Q 1 :

Compound A formed in the following reaction reacts with B gives the product C. Find out A and B.                 [2024]

  • A=CH3-CC-N+a,  B=CH3-CH2-CH2-Br

     

  • A=CH3-CH=CH2,  B=CH3-CH2-CH2-Br

     

  • A=CH3-CH2-CH3,  B=CH3-CCH

     

  • A=CH3-CC- +Na,  B=CH3-CH2-CH3

     

(1)

 



Q 2 :

The incorrect statement regarding ethyne is                            [2024]

  • Both carbons are sp hybridised

     

  • The C − C bonds in ethyne is shorter than that in ethene

     

  • Ethyne is linear

     

  • The carbon-carbon bonds in ethyne is weaker than that in ethene

     

(4)

More are number of bonds between two atoms, shorter and stronger is the bond. Thus, carbon-carbon bond in ethyne is shorter and stronger than that of ethene.

 

 



Q 3 :

The major product of the following reaction is P.                            

Number of oxygen atoms present in product ‘P’ is _______. (nearest integer)                     [2024]



(2)

 



Q 4 :

Given below are two statements:                                      [2025]

Statement I: One mole of propyne reacts with excess of sodium to liberate half a mole of H2 gas.

Statement II: Four gram of propyne reacts with NaNH2 to liberate NH3 gas which occupies 224 mL.

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

  • Statement I is correct but Statement II is incorrect

     

  • Both Statement I and Statement II are incorrect

     

  • Statement I is incorrect but Statement II is correct

     

  • Both Statement I and Statement II are correct

     

(1)

Statement I: One mole propyne gives half mole hydrogen gas as per the following stoichiometry:

CH3-CPropyneC-H Na (excess) CH3-CC(-)Na(+)+12H2

Statement II:

CH3-CPropyneC-H+NaNH2CH3-CC(-)Na(+)+NH3

Moles of propyne=Given massMolar mass=4 g40g mol-1=0.1 mol

By stoichiometry, moles of NH3 = moles of propyne = 0.1 mol.

Volume occupied by 0.1 mol gas at STP = 0.1×22400 mL = 2240 mL



Q 5 :

Identify product [A], [B] and [C] in the following reaction sequence.                        [2025]

CH3-CCHH2Pd/C[A](ii)Zn,H2O(i)O3[B]+[C]
 

  • [A]:CH3CH2CH3,[B]:CH3CHO,[C]:HCHO

     

  •  

  • [A]:CH3-CH=CH2,[B]:CH3CHO,[C]:CH3CH2OH

     

  • [A]:CH3-CH=CH2,[B]:CH3CHO,[C]:HCHO

     

(4)

Catalytic hydrogenation of alkyne with Pd/C reduces an alkyne to alkane. To stop the reduction at alkene stage poison such as BaSO4/CaCO3 is needed. However in this question as per given choices, we need to stop the reduction upto alkene stage.

CH3-CC-H+H2Pd/CCH3-CH[A]=CH2(ii)Zn+H2O(i)O3CH3-[A]CHO+O=[C]CH2

 



Q 6 :

The compound with molecular formula C6H6, which gives only one monobromo derivative and takes up four moles of hydrogen per mole for complete hydrogenation has ______ π electrons.                           [2025]



(8)

H3CCCCCCH3Br2+H3CCCCCCH2Br



Q 7 :

But-2-yne is reacted separately with one mole of Hydrogen as shown below;

[IMAGE 125]

(A) A is more soluble than B.

(B) The boiling point and melting point of A are higher and lower than B respectively.

(C) A is more polar than B because dipole moment of A is zero.

(D) Br2 adds easily to B than A.

Identify the incorrect statements from the options given below:             [2023]

  • B and C only

     

  • B, C and D only

     

  • A, C and D only

     

  • C and D only

     

(4)

Physical properties [IMAGE 126] Remarks
Dipole moment A > B

cis-isomer has resultant of dipoles while in trans-isomer dipole moments cancel out.

Boiling point A > B Molecules having higher dipole moment have higher boiling point due to larger intermolecular force of attraction.
Solubility (in H2O) A > B More polar molecules are more soluble in H2O.
Melting point B > A More symmetric isomers have higher melting points due to better packing in crystalline lattice & trans-isomers are more symmetric than cis.