Carbohydrates

Theory — Carbohydrates

1. What is a carbohydrate?

The general formula C_n(H₂O)_n reflects their historical name (from "hydrates of carbon"). Functionally, monosaccharides are polyhydroxy aldehydes (aldoses, ending in -ose with an aldehyde at C1) or polyhydroxy ketones (ketoses, with a ketone usually at C2). Disaccharides, oligosaccharides, and polysaccharides are linked monosaccharides.

2. Classification by carbon count

Name	Carbons	Aldose example	Ketose example
Triose	3	Glyceraldehyde	Dihydroxyacetone
Tetrose	4	Erythrose	Erythrulose
Pentose	5	Ribose, arabinose	Ribulose, xylulose
Hexose	6	Glucose, galactose, mannose	Fructose
Heptose	7	Sedoheptulose precursor	Sedoheptulose

3. Fischer projections and D/L convention

Fischer projections show 3D stereochemistry on a 2D plane: vertical bonds point AWAY from the viewer, horizontal bonds point TOWARD the viewer. The carbonyl carbon (C1 for aldoses) is drawn at the top.

D/L notation: determined by the configuration at the chiral carbon FARTHEST from the carbonyl (the highest-numbered chiral C). If the OH on this carbon points to the RIGHT in the Fischer projection \u2192 D-sugar. If LEFT \u2192 L-sugar. Almost all naturally occurring sugars are D. Note: D/L describes a single carbon\'s configuration and does NOT indicate optical rotation direction (use + or - for rotation).

D-Glucose Fischer projection: (top to bottom: CHO, H-C-OH, HO-C-H, H-C-OH, H-C-OH, CH₂OH). The C5 OH points right \u2192 D-sugar.

4. Cyclization: hemiacetal formation

Aldoses (and ketoses) with 5+ carbons spontaneously cyclize in solution. The OH at C5 (for aldohexoses) or C4 (for aldopentoses) attacks the carbonyl C, forming a HEMIACETAL: a 6-membered ring (PYRANOSE) for hexoses, or 5-membered ring (FURANOSE) for pentoses. The carbonyl C becomes a NEW STEREOCENTER \u2014 the ANOMERIC CARBON.

Anomers: at the anomeric carbon, the new OH can point either DOWN (\u03b1-anomer) or UP (\u03b2-anomer) in the Haworth projection. For D-glucose: \u03b1-D-glucose has axial OH at C1; \u03b2-D-glucose has equatorial OH at C1. The two anomers have different physical properties (\u03b1: [\u03b1] = +112\u00b0, \u03b2: [\u03b1] = +18.7\u00b0).

Mutarotation: when crystals of pure \u03b1- or \u03b2-D-glucose are dissolved in water, the optical rotation gradually changes until reaching equilibrium ([\u03b1] = +52.7\u00b0). This happens because the ring opens through the open-chain (free aldehyde) form and re-closes, giving a mixture of \u03b1 (~36%), \u03b2 (~64%), and trace open-chain (<0.01%).

5. Haworth projections

Haworth shows the cyclic form as a flat hexagon (or pentagon) with the oxygen typically at the upper right. Substituents that pointed RIGHT in the Fischer point DOWN in Haworth. The CH₂OH (C6) is up in D-sugars, down in L-sugars. The anomeric OH is down for \u03b1, up for \u03b2 (in D-sugars).

Chair conformations are more accurate than Haworth: \u03b2-D-glucopyranose has all substituents in the equatorial position \u2014 the lowest-energy hexose conformer. This stability explains why glucose is the most abundant sugar.

