Q.On the basis of which evidences D-glucose was assigned the following structure (the open chain structure of D-glucose)?

The key idea is Ring-Chain Tautomerism — the open-chain form of glucose is not the major species in solution, but its existence is proven indirectly.

Reasoning:

1. Reaction with HCN (cyanohydrin formation): Glucose adds HCN to form a cyanohydrin. This reaction is characteristic of an aldehyde ($-CHO$) group, confirming the presence of a free carbonyl carbon.

2. Reaction with hydroxylamine (oxime formation): Glucose reacts with hydroxylamine to give an oxime. This again requires a free $C=O$ group, supporting the open-chain aldehyde structure.

3. Reaction with Fehling’s solution / Tollen’s reagent: Glucose reduces these reagents, showing it behaves as a reducing sugar with a free aldehyde group (or its equivalent) under the test conditions. …

1. The core question: what does “structure” mean here?

When Emil Fischer worked out the structure of D‑glucose in the 1890s, he wasn’t just finding the molecular formula ($C_6H_{12}O_6$). He needed to determine:

• the functional group (aldehyde or ketone?),
• the carbon skeleton (straight chain or branched?),
• the stereochemistry at each chiral centre (which of the 16 possible aldohexoses?).

The open‑chain structure you see in textbooks — with the aldehyde at C1 and the four chiral centres (2R,3S,4R,5R) — is the result of a brilliant logical chain, not a single experiment. Let’s walk through the evidence piece by piece.

2. Step‑by‑step reasoning

1. It is an aldohexose, not a ketohexose.

• Glucose reduces Fehling’s solution and Tollens’ reagent — that tells us it has a free aldehyde or α‑hydroxyketone group.
• It forms a cyanohydrin with HCN, and the product can be hydrolysed and reduced to a heptanoic acid. This is the Kiliani–Fischer chain‑lengthening reaction, which only works with an aldehyde (ketones give tertiary cyanohydrins that are harder to handle).
• Crucially, glucose gives the same osazone as D‑mannose and D‑fructose. Osazone formation involves C1 and C2 only; if glucose were a ketose, its osazone would differ. The fact that three different sugars give the identical osazone proves that they differ only at C1 and C2 — and that glucose must be an aldose (C1 = CHO) while fructose is a ketose (C2 = C=O).

2. The carbon chain is unbranched.

• Glucose is reduced to a single hexitol (sorbitol) by NaBH₄ or by catalytic hydrogenation. If the chain were branched, reduction would give a mixture of diastereomeric polyols. A single product means the carbon skeleton is straight.
• Oxidation with nitric acid gives a dicarboxylic acid (saccharic acid) with the same number of carbons — again consistent with a straight chain.

3. The configuration at C5 is fixed by the “last chiral centre” rule.

• Fischer assigned the configuration of the highest‑numbered chiral centre (C5 in an aldohexose) by relating it to (+)-glyceraldehyde, the standard for D‑configuration.
• D‑glucose is dextrorotatory, but that alone doesn’t tell you the configuration. The key was the Kiliani–Fischer synthesis: starting from D‑arabinose (whose C4 configuration was known from D‑glyceraldehyde), adding HCN and reducing gives two epimeric heptonic acids. One of these, on degradation, yields D‑glucose. This chain of reactions ties the C5 configuration of glucose directly to the C2 configuration of (+)-glyceraldehyde.
• The result: C5 has the OH on the right in the Fischer projection — that’s what “D‑” means.

4. The configurations at C2, C3, and C4 are deduced from oxidation and degradation.

• Oxidation of D‑glucose with bromine water gives D‑gluconic acid (a monocarboxylic acid). This is optically active, confirming that C2 is chiral.
• Further oxidation with nitric acid gives D‑saccharic acid (a dicarboxylic acid). This compound is optically active. If C2 and C5 were related by a plane of symmetry (i.e., if the molecule were meso), the diacid would be optically inactive. Since it is active, C2 and C5 cannot be mirror images — the configurations at C2, C3, and C4 must be such that the molecule is not meso.
• Degradation of D‑glucose (via the Ruff degradation or Wohl degradation) shortens the chain by one carbon, giving D‑arabinose. The configuration of D‑arabinose was already known from independent work. This directly gives the configuration at C3 and C4.

5. The osazone evidence locks C3 and C4.

• D‑glucose, D‑mannose, and D‑fructose all give the same osazone. Since osazone formation destroys the chirality at C1 and C2, the three sugars must have identical configurations at C3, C4, and C5.
• D‑mannose is the C2 epimer of D‑glucose. That means glucose and mannose differ only at C2. Combined with the degradation to D‑arabinose, this fixes C2 as having the OH on the right (in the Fischer projection) for glucose.

6. The final structure is uniquely determined.

Putting it all together:

• C1 = CHO (aldehyde)
• C2 = OH on the right (R configuration) …

Concept: Structure Elucidation of D-Glucose (Open Chain Form)

The open chain structure of D-glucose was assigned based on a series of chemical evidences — each experiment narrowed down the possible arrangements of atoms. The key method used is chemical degradation and functional group analysis.

