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?

Telangana TsbieShort· 3mImportance★★★★★

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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.

TG EAPCET 2026Set eng-2026-05-10-AN1 markMCQ
$\beta$
›Reveal solutionSolution
For $\beta$ -D-glucopyranose the Haworth ring must read OH up at C-1, down at C-2, up at C-3, down at C-4, with $\mathrm{CH_2OH}$ up at C-5. Only option (C) shows that pattern.

The concept first

There are just three rules for turning the open-chain Fischer projection into a Haworth pyranose ring.

Rule 1 — the "right–down" rule. Any $-\mathrm{OH}$ written on the right in the Fischer projection is drawn below the ring in Haworth; anything on the left is drawn above.

Rule 2 — the D-configuration. In a D-sugar the terminal $\mathrm{CH_2OH}$ (C-6, attached at C-5) points up.

Rule 3 — $\alpha$ versus $\beta$ . The new stereocentre created on ring closure is C-1, the anomeric carbon.

$\boldsymbol\beta$ : the C-1 $\mathrm{OH}$ is on the same side as the $\mathrm{CH_2OH}$ — i.e. up.

$\boldsymbol\alpha$ : the C-1 $\mathrm{OH}$ is down (opposite to $\mathrm{CH_2OH}$ ).

Step-by-step

Step 1 — recall the Fischer projection of D-(+)-glucose. Reading down from C-1 (the $\mathrm{CHO}$ ):
$\text{C-2: OH right},\quad \text{C-3: OH \textbf{left}},\quad \text{C-4: OH right},\quad \text{C-5: OH right (D)}$
The memory aid many students use is "right–left–right–right", the lone left-hand OH being at C-3.

Step 2 — apply the right–down rule to C-2, C-3, C-4.

C-2: OH on the right $\Rightarrow$ OH below the ring, H above.

C-3: OH on the left $\Rightarrow$ OH above the ring, H below.

C-4: OH on the right $\Rightarrow$ OH below the ring, H above.

Step 3 — place the $\mathrm{CH_2OH}$ . D-sugar $\Rightarrow$ $\mathrm{CH_2OH}$ at C-5 points up. (Both the question's diagrams and every standard Haworth drawing agree on this.)

Step 4 — set the anomeric carbon. We want the $\beta$ anomer, so the C-1 $\mathrm{OH}$ must be cis to the $\mathrm{CH_2OH}$ — i.e. OH above, H below.

Step 5 — assemble the required pattern.

$\text{C-1: OH \textbf{up}},\quad \text{C-2: OH \textbf{down}},\quad \text{C-3: OH \textbf{up}},\quad \text{C-4: OH \textbf{down}},\quad \text{C-5: }\mathrm{CH_2OH}\ \textbf{up}$ …
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TG EAPCET 2026Set eng-2026-05-11-FN1 markMCQ
$Q. Match the following List-1 List-2 A. Antioxidant I. Antiseptic B. Food preservative II. Artificial sweetener C. Sucralose III. Butylated hydroxy toluene D. Bithionol IV. Sodium benzoate The correct answer is (A) A - III, B - I, C - IV, D - II (B) A - IV, B - III, C - II, D - I (C) A - III, B - IV, C - I, D - II (D) A - III, B - IV, C - II, D - I$
›Reveal solutionSolution

Match each compound with its function: antioxidants prevent oxidative rancidity, preservatives inhibit microbial growth in food, artificial sweeteners provide sweetness without calories, and antiseptics kill microbes on living tissue. The correct pairing is A–III, B–IV, C–II, D–I.

This question tests your knowledge of the classification and applications of important organic compounds in everyday life, particularly in the pharmaceutical and food industries. Each substance has a specific role determined by its chemical structure and properties.

Let's identify each compound and its function:

A. Antioxidant → III. Butylated hydroxy toluene (BHT)

Antioxidants prevent oxidative degradation of fats and oils in food by inhibiting free-radical chain reactions. BHT is a phenolic compound widely used in the food industry to prevent rancidity. The hydroxyl group on the aromatic ring donates hydrogen atoms to free radicals, breaking the oxidation chain and preserving freshness.

