MedStudy · Organic Chemistry
TMU MBBS 1st Year · Semester 2
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Chapter 9 · Organic Chemistry

Carbohydrates

Blood glucose, lactose intolerance, glycogen storage diseases, cellulose vs starch — carbohydrate chemistry is metabolic medicine. The subtlety of α vs β linkages determines whether humans can digest starch (yes) or cellulose (no), and the anomeric carbon at C-1 is the key to understanding reducing sugars and glycosidic bonds.

Aldoses & ketoses Fischer projection D/L configuration Haworth projection α/β anomers Reducing sugars Starch vs cellulose
9.1

Classification of Carbohydrates

Carbohydrates are polyhydroxy aldehydes or ketones, or compounds that hydrolyse to give them. The name comes from “carbon hydrate” — the empirical formula of many is (CH₂O)ₙ. They are classified by size: monosaccharides are the simplest units (glucose, fructose, galactose); disaccharides are two monosaccharides joined by a glycosidic bond (sucrose, lactose, maltose); oligosaccharides have a few units; and polysaccharides are long chains of thousands of monosaccharide units (starch, glycogen, cellulose).

Size Classification
Monosaccharide: single sugar unit; cannot be hydrolysed further (glucose, fructose, ribose)
Disaccharide: 2 monosaccharides + glycosidic bond; hydrolysed by disaccharidases (sucrose, lactose, maltose)
Polysaccharide: many (often thousands) monosaccharide units; structural (cellulose) or storage (starch, glycogen)
Test yourself — 9.1
• What is a monosaccharide chemically? → A polyhydroxy aldehyde (aldose) or polyhydroxy ketone (ketose); cannot be hydrolysed further.
• Give one example each of a monosaccharide, disaccharide, and polysaccharide. → Glucose / Lactose / Starch.
9.2

Aldoses vs Ketoses

Monosaccharides are further divided by where the carbonyl group sits. An aldose has the carbonyl at C-1 (an aldehyde). A ketose has the carbonyl at C-2 (a ketone). Glucose is an aldose; fructose is a ketose. The prefix tells you the number of carbons: triose (3C), tetrose (4C), pentose (5C), hexose (6C). Combine them: glucose is an aldohexose; fructose is a ketohexose; ribose is an aldopentose.

Key Monosaccharides
SugarTypeCarbonsClinical note
GlucoseAldohexose6Primary energy fuel; blood glucose
FructoseKetohexose6Fruit sugar; sweetest monosaccharide
GalactoseAldohexose6C-4 epimer of glucose; from lactose
RiboseAldopentose5Backbone of RNA, ATP, NAD⁺
DeoxyriboseAldopentose5Backbone of DNA (no C-2 OH)
Test yourself — 9.2
• What is an aldose? → Sugar with aldehyde at C-1.
• What is a ketose? → Sugar with ketone at C-2.
• Classify glucose and fructose. → Glucose = aldohexose; Fructose = ketohexose.
• What is galactose relative to glucose? → C-4 epimer (OH at C-4 is reversed).
9.3

Fischer Projections & D/L Configuration

Sugars are drawn in Fischer projections — a cross-like 2D representation where the carbon chain runs vertically (C-1 at top by convention), horizontal bonds point toward the viewer, and vertical bonds point away. The D/L designation is determined by the configuration of the highest-numbered chiral carbon (the one furthest from the carbonyl) in the Fischer projection. If the –OH on this carbon is on the right, the sugar is a D-sugar; if on the left, it is L. Almost all naturally occurring sugars are D-sugars.

D/L Rule in Fischer Projection
1. Draw the sugar in Fischer projection with C-1 (carbonyl) at top.
2. Find the highest-numbered chiral carbon (C-5 for hexoses, C-4 for pentoses).
3. –OH on the right = D-sugar; –OH on the left = L-sugar.

