Q.(a) Write structure of D-glucose and fructose. [2]

Showing the 12 most recent of 15 on this concept.

1. CBSE 2026Set 56/1/11 markMCQ
   Q.Assertion (A) : The pentaacetate of glucose does not react with $H_2N-OH$. Reason (R) : It indicates the presence of free $-CHO$ group in glucose. (A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A). (B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A). (C) Assertion (A) is true, but Reason (R) is false. (D) Assertion (A) is false, but Reason (R) is true.

   ›Reveal solutionSolution

   Glucose forms a cyclic hemiacetal, so its aldehyde group is locked in a ring and not free. The pentaacetate of glucose has all five –OH groups acetylated, but the ring remains closed — no free –CHO exists to react with hydroxylamine. Hence Assertion is true, Reason is false. The correct option is (C).

   Glucose is famously a reducing sugar — it reduces Tollens’ reagent, Fehling’s solution, and so on. That reducing behaviour comes from its aldehyde group. But here’s the twist: in solution, glucose exists almost entirely as a cyclic hemiacetal (a six-membered pyranose ring). The aldehyde group is not free; it’s tied up in the ring as a hemiacetal linkage. The open-chain aldehyde form is present only in trace amounts (about 0.02% at equilibrium). Yet glucose still behaves as a reducing sugar because the ring can open to regenerate the aldehyde under the reaction conditions.

   Now, the pentaacetate of glucose is made by acetylating all five –OH groups of glucose. That locks the ring structure completely. The ring cannot open because the anomeric –OH (the one at C1) is now acetylated — there’s no free –OH to participate in ring-opening. So the aldehyde group is permanently trapped in the cyclic form. Hydroxylamine ($H_2N-OH$) reacts with free carbonyl groups (aldehydes and ketones) to form oximes. Since no free –CHO exists in the pentaacetate, no reaction occurs. That makes Assertion (A) true.

   Reason (R) claims that this non-reactivity indicates the presence of a free –CHO group in glucose. That’s backwards. The non-reactivity of the pentaacetate actually shows that the –CHO group is not free in the cyclic form — it’s masked. The free –CHO is present only in the open-chain form, which is a tiny fraction. So Reason (R) is false.

   Let’s walk through the logic step by step.

   1. Glucose cyclizes to a hemiacetal.

      The –CHO group at C1 reacts with the –OH at C5 to form a six-membered ring (pyranose). The C1 carbon becomes a chiral centre (the anomeric carbon) and the oxygen of the original –CHO is now part of a C–O–C linkage. No free aldehyde remains in the cyclic form.

   2. Acetylation of glucose gives the pentaacetate.

      All five –OH groups (including the anomeric –OH at C1) are converted to acetate esters. The ring stays intact. The anomeric acetate is not a hemiacetal — it’s a full acetal (specifically a glycosidic bond analogue). Acetals do not equilibrate with the open-chain aldehyde under mild conditions.

   3. Hydroxylamine reacts only with free carbonyls. …

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2. CBSE 2026Set 56/2/11 markMCQ
   Q.Which of the following reactions is not explained by the open chain structure of glucose ? (A) Glucose on prolonged heating with HI forms n-hexane. (B) Glucose reacts with hydroxylamine to form an oxime. (C) Glucose gets oxidized to gluconic acid on reaction with bromine water. (D) Glucose exists in two different crystalline forms, alpha ($\alpha$) and beta ($\beta$).

   ›Reveal solutionSolution

   The open-chain structure of glucose (an aldohexose) explains its aldehyde chemistry—reduction to hexane, oxime formation, and oxidation to an acid—but cannot account for the existence of two distinct crystalline forms ($\alpha$ and $\beta$), which arise only from cyclic hemiacetal formation.

   Why cyclization matters

   Glucose was long thought to be a simple open-chain aldehyde with five hydroxyl groups. That structure does explain many reactions: the aldehyde group can be reduced, can form derivatives like oximes, and can be oxidized. But one experimental fact stubbornly refused to fit—glucose crystallizes in two forms with different melting points and optical rotations, and freshly dissolved samples show mutarotation (a slow change in rotation). An open-chain aldehyde has no mechanism to produce two distinct solid forms; the molecule would always be the same.

   The resolution came when it was recognized that glucose exists predominantly as a cyclic hemiacetal, formed by intramolecular attack of the C-5 hydroxyl on the C-1 aldehyde. This cyclization creates a new chiral center at C-1 (the anomeric carbon), giving rise to two stereoisomers—$\alpha$-D-glucose and $\beta$-D-glucose—that can be isolated as separate crystals.

