The Hook: From Sweet Sugar to Energy Storage
Every time you eat rice, bread, or fruits, you’re consuming carbohydrates! Glucose in your blood powers your brain (uses 20% of your daily energy). Athletes “carb-load” before marathons. Diabetics monitor blood sugar levels. Cotton clothes are made of cellulose (a carbohydrate!).
Here’s the JEE question: Why do glucose and fructose have the same molecular formula (C₆H₁₂O₆) but different properties? Why can we digest starch but not cellulose, even though both are glucose polymers?
The Core Concept
What are Carbohydrates?
Carbohydrates = “Hydrates of carbon”
General formula:
$$\boxed{\text{C}_n(\text{H}_2\text{O})_m}$$Or more commonly: Cₙ(H₂O)ₙ → CₙH₂ₙOₙ
Definition: Polyhydroxy aldehydes or ketones, or compounds that yield them on hydrolysis.
Etymology: Carbo (carbon) + hydrate (water)
Classification of Carbohydrates
Based on Hydrolysis
1. Monosaccharides (Cannot be hydrolyzed)
- Simplest sugars
- Examples: Glucose, fructose, ribose
2. Oligosaccharides (Yield 2-10 monosaccharides)
- Disaccharides: 2 monosaccharides
- Sucrose, maltose, lactose
- Trisaccharides: 3 monosaccharides
- Raffinose
3. Polysaccharides (Yield many monosaccharides)
- Examples: Starch, cellulose, glycogen
Based on Reducing Property
Reducing sugars:
- Have free aldehyde or ketone group
- Reduce Fehling’s/Benedict’s solution
- Examples: Glucose, fructose, maltose, lactose
Non-reducing sugars:
- No free carbonyl group
- Do NOT reduce Fehling’s solution
- Example: Sucrose
“Mono is One, Oligo is Occasionally, Poly is Plenty”
Reducing vs Non-reducing: “Reducing sugars have Real carbonyl (free CHO or C=O)” “Sucrose is Special - non-reducing!”
JEE Tip: If both anomeric carbons are involved in glycosidic linkage → non-reducing!
Monosaccharides
Classification of Monosaccharides
By Number of Carbons
Trioses: 3 carbons (C₃H₆O₃)
- Glyceraldehyde
Tetroses: 4 carbons (C₄H₈O₄)
Pentoses: 5 carbons (C₅H₁₀O₅)
- Ribose, arabinose, xylose
Hexoses: 6 carbons (C₆H₁₂O₆) - Most important
- Glucose, fructose, galactose
By Functional Group
Aldoses: Contain aldehyde group (-CHO)
- Examples: Glucose, ribose, galactose
Ketoses: Contain ketone group (C=O)
- Examples: Fructose, ribulose
Combined classification:
- Aldohexose: 6-carbon aldehyde sugar (glucose)
- Ketohexose: 6-carbon ketone sugar (fructose)
Glucose: The Most Important Monosaccharide
Molecular Formula and Structure
Molecular formula: C₆H₁₂O₆
Open chain structure (Fischer projection):
CHO (C1)
|
H-C-OH (C2)
|
HO-C-H (C3)
|
H-C-OH (C4)
|
H-C-OH (C5)
|
CH₂OH (C6)
D-(+)-Glucose
Key features:
- Aldohexose: 6 carbons + aldehyde group
- 4 chiral centers: C2, C3, C4, C5
- D-configuration: OH on C5 is on right (Fischer projection)
- Optically active: Rotates plane-polarized light
Cyclic Structure (Most Important for JEE)
In solution, glucose exists 99% in cyclic form!
Formation: Intramolecular hemiacetal formation
Step 1: C5-OH attacks C1=O
Open chain → Cyclic form (pyranose ring)
Two anomers form:
α-D-Glucose (36%)
CH₂OH
|
H—C—O—H
| |
HO—C C—OH ← OH at C1 is DOWN
| |
H—C C—H
| |
HO—C C—OH
|
H
(Haworth projection)
β-D-Glucose (64%)
CH₂OH
|
H—C—O—OH ← OH at C1 is UP
| |
HO—C C—H
| |
H—C C—OH
| |
HO—C C—H
|
H
Anomers = Two cyclic forms differing only at C1 (anomeric carbon)
Key differences:
| Property | α-Glucose | β-Glucose |
|---|---|---|
| OH at C1 | Down (axial) | Up (equatorial) |
| % in solution | 36% | 64% |
| Melting point | 146°C | 150°C |
| Specific rotation | +112° | +19° |
Why β is more stable:
- β-OH is equatorial (less steric repulsion)
- α-OH is axial (1,3-diaxial strain)
Mutarotation: Interconversion of α and β forms through open chain
- Final ratio: 36% α : 64% β
- Specific rotation changes: +112° → +52.5° (equilibrium)
JEE Tip: β-anomer is MORE stable (equatorial OH at C1)
Interactive Demo: Visualize Glucose Structure
Explore the cyclic forms of glucose and see how α and β anomers differ.
