How do you explain the presence of five —OH groups in glucose molecule?

NCERT Exemplar · Q33

Q. How do you explain the presence of five —OH groups in glucose molecule?

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Concept understanding — Glucose Cyclization

Glucose Cyclization: From a Straight Chain to a Ring

Imagine you have a long, flexible chain with a hook at one end and an eyelet at the other. If you swing that chain around, the hook can snap into the eyelet, forming a loop. That's the core idea behind glucose cyclization.

Glucose in its simplest written form is a straight chain of six carbon atoms with an aldehyde group ( $-CHO$ ) at one end. But in water (like in your blood), this chain doesn't stay straight. The aldehyde group reacts with the alcohol group on the fifth carbon, forming a stable six-membered ring.

Why does this happen?

The aldehyde carbon is electrophilic (electron-poor), and the oxygen on carbon-5 has lone pairs (nucleophilic). They attack each other, forming a new bond. This creates a hemiacetal — a carbon bonded to both an $-OH$ and an $-OR$ group. The ring is more stable than the open chain in solution.

Note

The open-chain form of glucose exists in equilibrium with the cyclic form, but at any given time, over 99% of glucose molecules are in the cyclic form.

The precise statement

Glucose cyclization is an intramolecular nucleophilic addition where the aldehyde group at C1 reacts with the hydroxyl group at C5, forming a six-membered pyranose ring (named after pyran, a six-membered oxygen heterocycle). This reaction creates a new chiral center at C1, giving two possible stereoisomers called anomers: $\alpha$ and $\beta$ .

The two anomers

When the ring forms, the oxygen from C5 becomes part of the ring. The carbon that was the aldehyde (now C1) becomes a new chiral center. The $-OH$ group at this new center can point:

Down (relative to the ring plane) → $\alpha$ -D-glucose
Up → $\beta$ -D-glucose

Important

The $\alpha$ and $\beta$ anomers are diastereomers, not enantiomers. They differ only at the anomeric carbon (C1). In solution, they interconvert through the open-chain form — a process called mutarotation.

How to draw it (Haworth projection)

Draw a hexagon with an oxygen atom at the top-right corner.
Number the carbons clockwise from the oxygen: C1 is the carbon to the right of oxygen, C2 next, and so on.
For $\alpha$ -D-glucose, the $-OH$ at C1 points down (opposite to the CH $_2$ OH group at C5).
For $\beta$ -D-glucose, the $-OH$ at C1 points up (same side as the CH $_2$ OH group). …

Why this formula?

Glucose Cyclization: Why the Ring Forms

Glucose cyclization is a classic example of an intramolecular reaction — a molecule reacting with itself. Let's build the understanding step by step.

1. The Starting Point: Open-Chain Glucose

Glucose ( $\text{C}_6\text{H}_{12}\text{O}_6$ ) in its open-chain form has:

An aldehyde group ( $-\text{CHO}$ ) at carbon 1 (C1)
A hydroxyl group ( $-\text{OH}$ ) at carbon 5 (C5)

The aldehyde is electrophilic (electron-deficient at the carbonyl carbon), and the hydroxyl is nucleophilic (electron-rich oxygen with lone pairs).

2. Why Cyclization Happens: Thermodynamic & Kinetic Favorability

The Key Insight

The molecule is flexible — it can bend so that the C5 hydroxyl oxygen approaches the C1 aldehyde carbon. This brings two reactive groups into close proximity.

Entropy is favorable: One molecule becomes one ring — no loss of translational entropy (unlike two separate molecules reacting).
Ring strain is manageable: A 5- or 6-membered ring (furanose or pyranose) has minimal angle strain (close to tetrahedral angles).

Result: The reaction is reversible but strongly favors the cyclic form — in aqueous solution, >99% of glucose exists as the ring.