   Examining each reaction

   1. Prolonged heating with HI → $n$-hexane

      Hydroiodic acid is a powerful reducing agent. The aldehyde group at C-1 is reduced to $\ce{-CH2OH}$, then all five hydroxyl groups (including the newly formed one) are replaced by iodine and subsequently reduced to hydrogen, yielding $\ce{CH3(CH2)4CH3}$. This is classic aldehyde reduction chemistry; the open-chain structure with an aldehyde at one end fully accounts for it.

   2. Reaction with hydroxylamine → oxime

      Aldehydes react with $\ce{NH2OH}$ to form oximes via nucleophilic addition-elimination:

   $\ce{R-CHO + NH2OH -> R-CH=N-OH + H2O}$

   Glucose, with its free (or equilibrium-accessible) aldehyde group, forms glucose oxime. Again, the open-chain aldehyde structure explains this perfectly.

   3. Oxidation with bromine water → gluconic acid Bromine water is a mild oxidizing agent that selectively oxidizes aldehydes to carboxylic acids without attacking alcohols: $\ce{CH2OH-(CHOH)4-CHO ->[Br2/H2O] CH2OH-(CHOH)4-COOH}$ …

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3. CBSE 2026Set 56/2/11 markMCQ
   Q.Assertion (A) : Glucose gets oxidized to six carbon gluconic acid on reaction with bromine water. Reason (R) : The carbonyl group is absent in the open chain structure of glucose.

   ›Reveal solutionSolution

   Glucose has an aldehyde group in its open-chain form, which is selectively oxidized by bromine water to a carboxylic acid, giving gluconic acid. The reason is false because the carbonyl group is present, not absent.

   The key to this question lies in understanding the structure of glucose and the specific action of bromine water as an oxidizing agent. Many students get confused because glucose usually exists as a cyclic hemiacetal, but in solution, a tiny amount of the open-chain aldehyde form is always present — and that’s what reacts.

   Let’s break it down.

   1. Glucose exists in equilibrium between cyclic and open-chain forms.

      In aqueous solution, glucose is predominantly in its cyclic pyranose form (about 99.9%). However, a very small fraction (roughly 0.1%) exists as the open-chain aldehyde. This equilibrium is dynamic — as the open-chain form is consumed in a reaction, more cyclic molecules open up to replenish it.

   2. Bromine water is a mild oxidizing agent.

      Unlike strong oxidizers like nitric acid (which can oxidize both ends of glucose to give saccharic acid), bromine water selectively oxidizes the aldehyde group ($-CHO$) to a carboxylic acid ($-COOH$). It does not attack the primary alcohol group at C-6 under these conditions.

   3. The reaction produces gluconic acid.

      When the open-chain aldehyde form of glucose reacts with bromine water, the aldehyde group at C-1 is oxidized to a carboxyl group. The product is gluconic acid, which still has six carbons — the chain length is preserved.

      $\text{Glucose (open-chain)} \xrightarrow{\text{Br}_2/\text{H}_2\text{O}} \text{Gluconic acid}$

      The reaction can be written as:

   $C_6H_{12}O_6 + Br_2 + H_2O \rightarrow C_6H_{12}O_7 + 2HBr$

   4. Now examine the Assertion and Reason. …

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4. CBSE 2026Set ANNUAL1 markMCQ
   Q.In the next two parts of Question No.-1, there are two statements labelled as Assertion (A) and Reason (R). From the following options (i), (ii),

   (iii) and (iv), select the correct answer. Assertion (A): All monosaccharides are reducing sugars. Reason (R): Monosaccharides either have an aldehyde group or an aldehyde group is formed in solution as a result of tautomerism.

   (a)

   (i) Both A and R are correct and R is the correct explanation of A.

   (b)

   (ii) Both A and R are correct but R is not the correct explanation of A.

   (c)

   (iii) A is correct but R is incorrect.

   (d)

   (iv) Both A and R are incorrect.

   ›Reveal solutionSolution

   A is true (all monosaccharides are reducing sugars) and R is true and is the correct explanation — aldoses carry a free –CHO, while ketoses form an aldehyde in solution via tautomerism, and it is this aldehyde group that is oxidised. Correct option: (i).

