Q 1 :

In the Claisen-Schmidt reaction to prepare 351 g of dibenzalacetone using 87 g of acetone, the amount of benzaldehyde required is _______ g. (Nearest integer)                      [2024]



(318)

The condensation of an aromatic aldehyde (without α hydrogen) with an aliphatic aldehyde or ketone (having α hydrogen) in the presence of a strong base (hydroxide or alkoxide) to give α, β unsaturated aldehydes or ketones is called Claisen-Schmidt condensation reaction.

Moles of dibenzalacetone=Mass of dibenzalacetonemolar mass

                                                   =351234mol=1.5mol

By stoichiometry of reaction, moles of benzaldehyde = 2 ×moles of dibenzalacetone = 2 ×1.5 = 3mol.

Mass of benzaldehyde = moles of benzaldehyde ×molar mass of benzaldehyde = 3mol ×106 gmol-1 = 318 g

 



Q 2 :

The final product A, formed in the following multistep reaction sequence is:                         [2024]

  •  

  •  

  •  

  •  

(2)

 



Q 3 :

The product of the following reaction is P.                                                 [2024]

The number of hydroxyl groups present in the product P is ______.         



(0)



Q 4 :

Major product B of the following reaction has ___________ π-bond.                           [2024]



(5)

 



Q 5 :

1,2-dibromocyclooctane(ii) NaNH2(iii) Hg2+/H+(iv) Zn-Hg/H+(i) KOH(alc.)P(Major Product)

'P'  is                                                                                         [2025]

  •  

  •  

  •  

  •  

(2)

 



Q 6 :

An organic compound (X) with molecular formula C3H6O is not readily oxidised. On reduction it gives C3H8O (Y) which reacts with HBr to give a bromide (Z) which is converted to Grignard reagent. This Grignard reagent on reaction with (X) followed by hydrolysis gives 2,3-dimethylbutan-2-ol.

Compounds (X), (Y) and (Z) respectively are:                              [2025]

  • CH3COCH3, CH3CH2CH2OH, CH3CH(Br)CH3

     

  • CH3COCH3, CH3CH(OH)CH3, CH3CH(Br)CH3

     

  • CH3CH2CHO, CH3CH2CH2OH, CH3CH2CH2Br

     

  • CH3CH2CHO, CH3CH=CH2, CH3CH(Br)CH2

     

(2)

 



Q 7 :

Consider the following reactions. From these reactions which reaction will give carboxylic acid as a major product?

Choose the correct answer from the options given below:      [2025]

  • A and D only

     

  • A, B and E only

     

  • B, C and E only

     

  • B and E only

     

(4)

 



Q 8 :

An optically active alkyl halide C4H9Br [A] reacts with hot KOH dissolved in ethanol and forms alkene [B] as major product which reacts with bromine to give dibromide [C]. The compound [C] is converted into a gas [D] upon reacting with alcoholic NaNH2. During hydration 18 gram of water is added to 1 mole of gas [D] on warming with mercuric sulphate and dilute acid at 333 K to form compound [E].

The IUPAC name of compound [E] is:                 [2025]

  • But-2-yne

     

  • Butan-2-ol

     

  • Butan-2-one

     

  • Butan-1-al

     

(3)

Out of four possible isomers of C4H9Br, only one is optically active, so this gives us structure of [A].

 



Q 9 :

The product (A) formed in the following reaction sequence is

CH3-CCH(iii) H2/Ni(i) Hg2+,H2SO4(ii) HCN(A)Product                           [2025]

  •  

  •  

  •  

  •  

(3)

 



Q 10 :

0.1 mole of compound ‘S’ will weigh _____ g.

