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Chemistry

Organic Compounds Containing Nitrogen — JEE Main PYQs

56 previous year questions

Q1JEE Main 2026 Apr 2 Shift 2Hard

Identify compounds A and E in the following reaction sequence.

[Figure: 1-ethyl-4-nitrobenzene (benzene ring with C2H5\mathrm{C_2H_5} and NO2\mathrm{NO_2} para to each other)] Br2/AlBr3\xrightarrow{\mathrm{Br_2/AlBr_3}} A HClSn\xrightarrow[\mathrm{HCl}]{\mathrm{Sn}} B 273-278 KNaNO2/HCl\xrightarrow[\text{273-278 K}]{\mathrm{NaNO_2/HCl}} C C2H5OH\xrightarrow{\mathrm{C_2H_5OH}} D (ii) H3O+(i) KMnO4/KOH\xrightarrow[\text{(ii) } \mathrm{H_3O^+}]{\text{(i) } \mathrm{KMnO_4/KOH}} E

  1. A

    [Figure: A = benzene ring with C2H5\mathrm{C_2H_5}, with Br ortho to NO2\mathrm{NO_2} (meta to C2H5\mathrm{C_2H_5}); E = 3-bromobenzoic acid]

  2. B

    [Figure: A = 2-bromo-1-ethyl-4-nitrobenzene (Br ortho to C2H5\mathrm{C_2H_5}); E = 2-bromobenzoic acid]

  3. C

    [Figure: A = 1-(2-bromoethyl)-4-nitrobenzene (CH2CH2Br\mathrm{CH_2CH_2Br} side chain, NO2\mathrm{NO_2} para); E = benzoic acid]

  4. D

    [Figure: A = 2-bromo-1-ethyl-4-nitrobenzene (Br ortho to C2H5\mathrm{C_2H_5}); E = 3-bromo-4-hydroxybenzoic acid (COOH with Br ortho and OH para)]

Show answer

Correct option: B

Q2JEE Main 2026 Apr 2 Shift 2HardNumerical

Consider the following reactions sequence

[Figure: 1-methyl-4-nitrobenzene (benzene ring with NO2\mathrm{NO_2} and H3C\mathrm{H_3C} para to each other)] (ii) (CH3CO)2O(iii) Br2/AlBr3(iv) H3O+(i) Sn/HCl; OH\xrightarrow[\substack{\text{(ii) } \mathrm{(CH_3CO)_2O} \\ \text{(iii) } \mathrm{Br_2/AlBr_3} \\ \text{(iv) } \mathrm{H_3O^+}}]{\text{(i) } \mathrm{Sn/HCl;\ OH^-}} Major Product (P)

When the product (P) is subjected to Carius analysis using AgNO3\mathrm{AgNO_3}, 1.0 g of the product (P) will produce __________ g of the precipitate of AgBr. (Nearest Integer)

(Given : molar mass in g mol1^{-1} C : 12, H : 1, O : 16, N : 14, Br : 80, Ag : 108)

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Answer: 1

Q3JEE Main 2026 Apr 4 Shift 1Medium

Product C of the following reaction sequence will be

[Figure: aniline as starting material]

Aniline Br2/water\xrightarrow{\mathrm{Br_2/water}} A (Major Product) 273278 KNaNO2/HCl\xrightarrow[273-278\ \mathrm{K}]{\mathrm{NaNO_2/HCl}} B (ii) NaNO2,Cu,Δ(i) HBF4\xrightarrow[\text{(ii) } \mathrm{NaNO_2, Cu, \Delta}]{\text{(i) } \mathrm{HBF_4}} C

  1. A

    1-Bromo-4-nitrobenzene

  2. B

    1, 3, 5-Tribromo-2-nitrobenzene

  3. C

    4-Bromo-1-nitrobenzene

  4. D

    1, 3, 5-Tribromobenzene

Show answer

Correct option: B

Q4JEE Main 2026 Apr 4 Shift 2Medium

The strongest conjugate acid will result from:

  1. A

    [Figure: aniline — benzene ring with NH2\mathrm{NH_2}]

  2. B

    [Figure: 4-methoxyaniline — benzene ring with NH2\mathrm{NH_2} and para OCH3\mathrm{OCH_3}]

  3. C

    [Figure: 4-nitroaniline — benzene ring with NH2\mathrm{NH_2} and para NO2\mathrm{NO_2}]

  4. D

    [Figure: 4-methylaniline — benzene ring with NH2\mathrm{NH_2} and para CH3\mathrm{CH_3}]

Show answer

Correct option: C

Q5JEE Main 2026 Apr 5 Shift 1Hard

Identify the incorrect statements.

A. [Figure: benzylamine] C6H5CH2NH2\mathrm{C_6H_5CH_2NH_2} is a stronger base than [Figure: aniline] C6H5NH2\mathrm{C_6H_5NH_2} B. [Figure: 4-methoxyaniline (p-anisidine)] can be synthesized by Gabriel phthalimide synthesis. C. [Figure: 2-phenylacetamide, C6H5CH2CONH2\mathrm{C_6H_5CH_2CONH_2}] Br2, NaOH\xrightarrow{\mathrm{Br_2},\ \mathrm{NaOH}} primary aromatic amine D. [Figure: 4-nitroaniline] (ii) Δ(i) NaNO2, HCl, 0C\xrightarrow[\text{(ii) } \Delta]{\text{(i) } \mathrm{NaNO_2},\ \mathrm{HCl},\ 0^\circ\mathrm{C}} product will dissolve in NaOH

Choose the correct answer from the options given below:

  1. A

    A and D Only

  2. B

    A and C Only

  3. C

    B and C Only

  4. D

    A and B Only

Show answer

Correct option: C

Q6JEE Main 2026 Apr 5 Shift 2Medium

Given below are two statements :

Statement I : Heating benzamide with bromine in an ethanolic solution of sodium hydroxide will give benzylamine.

Statement II : Nitration of aniline with HNO3/H2SO4\mathrm{HNO_3/H_2SO_4} at 288 K produces m-nitroaniline in higher amount than o-nitroaniline (pH adjusted).