B. Food preservative → IV. Sodium benzoate

Food preservatives extend shelf life by inhibiting microbial growth (bacteria, fungi, yeasts). Sodium benzoate ( $\ce{C6H5COONa}$ ) is one of the most common preservatives, especially effective in acidic foods like pickles, jams, and carbonated beverages. In acidic conditions it converts to benzoic acid, which penetrates microbial cell walls and disrupts their metabolism.

C. Sucralose → II. Artificial sweetener …
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TG EAPCET 2026Set ap-2026-05-05-FN1 markMCQ
$Q. Essential amino acids which are basic in nature are (A) Arg, Lys (B) His, Ser (C) Arg, Thr (D) Pro, Trp$
›Reveal solutionSolution
Basic amino acids have side chains that can accept protons (positive charge at neutral pH). Among the essential amino acids, arginine (Arg) and lysine (Lys) are basic, making option (A) correct.

The key to this question is understanding two classifications at once: which amino acids are essential (must come from diet) and which are basic (have a net positive charge at physiological pH). You need the overlap.

Amino acids are basic when their side chain contains an extra amino group or a guanidino group — something that can grab a proton. The three standard basic amino acids are arginine (Arg), lysine (Lys), and histidine (His). But not all of these are essential for adults.

Identify the basic amino acids.

At pH ~7.4, the side chains of Arg (guanidino group, pKa ~12.5), Lys (ε-amino group, pKa ~10.5), and His (imidazole group, pKa ~6.0) can carry a positive charge. So the set of basic amino acids is {Arg, Lys, His}.

Identify the essential amino acids.

The nine essential amino acids for humans are: histidine, isoleucine, leucine, lysine, methionine, phenylalanine, threonine, tryptophan, and valine. (Histidine is essential for infants and often considered conditionally essential for adults, but standard lists include it.)

Find the intersection.

From the basic set {Arg, Lys, His}, which are essential?

Lysine is essential.

Histidine is essential.

Arginine is not essential for adults (the body can synthesize it), though it is essential for children. In most exam contexts, arginine is classified as non-essential for adults.

So the essential basic amino acids are histidine and lysine. But look at the options — none say "His, Lys". Option (A) says "Arg, Lys". Why would that be correct?

Watch out

Many standard Indian exam lists (including NCERT) treat arginine as essential for children and sometimes include it in the essential set for general purposes. Also, histidine is sometimes omitted from the essential list in older or simplified versions. This mismatch is a known source of confusion.

Check each option against the standard NCERT/NEET classification. …
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TG EAPCET 2024Set eng-2024-05-09-FN1 markMCQ
$-\mathrm{CHO},\ -\mathrm{CHOH}$
›Reveal solutionSolution
Glucose is an aldohexose; mild oxidation (e.g., with bromine water) converts its aldehyde group ( $-\mathrm{CHO}$ ) to a carboxylic acid, giving gluconic acid. Stronger oxidation (e.g., with nitric acid) also oxidises the primary alcohol at the other end ( $-\mathrm{CH_2OH}$ ) to a carboxylic acid, yielding saccharic acid. Thus the functional groups involved are $-\mathrm{CHO}$ then $-\mathrm{CH_2OH}$ — option (B).

Concept & Intuition

Glucose is a six-carbon sugar with an aldehyde group at C1 and a primary alcohol group at C6 (the terminal $-\mathrm{CH_2OH}$ ). The other four carbons carry secondary alcohol groups ( $-\mathrm{CHOH}$ ).

Mild oxidation (like bromine water) selectively attacks the most easily oxidised group — the aldehyde — turning it into a carboxylic acid. This gives gluconic acid.