All common biological sugars are D-: D-glucose, D-fructose, D-galactose, D-ribose.
Epimers
Two sugars that differ only in configuration at ONE carbon (other than C-1) are epimers.
Glucose ↔ Galactose: C-4 epimers (Leloir pathway interconverts them; defect = galactosaemia)
Glucose ↔ Mannose: C-2 epimers
Test yourself — 9.3
• How do you determine D vs L for a sugar in Fischer projection? → Check the OH on the highest-numbered chiral C: right = D, left = L.
• Are most biological sugars D or L? → D.
• What are epimers? → Two sugars differing at exactly one chiral centre (not C-1).
• Glucose and galactose are epimers at which carbon? → C-4.
9.4

Ring Forms & Haworth Projections

In aqueous solution, glucose doesn't exist primarily as the open-chain aldehyde — it cyclises. The C-5 hydroxyl attacks the C-1 aldehyde (an intramolecular hemiacetal formation), forming a six-membered ring called a pyranose ring (named after pyran). Five-membered rings (from C-4 attacking C-2 ketone, as in fructose) are furanose rings.

The ring is drawn as a Haworth projection — a flat hexagon (or pentagon) viewed from the side. The ring oxygen is at the back-right. Groups that were on the right in the Fischer projection are drawn below the Haworth ring; groups on the left go above. The critical new centre formed by ring closure is C-1 for aldoses (the anomeric carbon) — more on this in §9.5.

Fischer → Haworth Conversion Rule (Aldohexose)
1. C-1 is at the right of the ring in Haworth.
2. Groups on the right in Fischer go below the ring in Haworth.
3. Groups on the left in Fischer go above the ring.
4. C-6 (–CH₂OH) is above the ring for D-sugars.
5. The OH at C-1 (new group formed by ring closure) can be α (below) or β (above) — see §9.5.
D-Glucose: Fischer Projection → Haworth Ring Fischer (open chain) CHO C-1 H OH C-2 HO H C-3 H OH C-4 H OH C-5★ CH₂OH C-6 ★ OH on right = D-sugar Conversion Rules Right in Fischer → below ring Left in Fischer → above ring C-6 CH₂OH: above (all D-sugars) C-1 OH (anomeric): α = below · β = above Haworth (α-D-glucose) O C-1 OH↓(α) OH↓ C-2 OH↑ C-3 OH↓ C-4 C-5 CH₂OH↑
D-glucose in Fischer: C-5 OH on the right confirms D-configuration. In Haworth: right-side groups go below the ring, left-side groups go above. C-6 CH₂OH is always above for D-sugars. The new C-1 OH (anomeric carbon) is α (below) in this drawing.
Test yourself — 9.4
• What intramolecular reaction forms the pyranose ring? → Hemiacetal formation — C-5 OH attacks C-1 aldehyde.
• What is a pyranose vs furanose? → Pyranose = 6-membered ring; furanose = 5-membered ring.
• In Haworth projection, right-side Fischer groups go where? → Below the ring.
9.5

Anomers & Mutarotation

When glucose cyclises, a new chiral centre forms at C-1 — the anomeric carbon. The two possible configurations of the C-1 OH are called anomers: in α-D-glucose, the C-1 OH is axial/below the ring; in β-D-glucose, the C-1 OH is equatorial/above the ring. These are not enantiomers (mirror images) or epimers (differ at a non-anomeric carbon) — they are specifically anomers (differ only at the anomeric carbon).

Mutarotation is the spontaneous interconversion of α and β anomers in solution, going through the open-chain aldehyde as an intermediate. If you dissolve pure α-D-glucose in water, its optical rotation gradually changes until equilibrium is reached (about 36% α and 64% β at equilibrium for glucose). This is mutarotation, and it is clinically relevant because the enzyme lactase, maltase, and other glucosidases are anomer-specific.

