaces the diazonium group by fluorine, giving 2,4,6-tribromofluorobenzene.

Teacher's Note:
1. Balz-Schiemann reaction introduces fluorine into the ring.

Answer: (a) Schiff's base

Explanation:
1. Condensation of a primary amine with an aldehyde gives an imine \(\ce{PhCH=NPh}\) (benzylideneaniline), called a Schiff base.
2. The other names are reagents, not products.

Teacher's Note:
1. Schiff bases (azomethines) contain the C=N group.

Answer: (c) 4-methylbenzonitrile

Explanation:
1. Diazotisation gives 4-methylbenzenediazonium chloride (D).
2. In the Sandmeyer reaction \(\ce{CuCN}\) replaces the diazonium group by -CN, giving 4-methylbenzonitrile (E).

Teacher's Note:
1. Sandmeyer reaction: CuCl, CuBr and CuCN replace the diazonium group by Cl, Br and CN.

Answer: (c) four moles of NaOH and one mole of \(\ce{Br2}\)

Explanation:
1. \(\ce{RCONH2 + Br2 + 4NaOH -> RNH2 + Na2CO3 + 2NaBr + 2H2O}\).
2. The balanced equation uses four moles of NaOH and one mole of bromine per mole of amine.

Teacher's Note:
1. Check the balance: 4 Na on the left equals 2 (Na2CO3) + 2 (NaBr) on the right.

Answer: (a) (iii) and (iv) only

Explanation:
1. The Hofmann rearrangement needs a primary amide, with two hydrogens on nitrogen.
2. \(\ce{RCONH2}\) and benzamide qualify; the N-methylamide and the ammonium salt do not.

Teacher's Note:
1. Primary amide only: RCONH2.

Answer: (c) 4

Explanation:
1. Balance bromine first: one Br2 gives 2 NaBr, and carbon ends up as \(\ce{Na2CO3}\), needing 2 more Na.
2. Total Na on the right = 2 + 2 = 4, so x = 4.
3. Balanced: \(\ce{RCONH2 + 4NaOH + Br2 -> RNH2 + 2NaBr + Na2CO3 + 2H2O}\).

Teacher's Note:
1. Balance Br first (two Br atoms need 2 NaBr), then Na.

Answer: (c) 3

Explanation:
1. \(\ce{C6H5NH2 + CHCl3 + 3KOH -> C6H5NC + 3KCl + 3H2O}\).
2. Three moles of KOH neutralise the three HCl units lost from chloroform and the amine.

Teacher's Note:
1. The carbylamine reaction consumes three moles of base per mole of amine.

Answer: (b) 5

Explanation:
1. Propylamine, guanidine, acetamidine and cyclohexylamine are strong bases (localised lone pair or resonance stabilised cation).
2. p-Anisidine is more basic than aniline because of the +R effect of the methoxy group.
3. Benzamide, p-nitroaniline, p-chloroaniline and p-aminobenzaldehyde are weaker.
4. Total = 5.

Teacher's Note:
1. Guanidine is one of the strongest organic bases because its conjugate acid is resonance stabilised.

Answer: (b) 3

Explanation:
1. Only primary amines (aliphatic or aromatic) give the carbylamine reaction.
2. Propylamine, aniline and 1-phenylethylamine are primary.
3. The others are secondary or tertiary.

Teacher's Note:
1. Carbylamine test: primary amines only.

Answer: (b) 5

Explanation:
1. Nitrile reduction (a), amide reduction (b), Hofmann degradation (c), Gabriel synthesis (e) and nitroalkane hydrogenation (h) give amines.
2. (d) is a Henry (nitro-aldol) reaction, (f) is Wolff-Kishner reduction giving an alkane and (g) dehydrates the amide to a nitrile.

Teacher's Note:
1. List the amine-forming reactions: reduction of nitriles, amides and nitro compounds; Hofmann; Gabriel; ammonolysis.

Answer: (b) 6

Explanation:
1. A secondary amine R-NH-R' has R + R' = 5 carbons.
2. Methyl with butyl: four butyl groups (n-, iso-, sec-, tert-) gives 4 isomers.
3. Ethyl with propyl: n-propyl and isopropyl give 2 isomers.
4. Total = 4 + 2 = 6.

Teacher's Note:
1. Split the five carbons as 1 + 4 and 2 + 3 and count the alkyl isomers for each part.

Answer: (b) 3

Explanation:
1. Acetic acid (pKa 4.76), p-nitrophenol (7.2) and o-nitrophenol (7.2) are stronger acids than phenol (pKa 10).
2. Ethanol, cyclohexanol and p-cresol are weaker.

Teacher's Note:
1. Order: carboxylic acids > nitrophenols > phenol > cresols > alcohols.

Answer: (a) 1

Explanation:
1. In basic medium 2-naphthol is present as the naphthoxide ion.
2. Attack occurs at the alpha position next to the carbon bearing O-, i.e. C-1, because the intermediate keeps the second benzene ring intact (more resonance stabilisation).
3. The product is 1-(phenylazo)-2-naphthol, an orange-red dye.

Teacher's Note:
1. In 2-naphthol the most reactive site for electrophiles is C-1.

Answer: (a) 14

Explanation:
1. The smallest ketone whose oxime has geometrical isomers is butan-2-one (acetone gives only one oxime).
2. The two oximes (syn and anti) rearrange to N-ethylacetamide and N-methylpropanamide.
3. Hydrolysis gives acetic acid (60 g/mol) in one case and propanoic acid (74 g/mol) in the other.
4. Difference = 74 - 60 = 14 g/mol.

Teacher's Note:
1. In the Beckmann rearrangement the group anti to the OH migrates to nitrogen.