Question 1

In Kjeldahl’s method, the nitrogen present in the sample is determined in the form of:

Accepted Answer

NH3. Kjeldahl’s method quantifies nitrogen by converting it into ammonia (NH3) through digestion with sulfuric acid, allowing for the measurement of nitrogen content in organic compounds.

Question 2

What product is formed when benzenediazonium chloride (C6H5N2+Cl−) is treated with CuCl?

Accepted Answer

C6H5Cl. When benzenediazonium chloride reacts with copper(I) chloride (CuCl), it undergoes a substitution reaction that replaces the diazonium group with a chlorine atom, resulting in the formation of chlorobenzene (C6H5Cl). This reaction is a classic method for synthesizing aryl halides from diazonium salts.

Question 3

When o-methoxybromobenzene is first reacted with sodamide and then treated with ammonia, which compound is obtained as the final product?

Accepted Answer

m-Methoxyaniline. NaNH2 abstracts the H ortho to Br (at C6, since C2 bears OMe) generating a benzyne between C1-C6. NH2- adds so the resulting carbanion sits on the carbon next to the electron-withdrawing (-I) OMe; this places the NH2 group meta to OMe, giving m-methoxyaniline.

Question 4

An organic compound A, when treated with NH3, forms B. On heating, B gives C, and C on treatment with Br2 in the presence of KOH yields ethylamine. The compound A is:

Accepted Answer

CH3CH2COOH. A + NH3 -> ammonium salt (B); heating -> amide (C); C + Br2/KOH (Hofmann bromamide) gives an amine with one fewer carbon. To get ethylamine (2 C) the amide must be propanamide, so A = propanoic acid, CH3CH2COOH.

Question 5

An organic nitrogen-containing compound gives an oily nitrosoamine when treated with aqueous nitrous acid at low temperature. Which compound is it?

Accepted Answer

CH3CH2NHCH2CH3. The correct option, CH3CH2NHCH2CH3, is a secondary amine that reacts with nitrous acid to form a nitrosoamine, which is typically oily in nature. Primary amines would form different products, while tertiary amines do not react with nitrous acid in this manner.

Question 6

Which of the following statements are true?
(i) In the Sandmeyer reaction, groups such as Cl, Br and CN are introduced into a benzene ring in the presence of Cu+ ions.
(ii) In the Gattermann reaction, substitution into the benzene ring occurs in the presence of copper powder and HCl.
(iii)

Accepted Answer

(i) and (ii). Sandmeyer (Cu+ salts, introducing Cl/Br/CN) and Gattermann (Cu powder + HX) are both correct descriptions. However the Sandmeyer reaction gives better yields than the Gattermann reaction, so (iii) is false and only (i) and (ii) are true.

Question 7

In the reaction sequence below, identify compound [A]:
[A] —(reduction)→ [B] —(HNO2)→ CH3CH2OH

Accepted Answer

CH3CN. HNO2 converts a primary amine to an alcohol, so [B] must be ethylamine (CH3CH2NH2), giving CH3CH2OH. Ethylamine is obtained by reduction of acetonitrile CH3CN, so [A] = CH3CN.

Question 8

Which reagent will transform p-methylbenzenediazonium chloride into p-cresol?

Accepted Answer

H2O. Warming an arenediazonium salt with water (steam/dilute acid) replaces the -N2+ group by -OH. Thus p-methylbenzenediazonium chloride is hydrolysed to p-cresol (4-methylphenol).

Question 9

Which reagent can be used to replace the diazonium group (–N2+Cl−) in benzene diazonium chloride with iodine?

Accepted Answer

KI. Iodobenzene is obtained simply by warming benzenediazonium chloride with potassium iodide (KI); the -N2+ group is replaced by iodine with loss of N2. No Cu catalyst is needed, unlike for Cl/Br (Sandmeyer).

Question 10

Nitrosamines of the type R2N–N=O are not soluble in water. When they are heated with concentrated H2SO4, they yield secondary amines. This transformation is known as the

Accepted Answer

Liebermann nitroso reaction. The conversion/colour reaction of N-nitrosamines (R2N-N=O) with concentrated H2SO4, used to detect secondary amines, is the Liebermann nitroso reaction.

Question 11

Which reagent is capable of reacting with primary, secondary, and tertiary amines alike?

Accepted Answer

HCl. HCl reacts with primary, secondary and tertiary amines alike, forming the corresponding ammonium chloride salts since all are basic. Acetic anhydride acylates only primary and secondary amines (no N-H in 3 amines).