Method: Chemical Evidence-Based Deduction

Step 1 — Molecular formula determination

• Elemental analysis and molecular mass determination gave the formula C₆H₁₂O₆.
• This tells us glucose is a hexose (6-carbon sugar).

Step 2 — Presence of an aldehyde group

• Glucose reduces Tollens’ reagent (silver mirror) and Fehling’s solution (red precipitate of Cu₂O).
• It also forms a cyanohydrin with HCN.
• Conclusion: Glucose contains an aldehydic group (–CHO).

Step 3 — Presence of five hydroxyl groups

• Glucose reacts with acetic anhydride to form a pentaacetate (5 –OH groups acetylated).
• Conclusion: Glucose has five hydroxyl groups (–OH).

Step 4 — Straight chain nature

• On reduction with HI and red phosphorus, glucose gives n-hexane.
• Conclusion: The carbon chain is straight (no branching).

Step 5 — Configuration of hydroxyl groups (stereochemistry)

• Using Ruff degradation (shortening the chain by one carbon) and Kiliani-Fischer synthesis (lengthening the chain), the relative positions of –OH groups were determined.
• The specific optical rotation and comparison with known sugars (like D-glyceraldehyde) established the D-configuration at the chiral carbon farthest from the aldehyde group (C5).

Step 6 — Final open chain structure

• Combining all evidence, the structure is: …

Here is a breakdown of the common mistakes students make when justifying the open-chain structure of D-glucose, along with how to avoid each.

The Core Concept: Why This Structure?

The open-chain structure of D-glucose (an aldohexose) was not guessed; it was deduced from a series of chemical reactions. The key is to connect each piece of evidence to a specific structural feature.

Common Mistake #1: Confusing "Evidence" with "Properties"

The Mistake: Students list properties of glucose (e.g., "it is sweet," "it is soluble in water") as evidence for its structure. This is incorrect. Evidence must be chemical reactions that reveal functional groups and carbon chain length.

How to Avoid:

• Rule of thumb: If the statement doesn't involve a chemical reaction (e.g., reduction, oxidation, reaction with HCN), it is not structural evidence.
• Focus on: Reactions that prove the presence of an aldehyde group ($-CHO$), a primary alcohol ($-CH_2OH$), and a straight chain of six carbon atoms.

Common Mistake #2: Forgetting the "Straight Chain" Proof

The Mistake: Students prove the aldehyde and alcohol groups but forget to prove that the carbon skeleton is unbranched (a straight chain). They assume it's a hexose and stop there.

How to Avoid:

• Remember the HI reduction experiment: When glucose is heated with hydrochloric acid (HI) in the presence of red phosphorus, it is reduced to n-hexane.
• The logic: If glucose had a branched chain, reduction would yield a branched alkane (e.g., isohexane). The fact that only n-hexane is obtained proves the carbon chain is straight (unbranched).

Key phrase to write: "Reduction with HI gives n-hexane, confirming a straight chain of six carbon atoms."

Common Mistake #3: Mixing Up the Evidence for the Aldehyde Group

The Mistake: Students say "glucose reduces Tollen's reagent" but don't explain why this proves an aldehyde. They also forget that a ketone can also reduce Tollen's reagent under certain conditions (though slowly).

How to Avoid:

• Be specific: Glucose reduces Tollen's reagent (ammoniacal silver nitrate) to give a silver mirror, and it reduces Fehling's solution to give a red precipitate of $Cu_2O$.
• The key distinction: These reactions are characteristic of aldehydes (and $\alpha$-hydroxy ketones). However, glucose also reacts with HCN to form a cyanohydrin, and with hydroxylamine ($NH_2OH$) to form an oxime. These are definitive proofs of a carbonyl group ($C=O$) .
• The clincher: Glucose does not give a positive Schiff's test (a test for aldehydes) under normal conditions because it exists predominantly as a cyclic hemiacetal. So, the best evidence is the HCN addition or oxime formation.

Correct approach: "Glucose adds one mole of HCN, forming a cyanohydrin. This proves the presence of a carbonyl group (aldehyde)."

Common Mistake #4: Ignoring the "Five Hydroxyl Groups" Proof

The Mistake: Students say "glucose has five -OH groups" but don't cite the experiment that proves this number.

How to Avoid:

• Use the acetylation reaction: Glucose reacts with excess acetic anhydride ($(CH_3CO)_2O$) in the presence of a catalyst (like pyridine) to form a pentaacetate derivative.
• The logic: The formation of a pentaacetate means that five hydroxyl groups have been acetylated. This proves the presence of five -OH groups in the molecule.

Key phrase: "Acetylation with excess acetic anhydride yields a pentaacetate, confirming five hydroxyl groups."

Common Mistake #5: Confusing the Evidence for Primary vs. Secondary Alcohol

The Mistake: Students say "glucose is oxidized to gluconic acid" but don't explain why this proves a primary alcohol ($-CH_2OH$) at one end.

How to Avoid: …