Reactions of Glucose
Reaction 1: Oxidation
Mild oxidation (Br₂ water or HNO₃):
$$\boxed{\text{Glucose} \xrightarrow{[\text{O}]} \text{Gluconic acid}}$$CHO COOH
| |
(CHOH)₄ → (CHOH)₄
| |
CH₂OH CH₂OH
Glucose Gluconic acid
Strong oxidation (conc. HNO₃):
$$\boxed{\text{Glucose} \xrightarrow{\text{HNO}_3} \text{Glucaric acid (Saccharic acid)}}$$CHO COOH
| |
(CHOH)₄ → (CHOH)₄
| |
CH₂OH COOH
Glucose Glucaric acid
Both C1 and C6 oxidized!
Reaction 2: Reduction
$$\boxed{\text{Glucose} \xrightarrow{\text{NaBH}_4 \text{ or } \text{H}_2/\text{Ni}} \text{Sorbitol}}$$CHO CH₂OH
| |
(CHOH)₄ → (CHOH)₄
| |
CH₂OH CH₂OH
Glucose Sorbitol (Glucitol)
Sorbitol:
- Sugar alcohol
- Used as sweetener (diabetic foods)
- Non-cariogenic (doesn’t cause tooth decay)
Reaction 3: Acetylation
With acetic anhydride:
$$\boxed{\text{Glucose} + 5(\text{CH}_3\text{CO})_2\text{O} \rightarrow \text{Glucose pentaacetate}}$$All 5 -OH groups get acetylated!
Products:
- α-Glucose pentaacetate
- β-Glucose pentaacetate
Evidence: Glucose has 5 OH groups (and 1 CHO in open form)
Reaction 4: Glycoside Formation
With methanol + HCl:
$$\boxed{\text{Glucose} + \text{CH}_3\text{OH} \xrightarrow{\text{HCl}} \text{Methyl glucoside} + \text{H}_2\text{O}}$$Products:
- Methyl α-D-glucoside
- Methyl β-D-glucoside
Glycosidic bond: C-O-C linkage between sugar and alcohol
Importance: This is how sugars link in disaccharides!
Reaction 5: Fehling’s Test (Reducing Sugar Test)
Glucose + Fehling’s solution:
$$\boxed{\text{Glucose} \xrightarrow{\text{Fehling's}} \text{Red ppt of Cu}_2\text{O}}$$Observation:
- Blue color → Brick red precipitate
Why positive?
- Glucose has free aldehyde group (in open form)
- Reduces Cu²⁺ → Cu⁺ (red Cu₂O)
All reducing sugars give this test!
Q: How will you prove that glucose contains: (a) 5 OH groups (b) Aldehyde group (c) Straight chain of 6 carbons
Solution:
(a) 5 OH groups:
- Acetylation: Forms glucose pentaacetate
- 5 molecules of (CH₃CO)₂O react
- Proves 5 OH groups present
(b) Aldehyde group:
- Fehling’s test: Positive (red ppt)
- Tollen’s test: Silver mirror forms
- Adds HCN to form cyanohydrin
- All prove CHO group
(c) Straight chain of 6 carbons:
- Prolonged HI treatment: Gives n-hexane
- Proves 6 carbons in straight chain
- No branching present
JEE Strategy: Know what each reaction proves about structure!
Fructose: The Ketohexose
Structure
Molecular formula: C₆H₁₂O₆ (same as glucose!)
Open chain structure:
CH₂OH (C1)
|
C=O (C2) ← Ketone group
|
HO-C-H (C3)
|
H-C-OH (C4)
|
H-C-OH (C5)
|
CH₂OH (C6)
D-Fructose
Key features:
- Ketohexose: 6 carbons + ketone at C2
- 3 chiral centers: C3, C4, C5 (one less than glucose)
- Sweetest natural sugar
Cyclic Structure
Forms 5-membered ring (furanose):
- α-D-Fructofuranose
- β-D-Fructofuranose
Properties:
- Reducing sugar (can open to show C=O)
- Gives positive Fehling’s test
- Sweeter than glucose
Disaccharides
What are Disaccharides?