Question 12

On treatment with lithium aluminium hydride, which of the following gives a secondary amine as the product?

Accepted Answer

Methyl isocyanide. Reduction of methyl isocyanide (CH3-NC) with LiAlH4 gives N,N-dimethylamine, CH3-NH-CH3, a secondary amine. Acetamide and methyl cyanide both reduce to primary amines.

Question 13

When aniline undergoes condensation with oil of bitter almonds (C6H5CHO), a benzal compound is produced. What is this reaction/product called?

Accepted Answer

Schiff's base. Condensation of a primary amine (aniline) with an aldehyde (benzaldehyde) gives an imine, R-CH=N-Ar, known as a Schiff's base.

Question 14

Which sequence of reactions can be used to synthesize 3-chloroaniline starting from benzene?

Accepted Answer

Nitration followed by chlorination and then reduction. To place Cl and NH2 in a meta relationship, the deactivating meta-director must be introduced before the second group. Nitrate benzene to nitrobenzene (NO2 directs meta), chlorinate to give m-chloronitrobenzene, then reduce the NO2 to NH2, giving 3-chloroaniline. Reducing first would give an amine (an o,p-director), leading to ortho/para products.

Question 15

Which of the following is the most preferred reagent system for reducing nitro compounds?

Accepted Answer

Sn with HCl. Sn with HCl is effective for reducing nitro compounds due to the strong reducing environment created by the combination of tin and hydrochloric acid, which facilitates the conversion of nitro groups to amines.

Question 16

Acetamide is subjected to each of the following reagents in separate reactions. Which treatment will produce methylamine as the product?

Accepted Answer

NaOH followed by Br2. Acetamide (CH3CONH2) on treatment with Br2 and NaOH undergoes the Hofmann bromamide degradation, losing the carbonyl carbon to give methylamine (CH3NH2). Soda lime would give methane, not methylamine.

Question 17

Which of the following transformations does not produce a primary amine?

Accepted Answer

CH3NC treated with LiAlH4. CH3-NC reduced by LiAlH4 gives the secondary amine dimethylamine (CH3)2NH, not a primary amine. The other three (amide+Br2/KOH, CH3CN+LiAlH4, amide+LiAlH4) all give primary amines.

Question 18

Toluene is first nitrated. The nitro compound obtained is then reduced using tin and hydrochloric acid. The amine formed is diazotized and the diazonium salt is subsequently heated with cuprous bromide. The final reaction mixture contains

Accepted Answer

a mixture of ortho- and para-bromotoluenes. The nitration of toluene introduces a nitro group, which directs further electrophilic substitution reactions to the ortho and para positions. The reduction of the nitro group to an amine followed by diazotization and reaction with cuprous bromide leads to bromination at these same positions, resulting in a mixture of ortho- and para-bromotoluenes.

Question 19

When an aliphatic primary amine is warmed with chloroform in the presence of ethanolic potassium hydroxide, which organic product is obtained?

Accepted Answer

an alkyl isocyanide. A primary amine warmed with CHCl3 and ethanolic KOH undergoes the carbylamine (isocyanide) test, forming a foul-smelling alkyl isocyanide R-NC. Answer: option 3.

Question 20

In water, which of the following amines is the strongest base, i.e., has the lowest pKb value?

Accepted Answer

(CH3)2NH. In aqueous solution basic strength of methylamines is (CH3)2NH > CH3NH2 > (CH3)3N, reflecting the balance of +I effect, solvation of the cation and steric hindrance. Aniline is far weaker (lone pair delocalised into the ring). So (CH3)2NH has the lowest pKb.

Question 21

In the Hofmann bromamide degradation, how many moles of NaOH and Br2 are required for each mole of amine formed?

Accepted Answer

4 moles of NaOH and 1 mole of Br2. Hofmann bromamide degradation: RCONH2 + Br2 + 4NaOH -> RNH2 + Na2CO3 + 2NaBr + 2H2O. So each mole of amine requires 4 moles of NaOH and 1 mole of Br2.

Question 22

On heating an aliphatic primary amine with chloroform and ethanolic potassium hydroxide, the organic compound formed is:

Accepted Answer

an alkyl isocyanide. Heating an aliphatic primary amine with chloroform and ethanolic potassium hydroxide leads to the formation of an alkyl isocyanide through the formation of an isocyanide functional group, which is characteristic of this reaction.

