Match List-I with List-II.
List-I
A. Sucrose
B. Lactose
C. Maltose
List-II
(i) beta-D-galactose and

Q: Why is sucrose considered a non-reducing sugar?

A sugar is reducing if it has a free anomeric carbon that can open to form an aldehyde or ketone. In sucrose, the glycosidic linkage is formed between the anomeric $C_1$ of glucose and the anomeric $C_2$ of fructose. Since both anomeric carbons are tied up in the bond, sucrose cannot act as a reducing agent.

Q: What is the difference between $\alpha$-D-glucose and $\beta$-D-glucose?

The difference lies in the spatial orientation of the hydroxyl (-OH) group at the anomeric carbon ($C_1$). In the $\alpha$-anomer, the -OH group is on the opposite side (trans) of the $CH_2OH$ group, while in the $\beta$-anomer, it is on the same side (cis).

Q: How can I easily remember the monomers of lactose?

A helpful mnemonic is that 'Lactose' sounds very similar to 'Galactose'. Lactose is simply made of $\beta$-D-galactose and $\beta$-D-glucose.

Q: Are all these disaccharides naturally occurring?

Yes! Sucrose is abundantly found in sugar cane and sugar beets. Lactose is the primary sugar in milk. Maltose is found in germinating seeds and is a product of starch digestion.

Q: What exactly is a glycosidic linkage?

A glycosidic linkage is a type of covalent ether bond that joins a carbohydrate (sugar) molecule to another group, which may or may not be another carbohydrate. It is formed via a dehydration (condensation) reaction where a water molecule is lost.

Question

Match List-I with List-II.

$\begin{array}{ll} 	extbf{List-I} & 	extbf{List-II} \ 	ext{A. Sucrose} & 	ext{(i) } \beta	ext{-D-galactose and } \beta	ext{-D-glucose} \ 	ext{B. Lactose} & 	ext{(ii) } \alpha	ext{-D-glucose and } \beta	ext{-D-fructose} \ 	ext{C. Maltose} & 	ext{(iii) } \alpha	ext{-D-glucose and } \alpha	ext{-D-glucose} \end{array}$

Choose the correct answer from the options given below.

Options:

(A) A $ightarrow$ (i), B $ightarrow$ (iii), C $ightarrow$ (ii)

(B) A $ightarrow$ (iii), B $ightarrow$ (i), C $ightarrow$ (iii)

(C) A $ightarrow$ (ii), B $ightarrow$ (i), C $ightarrow$ (iii)

(D) A $ightarrow$ (iii), B $ightarrow$ (ii), C $ightarrow$ (i)

Sigma Percentile · Accepted Answer

### The Architecture of Disaccharides

When we dive into the world of carbohydrates, disaccharides stand out as some of the most common and biologically significant molecules. A disaccharide is formed when two monosaccharide units undergo a condensation reaction. During this process, a molecule of water is eliminated, and the two sugar units are covalently bonded via an oxygen atom. This specific connection is known as a **glycosidic linkage**.

In this problem, we are tasked with matching three classic disaccharides—Sucrose, Lactose, and Maltose—with their respective monosaccharide building blocks. Let's break down the molecular architecture of each one.

### Analyzing Sucrose (Cane Sugar)

Sucrose is the common table sugar we use every day. It is a unique disaccharide because it is composed of an **$\alpha$-D-glucose** unit and a **$\beta$-D-fructose** unit.

The magic of sucrose lies in its linkage. The glycosidic bond forms between the $C_1$ carbon (the anomeric carbon) of the $\alpha$-D-glucose and the $C_2$ carbon (the anomeric carbon) of the $\beta$-D-fructose. Because both anomeric carbons are locked in this bond, neither ring can open to form a free aldehyde or ketone group. This structural feature makes sucrose a **non-reducing sugar**.

Therefore, Sucrose (A) perfectly matches with $\alpha$-D-glucose and $\beta$-D-fructose (ii).

### Analyzing Lactose (Milk Sugar)

Lactose is the primary carbohydrate found in mammalian milk. Its structure consists of a **$\beta$-D-galactose** unit linked to a **$\beta$-D-glucose** unit.

In lactose, the glycosidic linkage connects the $C_1$ carbon of galactose to the $C_4$ carbon of glucose. Because the anomeric carbon ($C_1$) of the glucose unit remains free and unbonded, the glucose ring can open in solution to expose a reactive aldehyde group. This makes lactose a **reducing sugar**, capable of reducing Tollens' or Fehling's reagents.

Therefore, Lactose (B) matches with $\beta$-D-galactose and $\beta$-D-glucose (i).

### Analyzing Maltose (Malt Sugar)

Maltose is produced during the breakdown of starch. It is structurally the simplest of the three because it is a homodisaccharide, meaning it is made of two identical monomer units: two **$\alpha$-D-glucose** molecules.

The linkage in maltose is an $\alpha(1 \rightarrow 4)$ glycosidic bond, connecting the $C_1$ of the first glucose unit to the $C_4$ of the second glucose unit. Just like in lactose, the second glucose unit retains a free anomeric carbon at $C_1$, making maltose a **reducing sugar**.

Therefore, Maltose (C) matches with $\alpha$-D-glucose and $\alpha$-D-glucose (iii).

### The Final Verdict

By systematically analyzing the monomers, we have established the following matches:
- **A (Sucrose)** $\rightarrow$ **(ii)**
- **B (Lactose)** $\rightarrow$ **(i)**
- **C (Maltose)** $\rightarrow$ **(iii)**

Scanning through the given choices, this sequence corresponds exactly to **Option (c)**. Memorizing these structures is crucial for JEE, but understanding *why* they link the way they do makes the chemistry truly intuitive!

Animated Solution for Chemistry - Organic Chemistry: 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.

Select Answer:

Visualized Solution

\text{Disaccharides and their Monomers}

\text{Composition of Sucrose}

\text{Composition of Lactose}

\text{Composition of Maltose}

\text{Final Matching}

The Sigma Insight: Biomolecules

The Architecture of Disaccharides

Analyzing Sucrose (Cane Sugar)

Analyzing Lactose (Milk Sugar)

Analyzing Maltose (Malt Sugar)

The Final Verdict

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