In the light of the above statements, choose the correct answer from the options given below :

  1. A

    Both Statement I and Statement II are true

  2. B

    Both Statement I and Statement II are false

  3. C

    Statement I is true but Statement II is false

  4. D

    Statement I is false but Statement II is true

Show answer

Correct option: D

Q7JEE Main 2026 Apr 6 Shift 1Easy

Arrange the following compounds according to increasing order of boiling points.

n-C4H9OH\mathrm{n\text{-}C_4H_9OH} (A), n-C4H9NH2\mathrm{n\text{-}C_4H_9NH_2} (B), n-C4H10\mathrm{n\text{-}C_4H_{10}} (C) and C2H5NHC2H5\mathrm{C_2H_5NHC_2H_5} (D).

  1. A

    C < B < A < D

  2. B

    D < C < B < A

  3. C

    C < D < B < A

  4. D

    D < B < A < C

Show answer

Correct option: C

Q8JEE Main 2026 Apr 6 Shift 1HardNumerical

Consider the following sequence of reactions.

[Figure: diazoaminobenzene, C6H5N=NNHC6H5\mathrm{C_6H_5{-}N{=}N{-}NH{-}C_6H_5}] (i) HCl (ii) Aniline (in excess) (iii) Warm 4045C (iv) Cool, acetic acid\xrightarrow{\text{(i) HCl (ii) Aniline (in excess) (iii) Warm } 40-45^{\circ}\mathrm{C} \text{ (iv) Cool, acetic acid}} Yellow Product (X)

The percentage of nitrogen in the yellow product (X) formed is ________ %. (Nearest Integer)

(Given Molar mass in g mol1^{-1} H:1, C:12, N:14)

Show answer

Answer: 21

Q9JEE Main 2026 Apr 6 Shift 2Medium

Consider the following organic reaction sequence. Choose the final product (X) from the following (consider the major product in all intermediate reactions)

[Figure: Propanoic acid, CH3CH2COOH\mathrm{CH_3CH_2COOH}, as starting material]

CH3CH2COOH(iv) C6H5N2Cl, 0C(i) NH3/Δ(ii) NaOH/Br2(iii) HNO2/H2O, 0C\mathrm{CH_3CH_2COOH} \xrightarrow[\text{(iv) } \mathrm{C_6H_5N_2Cl},\ 0^\circ\mathrm{C}]{\text{(i) } \mathrm{NH_3}/\Delta \quad \text{(ii) } \mathrm{NaOH/Br_2} \quad \text{(iii) } \mathrm{HNO_2/H_2O},\ 0^\circ\mathrm{C}} (X) Major Product

  1. A

    Benzene

  2. B

    Phenol

  3. C

    Propanol

  4. D

    Chlorobenzene

Show answer

Correct option: A

Q10JEE Main 2026 Apr 6 Shift 2Medium

The number of compounds from the following which can undergo reaction with Br2\mathrm{Br_2}/KOH (alcoholic) to give respective products and these respective products can also be obtained separately by Gabriel phthalimide reaction is :

[Figure: Six amide structures]

  1. C6H5CONH2\mathrm{C_6H_5{-}CO{-}NH_2} (benzamide)
  2. C6H5CH2CONH2\mathrm{C_6H_5{-}CH_2{-}CO{-}NH_2} (2-phenylacetamide)
  3. CH3CONH2\mathrm{CH_3{-}CO{-}NH_2} (acetamide)
  4. C6H11CONHCH2CH3\mathrm{C_6H_{11}{-}CO{-}NHCH_2CH_3} (N-ethylcyclohexanecarboxamide)
  5. (CH3)3CCONHCH3\mathrm{(CH_3)_3C{-}CO{-}NHCH_3} (N-methylpivalamide)
  6. C6H11CONH2\mathrm{C_6H_{11}{-}CO{-}NH_2} (cyclohexanecarboxamide)
  1. A

    5

  2. B

    4

  3. C

    3

  4. D

    6

Show answer

Correct option: C

Q11JEE Main 2026 Apr 8 Shift 2Hard

Consider the following reaction.

[Figure: 2-(2-oxopiperidin-1-yl)propanal — a piperidin-2-one (six-membered lactam) whose ring N bears the CH(CH3)CHO\mathrm{{-}CH(CH_3){-}CHO} group] (i) NaBH4/MeOH; (ii) NaOH(aq.), Δ; (iii) H3O+\xrightarrow{\text{(i) } \mathrm{NaBH_4}/\text{MeOH; (ii) NaOH(aq.), } \Delta \text{; (iii) } \mathrm{H_3O^+}} (P)

The major product (P) formed is :

  1. A

    [Figure: 2-(piperidin-1-yl)propan-1-ol — piperidine ring with N bonded to CH(CH3)CH2OH\mathrm{{-}CH(CH_3){-}CH_2OH}]

  2. B

    [Figure: 2-(2-hydroxypiperidin-1-yl)propanal — piperidine ring bearing OH at C-2, N bonded to CH(CH3)CHO\mathrm{{-}CH(CH_3){-}CHO}]

  3. C

    [Figure: open-chain amino acid HOOC(CH2)4NHCH(CH3)CH2OH\mathrm{HOOC{-}(CH_2)_4{-}NH{-}CH(CH_3){-}CH_2OH} (5-[(1-hydroxypropan-2-yl)amino]pentanoic acid)]

  4. D

    [Figure: open-chain amino diol HO(CH2)5NHCH(CH3)CH2OH\mathrm{HO{-}(CH_2)_5{-}NH{-}CH(CH_3){-}CH_2OH}]

Show answer

Correct option: C

Q12JEE Main 2026 Apr 8 Shift 2Medium

Which statements are True ?

A. In Hoffmann bromamide degradation, 4 moles of NaOH and 2 moles of Br2\mathrm{Br_2} are consumed per mole of an amide

B. Hoffmann bromamide reaction is not given by alkyl amides.

C. Primary amines can be synthesized by Hoffmann bromamide degradation.

D. Secondary amide on reaction with Br2\mathrm{Br_2} and NaOH will give secondary amine.

E. The by-products of Hoffmann degradation are Na2CO3\mathrm{Na_2CO_3}, NaBr and H2O\mathrm{H_2O}.

Choose the correct answer from the options given below :

  1. A

    A, C and E only

  2. B

    B, C and D only

  3. C

    C and E only

  4. D

    C, D and E only

Show answer

Correct option: C

Q13JEE Main 2025 Apr 2 Shift 1Medium

The correct order of basic nature in aqueous solution for the bases NH3\mathrm{NH_3}, H2NNH2\mathrm{H_2N-NH_2}, CH3CH2NH2\mathrm{CH_3CH_2NH_2}, (CH3CH2)2NH\mathrm{(CH_3CH_2)_2NH} and (CH3CH2)3N\mathrm{(CH_3CH_2)_3N} is :