6. Common monosaccharides

Sugar	Type	Notes
D-Glucose (dextrose, blood sugar)	Aldohexose	The body\'s primary energy substrate; \u03b2-form most stable; ALL OH equatorial in chair
D-Fructose (fruit sugar, levulose)	Ketohexose	Ketone at C2; sweetest natural sugar; furanose in solution; in HFCS
D-Galactose	Aldohexose	Differs from glucose at C4 (epimer); component of lactose
D-Mannose	Aldohexose	C2-epimer of glucose; component of glycoproteins
D-Ribose	Aldopentose	RNA backbone; furanose form; ATP, NADH
2-Deoxy-D-ribose	Aldopentose	DNA backbone; OH at C2 replaced by H
D-Glyceraldehyde	Aldotriose	Simplest aldose; ONE chiral center; D/L reference standard
Dihydroxyacetone (DHA)	Ketotriose	Simplest ketose; achiral (no chiral center)

7. Reducing vs non-reducing sugars

A reducing sugar has a free aldehyde (or hemiacetal that can open to an aldehyde, OR a ketose that can tautomerise to an aldose under basic conditions). It REDUCES Cu²⁺ or Ag⁺ to Cu₂O (red precipitate) or Ag⁰ (silver mirror) and gets oxidized itself. Non-reducing sugars have BOTH anomeric carbons locked in glycosidic bonds, so neither end is free to open and reduce.

Test	Reagent	Positive result	Notes
Tollens\' test	Ag(NH₃)₂⁺ in NaOH	SILVER MIRROR on tube wall	Most sensitive; aldehydes only (ketones don\'t react except via tautomerisation)
Benedict\'s test	Cu²⁺-citrate, alkaline	Brick-red Cu₂O precipitate (yellow \u2192 orange \u2192 red)	Quantitative for blood glucose; mild conditions; ketoses positive (tautomerise)
Fehling\'s test	Cu²⁺-tartrate, alkaline	Brick-red Cu₂O precipitate	Similar to Benedict\'s; less stable reagent
Bial\'s test (orcinol)	Orcinol / Fe³⁺ / HCl	Green colour for pentoses	Distinguishes pentoses from hexoses
Seliwanoff\'s test	Resorcinol / HCl	Red colour within 60s for ketoses; longer for aldoses	Distinguishes ketoses from aldoses

8. Glycosidic bonds and disaccharides

The hemiacetal OH on the anomeric carbon can be replaced by another alcohol or amine, forming a stable acetal called a GLYCOSIDIC BOND. The new C-O bond locks the anomeric configuration (\u03b1 or \u03b2).

Disaccharide	Components & linkage	Reducing?	Notes
Maltose	D-Glucose-\u03b1-1,4-D-Glucose	Yes (free C1 on second glucose)	Malt sugar; product of starch hydrolysis
Cellobiose	D-Glucose-\u03b2-1,4-D-Glucose	Yes	Product of cellulose hydrolysis
Lactose	D-Galactose-\u03b2-1,4-D-Glucose	Yes	Milk sugar; lactose intolerance = lactase deficiency
Sucrose	D-Glucose-\u03b1,\u03b2-1,2-D-Fructose	NO (both anomeric C linked)	Table sugar; non-reducing
Trehalose	D-Glucose-\u03b1,\u03b1-1,1-D-Glucose	NO	Insect blood sugar; non-reducing

9. Polysaccharides

Polymer	Linkage	Function
Starch (amylose, amylopectin)	\u03b1-1,4 + \u03b1-1,6 branches	Plant energy storage; helical coiling
Glycogen	\u03b1-1,4 + \u03b1-1,6 (more branches than starch)	Animal energy storage (liver, muscle)
Cellulose	\u03b2-1,4	Plant cell walls; flat, sheet-like; most abundant biomolecule
Chitin	\u03b2-1,4 of N-acetylglucosamine	Insect exoskeletons, fungal cell walls
Peptidoglycan	\u03b2-1,4 of NAM/NAG + peptide cross-links	Bacterial cell wall

Memorise these landmarks Aldoses: aldehyde at C1; CHO group at top of Fischer
Ketoses: ketone at C2; C=O at second carbon
D vs L: configuration at the highest-numbered chiral C (right = D, left = L)
Pyranose: 6-membered ring (most hexoses); Furanose: 5-membered ring (fructose, ribose)
\u03b1-anomer: anomeric OH down (axial in D-glucopyranose)
\u03b2-anomer: anomeric OH up (equatorial in D-glucopyranose, lowest energy)
\u03b2-D-glucopyranose: ALL OH equatorial \u2014 the most stable hexose
Reducing sugar: free anomeric C (or convertible to one); positive Tollens/Benedict
Sucrose, trehalose: non-reducing (both anomeric C locked)
DNA backbone: 2-deoxyribose; RNA backbone: ribose. The 2\u2032-OH of ribose makes RNA chemically labile; DNA is more stable for genome storage.

