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MCQMediumJEE 2023Carbohydrates (Glucose, Fructose, Sucrose…)

JEE Chemistry 2023 Question with Solution

Match items of Row I with those of Row II: Row I:

Structure labelled P showing a six-membered glucopyranose ring with CH2OH group up and anomeric OH below the plane.Structure labelled Q showing a six-membered glucopyranose ring with CH2OH group up and anomeric OH above the plane.Structure labelled R showing a five-membered fructofuranose ring with CH2OH substituents and anomeric OH below the plane.Structure labelled S showing a five-membered fructofuranose ring with CH2OH substituents and anomeric OH above the plane.

Row II:

  • A

    (1) P →\to iv, Q →\to iii, R →\to i, S →\to ii

  • B

    (2) P →\to i, Q →\to ii, R →\to iii, S →\to iv

  • C

    (3) P →\to iii, Q →\to iv, R →\to ii, S →\to i

  • D

    (4) P →\to iii, Q →\to iv, R →\to i, S →\to ii

Answer

Correct answer:D

Step-by-step solution

Standard Method

Given: Four carbohydrate ring structures P, Q, R, S are to be matched with α\alpha-D-(-)-Fructofuranose, β\beta-D-(-)-Fructofuranose, α\alpha-D-(+)-Glucopyranose, and β\beta-D-(+)-Glucopyranose.

Find: The correct matching option.

From the given identification in the solution:

  • P is α\alpha-D-(+)-Glucopyranose ⇒\Rightarrow (iii)
  • Q is β\beta-D-(+)-Glucopyranose ⇒\Rightarrow (iv)
  • R is α\alpha-D-(-)-Fructofuranose ⇒\Rightarrow (i)
  • S is β\beta-D-(-)-Fructofuranose ⇒\Rightarrow (ii)

This is based on the ring size and anomeric configuration:

  • Glucopyranose has a six-membered ring.
  • Fructofuranose has a five-membered ring.
  • α\alpha and β\beta forms differ by the position of the OH group at the anomeric carbon: C1 for glucose and C2 for fructose.

Therefore, the correct matching is P →\to iii, Q →\to iv, R →\to i, S →\to ii.

The correct option is D.

Ring Size and Anomer Analysis

Given: Structures P, Q, R, S.

Find: Which structures correspond to the named anomers.

  1. First identify the ring size.
  • P and Q are six-membered rings, so they are glucopyranose forms.
  • R and S are five-membered rings, so they are fructofuranose forms.
  1. Now identify the anomer.
  • In P, the anomeric OH is in the α\alpha-position, so P = α\alpha-D-(+)-Glucopyranose = (iii).
  • In Q, the anomeric OH is in the β\beta-position, so Q = β\beta-D-(+)-Glucopyranose = (iv).
  • In R, the anomeric OH at C2 is in the α\alpha-position, so R = α\alpha-D-(-)-Fructofuranose = (i).
  • In S, the anomeric OH at C2 is in the β\beta-position, so S = β\beta-D-(-)-Fructofuranose = (ii).

So the complete match is:

P→iii,Q→iv,R→i,S→ii\text{P} \to \text{iii}, \quad \text{Q} \to \text{iv}, \quad \text{R} \to \text{i}, \quad \text{S} \to \text{ii}

Hence, the correct option is D.

Common mistakes

  • Confusing pyranose with furanose. This is wrong because glucopyranose has a six-membered ring while fructofuranose has a five-membered ring. First count the ring members before assigning the sugar form.

  • Using the wrong anomeric carbon. This is incorrect because the anomeric carbon is C1 in glucose but C2 in fructose. Check the correct carbon before deciding between α\alpha and β\beta.

  • Reversing α\alpha and β\beta configurations. This leads to a wrong match because the distinction depends on the position of the OH group relative to the plane at the anomeric carbon. Compare the OH orientation carefully for each structure.

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