  1. A

    NH2NH2<NH3<CH3CH2NH2<(CH3CH2)3N<(CH3CH2)2NH\mathrm{NH_2-NH_2 < NH_3 < CH_3CH_2NH_2 < (CH_3CH_2)_3N < (CH_3CH_2)_2NH}

  2. B

    NH3<H2NNH2<(CH3CH2)3N<CH3CH2NH2<(CH3CH2)2NH\mathrm{NH_3 < H_2N-NH_2 < (CH_3CH_2)_3N < CH_3CH_2NH_2 < (CH_3CH_2)_2NH}

  3. C

    H2NNH2<NH3<(CH3CH2)3N<CH3CH2NH2<(CH3CH2)2NH\mathrm{H_2N-NH_2 < NH_3 < (CH_3CH_2)_3N < CH_3CH_2NH_2 < (CH_3CH_2)_2NH}

  4. D

    NH3<H2NNH2<CH3CH2NH2<(CH3CH2)2NH<(CH3CH2)3N\mathrm{NH_3 < H_2N-NH_2 < CH_3CH_2NH_2 < (CH_3CH_2)_2NH < (CH_3CH_2)_3N}

Show answer

Correct option: A

Q14JEE Main 2025 Apr 2 Shift 2Hard

When a concentrated solution of sulphanilic acid and 1-naphthylamine is treated with nitrous acid (273 K) and acidified with acetic acid, the mass (g) of 0.1 mole of product formed is :

(Given molar mass in g mol1\mathrm{g\ mol^{-1}} H : 1, C : 12, N : 14, O : 16, S : 32)

  1. A

    330

  2. B

    33

  3. C

    66

  4. D

    343

Show answer

Correct option: B

Q15JEE Main 2025 Apr 3 Shift 1Medium

Identify [A], [B] and [C], respectively in the following reaction sequence:

[A]273278 KNaNO2, HClC6H5N2+ClKI[B]Dry ether2Na[C][\mathrm{A}] \xrightarrow[273-278~\mathrm{K}]{\mathrm{NaNO_2,~HCl}} \mathrm{C_6H_5\overset{+}{N_2}Cl^-} \xrightarrow{\mathrm{KI}} [\mathrm{B}] \xrightarrow[\text{Dry ether}]{2\mathrm{Na}} [\mathrm{C}]

[Figure: the intermediate is drawn as benzenediazonium chloride]

  1. A

    [A] aniline (C6H5NH2\mathrm{C_6H_5NH_2}), [B] chlorobenzene (C6H5Cl\mathrm{C_6H_5Cl}), [C] benzene [structures shown]

  2. B

    [A] aniline (C6H5NH2\mathrm{C_6H_5NH_2}), [B] chlorobenzene (C6H5Cl\mathrm{C_6H_5Cl}), [C] iodobenzene (C6H5I\mathrm{C_6H_5I}) [structures shown]

  3. C

    [A] nitrobenzene (C6H5NO2\mathrm{C_6H_5NO_2}), [B] iodobenzene (C6H5I\mathrm{C_6H_5I}), [C] biphenyl (C6H5C6H5\mathrm{C_6H_5{-}C_6H_5}) [structures shown]

  4. D

    [A] aniline (C6H5NH2\mathrm{C_6H_5NH_2}), [B] iodobenzene (C6H5I\mathrm{C_6H_5I}), [C] biphenyl (C6H5C6H5\mathrm{C_6H_5{-}C_6H_5}) [structures shown]

Show answer

Correct option: D

Q16JEE Main 2025 Apr 3 Shift 1Medium

In the following reactions, which one is NOT correct?

[Figure: each option shows benzenediazonium chloride (C6H5N2+Cl\mathrm{C_6H_5\overset{+}{N_2}Cl^-}) converted to a product]

  1. A

    C6H5N2+ClEtOH\mathrm{C_6H_5\overset{+}{N_2}Cl^-} \xrightarrow{\mathrm{EtOH}} ethoxybenzene (C6H5OEt\mathrm{C_6H_5OEt})

  2. B

    C6H5N2+ClH3PO2/H2O\mathrm{C_6H_5\overset{+}{N_2}Cl^-} \xrightarrow{\mathrm{H_3PO_2/H_2O}} benzene

  3. C

    C6H5N2+ClCuCN/KCN\mathrm{C_6H_5\overset{+}{N_2}Cl^-} \xrightarrow{\mathrm{CuCN/KCN}} benzonitrile (C6H5CN\mathrm{C_6H_5CN})

  4. D

    C6H5N2+ClPotassium Iodide\mathrm{C_6H_5\overset{+}{N_2}Cl^-} \xrightarrow{\text{Potassium Iodide}} iodobenzene (C6H5I\mathrm{C_6H_5I})

Show answer

Correct option: A

Q17JEE Main 2025 Apr 3 Shift 2Hard

The sequence from the following that would result in giving predominantly 3, 4, 5-Tribromoaniline is :

  1. A

    [Figure] Aniline Br2, water\xrightarrow{\mathrm{Br_2},\ \text{water}}

  2. B

    [Figure] Nitrobenzene (i) Br2, acetic acid; (ii) Sn, HCl\xrightarrow{\text{(i) } \mathrm{Br_2},\ \text{acetic acid; (ii) } \mathrm{Sn,\ HCl}}

  3. C

    [Figure] 4-Nitroaniline (i) Br2(excess), acetic acid; (ii) NaNO2, HCl, CuBr; (iii) Sn, HCl\xrightarrow{\text{(i) } \mathrm{Br_2}\text{(excess), acetic acid; (ii) } \mathrm{NaNO_2,\ HCl,\ CuBr}\text{; (iii) } \mathrm{Sn,\ HCl}}

  4. D

    [Figure] Benzene (i) Br2, AlBr3; (ii) NH3\xrightarrow{\text{(i) } \mathrm{Br_2,\ AlBr_3}\text{; (ii) } \mathrm{NH_3}}

Show answer

Correct option: C

Q18JEE Main 2025 Apr 3 Shift 2MediumNumerical

X g of nitrobenzene on nitration gave 4.2 g of m-dinitrobenzene. X = __________ g. (nearest integer)

