MHT-CET Chemistry · Biomolecules

Carbohydrates: Classification, Structure and Reactions

Carbohydrates are polyhydroxy aldehydes or ketones (or what hydrolyses to them), classed by the number of sugar units; glucose is an aldohexose with four chiral carbons that closes into a six-membered pyranose ring between C-1 and C-5, fructose a ketohexose that closes into a five-membered furanose ring, and the reactions of the open-chain form prove each part of the structure.

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Why this matters

21 PYQs, one HARD. Fourteen are classification and structure — laevulose, aldohexose, the anomeric carbon, which carbons close the ring, hemiacetal versus hemiketal, the specific rotations, how many OH groups ribose has, which named sugar is a di-, tri- or tetrasaccharide; seven are the structure-proving reactions — bromine water for the aldehyde, acetic anhydride for the five OH, HI for the straight chain, saccharic acid, sorbitol, and the acetylation mass gain that fixes the number of OH groups. Two cards.

Concept 1 of 2

Classification, and the Structures of Glucose and Fructose

Intuition

Classify by units first: one sugar unit is a monosaccharide, two a disaccharide, three or four an oligosaccharide, many a polysaccharide. Then by the carbonyl: an aldose has –CHO, a ketose has C=O, and the carbon count gives triose, tetrose, pentose, hexose. Glucose is the aldohexose; fructose (laevulose, laevorotatory) is the ketohexose with the same formula. In water the open chain closes: the C-5 OH of glucose adds across the C-1 aldehyde to give a HEMIACETAL, a six-membered pyranose ring, and C-1 becomes a new stereocentre — the anomeric carbon — so there are two forms, alpha and beta. Fructose closes C-5 OH onto its C-2 ketone: a HEMIKETAL, a five-membered furanose ring.

Definition

By units: mono (glucose, fructose, galactose, ribose); di (sucrose, maltose, lactose); tri (raffinose = galactose + glucose + fructose); tetra (stachyose = two galactose + glucose + fructose); poly (starch, glycogen, cellulose).
By carbonyl and carbons: glucose = aldohexose, $\text{C}_6\text{H}_{12}\text{O}_6$ , 4 chiral carbons (C-2 to C-5); fructose (laevulose) = ketohexose, same formula; threose = aldotetrose, 2 chiral carbons; ribose = aldopentose, $\text{C}_5\text{H}_{10}\text{O}_5$ , 4 OH groups; glucose has 5 OH (one primary, at C-6).
Ring closure: glucose C-1 (aldehyde) + C-5 OH → six-membered pyranose, a hemiacetal; C-1 is the anomeric carbon; alpha and beta anomers differ only there. Fructose C-2 (ketone) + C-5 OH → five-membered furanose, a hemiketal.
Reducing or not: a free hemiacetal/hemiketal OH makes the sugar reducing (all monosaccharides, maltose, lactose). Sucrose is non-reducing — both anomeric carbons are tied in the glycosidic bond.
Specific rotation: glucose +52.7° (equilibrium), fructose −92.4°, sucrose +66.5°; hydrolysed sucrose (invert sugar) is laevorotatory, about −20°.

Glucose and fructose rings

\text{Glucose: C-1 (CHO)} + \text{C-5 OH} \to \text{pyranose (hemiacetal)};\quad \text{Fructose: C-2 (C=O)} + \text{C-5 OH} \to \text{furanose (hemiketal)}

Worked example

Erythrose is an aldotetrose. Give its molecular formula, the number of chiral carbons, the number of OH groups, and say which carbon becomes anomeric if it were to form a ring.

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 1BiomoleculesMODERATE

Which of following is NOT correct about fructose?

[Q69 · 11th May Shift 1 · 2024]

Hemiacetal for fructose, three chiral carbons for glucose

Glucose (aldehyde) gives a hemiACETAL and has FOUR chiral carbons; fructose (ketone) gives a hemiKETAL. Both wrong versions are offered as options in the same question.

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Concept 2 of 2

Reactions of Glucose: What Each One Proves

Intuition

Each reaction of the open-chain form is evidence for one feature. Bromine water, a mild oxidant, touches only the aldehyde: gluconic acid, so glucose has –CHO. Nitric acid oxidises both ends: saccharic acid, so there is a primary alcohol too. Five acetyl groups go on with acetic anhydride, so there are five OH. Prolonged HI reduces everything to n-hexane, so the six carbons are in a straight chain. Hydroxylamine gives an oxime and HCN a cyanohydrin (a carbonyl), and NaBH₄ reduces the aldehyde to sorbitol. What the open chain CANNOT explain is the existence of alpha and beta forms and mutarotation — those need the ring.

Definition

$\text{Br}_2$ water → gluconic acid (aldehyde → COOH): proves –CHO. Test for the aldehyde group; Tollens/Fehling only show a reducing sugar.
conc. $\text{HNO}_3$ → saccharic acid (glucaric acid): two COOH, four OH — proves the primary alcohol at C-6 as well.
$(\text{CH}_3\text{CO})_2\text{O}$ → glucose pentaacetate: five OH groups. Each OH acetylated adds 42 u ( $\text{OH} \to \text{OCOCH}_3$ ); a gain of 84 u means 2 OH → an aldotriose (glyceraldehyde).
HI, long heating → n-hexane: six carbons in a straight chain.
$\text{NH}_2\text{OH}$ → oxime; HCN → cyanohydrin; $\text{NaBH}_4$ → sorbitol (–CHO → –CH₂OH).
NOT explained by the open chain: alpha/beta anomers, mutarotation, no Schiff's test, no $\text{NaHSO}_3$ adduct, pentaacetate not reacting with $\text{NH}_2\text{OH}$ .

Mild and strong oxidation

\text{Glucose} \xrightarrow{\text{Br}_2/\text{H}_2\text{O}} \text{gluconic acid (1 COOH)};\quad \text{Glucose} \xrightarrow{\text{conc. HNO}_3} \text{saccharic acid (2 COOH)}

Worked example

A monosaccharide gains 126 u on complete acetylation. How many OH groups does it have, and is it a ketotetrose or an aldotetrose?

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 2BiomoleculesHARD

In which of the following carbohydrate, molecular mass increases 84 u after complete acetylation?

[Q74 · 14th May Shift 1 · 2024]

Counting the carbonyl carbon as an OH

An aldotriose has 3 carbons but only 2 OH — the CHO carbon carries none. 84 u is 2 × 42, so the answer is the aldotriose, not the tetrose.

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Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (2)

Classification, and the Structures of Glucose and Fructose
Glucose and fructose rings
$\text{Glucose: C-1 (CHO)} + \text{C-5 OH} \to \text{pyranose (hemiacetal)};\quad \text{Fructose: C-2 (C=O)} + \text{C-5 OH} \to \text{furanose (hemiketal)}$
Reactions of Glucose: What Each One Proves
Mild and strong oxidation
$\text{Glucose} \xrightarrow{\text{Br}_2/\text{H}_2\text{O}} \text{gluconic acid (1 COOH)};\quad \text{Glucose} \xrightarrow{\text{conc. HNO}_3} \text{saccharic acid (2 COOH)}$

Watch out for (2)

Hemiacetal for fructose, three chiral carbons for glucose
→ Classification, and the Structures of Glucose and Fructose
Counting the carbonyl carbon as an OH
→ Reactions of Glucose: What Each One Proves

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