Q.Which of the following statements is not true about glucose?

The key idea is Glucose Cyclization — glucose exists predominantly as a six-membered (pyranose) ring, not a five-membered (furanose) ring, and its cyclic hemiacetal form does not readily give a positive 2,4-DNP test.

• Glucose is an aldohexose (an aldehyde with 6 carbons) — statement (i) is true.
• Heating glucose with HI reduces it completely to n-hexane — statement (ii) is true.
• Glucose in solution forms a six-membered pyranose ring (glucopyranose), not a furanose form — statement (iii) is false. …

Glucose is the most abundant monosaccharide in nature, and its structure is a classic topic in organic chemistry. The key to this question lies in understanding how glucose cyclizes and what ring sizes are possible.

Glucose has an aldehyde group at C1 and hydroxyl groups on the other carbons. In solution, the aldehyde reacts with a hydroxyl group to form a cyclic hemiacetal. This cyclization can produce either a five-membered ring (furanose) or a six-membered ring (pyranose). The ring that actually forms depends on which hydroxyl group attacks the aldehyde.

Let's examine each statement carefully.

1. Statement (i): It is an aldohexose.

   Glucose has six carbon atoms (hexose) and contains an aldehyde group (aldo-). This is a fundamental fact. The open-chain structure of glucose confirms this: CHO at C1, with OH groups on C2 through C5, and CH2OH at C6. So (i) is true.

2. Statement (ii): On heating with HI it forms n-hexane.

   Hydroiodic acid (HI) is a strong reducing agent. When glucose is heated with HI, all the hydroxyl groups are reduced to hydrogen atoms, and the aldehyde group is also reduced to a methyl group. The result is a straight-chain alkane with six carbons — n-hexane. This reaction is a classic way to confirm the carbon skeleton of sugars. So (ii) is true.

3. Statement (iii): It is present in furanose form.

   Here's where the nuance lies. Glucose can theoretically form a furanose ring (by attack of the C4 hydroxyl on the aldehyde), but in reality, the predominant form in solution is the pyranose ring (by attack of the C5 hydroxyl). The pyranose ring is more stable because it has less angle strain and fewer steric clashes. While a small amount of furanose form exists in equilibrium (especially in certain derivatives like fructose), the statement says "it is present in furanose form" as if that were the main or exclusive form. That is incorrect. Glucose exists overwhelmingly as a pyranose. So (iii) is false. …

Method: Structural Analysis & Reaction-Based Elimination

This method uses the known structure and chemical behaviour of glucose to test each statement against established facts.

Step 1: Recall the basic structure of glucose

• Glucose is an aldohexose — it has 6 carbons and an aldehyde group ($-CHO$).
• Its open-chain form has the aldehyde at C1.

Step 2: Test each statement one by one

(A) "It is an aldohexose."

• True. Glucose is the classic example of an aldohexose.

(B) "On heating with HI it forms n-hexane."

• True. HI is a strong reducing agent. It reduces all oxygen-containing functional groups (including $-OH$ and $-CHO$) to $-H$, yielding straight-chain n-hexane ($CH_3(CH_2)_4CH_3$).

(C) "It is present in furanose form."

• False. Glucose predominantly exists as a pyranose (6-membered ring) form. The furanose (5-membered ring) form is characteristic of fructose, not glucose. …

Here are the common mistakes students make with this glucose cyclization question, and how to avoid each.

Mistake 1: Forgetting that glucose is primarily a pyranose, not a furanose

• The error: Students see "glucose" and "cyclic form" and assume both furanose and pyranose are equally common. They pick option (C) as true.
• Why it’s wrong: In solution, glucose exists overwhelmingly (>99%) in the pyranose (6-membered ring) form. The furanose (5-membered ring) form is a minor component (only about 0.5%).
• How to avoid: Memorise the key fact: Glucose → pyranose (6-membered ring). Fructose is the one that commonly forms a furanose ring. When you see "furanose form" for glucose, flag it as suspicious.

Mistake 2: Confusing the HI reduction product with other reduction reactions

• The error: Students think HI reduction gives sorbitol or gluconic acid, so they mark (B) as false.
• Why it’s wrong: HI is a strong reducing agent that removes all –OH groups, converting glucose into n-hexane (a straight-chain alkane). This is a classic test for the carbon skeleton.
• How to avoid: Remember the specific reagents:
  • HI + heat → removes all –OH → n-hexane
  • NaBH₄ / H₂ → reduces aldehyde to alcohol → sorbitol
  • Br₂ water → oxidises aldehyde to acid → gluconic acid
  • HNO₃ → oxidises both ends → saccharic acid

Mistake 3: Misunderstanding the 2,4-DNP test (Brady’s test)

• The error: Students think glucose has a free aldehyde group in solution, so it should give a positive 2,4-DNP test. They mark (D) as false.
• Why it’s wrong: In solution, glucose exists almost entirely as a cyclic hemiacetal. The aldehyde group is not free — it is tied up in the ring. Therefore, glucose does not give a positive 2,4-DNP test (no free carbonyl).
• How to avoid: Understand that cyclic hemiacetals do not behave like free aldehydes. Glucose only gives a weak, slow reaction with Fehling’s/Tollens’ (due to a tiny amount of open-chain form in equilibrium), but with 2,4-DNP, the equilibrium is too slow to give a visible precipitate under standard test conditions.

Mistake 4: Assuming "aldohexose" means it must have a free aldehyde …