[Given : molar mass (in g mol1\mathrm{g\ mol^{-1}}) C : 12, H : 1, O : 16, N : 14]

Show answer

Answer: 3

Q19JEE Main 2025 Apr 4 Shift 2Hard

The correct order of basicity for the following molecules is :

[Figure: three cyclic amide structures — (P) N-methylpiperidin-2-one (six-membered saturated lactam with N–Me); (Q) N-methyl-5,6-dihydropyridin-2(1H)-one (six-membered lactam with N–Me and a C=C conjugated with the C=O); (R) a bridged bicyclic lactam (1-azabicyclo[2.2.2]octan-2-one type) in which the nitrogen lone pair cannot conjugate with the C=O]

  1. A

    P > Q > R

  2. B

    Q > P > R

  3. C

    R > Q > P

  4. D

    R > P > Q

Show answer

Correct option: C

Q20JEE Main 2025 Apr 7 Shift 1Easy

Which of the following amine (s) show (s) positive carbylamine test?

A. [Figure: aniline — C6H5NH2\mathrm{C_6H_5NH_2}] B. (CH3)2NH\mathrm{(CH_3)_2NH} C. CH3NH2\mathrm{CH_3NH_2} D. (CH3)3N\mathrm{(CH_3)_3N} E. [Figure: N-methylaniline — C6H5NH(CH3)\mathrm{C_6H_5NH(CH_3)}]

Choose the correct answer from the options given below:

  1. A

    C Only

  2. B

    B, C and D Only

  3. C

    A and C Only

  4. D

    A and E Only

Show answer

Correct option: C

Q21JEE Main 2025 Apr 7 Shift 2Medium

The descending order of basicity of following amines is :

(A) aniline (C6H5NH2\mathrm{C_6H_5NH_2})

(B) 4-methoxyaniline (4-MeO-C6H4NH2\mathrm{4\text{-}MeO\text{-}C_6H_4NH_2})

(C) 4-nitroaniline (4-O2N-C6H4NH2\mathrm{4\text{-}O_2N\text{-}C_6H_4NH_2})

(D) CH3NH2\mathrm{CH_3NH_2}

(E) (CH3)2NH\mathrm{(CH_3)_2NH}

Choose the correct answer from the options given below :

  1. A

    E > D > B > A > C

  2. B

    E > D > A > B > C

  3. C

    E > A > D > C > B

  4. D

    B > E > D > A > C

Show answer

Correct option: A

Q22JEE Main 2025 Apr 8 Shift 2Medium

A(ii) H3O+(i) NaOHB(ii) H2SO4, Δ(i) EtOHC\mathrm{A \xrightarrow[(ii)\ H_3O^+]{(i)\ NaOH} B \xrightarrow[(ii)\ H_2SO_4,\ \Delta]{(i)\ EtOH} C}

'A' shows positive Lassaign's test for N and its molar mass is 121.

'B' gives effervescence with aq NaHCO3\mathrm{NaHCO_3}.

'C' gives fruity smell.

Identify A, B and C from the following.

  1. A

    [Figure: A = benzamide C6H5CONH2\mathrm{C_6H_5CONH_2}, B = benzoic acid C6H5CO2H\mathrm{C_6H_5CO_2H}, C = ethyl benzoate C6H5CO2Et\mathrm{C_6H_5CO_2Et}]

  2. B

    [Figure: A = benzene ring with CH2NHNH2\mathrm{CH_2NHNH_2} group (benzylhydrazine), B = benzoic acid C6H5CO2H\mathrm{C_6H_5CO_2H}, C = ethyl benzoate C6H5CO2Et\mathrm{C_6H_5CO_2Et}]

  3. C

    [Figure: A = 4-aminobenzaldehyde, B = benzoin-type condensation product of 4-aminobenzaldehyde (α-hydroxy ketone with two 4-aminophenyl rings), C = the corresponding diol with -OEt group (two 4-aminophenyl rings bearing OH, OH and OEt)]

  4. D

    [Figure: A = 2-aminobenzaldehyde (NH2\mathrm{NH_2} ortho to CHO), B = 2-aminobenzoic acid (NH2\mathrm{NH_2} ortho to CO2H\mathrm{CO_2H}), C = ethyl 2-(ethylamino)benzoate (NHEt ortho to CO2Et\mathrm{CO_2Et})]

Show answer

Correct option: A

Q23JEE Main 2025 Jan 22 Shift 1Medium

The products formed in the following reaction sequence are :

[Figure: benzene ring with NO2\mathrm{NO_2} group (nitrobenzene)]

C6H5NO2(i) Br2, AcOH; (ii) Sn, HCl; (iii) NaNO2, HCl, 273 K; (iv) C2H5OHA+B\mathrm{C_6H_5NO_2} \xrightarrow{\text{(i) } \mathrm{Br_2}, \text{ AcOH; (ii) Sn, HCl; (iii) } \mathrm{NaNO_2}, \text{ HCl, 273 K; (iv) } \mathrm{C_2H_5OH}} \mathrm{A} + \mathrm{B}

  1. A

    [Figure: 3-bromophenol (benzene ring with OH, Br at meta position) and 1-bromo-3-ethoxybenzene (benzene ring with OEt, Br at meta position)]

  2. B

    [Figure: bromobenzene (benzene ring with Br)] and CH3CHO\mathrm{CH_3-CHO}

  3. C

    [Figure: 3-bromophenol (benzene ring with OH, Br at meta position)] and CH3CHO\mathrm{CH_3-CHO}

  4. D

    [Figure: 1-bromo-3-ethoxybenzene (benzene ring with OEt, Br at meta position)] and CH3COOH\mathrm{CH_3-COOH}

Show answer

Correct option: B

Q24JEE Main 2025 Jan 22 Shift 1MediumNumerical

Consider the following sequence of reactions :

[Figure: benzene ring with NO2\mathrm{NO_2} group (nitrobenzene)]

C6H5NO2(i) Sn + HCl; (ii) NaNO2, HCl, 0°C; (iii) Cu2Cl2; (iv) Na, EtherA (Product)\mathrm{C_6H_5NO_2} \xrightarrow{\text{(i) Sn + HCl; (ii) } \mathrm{NaNO_2}, \text{ HCl, } 0°\mathrm{C}\text{; (iii) } \mathrm{Cu_2Cl_2}\text{; (iv) Na, Ether}} \mathrm{A}\ (\text{Product})

Molar mass of the product (A) formed is ________ g mol1\mathrm{g\ mol^{-1}}.

