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