Stronger oxidation (like hot nitric acid) is powerful enough to also oxidise the primary alcohol at C6 to a carboxylic acid. The result is a dicarboxylic acid called saccharic acid. The secondary alcohol groups ( $-\mathrm{CHOH}$ ) are not oxidised under these conditions (they would require even harsher reagents). So the two steps target different functional groups: first the aldehyde, then the primary alcohol.

Step-by-step reasoning

Identify the functional groups in glucose

Glucose has the formula $\mathrm{CHO(CHOH)_4CH_2OH}$ . The key groups are:

An aldehyde: $-\mathrm{CHO}$ at C1.

A primary alcohol: $-\mathrm{CH_2OH}$ at C6.

Four secondary alcohols: $-\mathrm{CHOH}$ at C2–C5.

First conversion: glucose → gluconic acid

Reagent: mild oxidising agent (e.g., $\mathrm{Br_2/H_2O}$ ).

Reaction: The aldehyde group is oxidised to a carboxylic acid:

$-\mathrm{CHO} \xrightarrow{[O]} -\mathrm{COOH}$
The product is gluconic acid: $\mathrm{HOOC(CHOH)_4CH_2OH}$ .

Functional group involved: $-\mathrm{CHO}$ .

Second conversion: gluconic acid → saccharic acid Reagent: strong oxidising agent (e.g., hot $\mathrm{HNO_3}$ ). Reaction: The primary alcohol at C6 is now oxidised to a carboxylic acid: …
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TG EAPCET 2023Set eng-2023-05-13-FN1 markMCQ
$\beta$
›Reveal solutionSolution
A six-membered ring rules out fructose; $\ce{CH2OH}$ up marks the D-series; the anomeric $\ce{OH}$ pointing down marks the $\alpha$ anomer. The sugar is $\alpha$ -D-(+)-glucopyranose — option (D).

The concept first: how to read a Haworth projection

A Haworth projection is the cyclic hemiacetal drawn as a flat ring seen edge-on. Three questions unlock it, in this order:

How many atoms in the ring? Six (5 C + O) = pyranose; five (4 C + O) = furanose. Glucose, an aldohexose, cyclises through C-1 and C-5 giving a six-membered pyranose. Fructose, a ketohexose, cyclises through C-2 and C-5 giving a five-membered furanose. So ring size alone separates glucose from fructose.

Which way does $\ce{CH2OH}$ (on C-5) point? Up ⇒ D-sugar; down ⇒ L-sugar.

Which way does the $\ce{OH}$ on the anomeric carbon (C-1 for glucose) point?

Down (i.e. trans / on the opposite side from $\ce{CH2OH}$ ) ⇒ $\alpha$ .

Up (cis to $\ce{CH2OH}$ ) ⇒ $\beta$ .

Step-by-step

Step 1 — Count the ring. The drawing is a hexagon with the oxygen at the top-right vertex: five carbons + one oxygen ⇒ pyranose. Immediately, both fructofuranose options fall away — fructose does not give a six-membered ring in these options.

Step 2 — Fix the D/L label. C-5 (top-left carbon, next to the ring O) carries $\ce{CH2OH}$ upward. ⇒ D-sugar. …
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TG EAPCET 2023Set ap-2023-05-11-AN1 markMCQ
$Q. Among the following, sulphur containing amino acids are (A) A & B only. (B) B & C only. (C) B & D only. (D) A & C only.$
›Reveal solutionSolution
The two standard amino acids that contain sulphur are Cysteine and Methionine. To answer the question, we need to identify which of the labels A, B, C, D correspond to these amino acids. Assuming B and D represent Cysteine and Methionine respectively, the correct option is (C).

Concept and Intuition

Amino acids are the fundamental building blocks of proteins. Each amino acid has a central carbon atom (the alpha-carbon) bonded to four different groups:

An amino group ( $-\text{NH}_2$ )

A carboxyl group ( $-\text{COOH}$ )

A hydrogen atom ( $-\text{H}$ )

A unique side chain, also known as the R-group

The R-group is what distinguishes one amino acid from another and determines its chemical properties. To identify sulphur-containing amino acids, we need to look for the presence of a sulphur atom within this R-group.

