Q.What are monosaccharides?

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.
2. Statement II: Lactose = $\beta$-D-galactose + $\beta$-D-glucose, linked by a $\beta(1\to4)$ glycosidic bond. This is also the standard textbook description. Correct.
3. Since both hold, the answer is that both statements are correct.

Common Mistakes

• Assuming both units in sucrose must be the same anomeric form ($\alpha$ or $\beta$) — they are actually different ($\alpha$-glucose, $\beta$-fructose).
• Forgetting that sucrose's non-reducing nature specifically requires both anomeric carbons to be tied up in the glycosidic bond.

ANSWER: A

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.
4. A free anomeric -OH can open to the aldehyde tautomer, which is oxidised by Fehling's/Tollens' reagents — the definition of a reducing sugar. True.

Common Mistakes

• Confusing lactose's components with maltose (glucose + glucose) — a very common mix-up.
• Assuming any disaccharide with a glycosidic bond is automatically non-reducing; what matters is whether both anomeric carbons are tied up (as in sucrose) or only one (as in lactose/maltose).

ANSWER: D

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).
3. Statement-II names the glucose unit as $\alpha$-D-glucose, which does not match the standard description of lactose (glucose unit is $\beta$, not $\alpha$) — incorrect.
4. Therefore Statement-I is correct, Statement-II is not correct — option (C).

Common Mistakes

• Mixing up the monosaccharide composition of sucrose (glucose + fructose) with lactose (galactose + glucose).
• Getting the $\alpha$/$\beta$ anomeric designation wrong for the glucose unit in lactose.

ANSWER: C

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)

   “Hydrolysis of sucrose results in change in optical rotation from dextro (+) to laevo (–)” — This is true, as shown by the calculation.

5. Evaluate Reason (R)

   “Both the products from the hydrolysis are laevorotatory” — This is false, because glucose is dextrorotatory. Only fructose is laevorotatory.

6. Determine the relationship

   Since A is correct but R is incorrect, the correct option is (C).

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).
2. Identify starch's bond types: amylose gives $\alpha$-1,4 bonds; amylopectin adds $\alpha$-1,6 bonds at branch points.
3. Combine in the order the question asks (cellulose, then starch): $\beta$-1,4 for cellulose; $\alpha$-1,4 and $\alpha$-1,6 for starch.
4. Match to the option that lists exactly this set — option (D).

Common Mistakes

• Assuming starch has only $\alpha$-1,4 bonds and forgetting amylopectin's branching ($\alpha$-1,6).
• Attributing a $\beta$-1,6 bond to cellulose — cellulose has no branching at all.

ANSWER: D

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).
5. Since $|{-92}| > |{+52.5}|$, the net optical rotation of the product mixture becomes negative overall, even though sucrose itself was positive — hence the classical name "inversion of sugar" and the product mixture "invert sugar."

Common Mistakes

• Assuming both products must have the same sign of rotation as the parent sucrose.
• Reversing which sugar is dextro- and which is laevo-rotatory (glucose is dextro, fructose is strongly laevo — remembered via the "inversion" name itself).

ANSWER: A

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.
4. Cellulose is a polysaccharide (not a disaccharide) of glucose units, hydrolysed by cellulase, not maltase.
5. So the disaccharide whose maltase-catalysed hydrolysis gives glucose only is maltose.

Common Mistakes

• Choosing cellulose because it is "all glucose" too, while missing that it is a polysaccharide, not a disaccharide, and is not acted on by maltase.

ANSWER: D

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: $C_{12}H_{22}O_{11} + H_2O \xrightarrow{H^+} C_6H_{12}O_6\,(\text{glucose}) + C_6H_{12}O_6\,(\text{fructose})$.
3. Since one mole of sucrose yields exactly one mole of glucose and one mole of fructose, the ratio formed is 1:1.

Common Mistakes

• Confusing sucrose (glucose + fructose) with other disaccharides like maltose (glucose + glucose) or lactose (glucose + galactose), which could mislead about the identity/ratio of hydrolysis products.
• Assuming some unequal stoichiometric ratio without recalling that sucrose contains exactly one unit each of glucose and fructose.

ANSWER: A

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.
4. Fructose, lactose, and maltose are not the hydrolysis products of starch (fructose/lactose come from different sugars; maltose is only the intermediate disaccharide, not the final complete-hydrolysis product).

Common Mistakes

• Stopping at the intermediate disaccharide maltose (partial hydrolysis product) instead of recognizing that full/complete hydrolysis (as implied by the harsh conditions given: acid, heat, pressure) goes all the way to glucose.

ANSWER: B