Disaccharides = Two monosaccharides joined by glycosidic linkage
General formula: C₁₂H₂₂O₁₁
Formation:
$$\text{Monosaccharide}_1 + \text{Monosaccharide}_2 \rightarrow \text{Disaccharide} + \text{H}_2\text{O}$$Sucrose (Table Sugar)
Molecular formula: C₁₂H₂₂O₁₁
Composition: Glucose + Fructose
Structure:
α-D-Glucose (C1) — O — β-D-Fructose (C2)
α(1→2) glycosidic linkage
Key features:
- Both anomeric carbons involved (C1 of glucose, C2 of fructose)
- Non-reducing sugar (no free CHO or C=O)
- Negative Fehling’s test
- Inverted by acids to glucose + fructose
Hydrolysis (Inversion):
$$\boxed{\text{Sucrose} + \text{H}_2\text{O} \xrightarrow{\text{H}^+ \text{ or invertase}} \text{Glucose} + \text{Fructose}}$$Invert sugar:
- Mixture of glucose + fructose
- Laevorotatory (rotates light left)
- Sweeter than sucrose
- Used in honey, soft drinks
Sucrose is non-reducing because:
Reducing sugar needs:
- Free anomeric carbon (C1 in aldoses, C2 in ketoses)
- Can open to show CHO or C=O
In sucrose:
- α-C1 of glucose linked to β-C2 of fructose
- Both anomeric carbons blocked
- Cannot open to free form
- No free carbonyl → non-reducing
Memory: “Sucrose has Special linkage - both anomeric carbons tied up!”
JEE Trap: “Sucrose contains glucose (reducing) so it’s reducing” - WRONG!
Maltose (Malt Sugar)
Molecular formula: C₁₂H₂₂O₁₁
Composition: Glucose + Glucose
Structure:
α-D-Glucose (C1) — O — β-D-Glucose (C4)
α(1→4) glycosidic linkage
Key features:
- Only one anomeric carbon involved (C1 of first glucose)
- Reducing sugar (C1 of second glucose is free)
- Positive Fehling’s test
Hydrolysis:
$$\boxed{\text{Maltose} + \text{H}_2\text{O} \xrightarrow{\text{maltase}} 2 \text{ Glucose}}$$Source:
- Starch hydrolysis
- Germinating cereals
- Malted barley (brewing)
Lactose (Milk Sugar)
Molecular formula: C₁₂H₂₂O₁₁
Composition: Glucose + Galactose
Structure:
β-D-Galactose (C1) — O — β-D-Glucose (C4)
β(1→4) glycosidic linkage
Key features:
- Reducing sugar (free C1 on glucose)
- Found only in milk
- Less sweet than sucrose
Hydrolysis:
$$\boxed{\text{Lactose} + \text{H}_2\text{O} \xrightarrow{\text{lactase}} \text{Glucose} + \text{Galactose}}$$Lactose intolerance:
- Lack of lactase enzyme
- Cannot digest lactose
- Causes bloating, diarrhea
Polysaccharides
What are Polysaccharides?
Polysaccharides = Long chains of monosaccharides (hundreds to thousands)
General formula: (C₆H₁₀O₅)ₙ
Examples:
- Starch (plants - storage)
- Cellulose (plants - structure)
- Glycogen (animals - storage)
Starch
Composition: Polymer of α-D-glucose
Two components:
1. Amylose (20-30%)
- Linear chain of glucose units
- α(1→4) glycosidic linkages
- Helical structure
- Water-soluble
- Blue color with iodine
2. Amylopectin (70-80%)
- Branched chain of glucose units
- α(1→4) in chains + α(1→6) at branches
- Water-swellable
- Purple/red color with iodine
Hydrolysis:
$$\text{Starch} \xrightarrow{\text{diastase}} \text{Maltose} \xrightarrow{\text{maltase}} \text{Glucose}$$Test for starch:
- Iodine test: Blue-black color
- Specific for starch
- Used to detect starch in food
Sources:
- Rice, wheat, potatoes
- Corn, tapioca
- Major energy source
Cellulose
Composition: Polymer of β-D-glucose
Structure:
- Linear chains of glucose
- β(1→4) glycosidic linkages
- Chains held by H-bonds (very strong)
- Fibrous structure
Properties:
- Insoluble in water
- Indigestible by humans (we lack cellulase enzyme)
- Structural material
Uses:
- Cotton, paper, wood
- Plant cell walls
- Textile industry
Why we can’t digest cellulose:
- Need cellulase enzyme to break β(1→4) linkages
- Humans don’t produce cellulase
- Herbivores have bacteria in gut that produce cellulase
Q: Why can humans digest starch but not cellulose, even though both are glucose polymers?