There are 20 standard amino acids commonly found in proteins. Among these, only two contain sulphur atoms in their side chains: Cysteine and Methionine.

Step-by-step Analysis

Identify Sulphur-Containing Amino Acids:

Out of the 20 standard amino acids, Cysteine and Methionine are the only two that contain sulphur in their R-groups.

Cysteine:

Cysteine has a thiol group ( $-\text{SH}$ ) in its side chain. This thiol group is highly reactive and can form disulfide bonds ( $-\text{S}-\text{S}-$ ) with another cysteine residue, which is crucial for stabilizing protein structures.

The structure of Cysteine is:

$\begin{array}{c} \text{H}_2\text{N}-\underset{\text{R-group}}{\underset{|}{\text{CH}}}-\text{COOH} \\ \text{CH}_2\text{SH} \end{array}$
Here, the sulphur atom is part of the $-\text{CH}_2\text{SH}$ (thiomethyl) group.

3. Methionine:

Methionine contains a thioether group ( $-\text{S}-\text{CH}_3$ ) in its side chain. Unlike cysteine, the sulphur in methionine is not directly bonded to a hydrogen atom, making it less reactive and unable to form disulfide bonds. Methionine is also unique as it is typically the first amino acid incorporated into a polypeptide chain during protein synthesis.

The structure of Methionine is:
$\begin{array}{c} \text{H}_2\text{N}-\underset{\text{R-group}}{\underset{|}{\text{CH}}}-\text{COOH} \\ \text{CH}_2\text{CH}_2\text{SCH}_3 \end{array}$
Here, the sulphur atom is part of the $-\text{CH}_2\text{CH}_2\text{SCH}_3$ (ethyl methyl thioether) group. > [!IMPORTANT] > Cysteine contains a thiol ($-\text{SH}$) group, while Methionine contains a thioether ($-\text{S}-\text{CH}_3$) group. Both contain sulphur.

4. Addressing the Options (A, B, C, D): …
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TG EAPCET 2023Set ap-2023-05-11-AN1 markMCQ
$Q. Match the following List - I (Example) A. Cimetidine B. Alitame C. Morphine D. Iodoform List-II (Type of chemical substance) I. Artificial sweetener II. Analgesic III. Antacid IV. Antifertility drug V. Antiseptic (A) A - I, B - III, C - IV, D - II (B) A - III, B - I, C - IV, D - II (C) A - V, B - II, C - I, D - III (D) A - III, B - I, C - II, D - V$
›Reveal solutionSolution
This question asks us to match common chemical substances with their respective categories based on their primary use or function. Cimetidine is an antacid, Alitame is an artificial sweetener, Morphine is an analgesic, and Iodoform is an antiseptic, leading to option (D).

The field of chemistry, particularly medicinal chemistry and food chemistry, involves a vast array of substances designed for specific purposes. Understanding the function of these chemicals is crucial. This problem tests your knowledge of common examples of drugs and food additives and their classifications. Each substance has a distinct chemical structure that dictates its biological activity or physical property, allowing it to serve a particular role.

Here's how we can match each substance to its correct category:

Analyze Cimetidine:

Cimetidine is a drug primarily used to treat heartburn and stomach ulcers. It works by blocking histamine H2 receptors in the stomach, which reduces the production of stomach acid. Substances that neutralize or reduce stomach acid are classified as antacids.

Therefore, Cimetidine matches with III. Antacid.

Analyze Alitame:

Alitame is a high-potency artificial sweetener. It is a dipeptide derivative, significantly sweeter than sucrose (table sugar), and is used as a sugar substitute in various food products.

Therefore, Alitame matches with I. Artificial sweetener.

Analyze Morphine:

Morphine is a powerful opioid drug used to relieve severe pain. It acts on the central nervous system to produce its pain-relieving effects. Drugs that alleviate pain are known as analgesics.

Therefore, Morphine matches with II. Analgesic.