10. Systematic carbohydrate analysis approach

Count the carbons. Triose, tetrose, pentose, hexose?
Identify the carbonyl. Aldehyde at C1 \u2192 aldose; ketone at C2 \u2192 ketose.
Read D vs L. Look at the highest-numbered chiral C. OH right = D; OH left = L.
Identify the ring form. Pyranose (6-ring) vs furanose (5-ring); \u03b1 vs \u03b2 anomer.
Predict reducing-sugar test. Free anomeric C \u2192 positive Tollens/Benedict.
For polymers/disaccharides: identify the linkage (e.g., \u03b1-1,4) and component sugars.

Instructions

The Simulation has four parts. Complete in order.

Section I — Sugar Library. 8 sugars. For each: D vs L? Aldose vs ketose? Pentose vs hexose vs triose?

Section II — Reducing-Sugar Bench. 6 sugar samples. Select a sample, run Tollens\' test, observe the result (silver mirror = reducing sugar; no change = non-reducing). Identify the sugar.

Section III — Carbohydrate Interpretation. 8 puzzles: anomers, mutarotation, glycosidic bonds, Fischer\u2194Haworth, polysaccharide identification.

Section IV — SDS & Microscale. SDS for 4 carbohydrate-lab materials (D-glucose, Tollens\' reagent, Benedict\'s reagent, conc. HCl) = 16 questions. Then 6 microscale sample-prep scenarios.

Prepare your lab notebook. Use the Example Report as your template.

Safety note: Tollens\' reagent must be FRESHLY PREPARED and used the same day. On standing, it forms silver fulminate (AgN(NH)₂) which is highly EXPLOSIVE on impact. Always destroy unused Tollens\' reagent by adding dilute HCl. Read the SDS carefully.

Simulation

Four interactive parts. Use the ↺ Reset Simulation button to clear all answers.

Eight monosaccharide identification cases. For each: (a) D vs L; (b) aldose vs ketose; (c) carbon count.

Score: 0 / 24

Six unknown sugar samples. Select a sample, click Run Tollens\' Test, observe the result, identify the sugar. Result hints appear after you answer.

Score: 0 / 6

Eight harder puzzles: anomers, mutarotation, glycosidic bonds, polysaccharide structure.

Score: 0 / 8

Round 1 — SDS interpretation

Four carbohydrate-lab materials. Each has 4 questions.

SDS score: 0 / 16

Round 2 — Microscale sample prep

Six sample-prep scenarios.

Microscale score: 0 / 6

Team Questions

Discuss with your team before answering.

Question 1 — D/L convention. A Fischer projection shows OH on the right at the highest-numbered chiral carbon. Is this a D or L sugar?

Question 2 — Anomer types. What is the difference between α-D-glucose and β-D-glucose?

Question 3 — Mutarotation. Why does the optical rotation of pure α-D-glucose change over time when dissolved in water?

Question 4 — Reducing sugar. Why is sucrose a non-reducing sugar but maltose is reducing?

Question 5 — Cellulose vs starch. Both are glucose polymers. Why can humans digest starch but not cellulose?

Question 6 — Tollens\' test. Why is Tollens\' reagent only freshly prepared and used immediately?

Example Lab Notebook Entry

Use the format below as a template.

Carbohydrates — Lab Notebook Entry

Submitted by: [Student Name]

Course: Organic Chemistry II · Section: 202-A · Date: May 11, 2026

Objective

To learn the systematic identification of monosaccharides (D/L, aldose/ketose, carbon count, ring form), the chemistry of cyclic hemiacetal formation and anomers, the principles of glycosidic bonds and disaccharides, and the diagnostic chemical tests (Tollens\', Benedict\'s) that distinguish reducing from non-reducing sugars.