Question 23

In the Hofmann bromamide degradation reaction, the number of moles of NaOH and Br2 used per mole of amine produced are:
(1) One mole of NaOH and one mole of Br2.
(2) Four moles of NaOH and two moles of Br2.
(3) Two moles of NaOH and two moles of Br2.
(4) Four moles of NaOH and one mole of

Accepted Answer

Four moles of NaOH and one mole of Br2.. The Hofmann bromamide degradation reaction requires four moles of NaOH to effectively convert the amide to an amine, while only one mole of Br2 is needed for the bromination step, making option D the correct choice.

Question 24

The increasing basicity order of the following compounds is: (A) CH3CH2NH2 (B) CH3CH2NHCH2CH2CH3 (C) CH3N(CH3)CH3 (D) Ph-NH-CH3

Accepted Answer

(D) < (C) < (A) < (B). In aqueous solution basicity order is 2deg > 1deg > 3deg amine, with the N-aryl amine (D) least basic. Here B is 2deg, A is 1deg, C is 3deg (Me3N), D is N-methylaniline. Increasing: (D) < (C) < (A) < (B).

Question 25

Ethylamine (C2H5NH2) can be obtained from N-ethylphthalimide on treatment with:

Accepted Answer

(4) NH2NH2. N-ethylphthalimide can be converted to ethylamine through a hydrazine (NH2NH2) reaction, which facilitates the cleavage of the phthalimide group, releasing ethylamine.

Question 26

The major product of the following reaction is: CH3CH(OH)CH2CH2NH2 + ethyl formate (1 equiv.) / triethylamine

Accepted Answer

CH3CH(OH)CH2CH2NHCHO. The reaction involves the nucleophilic attack of the amine on the carbonyl carbon of ethyl formate, leading to the formation of an amide. The product retains the hydroxyl group and incorporates the formyl group from ethyl formate, resulting in the major product being CH3CH(OH)CH2CH2NHCHO.

Question 27

Hinsberg reagent is:

Accepted Answer

C6H5SO2Cl. Hinsberg reagent, known as benzenesulfonyl chloride, is used to differentiate between primary, secondary, and tertiary amines based on their reactivity with the reagent, making option D the correct choice.

Question 28

Three isomers A, B and C (mol. formula C8H11N) give the following results:
A and C — Azotization → P + Q
(i) Hydrolysis
(ii) Oxidation (KMnO4/H+) → R (product of A)
→ S (product of C)
R has lower boiling point than S
B — C6H5SO2Cl → alkali-insoluble product
A, B and C, respectively are:

Accepted Answer

o-ethylaniline, N-ethylaniline, p-ethylaniline. A and C undergo diazotization, so they are primary aromatic amines; B gives an alkali-insoluble sulfonamide with PhSO2Cl, so it is a secondary amine (N-ethylaniline). The ortho product (from A) has intramolecular H-bonding and lower boiling point than the para product (from C). Thus A = o-ethylaniline, B = N-ethylaniline, C = p-ethylaniline.

Question 29

Which of the following will react with CHCl3 + alc. KOH ?
(1) Thymine and proline
(2) Adenine and lysine
(3) Adenine and thymine
(4) Adenine and proline

Accepted Answer

Adenine and lysine. Adenine and lysine are both amino compounds that can undergo reactions with chloroform in the presence of alcoholic KOH, leading to the formation of products through nucleophilic substitution. This reaction is characteristic of compounds containing amino groups, which are present in both adenine and lysine.

Question 30

The most appropriate reagent for conversion of C2H5CN into CH3CH2CH2NH2 is:

Accepted Answer

LiAlH4. LiAlH4 is a strong reducing agent that can effectively reduce nitriles like C2H5CN to primary amines, such as CH3CH2CH2NH2, by adding hydrogen across the carbon-nitrogen triple bond.

Question 31

Consider the following reactions,
(i) NaNO2/HCl, 0-5°C
(ii) β-naphthol/NaOH
[P] → Colored Solid
[P] → Br2/H2O → C7H6NBr3
The compound [P] is:

Accepted Answer

(2) A benzene ring with NH2 at position 1 and CH3 at position 3 (m-toluidine). P must be a toluidine (C7H9N): it couples with diazotised P + beta-naphthol to give an azo dye (colored solid). Bromination giving C7H6NBr3 needs three positions ortho/para to NH2 free, which only m-toluidine (NH2 at 1, CH3 at 3) provides (2,4,6-tribromo). So P is m-toluidine.