Analyze Iodoform: …
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TG EAPCET 2022Set ap-2022-07-30-FN1 markMCQ
$Q. Given sub-structure belongs to which amino acid (A) Proline (B) Arginine (C) Tyrosine (D) Histidine$
›Reveal solutionSolution
The sub-structure shown is the imidazole ring, which is the distinctive side chain of histidine; therefore the correct answer is histidine.

The question asks you to match a given sub-structure to its parent amino acid. The key is to recognize the unique chemical group that defines each amino acid’s side chain. Among the options, proline has a pyrrolidine ring, arginine has a guanidinium group, tyrosine has a phenol ring, and histidine has an imidazole ring. The sub-structure in question is a five-membered ring containing two nitrogen atoms — that is the imidazole ring, exclusive to histidine.

Let’s walk through the reasoning step by step:

Identify the sub-structure’s features.

The given sub-structure is a five-membered ring with two nitrogen atoms (one at position 1 and one at position 3) and three carbon atoms. This is the imidazole ring. It is aromatic and can exist in two tautomeric forms (with the hydrogen on either nitrogen).

Recall the side chains of the listed amino acids.

(A) Proline: Its side chain is a pyrrolidine ring — a five-membered ring with one nitrogen (a secondary amine). Not a match.

(B) Arginine: Its side chain is a straight chain ending in a guanidinium group (three nitrogens, no ring). Not a match.

(C) Tyrosine: Its side chain contains a phenol ring (a six-membered aromatic ring with one oxygen). Not a match.

(D) Histidine: Its side chain is exactly the imidazole ring. This is the only amino acid among the 20 standard ones that contains an imidazole group.

Confirm by elimination. …
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TG EAPCET 2022Set eng-2022-07-20-FN1 markMCQ
$_4$
›Reveal solutionSolution
Cold dilute KMnO $_4$ adds two –OH groups across each double bond (syn dihydroxylation). The starting compound has two double bonds (one in the ring, one in the allyl side chain), so the product gains four new –OH groups, plus the existing one, for a total of 5 hydroxyl groups.

The key is to recognise what cold, dilute, aqueous KMnO $_4$ does. This is Baeyer’s reagent — it performs syn dihydroxylation of alkenes, converting each C=C into a vicinal diol (two adjacent –OH groups). It does not cleave the double bond (that requires hot, concentrated KMnO $_4$ or ozonolysis). So every double bond in the molecule gets two –OH groups added across it.

Let’s look at the structure of 5-allylcyclohex-3-ene-1-ol.

Draw the molecule. The parent is cyclohex-3-ene-1-ol: a six-membered ring with a double bond between C3 and C4, and an –OH group at C1. At C5, there is an allyl substituent: –CH $_2$ –CH=CH $_2$ . So the molecule contains two separate double bonds — one in the ring (C3=C4) and one in the allyl side chain (the terminal CH $_2$ =CH–).

Apply cold KMnO $_4$ to each double bond.

The ring double bond (C3=C4) gets two –OH groups added, one on C3 and one on C4.

The allyl double bond (CH $_2$ =CH–) gets two –OH groups added, one on the terminal carbon and one on the middle carbon of the side chain.

Count the –OH groups in the product.

The original molecule already has one –OH at C1.

The ring dihydroxylation adds two new –OH groups.

The side-chain dihydroxylation adds two more –OH groups. …
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TG EAPCET 2021Set eng-2021-08-04-FN1 markMCQ
$\alpha$
›Reveal solutionSolution
α-D-Fructofuranose forms when C-2 (the ketone carbon) cyclizes with the C-5 hydroxyl, placing the anomeric OH below the ring plane in the α-anomer; the final structure has $\mathrm{CH_2OH}$ groups at both C-1 and C-6 positions flanking the ring oxygen.

Why fructose cyclizes differently from glucose

Fructose is a ketohexose — the carbonyl sits at C-2, not C-1. When it cyclizes into a five-membered furanose ring, the C-5 hydroxyl attacks the C-2 ketone, making C-2 the new anomeric center. This is fundamentally different from aldoses like glucose, where C-1 becomes anomeric.

