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Biomolecules question

2010 · Shift 1 · Q15
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Biomolecules question

2010 · Shift 1 · Q15

JEE AdvancedChemistryBiomoleculesMCQ+3 / −1
The correct statement about the following disaccharide is : IIT-JEE 2010 Paper 1 Offline Chemistry - Biomolecules Question 7 English
  1. A
    Ring (a) is pyranonse with α\alphaα-glycosidic link.
  2. B
    Ring (a) is furanonse with α\alphaα-glycosidic link.
  3. C
    Ring (b) is furanonse with α\alphaα-glycosidic link.
  4. D
    Ring (b) is pyranonse with α\alphaα-glycosidic link.
View written solutionFree

Correct answer: A

Step-by-step analysis of the disaccharide structure:

  1. Analyze Ring (a):

    • Ring Type: Ring (a) is a six-membered ring containing five carbon atoms and one oxygen atom. Such a six-membered ring structure for a monosaccharide is called a pyranose ring. (A five-membered ring is a furanose).
    • Glycosidic Linkage Configuration: The bond connecting the two monosaccharide units is called a glycosidic linkage. The nature of this linkage (\\{alpha\} or β\\{\beta\\}β) is determined by the configuration at the anomeric carbon of the first monosaccharide, which is C-1 of ring (a).
    • In the Haworth projection of a D-sugar, if the substituent on the anomeric carbon (C-1) is on the opposite side (trans) of the ring from the −CH2OH-CH_2OH−CH2​OH group (at C-5), the configuration is α\\{\alpha\\}α. If it's on the same side (cis), the configuration is β\\{\beta\\}β.
    • In ring (a), the −CH2OH-CH_2OH−CH2​OH group at C-5 is pointing upwards.
    • The glycosidic bond at the anomeric carbon (C-1) is pointing downwards.
    • Since the glycosidic bond and the −CH2OH-CH_2OH−CH2​OH group are on opposite sides (trans), the linkage is an α\\{\alpha\\}α-glycosidic link.
    • Therefore, ring (a) is a pyranose unit involved in an α\\{\alpha\\}α-glycosidic linkage.
  2. Analyze Ring (b):

    • Ring Type: Ring (b) is also a six-membered ring containing one oxygen atom, so it is also a pyranose ring.
    • Anomeric Configuration: The anomeric carbon of ring (b) (C-1') is not involved in the glycosidic linkage; it has a free hydroxyl (-OH) group. We can determine its configuration. The −CH2OH-CH_2OH−CH2​OH group at C-5' is pointing upwards. The -OH group at the anomeric carbon C-1' is also pointing upwards. Since they are on the same side (cis), the configuration of this anomeric center is β\\{\beta\\}β. Thus, ring (b) is a β\\{\beta\\}β-glucopyranose unit. The entire molecule shown is β\\{\beta\\}β-Maltose.
  3. Evaluate the Options:

    • A: Ring (a) is pyranonse with α\\{\alpha\\}α-glycosidic link.

      • As determined in step 1, ring (a) is a pyranose ring, and the glycosidic link it forms is α\\{\alpha\\}α. This statement is correct.
    • B: Ring (a) is furanonse with α\\{\alpha\\}α-glycosidic link.

      • Ring (a) is a pyranose, not a furanose. This statement is incorrect.
    • C: Ring (b) is furanonse with α\\{\alpha\\}α-glycosidic link.

      • Ring (b) is a pyranose, not a furanose. This statement is incorrect.
    • D: Ring (b) is pyranonse with α\\{\alpha\\}α-glycosidic link.

      • Ring (b) is a pyranose, which is correct. However, the glycosidic link is defined by ring (a). The anomeric center of ring (b) itself has a β\\{\beta\\}β-configuration. Therefore, this statement is incorrect.

Conclusion

The only correct statement is A.

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