3. The Reaction: Hemiacetal Formation

The nucleophilic oxygen of the C5 hydroxyl attacks the electrophilic carbonyl carbon of C1:

\text{R-CHO} + \text{R'-OH} \rightleftharpoons \text{R-CH(OH)(OR')}

This is a hemiacetal — a carbon bonded to both an $-\text{OH}$ and an $-\text{OR}$ group.

Mechanism (simplified)

Protonation of the carbonyl oxygen (acid-catalyzed) makes C1 more electrophilic.
Nucleophilic attack by the C5 oxygen.
Deprotonation yields the cyclic hemiacetal.

4. The Key Formula(e): Ring Size & Anomeric Carbon

Ring Size Determination

The ring size depends on which hydroxyl attacks:

C5 hydroxyl → pyranose (6-membered ring: 5 carbons + 1 oxygen)
C4 hydroxyl → furanose (5-membered ring: 4 carbons + 1 oxygen)

For D-glucose, the C5 attack is overwhelmingly favored, giving the pyranose form.

The Anomeric Carbon

The new chiral center formed at C1 is called the anomeric carbon. Two stereoisomers arise:

$\alpha$ -anomer: $-\text{OH}$ at C1 is trans to the $-\text{CH}_2\text{OH}$ group (axial in the chair conformation)
$\beta$ -anomer: $-\text{OH}$ at C1 is cis to the $-\text{CH}_2\text{OH}$ group (equatorial in the chair conformation)

Why two forms? The attack can occur from either face of the planar carbonyl group — leading to two possible configurations at the new stereocenter.

5. The Equilibrium Constant & Mutarotation

The interconversion between $\alpha$ and $\beta$ anomers is called mutarotation:

\alpha\text{-D-glucopyranose} \rightleftharpoons \text{open chain} \rightleftharpoons \beta\text{-D-glucopyranose}

At equilibrium (in water at 20°C):

$\beta$ -D-glucopyranose: ~64%
$\alpha$ -D-glucopyranose: ~36%
Open chain: <0.1% …

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The five —OH groups in glucose arise because its open-chain form is an aldohexose (a six-carbon aldehyde sugar), and when it cyclizes into a pyranose ring, the aldehyde group reacts with the C5 hydroxyl to form a hemiacetal, leaving all other hydroxyls intact — so the cyclic structure still shows five —OH groups, one of which is the anomeric —OH.

Glucose is an aldohexose — that means it has six carbons and an aldehyde group. In its open-chain (Fischer projection) form, each carbon except the first (the aldehyde carbon) carries a hydroxyl group. That gives you five —OH groups straight away: one on each of C2 through C6. The aldehyde group at C1 is a carbonyl, not an —OH.

But here’s the twist: glucose in solution is almost entirely cyclic, not open-chain. The cyclization happens when the aldehyde group at C1 reacts with the hydroxyl on C5, forming a six-membered ring (a pyranose ring). This reaction creates a new —OH group at C1 (the anomeric hydroxyl), while the —OH on C5 is now part of the ring oxygen bridge. So the cyclic structure still has five —OH groups: one at each of C1, C2, C3, C4, and C6. The C5 oxygen is now an ether linkage, not an —OH.

Let’s walk through it step by step.

Open-chain glucose has five —OH groups.

In the straight-chain form, glucose is written as:

$\text{CHO–(CHOH)}_4\text{–CH}_2\text{OH}$

Carbons 2, 3, 4, and 5 each carry one —OH, and carbon 6 carries a primary alcohol group (—CH $_2$ OH). That’s $4 + 1 = 5$ hydroxyls. The aldehyde at C1 has no —OH.
Cyclization creates a hemiacetal at C1.

The aldehyde group (C1) reacts with the —OH on C5. The oxygen of the C5—OH becomes part of the ring, and the hydrogen from that —OH moves to the carbonyl oxygen of the aldehyde. This forms a new —OH group at C1 (the anomeric —OH). The C5 oxygen is now an ether bridge, so C5 no longer has an —OH.
Count the —OH groups in the cyclic form.

