Q.Name the sugar present in milk. How many monosaccharide units are present in it? What are such oligosaccharides called?

Concept and Intuition

Disaccharides are named by which two monosaccharide units (and in which anomeric form) are joined by a glycosidic bond. Sucrose's non-reducing character comes specifically from the fact that BOTH anomeric carbons (C1 of glucose and C2 of fructose) are involved in the glycosidic bond, locking the ring forms as $\alpha$-D-glucose and $\beta$-D-fructose. Lactose, by contrast, is a reducing sugar because glucose's anomeric carbon is left free; the fixed unit is $\beta$-D-galactose joined via $\beta$-1,4 linkage to D-glucose, and standard descriptions state it as $\beta$-D-galactose and $\beta$-D-glucose.

Step-by-Step Solution

1. Statement I: Sucrose = $\alpha$-D-glucopyranose + $\beta$-D-fructofuranose, linked $C1(glucose) \to C2(fructose)$. This is the textbook description. Correct. …

Concept and Intuition

Disaccharides are named by which two monosaccharides are linked and through which carbons. Sucrose (glucose + fructose, both anomeric carbons involved) is the classic non-reducing sugar because neither free anomeric OH survives the glycosidic bond. Lactose and maltose are the classic reducing sugars because one anomeric carbon is left free.

Step-by-Step Solution

1. Lactose = $\beta$-D-galactose $+$ D-glucose, joined $\beta(1\to4)$ between galactose C1 and glucose C4.
2. Statement-I claims lactose is "$\alpha$-D-glucose $+$ $\beta$-D-glucose" — this describes maltose's/only-glucose composition, not lactose's actual galactose+glucose composition. False.
3. Because the glycosidic bond uses galactose's C1 and glucose's C4, glucose's own C1 (anomeric carbon) stays free with a free -OH. …

Concept and Intuition

Disaccharides are formed by a glycosidic linkage between two monosaccharide units, and the specific anomeric form ($\alpha$ or $\beta$) of each unit is a precise structural fact that must be remembered correctly, since sucrose and lactose have different compositions and linkages.

Step-by-Step Solution

1. Sucrose (cane sugar): formed by a glycosidic bond between C1 of $\alpha$-D-glucose and C2 of $\beta$-D-fructose. Statement-I matches this exactly — correct.
2. Lactose (milk sugar): formed by a glycosidic bond between C1 of $\beta$-D-galactose and C4 of $\beta$-D-glucose (glucose unit here is in $\beta$ form as it provides the free anomeric carbon, though the ring can open to $\alpha$/$\beta$ equilibrium — the standard textbook description names $\beta$-D-galactose and glucose, not $\alpha$-D-glucose). …

Concept & Intuition

Optical rotation measures how a substance rotates plane-polarized light. Sucrose is a disaccharide made of glucose and fructose. When hydrolyzed, it breaks into these two monosaccharides. The key is that sucrose itself is dextrorotatory (rotates light to the right, +), but the mixture of glucose and fructose after hydrolysis is laevorotatory (rotates light to the left, –). This phenomenon is called inversion of sucrose, and the product mixture is called invert sugar. The reason given claims both products are laevorotatory — that’s the trap. In reality, glucose is dextrorotatory, fructose is strongly laevorotatory, and the net effect is laevorotatory because fructose’s leftward rotation outweighs glucose’s rightward rotation.

Step-by-step reasoning

1. Identify the specific rotations

   • Sucrose: $[ \alpha ]_D = +66.5^\circ$ (dextrorotatory)
   • Glucose: $[ \alpha ]_D = +52.7^\circ$ (dextrorotatory)
   • Fructose: $[ \alpha ]_D = -92.4^\circ$ (laevorotatory)

2. Hydrolysis reaction

One molecule of sucrose yields one molecule each of glucose and fructose.

3. Net rotation after hydrolysis The observed rotation of the mixture is the weighted average of the rotations of the products. Since both are produced in equal molar amounts:

This is negative (laevorotatory). So the mixture is laevorotatory, even though glucose alone is dextrorotatory.

4. Evaluate Assertion (A) …

Concept and Intuition

Both cellulose and starch are glucose polymers, but the way the glucose units are joined determines their structure and digestibility. Cellulose is built entirely of $\beta$-D-glucose units connected end to end by $\beta$-1,4-glycosidic bonds. This linkage lets the chains lie flat and hydrogen-bond into rigid, fibrous microfibrils — ideal for cell walls, but not digestible by human enzymes (which only cleave $\alpha$ linkages).