Answer:
Starch:
- α(1→4) glycosidic linkages
- Humans have α-amylase enzyme
- Can hydrolyze α-linkages
- Digestible
Cellulose:
- β(1→4) glycosidic linkages
- Humans lack cellulase enzyme
- Cannot break β-linkages
- Indigestible (passes as fiber)
Key point: Different linkage geometry (α vs β) requires different enzymes!
Herbivores:
- Have bacteria/protozoa in gut
- These produce cellulase
- Can digest cellulose
JEE Tip: α vs β linkage determines digestibility!
Glycogen
Composition: Polymer of α-D-glucose (like amylopectin)
Structure:
- Highly branched
- More branched than amylopectin
- Compact, dense structure
Function:
- Animal starch (storage in liver and muscles)
- Readily mobilized for energy
- Provides glucose when needed
Hydrolysis:
$$\text{Glycogen} \rightarrow \text{Maltose} \rightarrow \text{Glucose}$$Common Mistakes to Avoid
Wrong: “Sucrose is reducing because it contains glucose”
Correct: Sucrose is non-reducing because both anomeric carbons are involved in glycosidic bond
Rule: Check if there’s a free anomeric carbon!
- Free anomeric C → Reducing
- Both involved → Non-reducing
JEE Tip: Draw the structure and check C1 (or C2 for ketoses)!
Wrong: Drawing all OH groups on same side
Correct: Glucose has specific configuration at each chiral center
Remember:
- C2, C4, C5: OH on right (Fischer)
- C3: OH on left (Fischer)
- For D-glucose
JEE Tip: Learn the Fischer projection for D-glucose by heart!
Wrong: “α has OH up, β has OH down”
Correct (Haworth projection):
- α-anomer: OH at C1 is DOWN (axial)
- β-anomer: OH at C1 is UP (equatorial)
Memory: “β is Better (up and equatorial)”
JEE Tip: β is more stable (64% in equilibrium)!
Practice Problems
Level 1: Foundation (NCERT)
Q: Classify the following as aldose or ketose: (a) Glucose (b) Fructose (c) Ribose
Solution:
(a) Glucose: Aldose
- Has CHO group at C1
- Aldohexose (6 carbons + aldehyde)
(b) Fructose: Ketose
- Has C=O group at C2
- Ketohexose (6 carbons + ketone)
(c) Ribose: Aldose
- Has CHO group at C1
- Aldopentose (5 carbons + aldehyde)
Q: Which of the following are reducing sugars? (a) Glucose (b) Sucrose (c) Maltose (d) Fructose
Solution:
Reducing: (a), (c), (d) Non-reducing: (b)
Explanation:
(a) Glucose: Reducing
- Free aldehyde group (in open form)
- Free C1
(b) Sucrose: Non-reducing
- Both anomeric carbons involved in bond
- No free carbonyl
(c) Maltose: Reducing
- One glucose has free C1
- Can show aldehyde group
(d) Fructose: Reducing
- Free ketone group (in open form)
- Free C2
Level 2: JEE Main
Q: The specific rotation of pure α-D-glucose is +112° and that of β-D-glucose is +19°. When either form is dissolved in water, the specific rotation changes to +52.5°. Explain.
Solution:
Phenomenon: Mutarotation
Explanation:
Initial state:
- Pure α-D-glucose: [α] = +112°
- OR Pure β-D-glucose: [α] = +19°
In solution:
- α and β forms interconvert through open chain
- Reach equilibrium: 36% α + 64% β
Final rotation:
$$[α] = 0.36 × (+112°) + 0.64 × (+19°)$$ $$[α] = +40.3° + 12.2° = +52.5°$$This constant value (+52.5°) proves equilibrium mixture!
JEE Concept: Mutarotation = change in optical rotation due to anomer interconversion
Q: What products are formed when sucrose is hydrolyzed? Why is the mixture called “invert sugar”?
Solution:
Hydrolysis:
$$\text{Sucrose} + \text{H}_2\text{O} \xrightarrow{\text{H}^+ \text{ or invertase}} \text{Glucose} + \text{Fructose}$$Products:
- D-(+)-Glucose: [α] = +52.5°
- D-(-)-Fructose: [α] = -92°
Why “invert sugar”?
Before hydrolysis:
- Sucrose: [α] = +66.5° (dextrorotatory)
After hydrolysis:
- Mixture: [α] = (52.5 - 92)/2 ≈ -20° (laevorotatory)
Sign of rotation inverts from + to - !
Hence called “invert sugar”
JEE Fact: Fructose is more laevorotatory than glucose is dextrorotatory!