After cyclization:
- C1: one —OH (anomeric)
- C2: one —OH
- C3: one —OH
- C4: one —OH
- C5: no —OH (now part of the ring)
- C6: one —OH (primary alcohol) That’s still five —OH groups. The ring oxygen is at C5, but the hydroxyl count remains unchanged.

Tip

A quick way to remember: an aldohexose always has five —OH groups in both open-chain and cyclic forms. The cyclization just relocates one —OH from C5 to C1 — it doesn’t change the total number.

Why doesn’t the ring reduce the count? …

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Common Mistakes: Explaining Five —OH Groups in Glucose

Students often struggle to connect the open-chain structure of glucose with its cyclic form. Here are the most frequent errors and how to avoid them.

✗ Mistake 1: Counting —OH groups from the open-chain formula only

The error:

Students write glucose as $C_6H_{12}O_6$ and say "there are 5 OH groups because the formula has 6 oxygens, one is in the aldehyde group, so 5 are OH." This is incomplete — it doesn't explain why the cyclic form also has 5 OH groups.

Why it's wrong:

In the cyclic (pyranose) form, the aldehyde carbon ( $C_1$ ) becomes a hemiacetal carbon and still carries an —OH group. So the total number of —OH groups remains 5, but the explanation must account for the ring closure.

✓ How to avoid:

Always draw both forms side-by-side:

Open chain:

$CHO$ at $C_1$ , then $CHOH$ at $C_2$ through $C_5$ , and $CH_2OH$ at $C_6$ → 5 OH groups (one on each of $C_2$ , $C_3$ , $C_4$ , $C_5$ , and $C_6$ ).
Cyclic (pyranose):

$C_1$ now has an —OH (hemiacetal), $C_2$ , $C_3$ , $C_4$ each have one —OH, and $C_6$ still has $CH_2OH$ → still 5 OH groups.

Key point: The —OH on $C_5$ is used to form the ring (oxygen bridge), so it disappears as an —OH, but a new —OH appears on $C_1$ . Count remains 5.

✗ Mistake 2: Forgetting the hemiacetal —OH

The error:

Students say "in the ring, $C_5$ loses its OH to form the ring, so there are only 4 OH groups left."

Why it's wrong:

The oxygen of the $C_5$ —OH becomes the ring oxygen, but the hydrogen from that —OH and the carbonyl oxygen of $C_1$ combine to form a new —OH on $C_1$ . So no net loss.

✓ How to avoid:

Memorise the hemiacetal formation mechanism:

\text{Aldehyde} + \text{Alcohol} \rightarrow \text{Hemiacetal}

In glucose:

$C_1$ (aldehyde) + $C_5$ —OH (alcohol) → ring closes
$C_1$ gets a new —OH (hemiacetal OH)
$C_5$ —OH is now part of the ring (O atom)

Visual trick: Write the open chain, then draw an arrow from $C_5$ —OH to $C_1$ =O. The =O becomes —OH, the —OH becomes O in ring.

✗ Mistake 3: Confusing —OH count with —CH $_2$ OH

The error:

Some students count the $CH_2OH$ group at $C_6$ as "one OH" but then also count the carbon it's attached to separately, leading to double-counting or missing it.

Why it's wrong:

$C_6$ has a $CH_2OH$ group — that's one —OH group (on a primary carbon). It is not two OH groups.

✓ How to avoid:

List the five carbons that carry —OH explicitly:

Carbon	Group	Type of —OH
$C_1$	—OH (hemiacetal)	Secondary (in ring)
$C_2$	—OH	Secondary
$C_3$	—OH	Secondary
$C_4$	—OH	Secondary
$C_6$	$CH_2OH$	Primary

Total = 5. Never count $C_5$ — it has no free —OH in the ring.

--- …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

KCET 2024Set B-21 markMCQ
$\alpha$
›Reveal solutionSolution
α- and β-D-glucose differ in configuration at exactly one carbon — C-1, the anomeric carbon created on ring closure — so by definition they are anomers.