Bench results (Section II) — Tollens\' test

Sample	Sugar	Type	Tollens\' result	Reducing?
1	D-Glucose	Aldohexose, free anomeric C	Silver mirror (positive)	YES
2	D-Fructose	Ketohexose, tautomerises	Silver mirror (positive, slightly slower)	YES (via tautomerization to glucose/mannose)
3	Sucrose	Glc-\u03b1,\u03b2-1,2-Fru, both anomeric C linked	No reaction	NO (non-reducing)
4	Lactose	Gal-\u03b2-1,4-Glc, free C1 on glucose	Silver mirror (positive)	YES
5	Maltose	Glc-\u03b1-1,4-Glc, free C1 on second glucose	Silver mirror (positive)	YES
6	D-Ribose	Aldopentose, free anomeric C	Silver mirror (positive)	YES

Discussion

Carbohydrates are polyhydroxy aldehydes or ketones. Their structural diversity comes from: chain length (3-7 carbons typically), aldose vs ketose, stereochemistry at each chiral carbon, ring size (pyranose vs furanose), and anomeric configuration (\u03b1 vs \u03b2). The systematic classification used in this lab \u2014 D/L, carbonyl type, carbon count \u2014 is essential for naming and predicting properties.

The Fischer projection elegantly shows monosaccharide stereochemistry. The convention assigns D/L based on the configuration at the highest-numbered chiral carbon (e.g., C5 for hexoses): if the OH points to the right in the Fischer projection, the sugar is D; if left, L. Almost all biological sugars are D \u2014 a remarkable conservation across all forms of life that suggests an early evolutionary preference. Note that D/L describes a SINGLE carbon\'s configuration and does NOT correlate directly with the direction of optical rotation (D-fructose, for example, is levorotatory).

In aqueous solution, sugars with 5+ carbons exist almost entirely in CYCLIC form (>99%) as hemiacetals. The C5 OH (for aldohexoses) or C4 OH (for aldopentoses) attacks the carbonyl C, forming a 6-membered (PYRANOSE) or 5-membered (FURANOSE) ring. The carbonyl C becomes a NEW STEREOCENTER \u2014 the anomeric carbon \u2014 and the new OH can be either down (\u03b1) or up (\u03b2) in the Haworth projection. The two anomers have different physical properties (specific rotation, melting point) but interconvert through the open-chain form (MUTAROTATION); equilibrium for D-glucose is ~36% \u03b1, ~64% \u03b2, with negligible open form. \u03b2-D-glucopyranose specifically has all substituents in EQUATORIAL positions in the chair conformation \u2014 the lowest-energy hexose conformer.

The instrument bench (Section II) demonstrated the reducing sugar test (Tollens\'). A reducing sugar has a free anomeric carbon that can open to expose an aldehyde, which reduces Ag⁺ to metallic Ag⁰ (the silver mirror). Glucose, fructose, lactose, maltose, and ribose are all reducing. Sucrose, by contrast, is NON-reducing because BOTH anomeric carbons are locked in the glycosidic bond (\u03b1-1,2 linkage between glucose C1 and fructose C2). Fructose is technically a ketose but tests positive for Tollens\' because under basic conditions it tautomerizes to glucose/mannose via the enediol intermediate. The Tollens\' silver mirror test is one of the most beautiful and dramatic chemical demonstrations: a freshly cleaned glass tube becomes mirror-coated as silver deposits.

For sample preparation, the key safety concern is Tollens\' reagent itself \u2014 if stored, it forms silver fulminate and silver azide which are EXPLOSIVE on impact. Standard lab practice: prepare Tollens\' fresh on the day of use; destroy any unused reagent immediately by adding dilute HCl (which precipitates AgCl, neutralising the explosive risk); wash glassware promptly. The silver mirror coating itself is safe but should be removed by dilute HNO₃ (not by harsh mechanical scrubbing, which can damage the glass).

