JEE Chemistry Amines MCQs Set 10

Answer: (b) nitrobenzene

Explanation:
1. Diazonium fluoroborate with \(\ce{NaNO2}\) and Cu powder on heating replaces \(\ce{N2+}\) by the NO2 group.
2. The product is nitrobenzene.

Teacher's Note:
1. Fluorobenzene is made from the fluoroborate by heating alone (Balz-Schiemann).
2. With NaNO2/Cu the nitro group is introduced.

Answer: (a) 2-methyl-2-nitropropane

Explanation:
1. Nitrous acid reacts with nitroalkanes that have an alpha hydrogen.
2. A primary nitroalkane gives a nitrolic acid (blood red with alkali) and a secondary one gives a pseudonitrole (blue).
3. A tertiary nitroalkane has no alpha hydrogen and does not react.

Teacher's Note:
1. Victor Meyer test: 1° nitro gives red colour, 2° gives blue, 3° gives no colour.

Answer: (d) nitrobenzene

Explanation:
1. C is p-hydroxyazobenzene \(\ce{C6H5N=NC6H4OH}\) (formula C12H10N2O), formed by coupling benzenediazonium chloride (B) with phenol.
2. B comes from aniline by diazotisation, and aniline is obtained from A by Sn/HCl reduction.
3. So A is nitrobenzene (aniline itself would not need Sn/HCl).

Teacher's Note:
1. Reduction with Sn/HCl followed by diazotisation points to a nitro compound as the starting material.

Answer: (c) Zn + \(\ce{NH4Cl}\)

Explanation:
1. Sn/HCl, Fe/HCl and H2/Pd fully reduce the nitro group to the amino group.
2. Zn with \(\ce{NH4Cl}\) (neutral medium) stops at N-phenylhydroxylamine.

Teacher's Note:
1. Choose the medium carefully when reducing nitrobenzene.

Answer: (a) 4-amino-2-methylbenzenesulphonic acid

Explanation:
1. Aniline forms the anilinium hydrogen sulphate, which on heating at about 200°C rearranges to the para sulphonic acid (sulphanilic acid type).
2. In m-toluidine the position para to NH2 is also ortho to CH3.
3. The product is 4-amino-2-methylbenzenesulphonic acid.

Teacher's Note:
1. Baking of the anilinium salt gives para-sulphonation.
2. Sulphanilic acid exists as a zwitterion.

Answer: (a) \(\ce{CH3-CH=CH-CH2-CH2-NH-CH3}\)

Explanation:
1. LiAlH4 reduces the C=N bond of an imine to a CH-NH bond.
2. It does not reduce an isolated C=C bond, so the double bond stays.
3. The product is the secondary amine in (a).

Teacher's Note:
1. LiAlH4 is a nucleophilic hydride donor and does not reduce isolated alkenes.

Answer: (b) Tollens' reagent, then \(\ce{Br2/KOH}\), then \(\ce{H2-Pd/C}\)

Explanation:
1. Tollens' reagent oxidises CHO to COOH without touching the C=C or amide.
2. \(\ce{Br2/KOH}\) converts CONH2 into NH2 (Hofmann).
3. Finally H2-Pd/C saturates the ring. Doing H2/Pd first would also reduce the aldehyde, and Br2/KOH first would give a different product with the aldehyde still present.

Teacher's Note:
1. Choose reagents so that each functional group is converted without affecting the others.
2. Tollens' reagent is selective for aldehydes.

Answer: (d) \(\ce{CH3CH2COOH}\)

Explanation:
1. Ethanamine is obtained by Hofmann degradation of propanamide (C).
2. Propanamide comes from heating the ammonium salt (B) of propanoic acid (A) with ammonia.
3. A is therefore \(\ce{CH3CH2COOH}\).

Teacher's Note:
1. Add one carbon to the amine to get the amide, and the amide comes from the acid with the same carbon count.

Answer: (b) A = 1° amine, B = methyl ketone, C = 1° amine, D = alcohol

Explanation:
1. Nitrile reduction gives a primary amine \(\ce{RCH2NH2}\).
2. Grignard reagent and then water give the methyl ketone \(\ce{RCOCH3}\).
3. Hydrolysis of an isocyanide gives primary amine \(\ce{RNH2}\) and formic acid.
4. A primary aliphatic amine with HNO2 gives an alcohol.

Teacher's Note:
1. Know the products of the three main reactions of nitriles: reduction, Grignard addition and hydrolysis.

Answer: (b) ethyl isocyanide, ethanamine and ethylmethylamine

Explanation:
1. AgCN is largely covalent, so it attacks through N and gives ethyl isocyanide (A).
2. Hydrolysis of isocyanide gives HCOOH and ethylamine (B).
3. The carbylamine reaction on B regenerates A, and reduction of A gives the secondary amine \(\ce{C2H5NHCH3}\) (C).

Teacher's Note:
1. Remember: AgCN gives isocyanides, KCN gives cyanides.
2. Isocyanide hydrolysis gives formic acid and 1° amine.