Show answer

Answer: 154

Q25JEE Main 2025 Jan 22 Shift 2Medium

Match the compounds (List-I) with the appropriate catalyst/reagents (List-II) for their reduction into corresponding amines :

List-I (Compound) List-II (Catalyst/Reagent)
(A) RCONH2\mathrm{R{-}\overset{\displaystyle O}{\overset{\|}{C}}{-}NH_2} (I) NaOH (aqueous)
(B) [Structure: nitrobenzene, C6H5NO2\mathrm{C_6H_5{-}NO_2}] (II) H2/Ni\mathrm{H_2/Ni}
(C) RCN\mathrm{R{-}C{\equiv}N} (III) LiAlH4, H2O\mathrm{LiAlH_4,\ H_2O}
(D) [Structure: N-alkyl phthalimide, N–R] (IV) Sn, HCl\mathrm{Sn,\ HCl}

Choose the correct answer from the options given below :

  1. A

    (A)-(II), (B)-(IV), (C)-(III), (D)-(I)

  2. B

    (A)-(III), (B)-(IV), (C)-(II), (D)-(I)

  3. C

    (A)-(III), (B)-(II), (C)-(IV), (D)-(I)

  4. D

    (A)-(II), (B)-(I), (C)-(III), (D)-(IV)

Show answer

Correct option: B

Q26JEE Main 2025 Jan 23 Shift 1Medium

Which among the following react with Hinsberg's reagent?

A. C6H5NH2\mathrm{C_6H_5{-}NH_2} (aniline)

B. C6H5N(CH3)2\mathrm{C_6H_5{-}N(CH_3)_2} (N,N-dimethylaniline)

C. CH3NH2\mathrm{CH_3{-}NH_2}

D. N(CH3)3\mathrm{N(CH_3)_3}

E. (C6H5)2NH\mathrm{(C_6H_5)_2NH} (diphenylamine)

Choose the correct answer from the options given below:

  1. A

    A, C and E Only

  2. B

    A, B and E Only

  3. C

    C and D Only

  4. D

    B and D Only

Show answer

Correct option: A

Q27JEE Main 2025 Jan 23 Shift 1MediumNumerical

Consider the following sequence of reactions to produce major product (A)

[Figure: 3-nitrotoluene (benzene ring with CH3\mathrm{CH_3} group and NO2\mathrm{NO_2} at the meta position)]

i) Br2, Feii) Sn,HCliii) NaNO2,HCl,273Kiv) H3PO2,H2O(A) Major Product\xrightarrow{\substack{\text{i) } \mathrm{Br_2},\ \mathrm{Fe} \\ \text{ii) } \mathrm{Sn}, \mathrm{HCl} \\ \text{iii) } \mathrm{NaNO_2}, \mathrm{HCl}, 273\,\mathrm{K} \\ \text{iv) } \mathrm{H_3PO_2}, \mathrm{H_2O}}} \text{(A) Major Product}

Molar mass of product (A) is _____ g mol1^{-1}.

(Given molar mass in g mol1^{-1} of C : 12, H: 1, O :16, Br : 80, N : 14, P :31)

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Answer: 171

Q28JEE Main 2025 Jan 23 Shift 2HardNumerical

Consider the following sequence of reactions.

[Figure: reaction scheme] 4-ethoxyaniline (p-ethoxyaniline) 0-5C(i) NaNO2, HCl\xrightarrow[\text{0-5}^{\circ}\text{C}]{\text{(i) NaNO}_2,\ \text{HCl}} (A, the diazonium salt) (ii) dil. HCl(i) phenol, NaOH\xrightarrow[\text{(ii) dil. HCl}]{\text{(i) phenol, NaOH}} (B) (Molecular formula C14H14N2O2\mathrm{C_{14}H_{14}N_2O_2}) (ii) H3CCH2Br(i) NaOH\xrightarrow[\text{(ii) } \mathrm{H_3CCH_2Br}]{\text{(i) NaOH}} (C) (Molecular formula C16H18N2O2\mathrm{C_{16}H_{18}N_2O_2})

Total number of sp3sp^3 hybridised carbon atoms in the major product C formed is ________.

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Answer: 4

Q29JEE Main 2025 Jan 24 Shift 2Medium

For reaction

[Figure: aniline (benzene ring with NH2\mathrm{-NH_2}) \rightarrow 2-bromoaniline (benzene ring with NH2\mathrm{-NH_2} and an ortho Br\mathrm{-Br}), labelled (Major)]

The correct order of set of reagents for the above conversion is :

  1. A

    Ac2O, Br2, H2O(Δ), NaOH\mathrm{Ac_2O},\ \mathrm{Br_2},\ \mathrm{H_2O(\Delta)},\ \mathrm{NaOH}

  2. B

    H2SO4, Ac2O, Br2, H2O(Δ), NaOH\mathrm{H_2SO_4},\ \mathrm{Ac_2O},\ \mathrm{Br_2},\ \mathrm{H_2O(\Delta)},\ \mathrm{NaOH}

  3. C

    Ac2O, H2SO4, Br2, NaOH\mathrm{Ac_2O},\ \mathrm{H_2SO_4},\ \mathrm{Br_2},\ \mathrm{NaOH}

  4. D

    Br2FeBr3, H2O(Δ), NaOH\mathrm{Br_2 | FeBr_3},\ \mathrm{H_2O(\Delta)},\ \mathrm{NaOH}

Show answer

Correct option: B

Q30JEE Main 2025 Jan 28 Shift 1Medium

[Stem unrecoverable: the PDF pages carrying this question's stem are corrupted in both the English and Hindi copies. The options indicate a five-statement (A-E) multiple-select question. Option texts below are recovered from the Hindi copy.]

  1. A

    A and C Only

  2. B

    A and B Only

  3. C

    A, B and E Only

  4. D

    A, C and D Only

Show answer

Correct option: D

Q31JEE Main 2025 Jan 28 Shift 1MediumNumerical

Consider the following sequence of reactions:

[Figure: benzene ring with Cl substituent (Chlorobenzene)]

Chlorobenzeneii) CO2,H3O+iii) NH3, Δi) Mg, dry etherABr2, NaOHB\text{Chlorobenzene} \xrightarrow[\substack{\text{ii) } \mathrm{CO_2, H_3O^+} \\ \text{iii) } \mathrm{NH_3},\ \Delta}]{\text{i) Mg, dry ether}} \mathrm{A} \xrightarrow{\mathrm{Br_2,\ NaOH}} \mathrm{B}

11.25 mg of chlorobenzene will produce ---- x101\mathrm{x}10^{-1} mg of product B.