Disaccharides and polysaccharides are formed by GLYCOSIDIC BONDS \u2014 the anomeric OH replaces the OH of another sugar, creating a stable acetal. The position of the bond (1,2; 1,4; 1,6) and the configuration (\u03b1 or \u03b2) determine the polymer\'s 3D structure. Cellulose (\u03b2-1,4-glucose) forms flat extended sheets (great structural strength but indigestible to mammals lacking cellulase enzymes); starch (\u03b1-1,4-glucose with \u03b1-1,6 branches) forms helical coils (digestible by amylase). Glycogen has more branching than starch \u2014 important because more branches means more chain ends for rapid hydrolysis when energy is needed quickly.

Conclusion

Carbohydrates exemplify how SUBTLE structural variations \u2014 a single epimer (glucose vs galactose), an \u03b1- vs \u03b2-glycosidic bond \u2014 can produce DRAMATICALLY different biological functions. The same monosaccharide unit (glucose) is the energy currency, the storage form (starch, glycogen), and the structural material (cellulose) of life on Earth, distinguished only by linkage chemistry. The next biomolecule labs (amino acids, lipids, nucleic acids) will continue this theme: chemical structure determines biological function.

References

1. Solomons, T. W. G.; Fryhle, C. B.; Snyder, S. A. Organic Chemistry, 12th ed., Wiley, 2016, Ch 22.
2. Lehninger, A. L.; Nelson, D. L.; Cox, M. M. Principles of Biochemistry, 7th ed., W.H. Freeman, 2017, Ch 7.
3. McMurry, J. Organic Chemistry, 9th ed., Cengage, 2016, Ch 25.
4. Berg, J. M.; Tymoczko, J. L.; Stryer, L. Biochemistry, 9th ed., W.H. Freeman, 2018, Ch 11.

Practice Questions

Work through each before peeking at the hint.

Practice 1 — Fischer to Haworth

Convert D-glucose Fischer projection to its Haworth projection (\u03b2-form). Where do the C2, C3, C4 OH groups point in the Haworth?

Hint: Substituents that point RIGHT in Fischer point DOWN in Haworth. C2 OH right \u2192 down. C3 OH left \u2192 up. C4 OH right \u2192 down. C5 OH right \u2192 D-sugar (CH₂OH up). For \u03b2-anomer: C1 OH up. So \u03b2-D-glucopyranose has: OH at C1 up; OH at C2 down; OH at C3 up; OH at C4 down; CH₂OH at C5 up.

Practice 2 — Identifying anomers

A hexose pyranose has the C1 OH in the equatorial position in its chair conformation. Is this an α or β anomer?

Hint: For D-glucopyranose: \u03b2-anomer has equatorial C1 OH; \u03b1-anomer has axial C1 OH. So equatorial = \u03b2. The \u03b2-form is more stable (all substituents equatorial), which is why \u03b2-D-glucose is the most stable hexose.

Practice 3 — Mutarotation kinetics

Pure \u03b1-D-glucose has [\u03b1] = +112\u00b0. Pure \u03b2-D-glucose has [\u03b1] = +18.7\u00b0. Equilibrium [\u03b1] = +52.7\u00b0. From these numbers, what is the α:β ratio at equilibrium?

Hint: Linear interpolation: f(\u03b1) \u00d7 112 + f(\u03b2) \u00d7 18.7 = 52.7, with f(\u03b1) + f(\u03b2) = 1. Solving: f(\u03b1)(112-18.7) = 52.7 - 18.7 \u2192 f(\u03b1) \u00d7 93.3 = 34 \u2192 f(\u03b1) = 0.36 (36%). So f(\u03b2) = 64%. The 36:64 ratio reflects the greater stability of the \u03b2-anomer (all-equatorial chair).

Practice 4 — Reducing test results

A student tests trehalose with Benedict\'s reagent and observes no colour change (clear blue). What does this tell you about trehalose\'s structure?

