For 'invert sugar', the correct statement(s) is (are)
(Given : specific rotations of (+)-sucrose, (+

Q: Why is the mixture called "invert" sugar?

Because the optical rotation of the solution "inverts" from dextrorotatory ($+66^\circ$ for sucrose) to levorotatory ($-20^\circ$ for the mixture) after hydrolysis.

Q: Why did we change the signs of the specific rotations given in the question?

The question deliberately provided the specific rotations for the L-isomers. However, the hydrolysis of natural sucrose yields D-isomers. Since D and L isomers are enantiomers, their specific rotations have equal magnitudes but opposite signs.

Q: Why doesn't fructose react with bromine water?

Bromine water is a mild oxidizing agent. It is strong enough to oxidize aldehydes (like the one in glucose) but not strong enough to oxidize ketones (like the one in fructose).

Q: How can we actually produce saccharic acid from glucose?

To form saccharic acid, you need to oxidize both the aldehyde group and the primary alcohol group of glucose. This requires a strong oxidizing agent like concentrated nitric acid ($\text{HNO}_3$).

Q: Does the hydrolysis of maltose also produce invert sugar?

No. Maltose is composed of two D-glucose units. Its hydrolysis yields only D-glucose, so the resulting solution remains dextrorotatory and is not called invert sugar.

Question

For 'invert sugar', the correct statement(s) is (are)
(Given : specific rotations of (+)-sucrose, (+)-maltose, L-(–)-glucose and L-(+)-fructose in aqueous solution are $+66^\circ$, $+140^\circ$, $-52^\circ$ and $+92^\circ$, respectively)

Options:

(A) 'invert sugar' is prepared by acid catalyzed hydrolysis of maltose

(B) 'invert sugar' is an equimolar mixture of D-(+) glucose and D-(–)-fructose

(C) specific rotation of 'invert surgar' is $-20^\circ$

(D) on reaction with $	ext{Br}_2$ water, 'invert sugar' forms saccharic acid as one of the products

Sigma Percentile · Accepted Answer

The journey into the world of carbohydrates often brings us face-to-face with fascinating phenomena, and the concept of "invert sugar" is undoubtedly one of the most intriguing. This problem from JEE Advanced beautifully weaves together structural chemistry, stereochemistry, and chemical reactivity. Let's embark on a detailed exploration of this sweet mystery.

### The Mystery of Invert Sugar

Imagine you have a beaker filled with a solution of pure sucrose, the common table sugar we use every day. If you pass plane-polarized light through this solution, you will observe that the light rotates to the right. Sucrose is dextrorotatory, with a specific rotation of $+66^\circ$.

However, if we add a few drops of acid and gently heat the solution, a chemical transformation occurs. The sucrose molecule undergoes hydrolysis, breaking its glycosidic bond to yield two simpler monosaccharides: glucose and fructose.

$$ 	ext{Sucrose} + 	ext{H}_2	ext{O} \xrightarrow{	ext{H}^+} 	ext{D-(+)-Glucose} + 	ext{D-(-)-Fructose} $$

This resulting equimolar mixture of D-glucose and D-fructose is what we call **invert sugar**. But why the name "invert"? To understand this, we must look at the optical properties of the products.

### The Enantiomer Trap

The problem presents us with a classic trap. It provides the specific rotations for L-(-)-glucose ($-52^\circ$) and L-(+)-fructose ($+92^\circ$).

**Here is the catch:** Naturally occurring sucrose does not break down into L-isomers; it exclusively yields D-isomers!

To find the specific rotations of the D-isomers, we must rely on a fundamental principle of stereochemistry: enantiomers have specific rotations of equal magnitude but opposite signs.

Therefore, we can easily calculate the specific rotations for our products:
$$ [\alpha]_{	ext{D-glucose}} = -([\alpha]_{	ext{L-glucose}}) = -(-52^\circ) = +52^\circ $$
$$ [\alpha]_{	ext{D-fructose}} = -([\alpha]_{	ext{L-fructose}}) = -(+92^\circ) = -92^\circ $$

### Calculating the Specific Rotation

Now that we have the correct specific rotations for D-glucose and D-fructose, we can determine the optical behavior of the invert sugar mixture.

Since the hydrolysis of one mole of sucrose produces exactly one mole of glucose and one mole of fructose, invert sugar is an equimolar mixture. The specific rotation of such a mixture is simply the arithmetic mean of the specific rotations of its individual components.

$$ [\alpha]_{	ext{mix}} = \frac{[\alpha]_{	ext{D-glucose}} + [\alpha]_{	ext{D-fructose}}}{2} $$

Substituting our values:

$$ [\alpha]_{	ext{mix}} = \frac{+52^\circ + (-92^\circ)}{2} = \frac{-40^\circ}{2} = -20^\circ $$

The final mixture has a specific rotation of $-20^\circ$. Notice what happened! We started with a dextrorotatory solution of sucrose ($+66^\circ$), and after hydrolysis, the solution became levorotatory ($-20^\circ$). The direction of optical rotation has literally *inverted*. This is the beautiful physical reality behind the name "invert sugar".

### The Bromine Water Test

The final piece of the puzzle involves chemical reactivity, specifically oxidation. Option D suggests that reacting invert sugar with bromine water ($	ext{Br}_2/	ext{H}_2	ext{O}$) yields saccharic acid. Let's test this claim.

Bromine water is a mild oxidizing agent. When it reacts with D-glucose, it selectively oxidizes the aldehyde group at the top of the Fischer projection to a carboxylic acid, leaving the rest of the molecule untouched. This reaction produces **gluconic acid**.

$$ 	ext{D-Glucose} \xrightarrow{	ext{Br}_2/	ext{H}_2	ext{O}} 	ext{D-Gluconic Acid} $$

D-fructose, on the other hand, is a ketose. Ketones are generally resistant to mild oxidation, so fructose does not react with bromine water.

To produce saccharic acid (also known as glucaric acid), which is a dicarboxylic acid, we would need a much stronger oxidizing agent like concentrated nitric acid ($	ext{HNO}_3$). Nitric acid is powerful enough to oxidize both the aldehyde group and the primary alcohol group at the bottom of the glucose molecule.

$$ 	ext{D-Glucose} \xrightarrow{	ext{HNO}_3} 	ext{Saccharic Acid} $$

Therefore, bromine water will never produce saccharic acid from invert sugar, making option D incorrect.

### Conclusion

By carefully navigating the stereochemical data and understanding the specific reactivity of carbohydrates, we can confidently conclude that invert sugar is indeed an equimolar mixture of D-(+)-glucose and D-(-)-fructose, and its specific rotation is $-20^\circ$. This makes options B and C the correct statements.

Select Answer:

Visualized Solution

\text{Sucrose} \xrightarrow{\text{H}^+} \text{D-Glucose} + \text{D-Fructose}

[\alpha]_{\text{D}} = -[\alpha]_{\text{L}}

[\alpha]_{\text{D-glucose}} = +52^\circ, [\alpha]_{\text{D-fructose}} = -92^\circ

[\alpha]_{\text{mix}} = -20^\circ

\text{D-Glucose} \xrightarrow{\text{Br}_2/\text{H}_2\text{O}} \text{D-Gluconic Acid}

\text{D-Glucose} \xrightarrow{\text{HNO}_3} \text{Saccharic Acid}

The Sigma Insight: Biomolecules

The Mystery of Invert Sugar

The Enantiomer Trap

Calculating the Specific Rotation

The Bromine Water Test

Conclusion

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