(Consider the reactions result in complete conversion.)

[Given molar mass of C, H, O, N and Cl as 12, 1, 16, 14 and 35.5 g mol1\mathrm{g\ mol^{-1}} respectively]

Show answer

Answer: 93

Q32JEE Main 2025 Jan 28 Shift 2Hard

The product B formed in the following reaction sequence is :

4-methyl-(prop-1-en-1-yl)benzene(Major)HCl(A)(Major)AgCN(B)\text{4-methyl-(prop-1-en-1-yl)benzene} \xrightarrow[\text{(Major)}]{\mathrm{HCl}} (A) \xrightarrow[\text{(Major)}]{\mathrm{AgCN}} (B)

[Figure: reactant is 1-(prop-1-en-1-yl)-4-methylbenzene, i.e. p-tolyl–CH=CH–CH3_3]

  1. A

    [Figure: p-tolyl–CH(NC)–CH2_2CH3_3; isocyanide (–NC) on the benzylic carbon]

  2. B

    [Figure: p-tolyl–CH2_2–CH(NC)–CH3_3; isocyanide (–NC) on the secondary non-benzylic carbon]

  3. C

    [Figure: p-tolyl–CH(CN)–CH3_3; nitrile (–CN) on the benzylic carbon]

  4. D

    [Figure: p-tolyl–CH2_2–CH(NC)–CH3_3; isocyanide (–NC) on the secondary carbon]

Show answer

Correct option: A

Q33JEE Main 2025 Jan 28 Shift 2Medium

Identify correct statements :

(A) Primary amines do not give diazonium salts when treated with NaNO2\mathrm{NaNO_2} in acidic condition. (B) Aliphatic and aromatic primary amines on heating with CHCl3\mathrm{CHCl_3} and ethanolic KOH form carbylamines. (C) Secondary and tertiary amines also give carbylamine test. (D) Benzenesulfonyl chloride is known as Hinsberg's reagent. (E) Tertiary amines reacts with benzenesulfonyl chloride very easily.

Choose the correct answer from the options given below :

  1. A

    (A) and (B) only

  2. B

    (B) and (C) only

  3. C

    (B) and (D) only

  4. D

    (D) and (E) only

Show answer

Correct option: C

Q34JEE Main 2024 Apr 4 Shift 1MediumNumerical

X g of ethylamine is subjected to reaction with NaNO2/HCl\mathrm{NaNO_2/HCl} followed by water; evolved dinitrogen gas which occupied 2.24 L volume at STP. X is ________ ×101\times 10^{-1} g.

Show answer

Answer: 45

Q35JEE Main 2024 Apr 4 Shift 2MediumNumerical

From 6.55 g of aniline, the maximum amount of acetanilide that can be prepared will be ________ ×101\times 10^{-1} g.

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Answer: 95

Q36JEE Main 2024 Apr 4 Shift 2MediumNumerical

Phthalimide is made to undergo following sequence of reactions.

Phthalimide(i) KOH (ii) Benzylchloride’P’\text{Phthalimide} \xrightarrow{\text{(i) KOH (ii) Benzylchloride}} \text{'P'}

Total number of π\pi bonds present in product 'P' is/are ________.

Show answer

Answer: 8

Q37JEE Main 2024 Apr 5 Shift 1MediumNumerical

9.3 g of pure aniline is treated with bromine water at room temperature to give a white precipitate of the product 'P'. The mass of product 'P' obtaind is 26.4 g. The percentage yield is ________ %.

Show answer

Answer: 80

Q38JEE Main 2024 Apr 5 Shift 2MediumNumerical

X g of ethanamine was subjected to reaction with NaNO2/HCl\mathrm{NaNO_2/HCl} followed by hydrolysis to liberate N2\mathrm{N_2} and HCl. The HCl generated was completely neutralised by 0.2 moles of NaOH. X is ________ g.

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Answer: 9

Q39JEE Main 2024 Apr 6 Shift 1MediumNumerical

9.3 g of pure aniline upon diazotisation followed by coupling with phenol gives an orange dye. The mass of orange dye produced (assume 100% yield/conversion) is _________ g. (nearest integer)

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Answer: 20

Q40JEE Main 2024 Apr 6 Shift 2Medium

Identify the product (A) in the following reaction.

[Figure: benzene ring with NH2\mathrm{NH_2} group (aniline)] (i) NaNO2+HCl(ii) Cu2Cl2(iii) NaOH, 623 K, 300 atm(iv) H+\xrightarrow{\substack{\text{(i) } \mathrm{NaNO_2 + HCl} \\ \text{(ii) } \mathrm{Cu_2Cl_2} \\ \text{(iii) } \mathrm{NaOH},\ 623\ \mathrm{K},\ 300\ \mathrm{atm} \\ \text{(iv) } \mathrm{H^+}}} (A)

  1. A

    [Figure: benzene ring with Cl and OH para to each other (4-chlorophenol)]

  2. B

    [Figure: benzene ring with NH2\mathrm{NH_2} and Cl para to each other (4-chloroaniline)]

  3. C

    [Figure: benzene ring with OH group (phenol)]

  4. D

    [Figure: benzene ring with NH2\mathrm{NH_2} and OH para to each other (4-aminophenol)]

Show answer

Correct option: C

Q41JEE Main 2024 Apr 6 Shift 2MediumNumerical

An amine (X) is prepared by ammonolysis of benzyl chloride. On adding p-toluenesulphonyl chloride to it the solution remains clear. Molar mass of the amine (X) formed is ________ g mol1\mathrm{g\ mol^{-1}}.

(Given molar mass in gmol1\mathrm{gmol^{-1}} C : 12, H : 1, O : 16, N : 14)

Show answer

Answer: 287

Q42JEE Main 2024 Apr 8 Shift 1MediumNumerical

If 279 g of aniline is reacted with one equivalent of benzenediazonium chloride, the maximum amount of aniline yellow formed will be _______ g. (nearest integer)

(consider complete conversion).