Answer: (b) ethylamine

Explanation:
1. Ethylamine with HNO2 gives ethanol (A).
2. \(\ce{PCl5}\) converts it into ethyl chloride (B).
3. Ammonolysis of ethyl chloride gives ethylamine again (and some secondary and tertiary amines), so C is mainly ethylamine.

Teacher's Note:
1. Ammonolysis of alkyl halide with excess ammonia gives the primary amine.

Answer: (c) carbylamine reaction

Explanation:
1. The carbylamine (isocyanide) test, with \(\ce{CHCl3 + KOH}\), gives a foul smell with primary amines only.
2. The iodoform test is for methyl ketones, Victor Meyer's test is for alcohols, and Liebermann's test is for phenols and secondary amines.

Teacher's Note:
1. Match each test with its functional group before the exam.

Answer: (a) \(\ce{C2H5COOH}\), \(\ce{CHI3}\), \(\ce{C6H5CH2NH2}\)

Explanation:
1. C2H5MgBr with ClCN gives \(\ce{C2H5CN}\) (A), and hydrolysis gives \(\ce{C2H5COOH}\) (B).
2. Acetone with iodine and NaOH gives iodoform \(\ce{CHI3}\) (X), and Ag removes iodine to give ethyne (Y).
3. Aniline gives the diazonium salt (P), then benzonitrile (Q), and Na/EtOH reduces it to benzylamine \(\ce{C6H5CH2NH2}\) (R).

Teacher's Note:
1. Iodoform test: methyl ketones give yellow CHI3.
2. Sandmeyer cyanation followed by reduction extends the chain by one carbon.

Answer: (b) \(\ce{Ph-NO2 -> Ph-NH2}\)

Explanation:
1. Fe/HCl is a metal-acid combination (acidic medium) and reduces the nitro group completely to the amino group.
2. Hydrazobenzene, azobenzene and hydroxylamine are formed in alkaline or neutral media.

Teacher's Note:
1. Acidic medium gives aniline; alkaline medium gives bimolecular products.

Answer: (b) \(\ce{-NO2}\)

Explanation:
1. Zn/NH4Cl reduces a nitro compound to N-arylhydroxylamine.
2. This reduces Tollens' reagent, giving silver as a black or grey precipitate (Mullikan-Barker test).
3. So A contains the nitro group.

Teacher's Note:
1. The Mullikan-Barker test detects the nitro group.

Answer: (d) Both Assertion and Reason are false.

Explanation:
1. Trimethylamine has no N-H, so it cannot form intermolecular hydrogen bonds and boils at 3°C.
2. n-Propylamine forms hydrogen bonds and boils at about 48°C.
3. The assertion is false and the reason is false.

Teacher's Note:
1. Boiling point order of isomeric amines is 1° > 2° > 3°.
2. Tertiary amines have no N-H bond for hydrogen bonding.

Answer: (d) Both Assertion and Reason are false.

Explanation:
1. Pyrrole is a five-membered aromatic heterocyclic compound, not alicyclic.
2. Its ring has six pi electrons, so it is aromatic and not aliphatic, making both statements false.

Teacher's Note:
1. Alicyclic means a ring of carbon atoms with aliphatic properties.
2. Heterocyclic rings contain atoms other than carbon.

Answer: (c) Assertion is true but the Reason is false.

Explanation:
1. Each of these nitroalkanes has an alpha hydrogen, so NaOH forms the soluble nitronate salt. The assertion is true.
2. Boiling points rise with molar mass, so these compounds boil higher than nitromethane (101°C). The reason is false.

Teacher's Note:
1. Nitroalkanes with alpha-H are acidic.
2. Boiling point increases with molecular mass in a homologous series.

Answer: (b) Both Assertion and Reason are true but the Reason is not a correct explanation of the Assertion.

Explanation:
1. \(\ce{CH3CN}\) is called acetonitrile in common nomenclature and ethanenitrile in IUPAC, so the assertion is true.
2. The alpha hydrogens are acidic due to the electron-withdrawing CN group, which is also true.
3. The acidity has nothing to do with the naming, so the reason does not explain the assertion.

Teacher's Note:
1. Both statements are true here, but one does not explain the other.
2. Do not pick option A just because both are true.

Answer: (c) Assertion is true but the Reason is false.

Explanation:
1. Ethanol reduces the diazonium salt, replacing N2+ by H and giving benzene, so the assertion is true.
2. Ethanol is oxidised (to ethanal) in this reaction, so it acts as a reducing agent, not an oxidising agent. The reason is false.

Teacher's Note:
1. Ethanol and H3PO2 are reducing agents for diazonium salts.

Answer: (c) Assertion is true but the Reason is false.

Explanation:
1. A primary amine with CHCl3 and KOH gives an isocyanide, so the assertion is true.
2. The intermediate is dichlorocarbene \(\ce{:CCl2}\), not methylene \(\ce{:CH2}\), so the reason is false.

Teacher's Note:
1. Dichlorocarbene is the electrophile in both the carbylamine and Reimer-Tiemann reactions.