1. Where the α/β difference comes from

Open-chain D-glucose is an aldohexose. Its C-5 –OH attacks the C-1 aldehyde carbon intramolecularly, forming a six-membered cyclic hemiacetal (the pyranose ring):
$\mathrm{CHO}\ \text{(C-1, planar, sp}^2) \longrightarrow \overset{\mathrm{OH}}{\underset{\mathrm{H}}{\mathrm{C}}}\text{(C-1, sp}^3\text{, a NEW stereocentre)}$
Because the carbonyl carbon is planar, the ring oxygen can close from either face, so the new –OH at C-1 ends up either:

down (trans to the $\mathrm{CH_2OH}$ ; on the same side as the C-2 OH in Fischer projection) ⇒ α-D-(+)-glucose, or

up ⇒ β-D-(+)-glucose.

The carbon whose configuration is fixed by this ring closure is called the anomeric carbon, and the two products are anomers. In solution they interconvert through the open-chain form — the phenomenon of mutarotation ( $[\alpha]_D$ drifting from $+111^\circ$ for pure α and $+19^\circ$ for pure β to the equilibrium $+52.7^\circ$ ).

2. Eliminate the other terms precisely

(A) Enantiomers — non-superimposable mirror images: they must differ at every stereocentre. α- and β-D-glucose share the same configuration at C-2, C-3, C-4 and C-5, so they are not mirror images (they are diastereomers). ✗ …
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COMEDK 2024Set 2024-A1 markMCQ
$\mathrm{HI}$
›Reveal solutionSolution
The question asks which statement about glucose is false. By checking each option against known glucose chemistry, we find that option (B) is incorrect because the reaction with HI and red P at 373 K yields n-hexane, not a mixture of cyclohexane and 1-iodohexane.

Concept & Intuition

Glucose is an aldohexose — a six-carbon sugar with an aldehyde group (in open-chain form) and multiple hydroxyl groups. Its reactions depend on these functional groups: the aldehyde gives typical carbonyl reactions (oxime formation, no Schiff’s test because it’s usually cyclic), the hydroxyls can be reduced, and the anomeric carbon allows mutarotation. To spot the false statement, we need to recall the specific outcome of each reaction, especially the drastic reduction with HI and red phosphorus, which removes all oxygen atoms.

Step-by-step reasoning

Option (A): Glucose does not give Schiff’s reagent test

Schiff’s reagent tests for aldehydes, but glucose exists predominantly in cyclic hemiacetal form, where the aldehyde group is masked. The equilibrium concentration of free aldehyde is too low to give a positive test under normal conditions. Thus, this statement is true.

Option (B): Glucose on reaction with HI and red P at 373 K gives a mixture of cyclohexane and 1-iodohexane

HI and red phosphorus are a powerful reducing agent that removes all hydroxyl groups (as water) and reduces any carbonyl to methylene. For glucose (C₆H₁₂O₆), the complete reduction yields straight-chain n-hexane (C₆H₁₄). No cyclohexane forms because the carbon skeleton remains unbranched and no ring closure occurs under these conditions. Also, 1-iodohexane would be an intermediate, but excess HI and red P drive reduction all the way to the alkane. So the product is n-hexane, not a mixture of cyclohexane and 1-iodohexane. This statement is false.

Option (C): Both α-D-glucose and β-D-glucose undergo mutarotation in aqueous solutions …
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COMEDK 2024Set 2024-E1 markMCQ
$\mathrm{Br}_2$
›Reveal solutionSolution

Statements A (Br $_2$ water → saccharic acid) and B (acetic anhydride → tetraacetate) are the incorrect ones; C and D correctly describe glucose chemistry.

A. Br $_2$ (aq) → Saccharic acid — INCORRECT. Bromine water is a mild oxidising agent; it oxidises only the –CHO group of glucose to –COOH, giving gluconic acid. Saccharic (glucaric) acid, a dicarboxylic acid, requires the stronger oxidant conc. HNO $_3$ .

