JEE Chemistry Amines MCQs Set 08

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

Explanation:
1. The amide is converted by Br2/KOH into the N-bromoamide \(\ce{RCONHBr}\), then to its anion, which loses Br- to give the isocyanate \(\ce{R-N=C=O}\).
2. Hydrolysis of the isocyanate gives \(\ce{RNH2}\) and carbonate.
3. The isocyanide \(\ce{RNC}\) is never an intermediate; it comes from the carbylamine reaction.

Teacher's Note:
1. Remember the order: amide, N-bromoamide, isocyanate, amine.
2. An isocyanide has the R-N=C skeleton, whereas an isocyanate has R-N=C=O, so do not confuse them.

Answer: (b) turn red litmus blue

Explanation:
1. Hofmann bromamide reaction removes the carbonyl carbon, so propanamide gives ethanamine \(\ce{CH3CH2NH2}\) (A).
2. Aliphatic amines are basic in water, \(\ce{RNH2 + H2O <=> RNH3+ + OH-}\).
3. The solution therefore turns red litmus blue.

Teacher's Note:
1. Hofmann degradation shortens the carbon chain by one carbon.
2. The product amine is basic, so it changes red litmus to blue.

Answer: (c) Mendius reaction

Explanation:
1. Dehydration of acetamide by \(\ce{P2O5}\) gives methyl cyanide \(\ce{CH3CN}\) (A).
2. Reduction of an alkyl cyanide to a primary amine with Na/C2H5OH is the Mendius reaction, so B is \(\ce{CH3CH2NH2}\).

Teacher's Note:
1. Stephen reduction uses SnCl2/HCl and stops at the aldehyde.
2. Mendius reaction takes nitrile all the way to the primary amine.

Answer: (a) a, b

Explanation:
1. Acetamide is first brominated to the N-bromoamide (a).
2. On loss of HBr and rearrangement of the methyl group to nitrogen it gives methyl isocyanate (b), which is hydrolysed to \(\ce{CH3NH2}\).
3. (c) and (d) are not formed, so the answer is a, b.

Teacher's Note:
1. The key step is the migration of the alkyl group from carbon to nitrogen.
2. The isocyanate formed in this step is hydrolysed to the amine.

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

Explanation:
1. Alkyl halide with ammonia (ammonolysis) gives amines.
2. Reduction of an oxime or of an amide with LiAlH4 gives a primary amine.
3. Acid hydrolysis of a nitrile gives the carboxylic acid RCOOH and ammonium ion, not an amine.

Teacher's Note:
1. Reduction of nitriles gives amines, but hydrolysis gives acids.
2. Always check whether the reagent is a reducing agent or water.

Answer: (a) Aniline treated with excess \(\ce{CH3I}\) in two successive steps

Explanation:
1. Aniline with one \(\ce{CH3I}\) gives N-methylaniline, a secondary amine.
2. A second \(\ce{CH3I}\) converts it into N,N-dimethylaniline \(\ce{C6H5N(CH3)2}\), a tertiary amine.
3. Nitrobenzene with Sn/HCl gives aniline, a primary amine.

Teacher's Note:
1. Each alkylation step replaces one N-H hydrogen.
2. Count the N-H bonds that remain to classify the amine.

Answer: (c) \(\ce{C2H5NC}\)

Explanation:
1. Nitroethane, acetaldoxime and methyl cyanide all contain two carbons in the chain and reduce to \(\ce{CH3CH2NH2}\).
2. Reduction of the isocyanide \(\ce{C2H5-NC}\) gives the secondary amine \(\ce{C2H5NHCH3}\) (the isocyanide carbon becomes a methyl group).
3. Hence ethanamine cannot be obtained from it.

Teacher's Note:
1. Isocyanides on reduction always give secondary amines (N-methyl derivatives).
2. Cyanides give primary amines with one more carbon than the alkyl group.

Answer: (c) imides

Explanation:
1. Phthalimide is an imide (two carbonyl groups on one N).
2. Its potassium salt is alkylated with an alkyl halide and the N-alkylphthalimide is hydrolysed to give a primary amine.

Teacher's Note:
1. Phthalimide has an acidic N-H, which is why KOH forms its potassium salt easily.

Answer: (c) primary amides

Explanation:
1. Primary amides have two N-H bonds and form extensive intermolecular hydrogen bonds.
2. Secondary amides have one N-H and tertiary amides have none.
3. Amines form weaker hydrogen bonds than amides, so primary amides melt highest.

Teacher's Note:
1. More hydrogen-bond donors give stronger attraction and higher melting point.
2. Primary amides are usually solids even for small molecules.

Answer: (c) formation of hydrogen bonds with water

Explanation:
1. The N-H bonds and the lone pair on nitrogen form hydrogen bonds with water molecules.
2. Low molar mass helps, but the actual cause of solubility is hydrogen bonding.

Teacher's Note:
1. Solubility of amines falls as the alkyl part gets larger.
2. Always link solubility in water with hydrogen bonding.

Answer: (b) b

Explanation:
1. A Schiff's base (imine) is formed when a primary amine condenses with an aldehyde or ketone.
2. So \(\ce{C2H5NH2 + C2H5CHO -> C2H5N=CHC2H5 + H2O}\) (b) is the Schiff's base.
3. The others give an alcohol (a), an alkyl chloride (c) and a sulphonamide (d).

