Question 1

Identify the reagent used to distinguish primary and secondary amines by Hinsberg test.

Accepted Answer

Hinsberg test is used to distinguish between primary, secondary, and tertiary amines. The reagent used in this test is benzenesulfonyl chloride, which has the chemical formula C _6 H _5 SO _2 Cl . Primary amines react with Hinsberg reagent to form an N-alkylbenzenesulfonamide, which is soluble in alkali due to the presence of an acidic hydrogen attached to nitrogen. Secondary amines react to form an N,N-dialkylbenzenesulfonamide, which is insoluble in alkali as it lacks an acidic hydrogen. Terti

Question 2

Aniline on treatment with NaNO _2 / HCl at about 0^ C gives a compound A, which on further treatment with CuCN / KCN gives a compound B, which when reacts with H _2 / Ni gives a compound C. The compou

Accepted Answer

Aniline ( C _6 H _5 NH _2) undergoes diazotization with NaNO _2 and HCl at 0^ C to form benzene diazonium chloride (A). C _6 H _5 NH _2 NaNO _2 / HCl , 0^ C C _6 H _5 N _2^+ Cl ^- (A) Benzene diazonium chloride reacts with CuCN / KCN to form benzonitrile (B) via the Sandmeyer reaction. C _6 H _5 N _2^+ Cl ^- CuCN / KCN C _6 H _5 CN (B) Benzonitrile is reduced by H _2 / Ni to form benzylamine (C). C _6 H _5 CN H _2 / Ni C _6 H _5 CH _2 NH _2 (C) Benzylamine, being an aliphatic primary amine, reac

Question 3

Identify the substrate 'A' in the following reaction. A [ ii) \ ] i) Moist Ag _2 O CH _3 CH _2 N(CH _3)_2 + CH _2 = CH _2

Accepted Answer

The given reaction is the Hofmann elimination of a quaternary ammonium salt. Substrate A reacts with moist Ag _2 O to form a quaternary ammonium hydroxide, which on heating undergoes -elimination to yield an alkene, a tertiary amine, and water. The products obtained are CH _3 CH _2 N(CH _3)_2 (N,N-dimethylethylamine) and CH _2= CH _2 (ethene). The quaternary ammonium ion must contain the alkyl groups present in the tertiary amine along with the alkyl group that underwent elimination to form the

Question 4

Which from following reactions occurs by removal of -CO- group from amide ?

Accepted Answer

Hofmann bromamide degradation reaction involves the conversion of a primary amide to a primary amine with one carbon atom less than the original amide. The general reaction is given by: R-CO-NH_2 + Br_2 + 4NaOH R-NH_2 + Na_2CO_3 + 2NaBr + 2H_2O In this reaction, the carbonyl group (-CO-) is removed from the amide as sodium carbonate. Answer: Hofmann degradation

Question 5

Which from following amines is obtained by Gabrial phthalimide synthesis?

Accepted Answer

Gabriel phthalimide synthesis is used exclusively for the preparation of primary aliphatic amines. Phthalimide on treatment with ethanolic potassium hydroxide forms the potassium salt of phthalimide, which on heating with an alkyl halide followed by alkaline hydrolysis produces the corresponding primary amine. Aromatic primary amines like C _6 H _5 NH _2 cannot be prepared by this method because aryl halides do not undergo nucleophilic substitution with the phthalimide anion. Secondary and terti

Question 6

Which from following compounds is obtained when propionamide is treated with Br _2 and concentrated aqueous KOH solution ?

Accepted Answer

Propionamide is CH _3 CH _2 CONH _2. The reaction of an amide with Br _2 and concentrated aqueous KOH is known as the Hoffmann bromamide degradation reaction. In this reaction, the amide is converted into a primary amine with one carbon atom less than the parent amide. CH _3 CH _2 CONH _2 + Br _2 + 4 KOH CH _3 CH _2 NH _2 + K _2 CO _3 + 2 KBr + 2 H _2 O The product obtained is ethylamine ( CH _3 CH _2 NH _2). Answer: CH _3 CH _2 NH _2

Question 7

Which from following amines when heated with ethanolic KOH and chloroform does NOT produce foul smell ?