B. Acetic anhydride → Glucose tetraacetate — INCORRECT. Glucose has five –OH groups, all acetylated, giving glucose pentaacetate, not tetraacetate. …
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COMEDK 2024Set 2024-M1 markMCQ
$\mathrm{HCN}$
›Reveal solutionSolution
Glucose’s open-chain aldehyde structure explains most of its reactions, but two tests—reaction with Schiff’s reagent and with sodium bisulphite—are not given by the open-chain form because in solution glucose exists predominantly as a cyclic hemiacetal, which lacks a free aldehyde group.

The key concept here is mutarotation and the equilibrium between open-chain and cyclic forms of glucose. In water, glucose is almost entirely (over 99%) in its cyclic pyranose form. Only a tiny fraction exists as the free aldehyde. Many aldehyde-specific tests require a high enough concentration of the free aldehyde to give a visible positive result. Some tests are so sensitive that even the trace amount of open-chain form suffices; others are not.

Let’s examine each pair of tests.

Reaction with Schiff’s reagent and with Sodium bisulphite

Schiff’s reagent (fuchsine decolorized by SO₂) gives a magenta color with aldehydes. However, glucose gives a very slow or no color change because the equilibrium concentration of the free aldehyde is too low to react quickly.

Sodium bisulphite (NaHSO₃) adds to aldehydes to form a crystalline bisulphite addition product. Glucose does not form such a solid adduct under normal conditions, again because the open-chain form is too scarce.

Conclusion: These two tests are not answered by the open-chain structure in practice.

Reaction with HCN and with HI

HCN adds to the carbonyl group of the open-chain form to give a cyanohydrin. This reaction does occur (though slowly) because the equilibrium shifts as the open-chain form is consumed.

HI reduces the aldehyde group (and also the alcohol groups) under drastic conditions, but the aldehyde is still the reactive site.

Conclusion: Both are possible via the open-chain form.

Reaction with HNO₃ and with Acetic anhydride

HNO₃ oxidizes the aldehyde group to a carboxylic acid (giving gluconic acid) and also the terminal CH₂OH to COOH (giving glucaric acid). This clearly involves the open-chain aldehyde.

Acetic anhydride reacts with all –OH groups (including the hemiacetal OH) to form an acetate. This does not require the open-chain form; it reacts with the cyclic form as well. So this test is not exclusive to the open-chain structure, but the question asks which tests are not answered by the open-chain structure — meaning which tests fail if only the open-chain form is considered. Acetic anhydride works fine, so it’s not a “not answered” case.

Conclusion: HNO₃ works via the open-chain; acetic anhydride works anyway.

Reaction with aqueous Bromine and with Hydroxylamine …
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COMEDK 2023Set 2023-E1 markMCQ
$Q. Structures of 3 Monosaccharides are given below. Two of them are Anomers. Identify the two Anomers. (A) II is the anomer of both I and II (B) I and II are Anomers. (C) II and III are Anomers. (D) I and III are Anomers$
›Reveal solutionSolution
[I] and [III] differ only at the anomeric carbon C1 (identical at C2–C4), so they are anomers; [II] also differs at C2 and is not an anomer of either.

Definition: anomers are a special pair of epimers (cyclic monosaccharides) that differ in configuration only at the anomeric carbon (C1, the former carbonyl carbon).

Compare the structures at C1–C4 (left/right substituents):

[I]: C1 = H/OH, C2 = H/OH, C3 = HO/H, C4 = H/OH

[III]: C1 = HO/H, C2 = H/OH, C3 = HO/H, C4 = H/OH

[II]: C1 = HO/H, C2 = HO/H, C3 = HO/H, C4 = H/OH …
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COMEDK 2023Set 2023-M1 markMCQ
$\alpha$
›Reveal solutionSolution
Statements 3 (anomers, not enantiomers) and 5 (amylopectin is branched) are the incorrect ones.

Evaluating each statement:

Glucose exists as two crystalline forms $\alpha$ - and $\beta$ -D-glucose — correct.

