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Chemistry · Ch 10 — Biomolecules

Glucose

10.1.2.1

Glucose

Glucose is an aldohexose, and its everyday name in commerce and physiology is dextrose. It occurs freely in nature — sweet fruits, honey and ripe grapes are all rich in it — and also occurs in the combined form, since it is the monomeric building block of larger carbohydrates such as starch and cellulose. It is, in fact, probably the single most abundant organic compound on Earth.

Preparation of Glucose

1. From sucrose (cane sugar)

Boiling sucrose with dilute HCl or H2SO4\text{H}_2\text{SO}_4 in an alcoholic solution hydrolyses it into equal amounts of glucose and fructose:

C12H22O11Sucrose+H2O→H+C6H12O6Glucose+C6H12O6Fructose\underset{\text{Sucrose}}{\text{C}_{12}\text{H}_{22}\text{O}_{11}} + \text{H}_2\text{O} \xrightarrow{\text{H}^+} \underset{\text{Glucose}}{\text{C}_6\text{H}_{12}\text{O}_6} + \underset{\text{Fructose}}{\text{C}_6\text{H}_{12}\text{O}_6}

2. From starch

This is the commercial route. Starch (or cellulose) is boiled with dilute H2SO4\text{H}_2\text{SO}_4 at 393 K under a pressure of 2–3 atmospheres, hydrolysing it fully down to glucose:

(C6H10O5)nStarch or cellulose+n H2O→  393 K; 2-3 atm  H+n C6H12O6Glucose\underset{\text{Starch or cellulose}}{(\text{C}_6\text{H}_{10}\text{O}_5)_n} + n\,\text{H}_2\text{O} \xrightarrow[\;393\,\text{K;}\ 2\text{-}3\ \text{atm}\;]{\text{H}^+} \underset{\text{Glucose}}{n\,\text{C}_6\text{H}_{12}\text{O}_6}

Working Out the Open-Chain Structure

Glucose's structure was not guessed — it was pieced together from a sequence of chemical evidence, each experiment ruling in or ruling out a structural feature.

1. Molecular formula. Elemental analysis and molecular-mass determination fixed the formula as C6H12O6\text{C}_6\text{H}_{12}\text{O}_6.

2. A straight chain of six carbons. Prolonged heating of glucose with HI reduces it all the way down to n-hexane:

CHO∣(CHOH)4∣CH2OH→HI, ΔCH3–CH2–CH2–CH2–CH2–CH3(n-Hexane)\begin{array}{c}\text{CHO}\\ \vert\\ (\text{CHOH})_4\\ \vert\\ \text{CH}_2\text{OH}\end{array} \xrightarrow{\text{HI},\ \Delta} \underset{(n\text{-Hexane})}{\text{CH}_3\text{–}\text{CH}_2\text{–}\text{CH}_2\text{–}\text{CH}_2\text{–}\text{CH}_2\text{–}\text{CH}_3}

Since exhaustive reduction collapses the molecule into an unbranched six-carbon alkane, all six carbon atoms of glucose must be linked in a single straight chain — none of them are branch points.

3. A carbonyl group is present. Glucose reacts with hydroxylamine to form an oxime (CH=N-OH\text{CH}{=}\text{N-OH} in place of CHO\text{CHO}), and it also adds one molecule of hydrogen cyanide to form a cyanohydrin. Both reactions are classic tests for a carbonyl group, > ⁣C=O>\!\text{C}{=}\text{O}, so both confirm its presence in glucose.

Glucose reacting with hydroxylamine to give the oxime and adding hydrogen cyanide to give the cyanohydrin, the two reactions that confirm the presence of a carbonyl group in glucose.
Glucose reacting with hydroxylamine to give the oxime and adding hydrogen cyanide to give the cyanohydrin, the two reactions that confirm the presence of a carbonyl group in glucose.

Drawn by us to help you understand the concept clearly, and verified to make sure it's accurate. For exams, practice from your textbook's own diagram.

Redrawn from the NCERT page with the structures, printed labels (CHO, (CHOH)4, CH2OH, CH=N–OH, CH, CN, OH) and reagent placement exactly as the textbook prints them. Every element of this display was checked against the printed page during the sweep's blind-judge verification pass, so …

4. The carbonyl is specifically an aldehyde, at the chain end. A mild oxidising agent — bromine water — oxidises glucose to a six-carbon carboxylic acid called gluconic acid. A ketone would not be oxidised this easily or this cleanly by a mild oxidant; that the terminal carbon is converted straight to a −COOH-\text{COOH} group shows the carbonyl is an aldehydic (−CHO-\text{CHO}) group sitting at the very end of the chain.

Mild oxidation of glucose by bromine water converting the aldehydic –CHO group to –COOH, giving the six-carbon gluconic acid.
Mild oxidation of glucose by bromine water converting the aldehydic –CHO group to –COOH, giving the six-carbon gluconic acid.

Drawn by us to help you understand the concept clearly, and verified to make sure it's accurate. For exams, practice from your textbook's own diagram.