α vs β Anomer (Haworth)
α-D-glucose: C-1 OH is below the ring (same side as ring oxygen in most depictions; trans to C-6 CH₂OH)
β-D-glucose: C-1 OH is above the ring (equatorial in chair form)

Memory trick: α = axial = down; β = both bonds up (equatorial)
Clinical — Anomer Specificity
Cellulose has β-1,4-glycosidic bonds; starch has α-1,4 bonds. Humans have α-amylase and α-glucosidases — we can digest starch but cannot hydrolyse the β-1,4 bonds of cellulose (no β-glucosidase). Termites and ruminants harbour gut bacteria with cellulase enzymes. This single α vs β difference at C-1 is the entire reason humans cannot digest wood.
Test yourself — 9.5
• What is the anomeric carbon? → C-1 in aldoses (C-2 in fructose) — the new chiral centre formed by ring closure.
• α-D-glucose: C-1 OH position? → Below the ring (axial in chair).
• β-D-glucose: C-1 OH position? → Above the ring (equatorial in chair).
• What is mutarotation? → Spontaneous interconversion of α and β anomers in solution through the open-chain form; equilibrium favours β (~64%).
9.6

Reducing vs Non-reducing Sugars

A reducing sugar is one that can reduce an oxidising agent (like Cu²⁺ in Fehling's or Tollens' reagent). For a sugar to do this, it needs a free aldehyde or a free hemiacetal at C-1 that can open to the aldehyde form. Any monosaccharide is a reducing sugar because it has a free anomeric OH (hemiacetal). In disaccharides, it depends on whether the glycosidic bond uses C-1 of both sugars.

In maltose (glucose + glucose, α-1,4), one glucose has its C-1 OH free — it can open to the aldehyde form, so maltose is a reducing sugar. In lactose (galactose + glucose, β-1,4), similarly, the glucose C-1 is free. But in sucrose (glucose + fructose, α,β-1,2), the glycosidic bond uses C-1 of glucose AND C-2 of fructose — both anomeric carbons are locked in the bond, leaving no free hemiacetal. Sucrose is therefore a non-reducing sugar.

Reducing Sugar Test Summary
SugarReducing?Reason
GlucoseYesFree C-1 hemiacetal
FructoseYesFree C-2 hemiketal
Maltose (α-1,4)YesOne free C-1 OH
Lactose (β-1,4)YesOne free C-1 OH (glucose end)
Sucrose (α,β-1,2)NoBoth anomeric C locked in bond
Test yourself — 9.6
• What structural feature makes a sugar reducing? → Free anomeric OH (hemiacetal/hemiketal) that can open to aldehyde/ketone form.
• Why is sucrose non-reducing? → The glycosidic bond uses C-1 of glucose and C-2 of fructose — both anomeric carbons are engaged, no free hemiacetal.
• Are all monosaccharides reducing sugars? → Yes — all have a free anomeric OH.
9.7

Important Disaccharides

Three disaccharides appear repeatedly in both exams and clinical medicine, each with a specific bond and specific significance. You need to know the monomers, the glycosidic bond type, whether it is reducing or not, and the enzyme that cleaves it.

The Three Key Disaccharides
DisaccharideMonomersBondReducing?Cleaving enzymeClinical note
MaltoseGlucose + Glucoseα-1,4YesMaltaseStarch hydrolysis product; malt/beer
LactoseGalactose + Glucoseβ-1,4YesLactaseMilk sugar; lactase deficiency = lactose intolerance
SucroseGlucose + Fructoseα,β-1,2NoSucrase/invertaseTable sugar; non-reducing
Clinical — Lactose Intolerance
Lactase (brush-border enzyme) hydrolyses lactose → glucose + galactose. Lactase activity declines after weaning in most human populations (lactase non-persistence). Undigested lactose reaches the colon → fermented by bacteria → gas (bloating, flatulence) + osmotic diarrhoea. Diagnosis: hydrogen breath test. Management: lactase supplements, lactose-free dairy, or fermented products (yoghurt — bacteria pre-digest lactose). Lactose intolerance is most prevalent in East Asian, African, and some Middle Eastern populations.
Test yourself — 9.7
• What are the monomers of lactose and what bond joins them? → Galactose + Glucose via β-1,4-glycosidic bond.
• Which disaccharide is non-reducing and why? → Sucrose — both anomeric carbons (C-1 glucose, C-2 fructose) are used in the glycosidic bond.
• What enzyme is deficient in lactose intolerance? → Lactase (intestinal brush-border β-galactosidase).
9.8

Polysaccharides

Polysaccharides are long chains of monosaccharide units joined by glycosidic bonds. The three most important are starch, glycogen, and cellulose — all polymers of glucose, yet with completely different properties due to the type of glycosidic bond (α vs β) and degree of branching.