Accepted Answer

The reaction of an amine with chloroform and ethanolic KOH to produce a foul-smelling compound is known as the carbylamine test or isocyanide test. This test is given only by primary aliphatic and primary aromatic amines. Secondary and tertiary amines do not respond to this test. Benzenamine is a primary aromatic amine. Propan-2-amine is a primary aliphatic amine. Phenylmethanamine is a primary aliphatic amine. N,N-Dimethyl aniline is a tertiary amine, so it does not give the carbylamine test an

Question 8

Identify 'B' in the following reaction Ethyl cyanide H _2 O A H _2 O / dil HCl B + NH _3

Accepted Answer

Ethyl cyanide is CH _3 CH _2 CN . Partial hydrolysis of ethyl cyanide yields propanamide (A). CH _3 CH _2 CN H _2 O CH _3 CH _2 CONH _2 Complete hydrolysis of propanamide in the presence of dilute HCl yields propanoic acid (B) and ammonia. CH _3 CH _2 CONH _2 H _2 O / dil HCl CH _3 CH _2 COOH + NH _3 Therefore, B is propanoic acid. Answer: Propanoic acid

Question 9

What is the number of moles of hydrogen atoms needed to obtain one mole ethanamine from acetamide using LiAl H _4 /ether as reducing agent.

Accepted Answer

The chemical reaction for the reduction of acetamide to ethanamine using LiAlH _4 is: CH _3 CONH _2 + 4[ H ] LiAlH _4/ ether CH _3 CH _2 NH _2 + H _2 O In this process, the carbonyl oxygen is removed as a water molecule, which consumes two hydrogen atoms. The carbonyl carbon is reduced to a methylene group ( CH _2), which consumes another two hydrogen atoms. Thus, a total of 4 moles of hydrogen atoms are required to reduce one mole of acetamide to one mole of ethanamine. Answer: 4

Question 10

Which from following statements is NOT correct regarding Hofmann's exhaustive alkylation?

Accepted Answer

Hofmann's exhaustive alkylation is the process of converting an amine into a quaternary ammonium salt by treating it with an excess of alkyl halide. When a primary amine (RNH_2) is heated with an alkyl halide (RX), it undergoes successive nucleophilic substitution reactions: RNH_2 + RX R_2NH + HX (Secondary amine) R_2NH + RX R_3N + HX (Tertiary amine) R_3N + RX R_4N^+X^- (Quaternary ammonium salt) To convert one mole of a primary amine into a tetraalkyl ammonium halide, three moles of alkyl hali

Question 11

Which from following amines does NOT form alkyl isocyanide when heated with alc. KOH and chloroform

Accepted Answer

The reaction of an amine with chloroform and alcoholic KOH to form an alkyl isocyanide is known as the carbylamine reaction. This test is given only by primary (aliphatic and aromatic) amines. Secondary and tertiary amines do not respond to this test. Methanamine, propan-2-amine, and ethanamine are primary amines, so they will form alkyl isocyanides. N, N-Diethylethanamine is a tertiary amine, hence it does not undergo the carbylamine reaction and will not form an alkyl isocyanide. Answer: N, N-

Question 12

Identify false statement from following regarding preparation of amines by Hofmann degradation method.

Accepted Answer

Hofmann bromamide degradation reaction is used for the conversion of primary amides to primary amines with one carbon atom less than the original amide. The general reaction is: R-CO-NH_2 + Br_2 + 4KOH R-NH_2 + K_2CO_3 + 2KBr + 2H_2O In this reaction, the carbonyl group (-CO-) is lost as potassium carbonate (K_2CO_3), and the alkyl or aryl group migrates from the carbonyl carbon to the nitrogen atom. The -CO- group is not reduced to a -CH_2- group. The reduction of an amide to an amine where the

Question 13

Which from following amines when heated with ethanolic KOH and chloroform forms aryl isocyanide ?