(Given molar mass in g mol-1 C : 12, H : 1, O : 16)               [2025]



(13)

Mass of 0.1 mol S = 0.1 × molar mass of S = 0.1 × 130 g = 13 g



Q 11 :

In the Claisen-Schmidt reaction to prepare dibenzalacetone from 5.3 g of benzaldehyde, a total of 3.51 g of product was obtained. The percentage yield in this reaction was _____%.                        [2025]



(60)

Moles of benzaldehyde initially taken (nC6H5CHO)

=Mass of C6H5CHOMolar mass of C6H5CHO=5.3106=0.05mol

For 100% yield of the reaction, 2 mol of benzaldehyde give 1 mol of dibenzalacetone, so 0.05 mol of benzaldehyde give 0.025 mol of dibenzalacetone.

Actual moles of dibenzalacetone obtained

= Mass of dibenzalacetone obtainedMolar mass of dibenzalacetone=3.51234=0.015mol

Percentage yield

= Actual moles of dibenzalacetone obtainedMoles of dibenzalacetone for 100% yield×100

=0.0150.025×100=60%



Q 12 :

Identify the structure of the final product (D) in the following sequence of the reactions:

Total number of sp2 hybridised carbon atoms in product D is _____.                [2025]



(7)

 



Q 13 :

Match List I with List II                          [2023]

  List I   List II
  Reaction   Reagents
A. Hoffmann Degradation I. Conc. KOH, Δ
B. Clemmensen reduction II. CHCl3,NaOH/H3O
C. Cannizzaro reaction III. Br2, NaOH
D. Reimer–Tiemann Reaction IV. Zn–Hg / HCl

Choose the correct answer from the options given below:

  • (A) – III, (B) – IV, (C) – II, (D) – I

     

  • (A) – III, (B) – IV, (C) – I, (D) – II

     

  • (A) – II, (B) – IV, (C) – I, (D) – III

     

  • (A) – II, (B) – I, (C) – III, (D) – IV

     

(2)

 



Q 14 :

Match List I with List II                      [2023]

  List I   List II
  Name of reaction   Reagent used
A. Hell–Volhard–Zelinsky reaction I. NaOH+I2
B. Iodoform reaction II. (i) CrO2Cl2,CS2 (ii) H2O
C. Etard reaction III. (i) Br2/red phosphorus  (ii) H2O
D. Gattermann–Koch reaction IV. CO,HCl,anhyd. AlCl3

Choose the correct answer from the options given below:

  • A–III, B–I, C–II, D–IV

     

  • A–I, B–II, C–III, D–IV

     

  • A–III, B–I, C–IV, D–II

     

  • A–III, B–II, C–I, D–IV

     

(1)



Q 15 :

The correct sequence of reagents for the above conversion of X to Y is:            [2026]

  • (i) Jones reagent (ii) NaOEt (iii) Hot KMnO4 / KOH

     

  • (i) B2H6/H2O2 (ii) NaOEt (iii) Jones reagent

     

  • (i) NaOH (aq) (ii) Jones reagent (iii) H3O+

     

  • (i) NaOEt (ii) B2H6/H2O2 (iii) Jones reagent

     

(4)



Q 16 :

A hydroxy compound (X) with molar mass 122 g mol-1 is acetylated with acetic anhydride, using a large excess of the reagent ensuring complete acetylation of all hydroxyl groups. The product obtained has a molar mass of 290 g mol-1. The number of hydroxyl groups present in compound (X) is:           [2026]

  • 5

     

  • 2

     

  • 3

     

  • 4

     

(4)

No. of OH groups=290-12242=4



Q 17 :

A student is given one compound among the following compounds that gives positive test with Tollen’s reagent.

The compound is:                   [2026]

  • D

     

  • B

     

  • C

     

  • A

     

(3)



Q 18 :

Consider the following two reactions A and B.

