A disaccharide X cannot be oxidised by bromine water. The acid hydrolysis of X leads to a laevorotat

Q: Why does bromine water only oxidize reducing sugars?

Bromine water is a mild oxidizing agent. It can only oxidize the free aldehyde group (or the hemiacetal form in equilibrium with the open-chain aldehyde) to an aldonic acid. Non-reducing sugars have their anomeric carbons locked in a glycosidic bond, preventing the ring from opening to expose the oxidizable aldehyde group.

Q: How do I quickly identify the anomeric carbon in a cyclic sugar structure?

Look for the carbon atom that is bonded to two oxygen atoms: the ring oxygen and a hydroxyl group (or a glycosidic linkage oxygen). In aldoses like glucose, it's C1. In ketoses like fructose, it's C2.

Q: Why is sucrose considered a non-reducing sugar even though it's made of glucose and fructose?

Because the glycosidic linkage in sucrose is formed between the anomeric carbon of glucose (C1) and the anomeric carbon of fructose (C2). Since both 'reactive' centers are tied up in the bond, neither ring can open to act as a reducing agent.

Q: What exactly does 'laevorotatory solution' mean in the context of hydrolysis?

It means the final mixture rotates plane-polarized light to the left (counter-clockwise). While sucrose itself is dextrorotatory (+66.5°), its hydrolysis yields glucose (+52.5°) and fructose (-92.4°). The strong laevorotation of fructose outweighs the dextrorotation of glucose, making the net mixture laevorotatory.

Q: Are all 1,2-linked disaccharides non-reducing?

If the linkage involves the anomeric carbons of both monosaccharides (like the 1,2 linkage between an aldose and a ketose in sucrose), then yes, it is non-reducing. However, if a linkage involves a non-anomeric carbon, the sugar remains reducing.

Question

A disaccharide X cannot be oxidised by bromine water. The acid hydrolysis of X leads to a laevorotatory solution. The disaccharide X is

Sigma Percentile · Accepted Answer

### Decoding the First Clue: Bromine Water Test

The problem begins with a crucial piece of chemical evidence: "A disaccharide X cannot be oxidised by bromine water." Bromine water ($Br_2/H_2O$) is a mild oxidizing agent. In carbohydrate chemistry, it is famously used to distinguish between reducing and non-reducing sugars.

When a sugar possesses a free hemiacetal or hemiketal group, it exists in equilibrium with its open-chain aldehyde or ketone form. Bromine water selectively oxidizes the free aldehyde group to a carboxylic acid (forming an aldonic acid). If a sugar fails this test, it means the ring cannot open. The anomeric carbon is locked, making it a **non-reducing sugar**.

### The Anatomy of a Non-Reducing Sugar

For a disaccharide to be non-reducing, the glycosidic linkage must connect the anomeric carbons of *both* constituent monosaccharides. If even one anomeric carbon is left free (as a hemiacetal), the sugar retains its reducing power.

Let's evaluate the given options by hunting for free anomeric carbons. In standard Haworth projections, the anomeric carbon is the one bonded to two oxygen atoms (the ring oxygen and a hydroxyl group).

### Evaluating the Suspects: Options B, C, and D

If we closely inspect the right-hand rings in options (B), (C), and (D), we notice a common structural feature. The anomeric carbon ($C_1$) of the rightmost pyranose ring is not involved in the glycosidic bond. It bears a free $-OH$ group.

Because this hemiacetal center is free to undergo mutarotation and open into an aldehyde, options (B), (C), and (D) are all reducing sugars. They would readily react with bromine water, eliminating them from our list of suspects.

### The Prime Suspect: Option A (Sucrose)

Now, let's turn our attention to option (A). The left ring is an $\alpha$-D-glucopyranose unit, and its anomeric carbon ($C_1$) is pointing down into the linkage. The right ring is a $\beta$-D-fructofuranose unit, and its anomeric carbon ($C_2$) is also directly tied into that exact same linkage.

This is an $\alpha$-D-glucopyranosyl-$(1\rightarrow 2)$-$\beta$-D-fructofuranoside bond. Because both reactive anomeric centers are locked together in an acetal/ketal linkage, the molecule has zero reducing power. This perfectly matches our first clue. This molecule is **Sucrose**.

### The Final Nail: Acid Hydrolysis and Optical Rotation

The problem provides a second, confirming clue: "The acid hydrolysis of X leads to a laevorotatory solution."

Sucrose itself is dextrorotatory, with a specific rotation of $[\alpha]_D = +66.5^\circ$. However, when we boil it with dilute acid ($H_3O^+$), the glycosidic bond breaks, yielding an equimolar mixture of D-glucose and D-fructose.

Let's look at the specific rotations of the products:
- D-Glucose: $+52.5^\circ$ (dextrorotatory)
- D-Fructose: $-92.4^\circ$ (strongly laevorotatory)

The massive negative rotation of fructose completely overwhelms the positive rotation of glucose. As a result, the net rotation of the hydrolyzed mixture flips from positive to negative. Because the sign of rotation "inverts" during the reaction, this mixture is famously known as **invert sugar**. This perfectly aligns with the problem's statement that the resulting solution is laevorotatory.

### Conclusion

Option (A) is the only structure that is both a non-reducing sugar and yields a laevorotatory mixture upon hydrolysis. Mastering the exact structural linkages of common disaccharides like sucrose, maltose, and lactose is an absolute must for conquering JEE biomolecules.

Animated Solution for Chemistry - Organic Chemistry: A disaccharide X cannot be oxidised by bromine water. The acid hydrolysis of X leads to a laevorotatory solution. The disaccharide X is

Select Answer:

Visualized Solution

$Oxidation by Bromine Water$

$Structural Condition$

$Analyzing Options B, C, D$

$Analyzing Option A$

$Hydrolysis of Sucrose$

$Conclusion$

The Sigma Insight: Biomolecules

Decoding the First Clue

The Anatomy of a Non-Reducing Sugar

Evaluating the Suspects

The Prime Suspect

The Final Nail

Conclusion

Similar Questions

Which of the following statement(s) is(are) true ?

A non-reducing sugar "A" hydrolyses to give two reducing monosaccharides. Sugar A is

For 'invert sugar', the correct statement(s) is (are) (Given : specific rotations of (+)-sucrose, (+)-maltose, L-(–)-glucose and L-(+)-fructose in aqueous solution are $+ 6 6^{\circ}$ , $+ 14 0^{\circ}$ , $- 5 2^{\circ}$ and $+ 9 2^{\circ}$ , respectively)

Compound A gives D-galactose and D-glucose on hydrolysis. The compound A is

Treatment of D-glucose with aqueous NaOH results in a mixture of monosaccharides, which are

The correct statement(s) regarding sugars is (are) Given : Specific rotations of L-(-)-glucose and L-(+)-fructose are -52.5° and +92.5°, respectively.

Seliwanoff test and xanthoproteic test are used for the identification of ......... and ......... respectively.

Match List-I with List-II. $List-I A. Sucrose B. Lactose C. Maltose List-II (i) β -D-galactose and β -D-glucose (ii) α -D-glucose and β -D-fructose (iii) α -D-glucose and α -D-glucose$ Choose the correct answer from the options given below.

Fructose and glucose can be distinguished by