Show answer

Answer: 591

Q43JEE Main 2024 Apr 8 Shift 1HardNumerical

Number of amine compounds from the following giving solids which are soluble in NaOH upon reaction with Hinsberg's reagent is ________.

[Figure: eleven structures — (1) aniline C6H5NH2\mathrm{C_6H_5NH_2}; (2) benzamide C6H5CONH2\mathrm{C_6H_5CONH_2}; (3) semicarbazide H2NNHCONH2\mathrm{H_2N{-}NH{-}CO{-}NH_2}; (4) N-methylaniline C6H5NHCH3\mathrm{C_6H_5{-}NH{-}CH_3}; (5) phenyl–NH– attached to a six-membered diene ring bearing NH2 (N-phenyl diamine structure); (6) urea H2NCONH2\mathrm{H_2N{-}CO{-}NH_2}; (7) 2-methoxyaniline (benzene ring with NH2 and OCH3 on adjacent carbons); (8) propan-1-amine CH3CH2CH2NH2\mathrm{CH_3CH_2CH_2NH_2}; (9) triethylamine (C2H5)3N\mathrm{(C_2H_5)_3N}; (10) tertiary amine — N bearing ethyl, methyl and cyclohexylmethyl groups; (11) cyclohexylamine]

Show answer

Answer: 5

Q44JEE Main 2024 Apr 8 Shift 2Medium

Given below are two statements :

Statement (I) : All the following compounds react with p-toluenesulfonyl chloride.

C6H5NH2\mathrm{C_6H_5NH_2} \quad (C6H5)2NH\mathrm{(C_6H_5)_2NH} \quad (C6H5)3N\mathrm{(C_6H_5)_3N}

Statement (II) : Their products in the above reaction are soluble is aqueous NaOH.

In the light of the above statements, choose the correct answer from the options given below :

  1. A

    Both Statement I and Statement II are true

  2. B

    Both Statement I and Statement II are false

  3. C

    Statement I is true but Statement II is false

  4. D

    Statement I is false but Statement II is true

Show answer

Correct option: B

Q45JEE Main 2024 Apr 9 Shift 1Easy

Correct order of basic strength of Pyrrole, Pyridine and Piperidine is :

[Figure: structures of pyrrole (five-membered aromatic ring with N-H), pyridine (six-membered aromatic ring with N) and piperidine (saturated six-membered ring with N-H)]

  1. A

    Piperidine > Pyridine > Pyrrole

  2. B

    Pyrrole > Pyridine > Piperidine

  3. C

    Pyrrole > Piperidine > Pyridine

  4. D

    Pyridine > Piperidine > Pyrrole

Show answer

Correct option: A

Q46JEE Main 2024 Feb 1 Shift 1Easy

Given below are two statements :

Statement (I) : Aminobenzene and aniline are same organic compounds.

Statement (II) : Aminobenzene and aniline are different organic compounds.

In the light of the above statements, choose the most appropriate answer from the options given below :

  1. A

    Both Statement I and Statement II are correct

  2. B

    Both Statement I and Statement II are incorrect

  3. C

    Statement I is correct but Statement II is incorrect

  4. D

    Statement I is incorrect but Statement II is correct

Show answer

Correct option: C

Q47JEE Main 2024 Feb 1 Shift 1Medium

Given below are two statements :

Statement (I) : The NH2\mathrm{NH_2} group in Aniline is ortho and para directing and a powerful activating group.

Statement (II) : Aniline does not undergo Friedel-Craft's reaction (alkylation and acylation).

In the light of the above statements, choose the most appropriate answer from the options given below :

  1. A

    Both Statement I and Statement II are correct

  2. B

    Both Statement I and Statement II are incorrect

  3. C

    Statement I is correct but Statement II is incorrect

  4. D

    Statement I is incorrect but Statement II is correct

Show answer

Correct option: A

Q48JEE Main 2024 Feb 1 Shift 2MediumNumerical

Number of compounds which give reaction with Hinsberg's reagent is ________.

[Figure: eleven structures — benzenediazonium chloride C6H5N2+Cl\mathrm{C_6H_5N_2^+Cl^-}; N-benzoylbenzamide (dibenzimide) (C6H5CO)2NH\mathrm{(C_6H_5CO)_2NH}; aniline C6H5NH2\mathrm{C_6H_5NH_2}; N-phenylpiperidine; N-methylbenzylamine C6H5CH2NHCH3\mathrm{C_6H_5CH_2NHCH_3}; trimethylamine (CH3)3N\mathrm{(CH_3)_3N}; ethane-1,2-diamine H2NCH2CH2NH2\mathrm{H_2NCH_2CH_2NH_2}; piperidine; pyridine; propan-1-amine CH3CH2CH2NH2\mathrm{CH_3CH_2CH_2NH_2}; urea H2NCONH2\mathrm{H_2N{-}CO{-}NH_2}]

Show answer

Answer: 5

Q49JEE Main 2024 Jan 27 Shift 1Medium

Which of the following is strongest Bronsted base ?

  1. A

    [Structure: aniline, C6H5NH2\mathrm{C_6H_5NH_2}]

  2. B

    [Structure: diphenylamine, (C6H5)2NH\mathrm{(C_6H_5)_2NH}]

  3. C

    [Structure: pyrrolidine (saturated five-membered ring with N-H)]

  4. D

    [Structure: pyrrole (aromatic five-membered ring with N-H)]

Show answer

Correct option: C

Q50JEE Main 2024 Jan 29 Shift 1Hard

The arenium ion which is not involved in the bromination of Aniline is ________.

