Q.What happens when D-glucose is treated with the following reagents?

Concept and Intuition

Carbohydrates are essential biomolecules, and disaccharides are a class of carbohydrates formed by the condensation of two monosaccharide units. This linkage between two monosaccharide units is called a glycosidic linkage. It's essentially an ether linkage ($-\text{O}-$) formed when a hydroxyl group from the anomeric carbon (C-1) of one monosaccharide reacts with a hydroxyl group from another carbon (often C-4 or C-6) of a second monosaccharide, with the elimination of a water molecule.

Maltose, commonly known as malt sugar, is a disaccharide. Understanding its structure requires knowing its constituent monosaccharides and the specific carbons involved in the glycosidic bond. Maltose is formed by two units of $\alpha$-D-glucose. The "$\alpha$" designation indicates the configuration of the hydroxyl group at the anomeric carbon (C-1) relative to the $-\text{CH}_2\text{OH}$ group at C-5 in the Haworth projection. For $\alpha$-D-glucose, the $-\text{OH}$ at C-1 is on the opposite side of the ring from the $-\text{CH}_2\text{OH}$ at C-5 (typically drawn below the plane of the ring).

Step-by-Step Explanation

1. Identify the constituent monosaccharides of maltose:

   Maltose is a disaccharide composed of two units of $\alpha$-D-glucose. This is a fundamental fact about maltose's structure.

2. Understand the nature of the glycosidic linkage:

   A glycosidic linkage is an ether bond formed between the anomeric carbon (C-1) of one monosaccharide and a hydroxyl group on another carbon of a second monosaccharide. During this reaction, a molecule of water is eliminated. The configuration of the anomeric carbon involved in the linkage (whether $\alpha$ or $\beta$) determines the type of glycosidic bond.

3. Determine the specific carbons involved in the linkage in maltose:

   In maltose, the glycosidic linkage is formed between the C-1 of one $\alpha$-D-glucose unit and the C-4 of the other $\alpha$-D-glucose unit. This specific connection is crucial for its structure and properties. …

Concept & Intuition

The problem describes a carbohydrate (A) that, under mild acidic conditions in alcohol, breaks into two isomeric sugars (B and C). Isomers here means they have the same molecular formula but different structures — typical of a disaccharide splitting into its two monosaccharide units. One of these (C) is explicitly a ketohexose (a six-carbon sugar with a ketone group, like fructose). The other (B) reacts with bromine water to give a monocarboxylic acid — bromine water oxidizes only aldoses (sugars with an aldehyde group) to their corresponding aldonic acids. So B must be an aldose. Therefore, A is a disaccharide composed of one aldose and one ketohexose. Among the options, only sucrose fits: it is made of glucose (an aldose) and fructose (a ketohexose). Starch is a polysaccharide, maltose is two glucose units (both aldoses), and lactose is glucose + galactose (both aldoses). Let’s verify step by step.

Step-by-step reasoning

1. Identify the reaction type

   Dilute HCl in alcoholic solution is a classic condition for hydrolyzing glycosidic bonds in disaccharides. The products are the constituent monosaccharides. So A is a disaccharide that yields two monosaccharides (B and C).

2. Characterize product C

   The problem states C is a ketohexose. Ketohexoses have a ketone group (e.g., fructose). This immediately rules out disaccharides that yield only aldoses (like maltose and lactose, which give two aldoses each).

3. Characterize product B

   B reacts with bromine water to give a monocarboxylic acid. Bromine water is a mild oxidizing agent that specifically oxidizes the aldehyde group (–CHO) of an aldose to a carboxylic acid (–COOH), producing an aldonic acid. This reaction does not occur with ketoses under these conditions. Therefore, B must be an aldose.

4. Combine the clues

   A must be a disaccharide that, upon hydrolysis, gives one aldose (B) and one ketohexose (C). Let’s check the options:

   • (A) Starch: A polysaccharide, not a disaccharide; hydrolysis gives many glucose units, not just two isomers. Incorrect. …

Concept & Intuition

Maltose is produced by the partial hydrolysis of starch. Its structure is well‑known: two D‑glucose molecules joined by an α‑1,4‑glycosidic bond. Because the anomeric carbon of the second glucose is free (not involved in the linkage), maltose is a reducing sugar. Any statement that contradicts this structure — especially one that swaps a glucose for a fructose — must be false.

1. Identify the monosaccharides from maltose hydrolysis Maltose (C₁₂H₂₂O₁₁) hydrolyses in the presence of dilute acid or the enzyme maltase to give two molecules of D‑glucose.

Both units are glucose; no fructose is produced.

2. Check the anomeric form of each glucose

   The glycosidic bond in maltose is formed between the α‑anomeric carbon (C1) of the first glucose and the C4 hydroxyl of the second glucose. The first glucose is locked in the α‑configuration at C1. The second glucose retains a free anomeric carbon, which can mutarotate between α and β forms, but in the intact disaccharide the second unit is also α‑D‑glucose (the bond does not alter its ring form). Thus both monosaccharide units are α‑D‑glucose.

3. Evaluate each option

   • (A) “Both are α‑D‑glucose units only” — True, as explained.
   • (B) “One is α‑D‑glucose and second one is β‑D‑fructose” — False. Fructose is a ketohexose; maltose contains only aldohexoses (glucose). This statement is incorrect. …

Concept & Intuition

Dehydration synthesis (also called condensation) is a chemical reaction where two molecules join by removing a water molecule (H₂O). This happens when a hydroxyl group (–OH) from one molecule and a hydrogen atom (–H) from another combine to form water, leaving a covalent bond between the two residues. In biology, this is how many key polymers are built. Hydrogen bonds, however, are weak electrostatic attractions between a partially positive hydrogen and a partially negative atom (like oxygen or nitrogen); they form spontaneously without any water molecule being released. So the question asks: which bond types are formed by a dehydration reaction?

