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Topic Question Set

Q 1 :    

Sugar which does not give reddish brown precipitate with Fehling’s reagent, is:              [2024]

  • Sucrose

     

  • Lactose

     

  • Glucose

     

  • Maltose

     

(A)

Sucrose does not have an aldehydic or ketonic group. Hence sucrose is a non-reducing sugar and does not give reddish brown precipitate with Fehling’s reagent. All monosaccharides (such as glucose) are reducing sugars.

 

Q 2 :    

Match List I with List II

  List I   List II
A. α-Glucose and α-Galactose I. Functional isomers
B. α-Glucose and β-Glucose II. Homologous
C. α-Glucose and α-Fructose III. Anomers
D. α-Glucose and α-Ribose IV. Epimers

 

Choose the correct answer from the options given below:              [2024]

  • A-III, B-IV, C-II, D-I

     

  • A-III, B-IV, C-I, D-II

     

  • A-IV, B-III, C-II, D-I

     

  • A-IV, B-III, C-I, D-II

     

(D)

    A. Epimers are diastereomers that differ in configuration at only one chiral centre. α-Glucose and α-Galactose are C-4 epimers.

   B. Anomers are epimers which differ in configuration at acetal or hemiacetal carbon. α-D-glucose and β-D-glucose are anomers.

   C. Glucose is an aldohexose and fructose is a ketohexose. These are thus functional isomers.

   D. Formula of glucose (C6H12O6) differs by — CH2 unit with the formula of ribose (C5H10O5). So, Homologns

Q 3 :    

The incorrect statement about Glucose is:                   [2024]

  • Glucose remains in multiple isomeric form in its aqueous solution

     

  • Glucose is one of the monomer unit in sucrose

     

  • Glucose is an aldohexose

     

  • Glucose is soluble in water because of having aldehyde functional group.

     

(D)

        1.  Glucose exists as equilibrium mixture of α-D-glucose, open chain structure and β-D-glucose.

        2.  Sucrose is a disaccharide formed by glycosidic linkage between C1 of α-D-glucose and C2 of β-D-fructose.

        3.  Glucose is an aldohexose.

        4.  Water solubility of glucose is because it forms extensive hydrogen bonding with water.

 

Q 4 :    

Match List I with List II                                                             [2024]

LIST I LIST II
A. Glucose/NaHSO3/Δ I. Gluconic acid
B. Glucose/HNO3 II. No reaction
C. Glucose/HI/Δ III. n-hexane
D. Glucose/Bromine water IV. Saccharic acid

 

Choose the correct answer from the options given below:

  • A-IV, B-I, C-III, D-II

     

  • A-I, B-IV, C-III, D-II

     

  • A-II, B-IV, C-III, D-I

     

  • A-III, B-II, C-I, D-IV

     

(3)

(A) Despite having the aldehyde group, glucose does not give Schiff’s test and it does not form the hydrogensulphite addition product with NaHSO3. This shows glucose exists in cyclic structure.

[IMAGE 563]

Note: In question, NaHCO3 should be replaced by NaHSO3.

 

Q 5 :    

Which of the following is the correct structure of L-Glucose?                       [2024]

  • [IMAGE 564]

     

  • [IMAGE 565]

     

  • [IMAGE 566]

     

  • [IMAGE 567]

     

(2)

[IMAGE 568]

Q 6 :    

[IMAGE 569]

The incorrect statement regarding the given structure is:                          [2024]

  • has 4-asymmetric carbon atom

     

  • will coexist in equilibrium with 2 other cyclic structures

     

  • can be oxidized to a dicarboxylic acid with Br2 water

     

  • despite the presence of –CHO does not give Schiff’s test

     

(3)

1.    [IMAGE 570]

2.    Two cyclic forms of glucose exist in equilibrium with its open chain structure.

3.    Bromine water is a mild oxidizing agent, it oxidizes easily oxidisable aldehyde group to carboxylic acid, but cannot oxidize alcoholic group.

4.    Despite having the presence of –CHO group, glucose does not give Schiff’s test because under the reaction conditions, it exists as cyclic form and free –CHO group is thus not available.