   Concept. A reducing sugar is one that can reduce Tollens' reagent (silver mirror) or Fehling's/Benedict's solution (red $\text{Cu}_2\text{O}$). Reduction requires a free (or potentially free) aldehyde/keto group at the anomeric carbon.

   Why the Assertion is true. Every monosaccharide — whether an aldose (e.g. glucose) or a ketose (e.g. fructose) — is a reducing sugar because its anomeric carbon is not locked as a glycoside.

   Why the Reason is the correct explanation.

   • Aldoses already possess a free $-\text{CHO}$ group, which is directly oxidised. …

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5. CBSE 2026Set ANNUAL1 markMCQ
   Q.Which of the following characters of D-(+)-Glucose CANNOT be explained by the open chain structure?

   (a) D- and L- forms

   (b) (+) and (–) forms

   (c) α- and β- forms

   (d) pentaacetate formation

   ›Reveal solutionSolution

   The open-chain structure lacks the cyclic hemiacetal (anomeric) carbon, so it cannot account for the α- and β-anomers of glucose — option (C).

   The open-chain structure of D-(+)-glucose (an aldohexose, $CHO-(CHOH)_4-CH_2OH$) explains most of its reactions: aldehyde reactions, formation of the pentaacetate (five –OH groups), and its optical activity/D–L designation.

   However, some observations cannot be explained by the open chain:

   • Glucose does not give certain characteristic aldehyde tests (e.g. it fails to react with Schiff's reagent / $NaHSO_3$ readily). …

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6. CBSE 2025Set 56/4/11 markMCQ
   Q.In the Haworth structure of the following carbohydrate, various carbon atoms have been numbered. The anomeric carbon is numbered as : (Drawn: the Haworth ring of beta-D-glucopyranose with carbons numbered 1-5 - ring oxygen at the top right; C1 at the right bearing OH above and H below; C5 at the top bearing the CH_2OH group; OH below at C2, OH above at C3, HO at C4.) (A) 1 (B) 2 (C) 3 (D) 5

   ›Reveal solutionSolution

   The anomeric carbon is the new chiral centre created when a linear sugar cyclizes — in $\beta$-D-glucopyranose, this is the carbon that becomes attached to both the ring oxygen and a hemiacetal OH group, which is C1.

   The question shows a Haworth projection of $\beta$-D-glucopyranose with carbons numbered 1 through 5 around the ring. You are asked to identify which of these is the anomeric carbon.

   The term "anomeric carbon" comes directly from the cyclization of glucose. In the open-chain form, glucose has an aldehyde group at C1. When the ring closes, the OH group on C5 attacks that aldehyde carbon, forming a hemiacetal. That carbon — originally the aldehyde carbon — becomes a new stereocentre. It is bonded to the ring oxygen, to a hydrogen, to an OH group, and to the rest of the ring. This carbon is called the anomeric carbon, and the two possible stereochemical arrangements at this centre are called the $\alpha$ and $\beta$ anomers.

   In the drawn structure, the ring oxygen is at the top right. The carbon immediately to the right of that oxygen, bearing an OH group above the ring and an H below, is C1. That is the carbon that was the aldehyde carbon in the open chain. It is the only carbon in the ring that is attached to two oxygens — one from the ring and one from the OH group. No other ring carbon has this feature.

   Let’s walk through the numbering systematically.

   1. Identify the ring oxygen. In the standard Haworth drawing of $\beta$-D-glucopyranose, the oxygen is placed at the top right corner of the hexagon. This oxygen is not numbered — it is the bridging atom from the cyclization.

   2. Locate C1. The carbon immediately clockwise from the ring oxygen (at the rightmost position of the ring) is C1. In the $\beta$ anomer, the OH at C1 points upward (on the same side as the CH$_2$OH group at C5). This carbon is the hemiacetal carbon.

   3. Check the other carbons. Moving clockwise around the ring: the next carbon (at the bottom right) is C2, with an OH below. Then C3 at the bottom left, with OH above. Then C4 at the top left, with OH below. Finally, C5 at the top, bearing the CH$_2$OH group. None of these carbons are attached to two oxygens — they each have only one OH group and are part of the ring. …

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7. CBSE 2025Set 56/6/11 markMCQ
   Q.Pyranose ring of glucose is formed due to the reaction between : (A) $C_1$ and $C_3$ (B) $C_1$ and $C_5$ (C) $C_1$ and $C_4$ (D) $C_1$ and $C_2$

   ›Reveal solutionSolution

   Glucose cyclizes when its aldehyde group ($C_1$) reacts with the hydroxyl on $C_5$, forming a six-membered pyranose ring. The answer is (B).