$\alpha$ - and $\beta$ -D-glucose are anomers (differ only in configuration at the anomeric C-1) — correct.

$\alpha$ - and $\beta$ -D-glucose are enantiomers — incorrect; they are anomers/diastereomers, not mirror images.

Cellulose is a straight (unbranched) chain of only $\beta$ -D-glucose units — correct. …
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KCET 2020Set A-11 markMCQ
$C_1 - C_4$
›Reveal solutionSolution
A $C_1 - C_4$ glycosidic bond links carbon-1 of one sugar to carbon-4 of another. Sucrose has a $C_1 - C_2$ bond instead, so it is the one that does not contain a $C_1 - C_4$ linkage. The correct option is (C).

The key to this question is understanding what a glycosidic bond is and how the numbering of carbon atoms works in sugars. A glycosidic bond is a covalent bond that joins a carbohydrate molecule to another group, which could be another carbohydrate. The notation $C_1 - C_4$ tells us exactly which carbon atoms are involved: the anomeric carbon (C1) of one monosaccharide is linked to the C4 carbon of another.

Let's check each option.

Starch is a polymer of glucose. The main linkage in starch (both amylose and amylopectin) is an $\alpha$ - $1,4$ -glycosidic bond — that is, a $C_1 - C_4$ bond. So starch definitely contains this bond.

Maltose is a disaccharide made of two glucose units. They are joined by an $\alpha$ - $1,4$ -glycosidic bond. Again, this is a $C_1 - C_4$ linkage.

Sucrose is a disaccharide of glucose and fructose. The glycosidic bond here is between the C1 of glucose and the C2 of fructose. This is a $C_1 - C_2$ bond, not $C_1 - C_4$ . This is the odd one out.

Lactose is a disaccharide of galactose and glucose. The bond is a $\beta$ - $1,4$ -glycosidic bond — the C1 of galactose links to the C4 of glucose. So it does have a $C_1 - C_4$ linkage. …
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KCET 2020Set A-11 markMCQ
$Q. Which of the following monomers can undergo condensation polymerization ? (A) Propene (B) Styrene (C) Glycine (D) Isoprene$
›Reveal solutionSolution
Condensation polymerisation requires a bifunctional monomer; only glycine carries two functional groups ( $-NH_2$ and $-COOH$ ) and can polymerise with loss of water.

Step 1 — The two modes of polymerisation (the concept).

Addition polymerisation: monomers must contain a multiple bond (usually $C{=}C$ ). The $\pi$ bond opens and the units simply add on; nothing is eliminated, so the polymer's empirical formula equals the monomer's. (Polythene, PVC, polystyrene, rubber.)

Condensation polymerisation: monomers must be bi- or poly-functional — carrying two reactive groups such as $-OH$ , $-COOH$ , $-NH_2$ . Repeated condensation between these groups builds the chain and eliminates a small molecule (usually $H_2O$ , sometimes $HCl$ or $CH_3OH$ ). (Nylon-6,6, Terylene, Bakelite.)

So the question is really: which monomer has two functional groups rather than a double bond?

Step 2 — Examine each option.

(A) Propene, $CH_3-CH{=}CH_2$ — a single $C{=}C$ , no functional groups → addition polymerisation → polypropylene.

(B) Styrene, $C_6H_5-CH{=}CH_2$ — vinyl double bond → addition → polystyrene.

(C) Glycine, $H_2N-CH_2-COOH$ — an $\alpha$ -amino acid carrying two different functional groups: an amino group $-NH_2$ and a carboxyl group $-COOH$ . ✓ …
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KCET 2018Set A-11 markMCQ
$Q. The two forms of D-Glucopyranose are called (A) Diastereomers (B) Anomers (C) Epimers (D) Enantiomers$
›Reveal solutionSolution
The two forms of D-glucopyranose differ only at the anomeric carbon (C1) when the ring closes, making them anomers — the correct answer is (B).