Question 32

Ammolysis of Alkyl halides followed by treatment with NaOH solution can be used to prepare primary, secondary and tertiary amines.
The purpose of NaOH in the reaction is:

Accepted Answer

to remove acidic impurities. Ammonolysis gives the amine as its hydrohalide salt (R-NH3+ X-). NaOH neutralises the HX / acidic by-product, freeing the neutral amine. So its purpose is to remove acidic impurities.

Question 33

Which of the following is least basic ?

Accepted Answer

(CH3CO)2NH. (CH3CO)2NH is least basic because the presence of two acetyl groups (CH3CO) significantly stabilizes the lone pair of electrons on the nitrogen through resonance, making it less available to accept protons compared to the other options.

Question 34

Compound A is converted to B on reaction with CHCl3 and KOH. The compound B is toxic and can be decomposed by C. A, B and C respectively are:

Accepted Answer

primary amine, isonitrile compound, conc. HCl. A primary amine (A) reacts with CHCl3 + KOH (carbylamine reaction) to give a foul-smelling, toxic isonitrile (B). Isonitriles are hydrolysed/decomposed by conc. HCl (C) back to the amine and formic acid. So A = primary amine, B = isonitrile, C = conc. HCl.

Question 35

A. Phenyl methanamine
B. N,N-Dimethylaniline
C. N-Methylaniline
D. Benzenamine

Choose the correct order of basic nature of the above amines:

Accepted Answer

A > B > C > D. Benzylamine (A) is an aliphatic-type amine and most basic. Among the anilines, more N-alkyl groups raise basicity in water: N,N-dimethylaniline (B) > N-methylaniline (C) > aniline (D). Hence A > B > C > D.

Question 36

Arrange the following in decreasing order of basic strength:
(A) C6H5CH2NH2
(B) C6H5N(CH3)2
(C) C6H5NHCH3
(D) C6H5NH2

Accepted Answer

(A) > (B) > (C) > (D). The order of basic strength is determined by the electron-donating ability of substituents on the nitrogen atom. In this case, the alkyl group in (A) enhances basicity more than the other groups in (B), (C), and (D), leading to the arrangement (A) > (B) > (C) > (D).

Question 37

Which of the following is not a correct statement for primary aliphatic amines?

Accepted Answer

The intermolecular association in primary amines is less than the intermolecular association in secondary amines.. A primary amine has two N-H bonds available for hydrogen bonding versus one in a secondary amine, so primary amines show greater intermolecular association. Hence the claim that their association is less than that of secondary amines is the incorrect statement.

Question 38

Given below are two statements:
Statement I: Aniline is less basic than acetamide.
Statement II: In aniline, the lone pair of electrons on nitrogen atom is delocalised over benzene ring due to resonance and hence less available to a proton.
Choose the most appropriate option:

Accepted Answer

Statement I is false but statement II is true.. Statement I is false because aniline is actually more basic than acetamide due to the presence of the electron-donating amino group, which increases the availability of the lone pair on nitrogen. Statement II is true as it correctly explains that the lone pair in aniline is delocalized into the benzene ring, making it less available to bond with protons.

Question 39

Given below are two statements, one is labelled as Assertion (A) and other is labelled as Reason (R).
Assertion (A): Gabriel phthalimide synthesis cannot be used to prepare aromatic primary amines.
Reason (R): Aryl halides do not undergo nucleophilic substitution reaction.
In the light of

Accepted Answer

Both (A) and (R) true and (R) is correct explanation of (A).. Gabriel phthalimide synthesis fails for aromatic primary amines precisely because aryl halides do not undergo the nucleophilic substitution needed to react with potassium phthalimide. So both statements are true and the Reason is the correct explanation of the Assertion.

Question 40

The conversion of propan-1-ol to n-butylamine involves the sequential addition of reagents. The correct sequential order of reagents is

Accepted Answer

(i) SOCl2 (ii) KCN (iii) H2/Ni, Na(Hg)/C2H5OH. Propan-1-ol -> (i) SOCl2 gives propyl chloride; (ii) KCN gives butanenitrile (adds one C); (iii) H2/Ni or Na(Hg)/C2H5OH reduces the nitrile to CH3CH2CH2CH2NH2 (n-butylamine). So the order is SOCl2, KCN, then reduction.

Question 41

The correct sequential order of the reagents for the given reaction is

NO2-substituted benzene with NH2 group converted to OH-substituted benzene with I group as shown.

