Question

Which of the following reactions is not explained by the open chain structure of glucose?

• A. Glucose on prolonged heating with HI forms n-hexane.
• B. Glucose reacts with hydroxylamine to form an oxime.
• C. Glucose gets oxidized to gluconic acid on reaction with bromine water.
• D. \( \text{Glucose exists in two different crystalline forms, alpha (}\alpha\text{) and beta (}\beta\text{).} \)

Hint:

The cyclic structure of glucose is a six-membered ring called a pyranose ring. The two forms (\(\alpha\) and \(\beta\)) differ only in the configuration of the hydroxyl group at the hemiacetal carbon (C-1).

Answer 1

The Correct Option is D

Concept: The structure of glucose was initially proposed as an open-chain polyhydroxy aldehyde. While this model successfully accounts for many properties, it fails to explain several experimental observations:

• Aldehyde tests: Despite having an aldehyde group, glucose does not give Schiff's test and does not form the hydrogen sulphite addition product with \(\text{NaHSO}_3\).

• Pentaacetate reactivity: The pentaacetate of glucose does not react with hydroxylamine, indicating the absence of a free \(-\text{CHO}\) group.

• Anomerism: Glucose exists in two distinct crystalline forms (\(\alpha\) and \(\beta\)) which exhibit mutarotation.
Step 1: Analyzing the validity of the open-chain model for each option.
(A) Prolonged heating with HI reduces all carbons to a straight chain of six carbons (n-hexane). This confirms the 6-carbon straight chain skeleton, which is consistent with an open chain. (B) Reaction with \(\text{NH}_2\text{OH}\) forms an oxime. This is a standard reaction for carbonyl groups (C=O) and is explained by the open-chain aldehyde structure. (C) Bromine water is a mild oxidizing agent that converts the aldehyde group (\(-\text{CHO}\)) to a carboxylic acid group (\(-\text{COOH}\)), forming gluconic acid. This also confirms the presence of an aldehyde group. (D) The existence of \(\alpha\)- and \(\beta\)-anomers implies that the \(-\text{OH}\) group at C-5 adds to the \(-\text{CHO}\) group to form a cyclic hemiacetal. This creates a new chiral center at C-1. The open-chain structure cannot explain this because it only shows one form of the aldehyde.