   Why glucose forms a ring

   Glucose exists predominantly as a cyclic structure in solution, not as the open-chain aldehyde you might first draw. This happens because the hydroxyl groups within the same molecule can attack the carbonyl carbon, forming a stable ring through intramolecular hemiacetal formation.

   The name "pyranose" tells you the ring size: it comes from pyran, a six-membered ring containing five carbons and one oxygen. When glucose forms this ring, it creates a structure analogous to pyran.

   Understanding the cyclization mechanism

   In the open-chain form of D-glucose, you have:

   • An aldehyde group at $C_1$ (the carbonyl carbon)
   • Hydroxyl groups at $C_2$, $C_3$, $C_4$, and $C_5$

   For a stable ring to form, the hydroxyl oxygen needs to be positioned close enough in space to attack the electrophilic carbonyl carbon. The question is: which hydroxyl?

   Step-by-step ring formation

   1. The nucleophilic attack

      The hydroxyl group on $C_5$ acts as a nucleophile and attacks the carbonyl carbon at $C_1$. This is geometrically favorable because when you draw the chain in its extended zigzag form and allow rotation around single bonds, the $C_5$ hydroxyl can easily reach $C_1$.

   2. Hemiacetal formation

      The attack converts the aldehyde into a hemiacetal:

   $\ce{R-CHO + R'-OH -> R-CH(OH)-O-R'}$

   Here, the oxygen from the $C_5$ hydroxyl becomes part of the ring, and the former carbonyl carbon ($C_1$) now bears both an $\ce{-OH}$ group and is bonded to the ring oxygen.

   3. The six-membered ring

      Count the atoms in the ring: $C_1$, $C_2$, $C_3$, $C_4$, $C_5$, and the oxygen (originally from the $C_5$ hydroxyl). That's six atoms total—a pyranose ring.

   4. The anomeric carbon

      $C_1$ becomes the anomeric carbon, a new chiral center. The newly formed hydroxyl can be either axial (α-anomer) or equatorial (β-anomer) in the chair conformation. …

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8. CBSE 2025Set A1 mark
   Q.Fill in the blank: Glucose occurs freely in nature as well as in the ______ form.

   ›Reveal solutionSolution

   Glucose is found both as a free monosaccharide and combined (bonded via glycosidic linkages) within larger carbohydrates.

   Glucose occurs freely in ripe fruits and in honey. It also occurs in the combined form, i.e. joined to other sugar units through glycosidic bonds, as a building block of larger carbohydrates — for example, in sucrose (glucose + fructose), in the disaccharide m …

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9. CBSE 2024Set 56/1/11 markMCQ
   Q.Which functional groups of glucose interact to form cyclic hemiacetal leading to pyranose structure? (A) Aldehyde group and hydroxyl group at C-4 (B) Aldehyde group and hydroxyl group at C-5 (C) Ketone group and hydroxyl group at C-4 (D) Ketone group and hydroxyl group at C-5

   ›Reveal solutionSolution

   Glucose cyclizes when its aldehyde group (C-1) reacts with the hydroxyl group on C-5, forming a six-membered pyranose ring via a hemiacetal linkage. The correct option is (B).

   Glucose is an aldohexose — it has an aldehyde group at C-1 and hydroxyl groups on every other carbon. In solution, it doesn't stay as a straight chain. Instead, the aldehyde reacts with one of its own hydroxyl groups to form a cyclic hemiacetal. The key question is: which hydroxyl group attacks?

   The ring size depends entirely on which carbon's OH does the attacking. If the OH at C-4 attacks, you get a five-membered ring (furanose). If the OH at C-5 attacks, you get a six-membered ring (pyranose). Glucose overwhelmingly prefers the six-membered pyranose form — and that means the attacking group is the hydroxyl on C-5.

   Let's walk through the reasoning step by step.

   1. Identify the reactive groups. Glucose has an aldehyde group at C-1. In the open-chain form, this aldehyde carbon is electrophilic. Any nearby alcohol (OH) can act as a nucleophile and attack it. The product is a hemiacetal — a carbon bonded to both an OH and an OR group.