Accepted Answer

HNO2, KI, Fe/H+, HNO2, H2O/warm. Diazotize the existing -NH2 (HNO2) and replace with iodine (KI); then reduce -NO2 to -NH2 (Fe/H+), diazotize it (HNO2), and hydrolyze with H2O/warm to -OH. Order: HNO2, KI, Fe/H+, HNO2, H2O/warm.

Question 42

Decarboxylation of all six possible forms of diaminobenzoic acid C6H3(NH2)2COOH yields three products A, B and C. Three acids give a product 'A', two acids give a product 'B' and one acid gives a product 'C'. The melting point of product 'C' is

Accepted Answer

142°C. Product 'C' is derived from the unique structure of one specific form of diaminobenzoic acid, which upon decarboxylation results in a compound with a higher melting point due to its distinct molecular interactions and stability compared to the other products.

Question 43

A primary aliphatic amine with nitrous acid in cold (273 K) and thereafter raising temperature of reaction mixture to room temperature (298 K), gives

Accepted Answer

alcohol. When a primary aliphatic amine reacts with nitrous acid at low temperatures, it forms a diazonium salt, which is unstable. Upon raising the temperature, this intermediate decomposes to produce an alcohol through hydrolysis.

Question 44

The Hinsberg reagent is:

Accepted Answer

C6H5SO2Cl. Hinsberg's reagent, used to distinguish primary, secondary, and tertiary amines, is benzenesulfonyl chloride, C6H5SO2Cl. The other options (diazonium salt, phthalimide salt, an azo dye) are not Hinsberg's reagent.

Question 45

In Friedel-Crafts alkylation of aniline, one gets

Accepted Answer

Positively charged nitrogen at benzene ring.. In Friedel-Crafts alkylation, the nitrogen atom of aniline can become positively charged due to the formation of a complex with the electrophile, which enhances the reactivity of the aromatic ring but also leads to the deactivation of the amine group, making it less nucleophilic.

Question 46

The correct order in aqueous medium of basic strength in case of methyl substituted amines is:

Accepted Answer

(1) Me2NH > MeNH2 > Me3N > NH3. The correct order reflects the increasing steric hindrance and decreasing availability of the lone pair on nitrogen for protonation in the methyl-substituted amines. Dimethylamine (Me2NH) has less steric hindrance compared to trimethylamine (Me3N), allowing it to better accept protons, while methylamine (MeNH2) is less basic than dimethylamine due to its single methyl group.

Question 47

Reaction of propanamide with Br2/KOH (aq) produces:

Accepted Answer

(3) Ethylamine. Propanamide undergoes a reaction known as the Hofmann degradation when treated with bromine and aqueous potassium hydroxide, resulting in the formation of ethylamine through the cleavage of the carbon chain and the release of nitrogen.

Question 48

Number of isomeric aromatic amines with molecular formula C8H11N, which can be synthesized by Gabriel phthalimide synthesis is _________.

Accepted Answer

[5]. The Gabriel phthalimide synthesis allows for the formation of primary amines from phthalimide, and in the case of C8H11N, five distinct isomeric aromatic amines can be generated based on the different arrangements of the aromatic ring and the alkyl substituents.

Question 49

Identify correct statements:
(A) Primary amines do not give diazonium salts when treated with NaNO2 in acidic condition.
(B) Aliphatic and aromatic primary amines on heating with CHCl3 and ethanolic KOH form carbylamines.
(C) Secondary and tertiary amines also give carbylamine test.
(D) Be

Accepted Answer

(B) and (D) only. Option (B) is correct because both aliphatic and aromatic primary amines react with CHCl3 and ethanolic KOH to form carbylamines, which is a characteristic reaction. Option (D) is also correct as benzenesulfonyl chloride is indeed known as Hinsberg's reagent, used to differentiate between primary, secondary, and tertiary amines.

Question 50

Choose the correct option for structures of A and B, respectively

CH(CH3)2
|
H2N — CH — COOH

pH = 2 → A
pH = 10 → B

Accepted Answer

H3N+ — CH — COOH
|
CH(CH3)2 and H2N — CH — COO−
|
CH(CH3)2. At pH 2, the amino group is protonated (H3N+) and the carboxylic acid group is still in its acidic form (COOH), while at pH 10, the carboxylic acid group loses a proton to become deprotonated (COO−), resulting in the correct structures for A and B.

JEE Main Chemistry: Amines questions with solutions

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