  1. A

    Cyclohexadienyl cation with an exocyclic C=NH2\mathrm{C{=}NH_2^{\oplus}} group and H, Br together on the carbon ortho to it

  2. B

    Cyclohexadienyl cation with an exocyclic C=NH2\mathrm{C{=}NH_2^{\oplus}} group and H, Br together on the carbon para to it

  3. C

    Ring with intact NH2\mathrm{NH_2} group, positive charge on the carbon ortho to NH2\mathrm{NH_2}, and H, Br together on the adjacent (meta) carbon

  4. D

    Ring with intact NH2\mathrm{NH_2} group, H, Br together on the carbon ortho to NH2\mathrm{NH_2}, and positive charge on the adjacent (meta) carbon

Show answer

Correct option: C

Q51JEE Main 2024 Jan 30 Shift 1Medium

The final product A, formed in the following multistep reaction sequence is :

C6H5Br(i) Mg, ether then CO2, H+(ii) NH3, Δ(iii) Br2, NaOHA\mathrm{C_6H_5Br} \xrightarrow{\substack{\text{(i) Mg, ether then } \mathrm{CO_2},\ \mathrm{H^+} \\ \text{(ii) } \mathrm{NH_3},\ \Delta \\ \text{(iii) } \mathrm{Br_2},\ \mathrm{NaOH}}} A

  1. A

    C6H5CONH2\mathrm{C_6H_5CONH_2} (benzamide structure)

  2. B

    C6H5COOH\mathrm{C_6H_5COOH} (benzoic acid structure)

  3. C

    C6H5COBr\mathrm{C_6H_5COBr} (benzoyl bromide structure)

  4. D

    C6H5NH2\mathrm{C_6H_5NH_2} (aniline structure)

Show answer

Correct option: D

Q52JEE Main 2024 Jan 30 Shift 1Medium

Following is a confirmatory test for aromatic primary amines. Identify reagent (A) and (B).

C6H5NH2(A)C6H5N2ClNaOH(B)Scarlet red dye\mathrm{C_6H_5NH_2 \xrightarrow{(A)} C_6H_5\overset{\oplus}{N_2}\overset{\ominus}{Cl} \xrightarrow[\text{NaOH}]{(B)} \text{Scarlet red dye}}

  1. A

    A=NaNO2+HCl, 05°C\mathrm{A = NaNO_2 + HCl,\ 0{-}5°C} ; B = [Structure: naphthalen-2-ol (β\beta-naphthol)]

  2. B

    A=NaNO2+HCl, 05°C\mathrm{A = NaNO_2 + HCl,\ 0{-}5°C} ; B = [Structure: naphthalen-1-ol (α\alpha-naphthol)]

  3. C

    A=HNO3/H2SO4\mathrm{A = HNO_3/H_2SO_4} ; B = [Structure: phenol]

  4. D

    A=NaNO2+HCl, 05°C\mathrm{A = NaNO_2 + HCl,\ 0{-}5°C} ; B = [Structure: aniline]

Show answer

Correct option: A

Q53JEE Main 2024 Jan 30 Shift 2Medium

The products A and B formed in the following reaction scheme are respectively

[Figure: benzene ii) Sn / HCli) con. HNO3/Con. H2SO4, 323333 K\xrightarrow[\text{ii) Sn / HCl}]{\text{i) con. } \mathrm{HNO_3}/\text{Con. } \mathrm{H_2SO_4},\ 323-333\ \mathrm{K}} A ii) Phenoli) NaNO2, HCl, 273278 K\xrightarrow[\text{ii) Phenol}]{\text{i) } \mathrm{NaNO_2},\ \mathrm{HCl},\ 273-278\ \mathrm{K}} B]

  1. A

    [Figure: A = aniline (NH2\mathrm{NH_2} on ring); B = a biphenyl with NH2\mathrm{NH_2} on one ring and OH on the other (C–C linked)]

  2. B

    [Figure: A = 1-chloro-3-nitrobenzene (Cl and NO2\mathrm{NO_2} on ring); B = a biphenyl bearing HO on one ring and NO2\mathrm{NO_2} on the other]

  3. C

    [Figure: A = aniline (NH2\mathrm{NH_2} on ring); B = p-hydroxyazobenzene, HO–C6_6H4_4–N=N–C6_6H5_5]

  4. D

    [Figure: A = aniline (NH2\mathrm{NH_2} on ring); B = o-hydroxyazobenzene, ring with N=N–C6_6H5_5 and OH at the ortho position]

Show answer

Correct option: C

Q54JEE Main 2024 Jan 31 Shift 2Medium

Given below are two statements :

Statement I : Aniline reacts with con. H2SO4\mathrm{H_2SO_4}, followed by heating at 453 - 473 K gives p-aminobenzene sulphonic acid, which gives blood red colour in the 'Lassaigne's test'.

Statement II : In Friedel - Craft's alkylation and acylation reactions, aniline forms salt with the AlCl3\mathrm{AlCl_3} catalyst. Due to this, nitrogen of aniline aquires a positive charge and acts as deactivating group.

In the light of the above statements, choose the correct answer from the options given below :

  1. A

    Both statement I and statement II are true

  2. B

    Both statement I and statement II are false

  3. C

    Statement I is true but statement II is false

  4. D

    Statement I is false but statement II is true

Show answer

Correct option: A

Q55JEE Main 2024 Jan 31 Shift 2Medium

The azo-dye (YY) formed in the following reactions is

Sulphanilic acid + NaNO2\mathrm{NaNO_2} + CH3COOH\mathrm{CH_3COOH} → X.

X+naphthalen-1-amineY\mathrm{X} + \text{naphthalen-1-amine} \longrightarrow \mathrm{Y}

[Figure: X reacting with 1-aminonaphthalene (naphthalene with NH2\mathrm{NH_2} at position 1)]

  1. A

    [Figure] HSO3C6H4N=N\mathrm{HSO_3-C_6H_4-N{=}N-}naphthalene with NH2\mathrm{NH_2} on the same ring as the azo linkage (azo group and NH2\mathrm{NH_2} both on the top ring, at positions 1 and 4)

  2. B

    [Figure] HSO3C6H4N=N\mathrm{HSO_3-C_6H_4-N{=}N-}naphthalene with NH2\mathrm{NH_2} on the other ring (azo group at position 1, NH2\mathrm{NH_2} at position 5)

  3. C

    [Figure] Bis-azo structure: two HO3S\mathrm{HO_3S}-bearing naphthalene units each linked through N=N\mathrm{N{=}N}, one unit carrying NH2\mathrm{NH_2}

  4. D

    [Figure] HSO3C6H4N=N\mathrm{HSO_3-C_6H_4-N{=}N-}naphthaleneN=NC6H4SO3H\mathrm{-N{=}N-C_6H_4-SO_3H} (naphthalene bearing two azo linkages, one on each ring)

Show answer

Correct option: A

Q56JEE Main 2024 Jan 31 Shift 2MediumNumerical

A compound (xx) with molar mass 108 g mol1108\ \mathrm{g\ mol^{-1}} undergoes acetylation to give product with molar mass 192 g mol1192\ \mathrm{g\ mol^{-1}}. The number of amino groups in the compund (xx) is _______.

Show answer

Answer: 2