Numerical value of [molar mass of x+ molar mass of y] is ______.           [2026]

  • 88

     

  • 160

     

  • 4

     

  • 46

     

(4)

x=H2 (gas),  y=CO2 (gas)

Sum of molar mass=2+44=46



Q 19 :

Match the LIST-I with LIST-II                        [2026]

  List-I   List-II
  Reagents   Name of reaction involving carbonyl compounds
A. NH2NH2, KOH I. Tollen’s Test
B. Ag(NH3)2OH II. Clemmensen Reduction
C. Aqueous CuSO4, Sodium potassium tartrate, KOH III. Wolff–Kishner Reduction
D. ZnHg, HCl IV. Fehling’s Test

 

Choose the correct answer from the options given below:

  • A–III, B–IV, C–I, D–II

     

  • A–IV, B–III, C–II, D–I

     

  • A–III, B–I, C–IV, D–II

     

  • A–II, B–I, C–IV, D–III

     

(3)

Theoretical (NCERT Based)



Q 20 :

Which of the following statements are TRUE about Haloform reaction?

A. Sodium hypochlorite reacts with KI to give KOI.
B. KOI is a reducing agent.
C. α,β-unsaturated methylketone  will give iodoform reaction.
D. Isopropyl alcohol will not give iodoform test.
E. Methanoic acid will give positive iodoform test.

Choose the correct answer from the options given below:    [2026]

  • A, B & C Only

     

  • A, C & E Only

     

  • A & C Only

     

  • B, D & E Only

     

(3)

(A) NaOCl + KI → NaCl + KOI
(B) Incorrect statement

(C)  gives iodoform reaction.
(D) Incorrect statement
(E) Incorrect statement



Q 21 :

Iodoform test can differentiate between

A. Methanol and Ethanol
B. CH3COOH and CH3CH2COOH
C. Cyclohexene and cyclohexanone
D. Diethyl ether and Pentan-3-one
E. Anisole and acetone

Choose the correct answer from the options given below:   [2026]

  • A & D Only

     

  • B, C & E Only

     

  • A, B & E Only

     

  • A & E Only

     

(4)



Q 22 :

Given below are two statements :

Statement I : Compound (X), shown below, dissolves in NaHCO solution and has two chiral carbon atoms
.

Statement II : Compound (Y), shown below, has two carbons with sp³ hybridization, one carbon with sp² and one carbon with sp hybridization 

In the light of the above statements, choose the correct answer from the options given below : [2026]

  • Both Statement I and Statement II are false

     

  • Statement I is true but Statement II is false

     

  • Both Statement I and Statement II are true

     

  • Statement I is false but Statement II is true

     

(3)

Two chiral centre and due to presence of –COOH compound dissolves in NaHCO3.



Q 23 :

Match List – I with List – II

 

List – I

 

List – II

 

Reagents

 

Reaction Name (Involving aldehydes)

A.

H2, PdBaSO4

I.

Etard Reaction

B.

SnCl2, HCl

II.

Rosenmund Reduction

C.

CrO2Cl2, CS2

III.

Gatterman – Koch Reaction

D.

CO, HCl, Anhyd. AlCl3

IV.

Stephen Reaction

Choose the correct answer from the options given below:       [2026]

  • A–II, B–IV, C–I, D–III

     

  • A–IV, B–I, C–II, D–III

     

  • A–II, B–III, C–IV, D–I

     

  • A–IV, B–III, C–I, D–II

     

(1)

NCERT Name reaction theory based



Q 24 :

Given below are the four isomeric compounds (P, Q, R, S)

Identify correct statements from below.

A. Q, R and S will give precipitate with 2, 4 – DNP.
B. P and Q will give positive Baeyer’s test.
C. Q and R will give sooty flame.
D. R and S will give yellow precipitate with I2 / NaOH.
E. Q alone will deposit silver with Tollen’s reagent

Choose the correct option.    [2026]

  • A, B, D and E only

     

  • A and E only

     

  • A, C and E only

     

  • C and E only

     

(3)

(A) Q, R, S all three give 2, 4 DNP test as they have Aldehyde/ketone group

(C) Q & R gives sooty flame

(E) Q gives Tollens reagent test