The concept: why the ring creates a special pair

D-Glucose exists predominantly as a six-membered pyranose ring. When the open-chain aldehyde form cyclises, the carbonyl carbon (C1) becomes a new chiral centre — the anomeric carbon. The two possible configurations at this carbon (OH pointing down or up in the standard Haworth projection) give rise to $\alpha$ and $\beta$ forms. These are not mirror images, nor do they differ at any other single carbon; they are a specific subclass of diastereomers called anomers.

Watch out

A common mistake is to confuse anomers with epimers. Epimers differ at any one chiral carbon except the anomeric carbon. Anomers differ only at the anomeric carbon — that is the defining feature.

Step-by-step reasoning

Identify the two forms.

D-Glucopyranose exists as $\alpha$ -D-glucopyranose and $\beta$ -D-glucopyranose. In the $\alpha$ form, the OH at C1 is trans to the CH $_2$ OH group (axial in the chair conformation); in the $\beta$ form, it is cis (equatorial).

Check the relationship.

Are they mirror images? No — they are not enantiomers (option D).

Do they differ at a single chiral carbon other than C1? No — they differ only at C1, so they are not epimers (option C). …
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KCET 2018Set A-11 markMCQ
$C_1 - C_4$
›Reveal solutionSolution
Amylose is a linear polymer of glucose linked by $C_1 - C_4$ $\alpha$ -glycosidic bonds. The correct option is (D).

The question is about the specific type of glycosidic bond that joins glucose units in amylose, which is the linear component of starch. Starch is a mixture of two polysaccharides: amylose (unbranched) and amylopectin (branched). The key is to recall the stereochemistry and position of the linkage.

Glucose can exist in two anomeric forms: $\alpha$ and $\beta$ , depending on the orientation of the -OH group at carbon 1. In starch, the glucose units are all in the $\alpha$ -D-glucopyranose form. The linkage between them always involves carbon 1 of one glucose and a specific carbon of the next glucose.

Identify the structure of amylose. Amylose is a long, unbranched chain of glucose molecules. Each glucose is connected to the next by a bond that forms between the anomeric carbon (C1) of one unit and the hydroxyl group on carbon 4 of the next unit. This is a $C_1 - C_4$ linkage.

Determine the stereochemistry of the bond. Because the glucose units in starch are $\alpha$ -D-glucose, the glycosidic bond is formed with the $\alpha$ configuration at C1. This means the oxygen bridge is oriented downward (in the standard Haworth projection) from C1. This is called an $\alpha$ -linkage. …
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Q.How do you explain the presence of five —OH groups in glucose molecule?

Concept understanding — Glucose Cyclization

Glucose Cyclization: From a Straight Chain to a Ring

Why does this happen?

The precise statement

The two anomers

How to draw it (Haworth projection)

Why this formula?

Glucose Cyclization: Why the Ring Forms

1. The Starting Point: Open-Chain Glucose

2. Why Cyclization Happens: Thermodynamic & Kinetic Favorability

The Key Insight

3. The Reaction: Hemiacetal Formation

Mechanism (simplified)

4. The Key Formula(e): Ring Size & Anomeric Carbon

Ring Size Determination

The Anomeric Carbon

5. The Equilibrium Constant & Mutarotation

Method: Open-Chain to Cyclic Conversion (Hemiacetal Formation)

Step 1 – Recall the open-chain structure

Step 2 – Understand why cyclization happens

Step 3 – Count the —OH groups in the cyclic form

Step 4 – Key exam point …

Common Mistakes: Explaining Five —OH Groups in Glucose

✗ Mistake 1: Counting —OH groups from the open-chain formula only

✗ Mistake 2: Forgetting the hemiacetal —OH

✗ Mistake 3: Confusing —OH count with —CH $_2$ OH

1. Where the α/β difference comes from

2. Eliminate the other terms precisely

The concept: why the ring creates a special pair

Step-by-step reasoning

Q.How do you explain the presence of five —OH groups in glucose molecule?

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