Step-by-step reasoning

1. Peptide bond (I) – A peptide bond links amino acids in proteins. It forms when the carboxyl group (–COOH) of one amino acid reacts with the amino group (–NH₂) of another, releasing a water molecule. This is a classic dehydration reaction.

   → Yes, dehydration occurs.

2. Hydrogen bond (II) – A hydrogen bond is an intermolecular force, not a covalent bond. It arises from the attraction between a hydrogen atom (bonded to an electronegative atom like O or N) and another electronegative atom. No atoms are removed or added; no water is produced.

   → No dehydration occurs.

3. Glycosidic bond (III) – This bond joins monosaccharides to form disaccharides or polysaccharides (e.g., maltose, sucrose). It forms when the hydroxyl group of one sugar reacts with the anomeric carbon of another, eliminating a water molecule.

   → Yes, dehydration occurs. …

Maltose is a common disaccharide, and its structure is a classic example in carbohydrate chemistry. The key to this question is knowing exactly which monosaccharides make up maltose and how they are linked. Many students confuse maltose with lactose (milk sugar), which does contain galactose. That mix-up is the trap here.

Let’s examine each statement:

1. Statement (A): "It is a reducing sugar"

   A reducing sugar has a free anomeric carbon (the carbonyl carbon of the open-chain form) that can reduce Cu²⁺ or Ag⁺ ions. In maltose, the glycosidic bond is formed between the anomeric carbon of one glucose unit and the C-4 hydroxyl of the other. This leaves the anomeric carbon of the second glucose unit free (it can open to an aldehyde). Therefore, maltose is a reducing sugar. This statement is correct.

2. Statement (B): "It is composed of two α-D-glucose units"

   Maltose is formed from two molecules of α-D-glucose joined by an α(1→4) glycosidic linkage. This is a standard fact. This statement is correct.

3. Statement (C): "It is composed of one β-D-glucose and one β-D-galactose unit" …

The Cori cycle is a beautiful example of metabolic cooperation between tissues. When your muscles work hard — say, during a sprint — they rely on glycolysis for quick energy, even when oxygen is limited. That process produces lactate as a byproduct. But lactate isn't just waste; it's a valuable fuel that gets shipped to the liver, which converts it back into glucose. That glucose then returns to the muscle, completing the cycle.

The key insight is that the liver has the enzymes for gluconeogenesis (making new glucose from lactate), while muscle does not. Muscle can only produce lactate; it cannot recycle it. So the cycle must involve both tissues — one that generates lactate and one that consumes it to remake glucose.

1. In muscle (during intense exercise): Glycolysis breaks down glucose to pyruvate, which is then reduced to lactate (by lactate dehydrogenase) to regenerate NAD⁺. This lactate is released into the bloodstream.

2. In the liver: The liver takes up lactate from the blood and converts it back to glucose via gluconeogenesis — a process that requires ATP. The newly made glucose is then released back into circulation. …

The key here is the stoichiometry of the reaction. Sucrose is a disaccharide made of one glucose unit and one fructose unit linked together. When you hydrolyse it in the presence of an acid, that bond breaks, and you get one molecule of glucose and one molecule of fructose.

So the reaction is simply:

Notice that one mole of sucrose gives exactly one mole of glucose. That’s the whole story — no coefficients to balance, no side products. The water is in excess, so it doesn’t limit anything.

Now, to find the mass of sucrose needed for 1 mole of glucose, you just need the molar mass of sucrose.

1. Molar mass of sucrose ($\text{C}_{12}\text{H}_{22}\text{O}_{11}$):
   • Carbon: $12 \times 12 = 144$ g/mol
   • Hydrogen: $22 \times 1 = 22$ g/mol
   • Oxygen: $11 \times 16 = 176$ g/mol
   • Total: $144 + 22 + 176 = 342$ g/mol …

The key is to recall the specific step in glycolysis where each enzyme acts and which molecule is its substrate. Instead of memorizing blindly, think of the metabolic pathway as a story: each enzyme recognizes a particular molecule and transforms it into the next.

1. Phosphoenol pyruvate (PEP) → Enolase (II)

   Enolase catalyzes the dehydration of 2-phosphoglycerate to PEP. But wait — PEP is the product of enolase, not its substrate. However, the question asks for the enzyme that acts on the given molecule. In the reverse direction (gluconeogenesis) or in the forward direction, enolase is the enzyme that interconverts 2-phosphoglycerate and PEP. Since PEP is directly linked to enolase’s action, the match is A–II.

2. Pyruvic acid → Pyruvic kinase (III)

   Pyruvic kinase catalyzes the final step of glycolysis: transfer of a phosphate from PEP to ADP, producing pyruvic acid (pyruvate) and ATP. So pyruvic acid is the product of pyruvic kinase. Again, the enzyme is named for its role in forming pyruvate. Thus B–III.

3. Triose phosphate → Aldolase (IV)

   Aldolase splits fructose-1,6-bisphosphate into two triose phosphates: dihydroxyacetone phosphate (DHAP) and glyceraldehyde-3-phosphate (G3P). So triose phosphates are the products of aldolase. Hence C–IV.

4. Glucose-6-phosphate → Hexokinase (I) …