   2. Which OH is close enough? For a stable ring to form, the attacking OH must be able to reach the aldehyde without excessive strain. In glucose, the OH on C-5 is perfectly positioned to form a six-membered ring (atoms: C-1 through C-5 plus the oxygen bridge). This is the pyranose ring, named after pyran (a six-membered oxygen heterocycle).

   3. What about C-4? The OH on C-4 can also attack, but that gives a five-membered furanose ring. While glucose can form a furanose in small amounts, the pyranose form is far more stable and predominant (over 99% in solution). The question specifically asks about the pyranose structure, so we need the C-5 OH.

   4. Check the options. …

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10. CBSE 2023Set 56/1/11 markMCQ
    Q.The glycosidic linkage involved in linking the glucose units in amylose part of starch is : (A) $\alpha$-$C_1-C_6$ linkage (B) $\beta$-$C_1-C_6$ linkage (C) $\alpha$-$C_1-C_4$ linkage (D) $\beta$-$C_1-C_4$ linkage

    ›Reveal solutionSolution

    Amylose is a linear polymer of glucose units joined by $\alpha$-$C_1$–$C_4$ glycosidic linkages. The correct answer is (C).

    Understanding Glycosidic Linkages in Polysaccharides

    When glucose molecules polymerize to form polysaccharides like starch, they connect through glycosidic bonds — covalent linkages formed between the anomeric carbon (usually $C_1$) of one sugar and a hydroxyl group on another. The nature of this linkage determines the structure and digestibility of the resulting polymer.

    Two features define any glycosidic bond:

    • The stereochemistry at $C_1$: $\alpha$ (hydroxyl below the ring plane) or $\beta$ (hydroxyl above)
    • The position of attachment: which carbon on the second glucose receives the bond

    Starch, the storage polysaccharide in plants, has two components: amylose (linear) and amylopectin (branched). The question asks specifically about amylose.

    Step-by-Step Analysis

    1. Glucose cyclization and the anomeric carbon

       When glucose cyclizes into its pyranose (six-membered ring) form, $C_1$ becomes the anomeric carbon. In $\alpha$-D-glucose, the $-OH$ on $C_1$ points downward (same side as $C_6$); in $\beta$-D-glucose, it points upward.

    2. Amylose structure

       Amylose is an unbranched chain of glucose units. Each glucose connects to the next through its anomeric carbon ($C_1$) linking to the $C_4$ hydroxyl of the adjacent glucose. This creates a linear polymer.

    3. The $\alpha$ configuration

       In amylose, all glycosidic bonds have the $\alpha$ configuration at $C_1$. This means the oxygen bridge connecting two glucose units lies below the plane of the ring at the anomeric carbon. This $\alpha$-linkage allows the chain to adopt a helical structure.

    4. Why not $C_1$–$C_6$?

       A $C_1$–$C_6$ linkage creates a branch point, not a linear extension. This type of bond appears in amylopectin (the branched component of starch) and glycogen, where side chains sprout from the main backbone. Amylose has no branches, so it uses only $C_1$–$C_4$ linkages. …

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11. CBSE 2023Set A1 mark
    Q.Match the following. Column A item: 'Aldohexose'. Choose its correct match from Column B:

    (a) Ether

    (b) Primary amine

    (c) Lactose

    (d) C12H22O11

    (e) Glucose

    (f) Negative ions

    (g) C6H5SO2Cl

    (h) +7

    ›Reveal solutionSolution

    Glucose is the classic example of an aldohexose — a six-carbon sugar bearing an aldehyde group.

    …

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12. CBSE 2023Set ANNUAL1 mark
    Q.Write the names of the anomers of glucose.

    ›Reveal solutionSolution

    When open-chain glucose cyclises to its 6-membered (pyranose) ring form, C-1 becomes a new stereocentre called the anomeric carbon, giving rise to two anomers: $\alpha$-D-glucose and $\beta$-D-glucose.

    Glucose exists predominantly in a cyclic hemiacetal (pyranose) form, formed by intramolecular reaction between the $-CHO$ at C-1 and the $-OH$ at C-5. This converts C-1 into a new chiral centre — the anomeric carbon, bearing an $-OH$ group whose orientation can be either of two ways relative to the reference $-CH_2OH$ group (C-6):

    • $\alpha$-D-glucose: the $-OH$ at C-1 lies on the same side (below the ring plane, cis) as the $-OH$ used to define the D-configuration (i.e. trans to the $CH_2OH$ at C-5 in the Haworth structure), melting point $419\ K$, $[\alpha]_D = +111^\circ$. …

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