The furanose ring of D-fructose therefore contains carbons 2, 3, 4, 5, and the ring oxygen. Carbon-1 and carbon-6 both end up as $\mathrm{CH_2OH}$ groups outside the ring — C-1 attached to C-2 (the anomeric carbon) and C-6 attached to C-5.

Building the Haworth projection step by step

Identify the ring atoms and numbering.

The furanose ring runs C-2 (anomeric) → C-3 → C-4 → C-5 → O → back to C-2. In a standard Haworth projection, we draw the ring with oxygen at the back (top), C-2 to the right of oxygen, C-3 at the bottom right, C-4 at the bottom left, and C-5 to the left of oxygen.

Place the exocyclic $\mathrm{CH_2OH}$ groups.

C-1 is bonded to C-2 (the anomeric carbon, upper right position). So $\mathrm{CH_2OH}$ from C-1 projects from the anomeric carbon.

C-6 is bonded to C-5 (upper left position). So $\mathrm{HOH_2C}$ (or $\mathrm{CH_2OH}$ ) projects from C-5.

This immediately tells us the upper-left and upper-right substituents are both $\mathrm{CH_2OH}$ groups.

Determine the anomeric configuration at C-2.

In D-sugars, the α-anomer has the anomeric OH on the opposite side of the ring from the C-6 $\mathrm{CH_2OH}$ group (which defines the D-configuration). Since C-6 projects upward from C-5 in D-fructose, the α-OH at C-2 must point downward (below the ring plane).

The anomeric carbon (C-2) therefore carries the C-1 $\mathrm{CH_2OH}$ substituent and an OH group; in α-D-fructofuranose, that OH is down.

Assign the remaining hydroxyls using Fischer → Haworth rules.

In the D-fructose Fischer projection:

C-3: OH on the right → projects down in Haworth (below the ring). …
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TG EAPCET 2021Set eng-2021-08-05-AN1 markMCQ
$\mathrm{CH_2I} \\ | \\ (\mathrm{CHOH})_4 \\ | \\ \mathrm{CH_2OH}$
›Reveal solutionSolution
Prolonged heating of glucose with HI causes exhaustive reduction — all hydroxyl (–OH) groups are replaced by hydrogen, and the aldehyde group is also reduced to a methyl group. The final product is n-hexane, which matches option (C).

The key here is understanding what HI does to a sugar like glucose. HI is a powerful reducing agent, especially under prolonged heating. It doesn’t just replace one or two –OH groups — it goes all the way. Every hydroxyl group in glucose gets replaced by hydrogen, and the aldehyde group at C1 gets reduced to a methyl group as well. So the entire carbon skeleton becomes a straight-chain alkane.

Let’s walk through it step by step.

Glucose’s structure Glucose is an aldohexose: a six-carbon straight chain with an aldehyde group at C1 and a hydroxyl group on every other carbon. Its open-chain form is:

$\mathrm{CHO} - (\mathrm{CHOH})_4 - \mathrm{CH_2OH}$

What HI does

HI replaces each –OH group with –I in the first step, but under prolonged heating, the C–I bonds are further reduced to C–H bonds. The aldehyde group (–CHO) is also reduced: first to –CH₂OH, then to –CH₃. So every oxygen atom is removed and replaced by hydrogen.

The result on each carbon

C1 (originally –CHO) becomes –CH₃.

C2 through C5 (originally –CHOH–) each become –CH₂–.

C6 (originally –CH₂OH) becomes –CH₃.

So the product is a straight chain of six carbons with no functional groups:

$\mathrm{CH_3} - (\mathrm{CH_2})_4 - \mathrm{CH_3}$

That’s n-hexane …
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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

The concept first

Step-by-step

The concept first: how to read a Haworth projection

Step-by-step

Concept and Intuition

Step-by-step Analysis

Why fructose cyclizes differently from glucose

Building the Haworth projection step by step

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

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