Starch, Glycogen, Cellulose Comparison
PolysaccharideMonomerBondBranchingFunctionDigestible?
Amylose (starch)α-D-Glucoseα-1,4None (linear)Plant energy storageYes (α-amylase)
Amylopectin (starch)α-D-Glucoseα-1,4 + α-1,6Every ~24–30 unitsPlant energy storageYes
Glycogenα-D-Glucoseα-1,4 + α-1,6Every ~8–12 units (more than starch)Animal energy storage (liver, muscle)Yes (glycogenolysis)
Celluloseβ-D-Glucoseβ-1,4NonePlant cell wall structureNo (no cellulase)
Clinical — Glycogen Storage Diseases
Glycogen storage diseases (GSDs) result from enzyme deficiencies in glycogen synthesis or breakdown. von Gierke disease (GSD I): glucose-6-phosphatase deficiency → cannot release glucose from glycogen → hypoglycaemia + hepatomegaly + hyperlipidaemia. Pompe disease (GSD II): lysosomal α-glucosidase deficiency → glycogen accumulates in lysosomes → cardiomyopathy and muscle weakness. Treatment for Pompe: alglucosidase alfa (enzyme replacement therapy — recombinant α-glucosidase).
Test yourself — 9.8
• What bond type is in starch vs cellulose? → Starch: α-1,4 (and α-1,6 for branches); Cellulose: β-1,4.
• Why can't humans digest cellulose? → Humans lack β-glucosidase (cellulase) to hydrolyse β-1,4 bonds.
• How does glycogen differ from amylopectin? → Glycogen is more highly branched (branch every 8–12 units vs 24–30 in amylopectin) — allows faster glucose release.
• Enzyme deficient in von Gierke disease? → Glucose-6-phosphatase.
🎓

Past-paper Drill

1. Explain why β-D-glucose is more stable than α-D-glucose at equilibrium in solution.
In the chair conformation of β-D-glucose, the C-1 OH is equatorial (less steric strain) whereas in α-D-glucose it is axial. Equatorial substituents are lower in energy. Therefore β is more stable and predominates at equilibrium (~64% β vs ~36% α).
2. A patient presents with bloating, flatulence, and osmotic diarrhoea after drinking milk. Which enzyme is deficient and what is the diagnosis?
Lactase (intestinal β-galactosidase) deficiency → lactose intolerance. Undigested lactose osmotically draws water into the colon and is fermented by colonic bacteria to produce gas (H₂, CO₂, CH₄).
3. Benedict's test is positive for glucose and lactose but negative for sucrose. Explain each result.
Glucose: free aldehyde (C-1 hemiacetal can open) → reduces Cu²⁺ → positive. Lactose: glucose end has free C-1 OH (hemiacetal) → can open to aldehyde → reduces Cu²⁺ → positive. Sucrose: both anomeric carbons (C-1 glucose, C-2 fructose) locked in the glycosidic bond — no free hemiacetal, cannot open to aldehyde — negative.
4. Starch and cellulose are both polymers of glucose. Explain their different digestibilities based on organic chemistry.
Starch uses α-1,4 (and α-1,6) glycosidic bonds. Human α-amylase (salivary and pancreatic) specifically hydrolyses α-1,4 bonds. Cellulose uses β-1,4 bonds. Humans lack β-glucosidase (cellulase) — the active-site geometry of our digestive enzymes cannot accommodate the different orientation of the β bond. This single stereochemical difference (α vs β at C-1) determines digestibility.
Chapter 9 Master Summary
D/L: highest-numbered chiral C → OH right = D; left = L (all biological sugars = D)
Anomers: α = OH down at C-1; β = OH up; mutarotation interconverts them
Reducing sugar: free anomeric OH → all monosaccharides YES; sucrose NO
Disaccharides: Maltose (α-1,4, reducing) · Lactose (β-1,4, reducing) · Sucrose (αβ-1,2, non-reducing)
Polysaccharides: α-1,4 = digestible (starch/glycogen); β-1,4 = indigestible (cellulose)