Starch, by contrast, is made of $\alpha$-D-glucose units. Its two components are amylose (a straight chain held together by $\alpha$-1,4 bonds) and amylopectin (a branched molecule with an $\alpha$-1,4 backbone plus $\alpha$-1,6 bonds at branch points, roughly every 24–30 residues). Because starch as a storage polysaccharide is really this combination, both $\alpha$-1,4 and $\alpha$-1,6 bonds are correctly attributed to it.

Step-by-Step Solution

1. Identify cellulose's bond type: only $\beta$-1,4-glycosidic bonds (no branching, no $\alpha$ bonds). …

Concept and Intuition

Sucrose is built from $\alpha$-D-glucopyranose and $\beta$-D-fructofuranose joined C1–C2 through their anomeric carbons, which locks both anomeric centres and makes sucrose a non-reducing sugar with no free aldehyde/ketone. Hydrolysing this glycosidic bond liberates both monosaccharides in their free, mutarotating forms, each with its own intrinsic optical rotation.

Step-by-Step Solution

1. Sucrose itself is dextrorotatory, $[\alpha]_D = +66.5^\circ$.
2. Acid hydrolysis (or the enzyme invertase) breaks the glycosidic bond: $\text{Sucrose} + H_2O \rightarrow \text{Glucose} + \text{Fructose}$.
3. Free D-glucose is dextrorotatory, $[\alpha]_D = +52.5^\circ$.
4. Free D-fructose is strongly laevorotatory, $[\alpha]_D = -92^\circ$ (fructose's rotation is large and negative — it is sometimes called laevulose for this reason). …

Concept and Intuition

Disaccharides hydrolyse into their two constituent monosaccharides, and the specific enzyme named must match the specific glycosidic bond being cleaved. Maltase is the enzyme that hydrolyses the $\alpha(1\to4)$ bond in maltose; since maltose's two building blocks are both glucose, hydrolysis gives only glucose as product.

Step-by-Step Solution

1. Maltose = glucose + glucose (joined by an $\alpha(1\to4)$ glycosidic linkage). Enzyme maltase hydrolyses this bond $\Rightarrow$ 2 glucose molecules only.
2. Sucrose = glucose + fructose, hydrolysed by sucrase/invertase $\Rightarrow$ gives glucose and fructose, not glucose alone.
3. Lactose = glucose + galactose, hydrolysed by lactase $\Rightarrow$ gives glucose and galactose, not glucose alone. …

Concept and Intuition

Sucrose is a disaccharide formed by the condensation of one molecule of alpha-D-glucose and one molecule of beta-D-fructose, joined through a glycosidic linkage between C1 of glucose and C2 of fructose. Because this glycosidic bond is the only bond joining the two monosaccharide units, hydrolyzing it (by boiling with dilute acid, a reaction historically called 'inversion' because the optical rotation changes sign) breaks sucrose into exactly one glucose unit and one fructose unit — there is no possibility of an unequal split, since each sucrose molecule contains precisely one of each monosaccharide.

Step-by-Step Solution

1. Recall the structure of sucrose: glucose + fructose joined by one glycosidic bond (1→2 linkage), with the molecular formula $C_{12}H_{22}O_{11}$.
2. Acid hydrolysis reaction: $C_{12}H_{22}O_{11} + H_2O \xrightarrow{H^+} C_6H_{12}O_6\,(\text{glucose}) + C_6H_{12}O_6\,(\text{fructose})$. …

Concept and Intuition

Starch is a polysaccharide made of many glucose units linked by glycosidic bonds ($\alpha$-1,4 and $\alpha$-1,6 linkages in amylose/amylopectin). Acid-catalyzed hydrolysis under heat and elevated pressure cleaves all these glycosidic bonds completely, releasing the individual glucose monomer units — this is the industrial process used to make glucose syrup from starch.

Step-by-Step Solution

1. Starch's repeating unit formula is $(C_6H_{10}O_5)_n$.
2. Complete hydrolysis adds one water molecule per glycosidic bond broken: $(C_6H_{10}O_5)_n + nH_2O \xrightarrow[2\text{-}3\ atm]{H^+, 393K} nC_6H_{12}O_6$.
3. The product, $C_6H_{12}O_6$, is glucose — the single repeating monosaccharide unit of starch. …