Hint: No reaction = NON-REDUCING sugar. Trehalose is a disaccharide with both anomeric carbons (C1 of each glucose) involved in the \u03b1,\u03b1-1,1 glycosidic bond. Neither sugar can open to expose an aldehyde. Like sucrose, trehalose is non-reducing. Found in insects, fungi, dehydration-tolerant organisms.

Practice 5 — Cellulose digestion

Why can termites digest cellulose but humans cannot, despite both being heterotrophs that derive energy from glucose?

Hint: Cellulose has \u03b2-1,4 glycosidic bonds, requiring CELLULASE enzymes. Humans don\'t produce cellulase. Termites have CELLULASE-PRODUCING SYMBIONTS (gut protozoa, bacteria) that hydrolyze cellulose to glucose for the host to absorb. Cows, goats etc. have similar microbial fermentation in the rumen. Without these symbionts, mammals can only digest \u03b1-1,4 (starch) and \u03b1-1,6 bonds of glucose polymers.

Practice 6 — Lactose intolerance

Lactose intolerance is caused by deficiency of which enzyme? What happens to undigested lactose?

Hint: LACTASE deficiency. Lactase normally hydrolyzes the \u03b2-1,4 glycosidic bond of lactose into galactose + glucose. Without lactase, lactose passes to the colon undigested, where bacteria ferment it producing gas (CO₂, H₂, methane), short-chain fatty acids, and lactate. Symptoms: bloating, gas, diarrhoea. Most adults worldwide have decreased lactase expression after weaning ("lactase non-persistence"); persistence is genetic, common in northern European populations.

Practice 7 — Sucrose hydrolysis

Sucrose is hydrolyzed by acid (or by sucrase enzyme) to give what products? How would you test for the products?

Hint: Sucrose + H₂O \u2192 D-glucose + D-fructose (called "invert sugar" or "inverted syrup" because the optical rotation reverses sign during hydrolysis). Both products are reducing sugars. After hydrolysis, the mixture would test POSITIVE for Tollens\'/Benedict\'s (sucrose itself is negative). The fructose can be detected by Seliwanoff\'s test (red color within 60 sec for ketoses).

Practice 8 — DNA vs RNA

DNA and RNA backbones differ only in one C-H/C-OH bond. Which carbon is involved, and why is DNA more chemically stable?

Hint: 2\u2032-position (C2 of the sugar). DNA = 2\u2032-deoxyribose (H at C2); RNA = ribose (OH at C2). The 2\u2032-OH of RNA can attack the adjacent phosphate diester bond intramolecularly, hydrolyzing the chain. This makes RNA chemically labile (half-life seconds-minutes in cells without protection). DNA, lacking the 2\u2032-OH, is much more stable (half-life years-millennia) and is therefore the preferred molecule for long-term genetic information storage.

Practice 9 — Glycogen branching

Glycogen has more branches (\u03b1-1,6) than starch. Why is this advantageous for animals?

Hint: More branches = more CHAIN ENDS = more sites for simultaneous enzymatic hydrolysis (by glycogen phosphorylase). When animals need glucose RAPIDLY (fight-or-flight, intense exercise), glycogen can be hydrolyzed fast. Plants, by contrast, have slower energy demands, so starch with fewer branches works fine. Glycogen also has a more compact, dense structure (better storage per volume).

Practice 10 — Combining technqiues

An unknown sugar gives a positive Tollens\' (silver mirror), positive Benedict\'s (red precipitate), and a positive Bial\'s (green colour). What is the carbon count and aldehyde/ketone status?

Hint: Tollens\' positive + Benedict\'s positive = REDUCING sugar (free anomeric C). Bial\'s test (green colour) is specific for PENTOSES (5-carbon sugars). So the unknown is an aldopentose: D-ribose or D-arabinose or D-xylose, etc. The aldopentoses give silver mirror and red Cu₂O ppt because their anomeric C is free; Bial\'s confirms the 5-carbon count specifically. Confirmation by NMR or HPLC would distinguish between the four aldopentoses.