Glucose on prolonged heating with HI gives
(a) n-hexane
(b) 1-hexene
(c) Hexanoic acid
(d) 6-iodohex

Q: Why doesn't the reaction form a branched alkane like isohexane?

The formation of a straight-chain alkane ($n$-hexane) proves that the original carbon skeleton of glucose is completely unbranched. If glucose had a branched structure, the reduction would have yielded a branched alkane.

Q: What would happen if we used a mild reducing agent like Sodium Borohydride (NaBH4) instead?

A mild reducing agent like $\text{NaBH}_4$ is only strong enough to reduce the aldehyde group to a primary alcohol. It would leave the other hydroxyl groups untouched, resulting in a sugar alcohol called sorbitol (or glucitol), not an alkane.

Q: Why is 'prolonged heating' specifically mentioned?

Reducing all five hydroxyl groups and the aldehyde group requires a significant amount of energy and time. Prolonged heating ensures the reaction goes to completion, fully reducing the molecule to $n$-hexane.

Q: Would fructose give the same product under these conditions?

Yes! Fructose is a ketohexose, but it also has a straight chain of six carbon atoms. Prolonged heating of fructose with HI also yields $n$-hexane, proving its straight-chain nature as well.

Q: What is the historical significance of this specific reaction?

Before the advent of X-ray crystallography and NMR spectroscopy, chemists used this exact reaction to deduce the fundamental carbon backbone of glucose, establishing that it was a straight six-carbon chain.

Question

Glucose on prolonged heating with HI gives

Options:

(A) $n$-hexane

(B) 1-hexene

(C) Hexanoic acid

(D) 6-iodohexanal

Sigma Percentile · Accepted Answer

### Unveiling the Skeleton of Glucose

Have you ever wondered how early chemists figured out the exact structure of complex molecules like glucose without modern spectroscopic tools? They relied on clever chemical reactions that acted like molecular interrogators. One of the most famous of these is the reaction of glucose with hydrogen iodide (HI).

Let's look at the open-chain structure of glucose. It is an aldohexose, meaning it has a six-carbon backbone. At the very top sits an aldehyde group ($	ext{CHO}$). Below it, there are four secondary alcohol groups ($	ext{CHOH}$), and at the very bottom, a primary alcohol group ($	ext{CH}_2	ext{OH}$).

### The Power of Hydrogen Iodide

Now, what happens when we treat this molecule with hydrogen iodide and heat it for a prolonged period? Hydrogen iodide is not just any reagent; it is a remarkably strong reducing agent.

Because HI is such a powerful reducing agent, it doesn't just stop at modifying one functional group. It aggressively attacks all the oxygen-containing groups in the molecule. It reduces the carbonyl group of the aldehyde, the secondary alcohols, and the primary alcohol all the way down to simple alkane linkages.

### The Final Proof

During this intense reduction process, all six carbon atoms in the straight chain of glucose are completely stripped of their oxygen atoms. Hydrogen atoms take their place, resulting in a fully saturated, straight chain of six carbon atoms.

$$ 	ext{CHO}-(	ext{CHOH})_4-	ext{CH}_2	ext{OH} \xrightarrow{	ext{HI, } \Delta} 	ext{CH}_3-(	ext{CH}_2)_4-	ext{CH}_3 $$

This straight-chain alkane with six carbons is called **$n$-hexane**.

This reaction is incredibly significant in the history of chemistry. It serves as the classic, undeniable proof that all six carbon atoms in glucose are linked together in a continuous, unbranched straight chain! If there had been any branching in the carbon skeleton of glucose, we would have obtained a branched alkane instead. Therefore, the correct answer is $n$-hexane.

Animated Solution for Chemistry - Organic Chemistry: Glucose on prolonged heating with HI gives

Select Answer:

Visualized Solution

$Structure of Glucose$

$The Reagent: HI and Δ$

$Reduction Process$

$Formation of Product$

$Conclusion$

$Food for Thought$

The Sigma Insight: Biomolecules

Unveiling the Skeleton of Glucose

The Power of Hydrogen Iodide

The Final Proof

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