Accepted Answer

The reaction of an amine with chloroform and ethanolic KOH to form an isocyanide is known as the carbylamine reaction. This reaction is a characteristic test given only by primary aliphatic and primary aromatic amines. Among the given options, benzenamine (aniline) is a primary aromatic amine, while N-methylaniline and ethylethanamine are secondary amines, and N,N-dimethylaniline is a tertiary amine. Therefore, benzenamine reacts with chloroform and ethanolic KOH to form phenyl isocyanide (an ar

Question 14

Identify correct decreasing order of basic strength of amines

Accepted Answer

Basic strength of amines depends on the availability of the lone pair of electrons on the nitrogen atom for protonation. In aliphatic amines, the electron-donating +I effect of alkyl groups increases the electron density on nitrogen, making them more basic than ammonia. For methylamines in aqueous solution, the combined effect of the +I effect, steric hindrance, and hydration energy results in the basic strength order: secondary > primary > tertiary. Thus, ( CH _3)_2 NH > CH _3 NH _2 > NH _3. In

Question 15

Identify the major product obtained when ethyl amine is reacted with excess of methyl iodide ?

Accepted Answer

Ethyl amine reacts with excess of methyl iodide to undergo exhaustive methylation, forming a quaternary ammonium salt. C_2H_5NH_2 + CH_3I C_2H_5NHCH_3 + HI C_2H_5NHCH_3 + CH_3I C_2H_5N(CH_3)_2 + HI C_2H_5N(CH_3)_2 + CH_3I [C_2H_5N(CH_3)_3]^+ I^- Since methyl iodide is in excess, the reaction does not stop at the secondary or tertiary amine stage but proceeds to the final quaternary ammonium salt. The major product obtained is ethyltrimethylammonium iodide. Answer: Ethyltrimethylammonium iodide

Question 16

Tertiary amines have the lowest boiling points because ,

Accepted Answer

Tertiary amines have the general formula R_3N. Since there are no hydrogen atoms directly attached to the highly electronegative nitrogen atom, they cannot form intermolecular hydrogen bonds. In contrast, primary and secondary amines have hydrogen atoms attached to nitrogen and can form intermolecular hydrogen bonds, which leads to higher boiling points. Due to the absence of intermolecular hydrogen bonding, tertiary amines have the lowest boiling points among isomeric amines.

Question 17

Which of the following statements is NOT correct?

Accepted Answer

Primary amines have the general formula R-NH_2. Since they have two hydrogen atoms attached to the highly electronegative nitrogen atom, they can form intermolecular hydrogen bonds. Tertiary butylamine has the structure (CH_3)_3C-NH_2. The nitrogen atom is attached to only one carbon atom, making it a primary amine. Tertiary amines have the general formula R_3N. They do not have any hydrogen atoms attached to the nitrogen atom, so they cannot form intermolecular hydrogen bonds with each other. i

Question 18

Which of the following reactions uses diazonium salt ?

Accepted Answer

The Sandmeyer reaction involves the conversion of an aryl diazonium salt into an aryl halide using copper(I) salts, such as CuCl or CuBr. The Mendius reaction is the reduction of nitriles to primary amines using sodium and ethanol. The Hofmann rearrangement reaction involves the conversion of a primary amide to a primary amine with one fewer carbon atom using bromine and a strong base. The carbylamine reaction is a test for primary amines, where a primary amine reacts with chloroform and a base

Question 19

Which of the following amines is not represented by formula C _3 H _9 N ?

Accepted Answer

The molecular formula C _3 H _9 N corresponds to saturated acyclic amines. The possible structural isomers are: Primary (1^ ) amines: CH _3 CH _2 CH _2 NH _2 (propan-1-amine) and CH _3 CH ( NH _2) CH _3 (propan-2-amine). Secondary (2^ ) amine: CH _3 CH _2 NHCH _3 (N-methylethanamine). Tertiary (3^ ) amine: ( CH _3)_3 N (N,N-dimethylmethanamine). A quaternary ammonium salt requires the nitrogen atom to be bonded to four alkyl groups. The simplest quaternary ammonium salt is the tetramethylammoniu

Question 20

Identify Hinsberg's reagent from following.

Accepted Answer

Hinsberg's reagent is benzene sulphonyl chloride (C_6H_5SO_2Cl). It is used to distinguish between primary, secondary, and tertiary amines. Primary amines react with it to form a precipitate that is soluble in alkali. Secondary amines react to form a precipitate that is insoluble in alkali. Tertiary amines do not react with it.

Amines — Previous Year Questions & Solutions