Teacher's Note:
1. Schiff's bases contain the C=N group and are also called imines or anils.
2. They form from 1° amines only.

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

Explanation:
1. Primary amines react with nitrosyl chloride to give the alkyl chloride: \(\ce{RNH2 + NOCl -> RCl + N2 + H2O}\).
2. \(\ce{SOCl2}\) is used for alcohols and acids, not for converting the NH2 group directly.

Teacher's Note:
1. NOCl converts the NH2 group into Cl with loss of nitrogen.
2. Do not confuse it with SOCl2 used on OH groups.

Answer: (c) both of the above

Explanation:
1. \(\ce{RCONH2}\) is acidic enough for \(\ce{NaNH2}\) to remove the N-H hydrogen (as \(\ce{NH3}\)), giving \(\ce{RCONHNa}\).
2. \(\ce{RNH2}\) reacts with sodium to give \(\ce{RNHNa}\) and \(\ce{H2}\).
3. So an N-H hydrogen is lost in both.

Teacher's Note:
1. An N-H hydrogen can be removed as H+ equivalent by a strong base or active metal.

Answer: (d) a Schiff's base

Explanation:
1. A primary amine condenses with a carbonyl compound with loss of water: \(\ce{R2C=O + R'NH2 -> R2C=NR' + H2O}\).
2. The product containing the C=N bond is a Schiff's base (imine).

Teacher's Note:
1. Only primary amines give stable Schiff's bases.
2. Secondary amines give enamines with suitable ketones.

Answer: (c) acetyl chloride with dimethylamine

Explanation:
1. N,N-dimethylacetamide is \(\ce{CH3CON(CH3)2}\).
2. It is made by acylation of dimethylamine \(\ce{(CH3)2NH}\) with acetyl chloride, with loss of HCl.

Teacher's Note:
1. The substituents on nitrogen in the product come from the amine used.
2. The acyl group comes from the acid chloride.

Answer: (c) A-b, B-c, C-d, D-a

Explanation:
1. Ammoniacal AgNO3 (Tollens' reagent) tests for an aldehyde: A-b.
2. Periodic acid HIO4 cleaves vicinal diols: B-c.
3. Cold dilute alkaline KMnO4 (Baeyer's reagent) tests for a double bond: C-d.
4. Chloroform with alkali (carbylamine) tests for a primary amine: D-a.

Teacher's Note:
1. Match each reagent with its characteristic observation.
2. Baeyer's reagent is decolourised by C=C bonds.

Answer: (b) benzylamine

Explanation:
1. An aliphatic primary amino group gives an alcohol with HNO2 through an unstable diazonium salt.
2. In benzylamine \(\ce{C6H5CH2NH2}\) the NH2 is on an aliphatic carbon, so \(\ce{C6H5CH2OH}\) is formed.
3. Aniline gives a stable diazonium salt at 0-5°C, and 2 and 3° amines do not give alcohols.

Teacher's Note:
1. Benzylamine behaves like an aliphatic amine because NH2 is not attached to the ring.
2. Diazonium salts are stable only when NH2 is directly on the ring.

Answer: (a) aniline

Explanation:
1. In weakly acidic or neutral medium, electrolytic reduction goes to the amine stage, giving aniline.
2. In strongly acidic medium N-phenylhydroxylamine forms and rearranges to p-aminophenol.

Teacher's Note:
1. Medium decides the product of reduction of nitrobenzene: weakly acidic gives aniline, strongly acidic gives p-aminophenol.
2. Zn/NH4Cl gives N-phenylhydroxylamine.

Answer: (b) acetylation

Explanation:
1. Direct nitration oxidises aniline and also forms the anilinium ion, giving meta product.
2. Acetylation converts \(\ce{NH2}\) into \(\ce{NHCOCH3}\), which is still an ortho/para director but less activating, and protects it from oxidation.

Teacher's Note:
1. Protection by acetylation allows clean para-nitration.
2. After nitration, hydrolysis restores the amino group.

Answer: (c) 2,4,6-tribromoaniline

Explanation:
1. The NH2 group strongly activates the ring at the ortho and para positions.
2. With bromine water all three positions are brominated at once, giving 2,4,6-tribromoaniline as a white precipitate.

Teacher's Note:
1. This reaction is used as a test for aniline.
2. To get a monobromo product, the amino group must first be acetylated.

Answer: (d) benzylamine

Explanation:
1. Diazotisation needs a primary aromatic amine, with NH2 attached directly to the benzene ring.
2. Aniline, p-toluidine and p-nitroaniline satisfy this.
3. Benzylamine has NH2 on a CH2 group, so it behaves as an aliphatic amine.

Teacher's Note:
1. Aromatic primary amines give stable diazonium salts at 0-5°C.
2. Aliphatic ones give unstable salts that decompose to alcohols.

Answer: (c) \(\ce{C6H5NHR}\) (N-alkylaniline)

Explanation:
1. Secondary amines react with nitrous acid to give yellow oily N-nitrosoamines \(\ce{R2N-N=O}\).
2. The N-alkylaniline is the secondary amine in the list.
3. Primary amines give diazonium salts or alcohols.

Teacher's Note:
1. 1° gives diazonium or alcohol, 2° gives N-nitrosoamine, 3° gives a salt (with aliphatic) or a C-nitroso compound (with aromatic).
2. The Liebermann nitroso test uses the yellow nitrosoamine.