Level 3: JEE Advanced
Q: Glucose reacts with acetic anhydride to form glucose pentaacetate. What does this prove about glucose structure?
Solution:
Reaction:
$$\text{Glucose} + 5(\text{CH}_3\text{CO})_2\text{O} \rightarrow \text{Glucose pentaacetate}$$Proves:
Five -OH groups present
- 5 molecules of acetic anhydride react
- Each acetylates one OH
One carbonyl group
- In open form: CHO group (doesn’t acetylate)
- In cyclic form: Anomeric OH (acetylates)
Structure conclusion:
CHO or CH(OAc)-O-
| |
(CHOH)₄ → (CHOAc)₄
| |
CH₂OH CH₂OAc
(5 OH groups) (all acetylated)
Additional info:
- Two products: α and β pentaacetate
- Proves existence of cyclic forms!
JEE Strategy: Counting acetyl groups = counting OH groups!
Q: Explain why: (a) Glucose gives Fehling’s test but sucrose doesn’t (b) Cellulose is indigestible but starch is digestible
Solutions:
(a) Fehling’s Test:
Glucose: Positive
- Free aldehyde group (in open form)
- Reduces Cu²⁺ to Cu₂O (red ppt)
- Free anomeric carbon (C1)
Sucrose: Negative
- Both anomeric carbons tied in glycosidic bond
- Cannot open to show free carbonyl
- No reduction of Cu²⁺
(b) Digestibility:
Starch: Digestible
- α(1→4) linkages
- Humans have α-amylase enzyme
- Hydrolyzes to glucose
Cellulose: Indigestible
- β(1→4) linkages
- Humans lack cellulase enzyme
- Cannot be broken down
- Passes as dietary fiber
Key: Enzyme specificity for linkage geometry (α vs β)!
Quick Revision Box
| Carbohydrate | Type | Composition | Reducing? | Key Feature |
|---|---|---|---|---|
| Glucose | Monosaccharide | C₆H₁₂O₆ | Yes | Aldohexose, 4 chiral centers |
| Fructose | Monosaccharide | C₆H₁₂O₆ | Yes | Ketohexose, sweetest |
| Sucrose | Disaccharide | Glu + Fru | No | Both anomeric C involved |
| Maltose | Disaccharide | Glu + Glu | Yes | α(1→4) linkage |
| Lactose | Disaccharide | Gal + Glu | Yes | Milk sugar |
| Starch | Polysaccharide | (Glucose)ₙ | - | α-linkage, digestible |
| Cellulose | Polysaccharide | (Glucose)ₙ | - | β-linkage, indigestible |
Connection to Other Topics
Prerequisites:
- Aldehydes and Ketones - Carbonyl chemistry
- Stereochemistry - D/L, R/S configuration
- Organic Principles - Functional groups
Related Topics:
- Amino Acids and Proteins - Other biomolecules
- Nucleic Acids - Ribose, deoxyribose sugars
- Enzymes - Carbohydrate metabolism
Applications:
- Biochemistry - Energy metabolism
- Food Chemistry - Nutrition, sweeteners
Teacher’s Summary
1. Classification (MASTER THIS):
- Monosaccharides: Cannot hydrolyze (glucose, fructose)
- Disaccharides: 2 units (sucrose, maltose, lactose)
- Polysaccharides: Many units (starch, cellulose)
2. Glucose Structure (HIGH-YIELD):
- Aldohexose: CHO + 5 carbons + CH₂OH
- 4 chiral centers: C2, C3, C4, C5
- Cyclic forms: α (36%) and β (64%) - anomers
- Mutarotation: Interconversion via open chain
3. Reducing vs Non-reducing:
- Reducing: Free anomeric carbon (glucose, maltose, lactose, fructose)
- Non-reducing: Both anomeric carbons involved (sucrose only)
4. Important Reactions:
- Oxidation: → Gluconic acid (mild) or Glucaric acid (strong)
- Reduction: → Sorbitol
- Acetylation: → Pentaacetate (proves 5 OH)
- Fehling’s test: Red ppt for reducing sugars
5. Disaccharides:
- Sucrose: Glu-Fru, α(1→2), non-reducing
- Maltose: Glu-Glu, α(1→4), reducing
- Lactose: Gal-Glu, β(1→4), reducing
6. Polysaccharides:
- Starch: α-glucose polymer, digestible (we have α-amylase)
- Cellulose: β-glucose polymer, indigestible (we lack cellulase)
- Glycogen: Animal starch, highly branched
“Carbohydrates are life’s fuel - from the glucose in your blood to the starch in your food!”
Next: Study Proteins to understand amino acids, peptide bonds, and protein structure!