Question

Which of the following statement(s) is/are correct about the given compound?

• A. It exhibits ring-chain tautomerism.
• B. It forms osazone with phenylhydrazine.
• C. It gives eight (8) stereoisomers.
• D. It responds to Tollen's reagent.

Hint:

Remember that all aldohexoses like glucose, galactose, and mannose have 4 chiral carbons, yielding 16 stereoisomers in total. Since they are reducing sugars, they all give positive Tollen's, Fehling's, and Benedict's tests, and undergo osazone formation.

Answer 1

The Correct Option is B, C, D

Step 1: Understanding the Question:
The given compound is D-glucose (an aldohexose). We need to analyze its chemical properties and stereochemistry to determine which of the statements are correct.


Step 2: Detailed Explanation:
Let us evaluate each statement carefully:
- Statement (A): Glucose exhibits ring-chain tautomerism because in aqueous solution, it exists in dynamic equilibrium between its open-chain form and cyclic hemiacetal forms ($\alpha$-D-glucopyranose and $\beta$-D-glucopyranose). Thus, statement (A) is correct.
- Statement (B): Glucose has an aldehyde group on C1 and a secondary alcohol group on C2. It reacts with three equivalents of phenylhydrazine to form a crystalline phenylosazone (glucosazone). Thus, statement (B) is correct.
- Statement (C): The open-chain form of glucose has 4 asymmetric (chiral) carbon centers (C2, C3, C4, and C5).
The total number of stereoisomers is given by:
\[ 2^n = 2^4 = 16 \text{ stereoisomers} \]
This consists of 8 D-isomers and 8 L-isomers. Therefore, statement (C) is incorrect.
- Statement (D): Glucose is a reducing sugar because its open-chain form contains a free, oxidizable aldehyde group ($-\text{CHO}$). It readily reduces ammoniacal silver nitrate (Tollen's reagent) to form a shiny silver mirror. Thus, statement (D) is correct.


Step 3: Final Answer:
The correct statements are (A), (B), and (D).