Redrawn from the NCERT page with the structures, printed labels (CHO, (CHOH)4, CH2OH, COOH) and reagent placement exactly as the textbook prints them. Every element of this display was checked against the printed page during the sweep's blind-judge verification pass, so w …

5. Five –OH groups, each on a different carbon. Treating glucose with acetic anhydride acetylates it fully, giving glucose pentaacetate — five acetate groups are introduced. Since this pentaacetate is a stable compound, the five −OH-\text{OH} groups being esterified must all sit on different carbon atoms (a single carbon cannot stably carry more than one −OH-\text{OH}).

Acetylation of glucose with acetic anhydride giving glucose pentaacetate, whose five acetate groups confirm five –OH groups on different carbon atoms.
Acetylation of glucose with acetic anhydride giving glucose pentaacetate, whose five acetate groups confirm five –OH groups on different carbon atoms.

Drawn by us to help you understand the concept clearly, and verified to make sure it's accurate. For exams, practice from your textbook's own diagram.

Redrawn from the NCERT page with the structures, printed labels (CHO, (CHOH)4, CH2OH, (CH–O–C–CH3)4, CH2–O–C–CH3) and reagent placement exactly as the textbook prints them. Every element of this display was checked against the printed page during the sweep's blind-judge verification pass, s …

6. A primary alcohol group at the other end. Oxidising glucose with the stronger oxidant nitric acid converts it to a dicarboxylic acid, saccharic acid — and oxidising gluconic acid (from step 4) under the same conditions gives the same saccharic acid. Since gluconic acid already has a −COOH-\text{COOH} at C1, the only way both routes can land on a dicarboxylic acid is if the −CH2OH-\text{CH}_2\text{OH} at the opposite end of the chain (C6) is a primary alcohol that nitric acid also oxidises to −COOH-\text{COOH}.

Nitric-acid oxidation converging from both glucose and gluconic acid onto the same dicarboxylic saccharic acid, proving a primary alcoholic –OH at C6 of glucose.
Nitric-acid oxidation converging from both glucose and gluconic acid onto the same dicarboxylic saccharic acid, proving a primary alcoholic –OH at C6 of glucose.

Drawn by us to help you understand the concept clearly, and verified to make sure it's accurate. For exams, practice from your textbook's own diagram.

Redrawn from the NCERT page with the structures, printed labels (CHO, (CHOH)4, CH2OH, COOH, Oxidation) and reagent placement exactly as the textbook prints them. Every element of this display was checked against the printed page during the sweep's blind-judge verification pass, so …

Put together, these six results fix glucose as a straight, unbranched six-carbon chain with an aldehyde group at C1, hydroxyl groups at C2 through C5, and a primary alcohol (−CH2OH-\text{CH}_2\text{OH}) at C6 — written in Fischer-projection style as:

CHO∣(CHOH)4∣CH2OHGlucose\underset{\text{Glucose}}{\begin{array}{c}\text{CHO}\\ \vert\\ (\text{CHOH})_4\\ \vert\\ \text{CH}_2\text{OH}\end{array}}

Fixing the Stereochemistry — Fischer's Configuration

Knowing the connectivity is not the same as knowing the exact three-dimensional arrangement of each −OH-\text{OH} group. That precise spatial arrangement — the configuration — was worked out by Emil Fischer from a wider study of glucose's properties, and is written (top carbon = C1, most oxidised carbon drawn at the top) as:

Fischer projections I, II and III showing the configurations of glucose, gluconic acid and saccharic acid with the –OH groups on the same sides in all three.
Fischer projections I, II and III showing the configurations of glucose, gluconic acid and saccharic acid with the –OH groups on the same sides in all three.

Drawn by us to help you understand the concept clearly, and verified to make sure it's accurate. For exams, practice from your textbook's own diagram.

Redrawn from the NCERT page with the structures, printed labels (CHO, OH, HO, CH2OH, COOH) and reagent placement exactly as the textbook prints them. Every element of this display was checked against the printed page during the sweep's blind-judge verification pass, so wh …

CarbonConfiguration I — GlucoseConfiguration II — Gluconic acidConfiguration III — Saccharic acid
C1CHOCOOHCOOH
C2H–OH (OH right)H–OH (OH right)H–OH (OH right)
C3HO–H (OH left)HO–H (OH left)HO–H (OH left)
C4H–OH (OH right)H–OH (OH right)H–OH (OH right)
C5H–OH (OH right)H–OH (OH right)H–OH (OH right)
C6CH₂OHCH₂OHCOOH

Configuration I is glucose itself; oxidising C1 alone (bromine water) gives configuration II, gluconic acid; oxidising both ends (nitric acid) gives configuration III, saccharic acid — exactly matching the oxidation evidence above.

D and L notation

Glucose's full, correct name is D-(+)-glucose. The two symbols carry independent meanings that are easy to conflate:

  • 'D' describes the molecule's configuration — the spatial arrangement of its groups.
  • '(+)' describes its optical activity — here, that it rotates plane-polarised light to the right (dextrorotatory).
Watch out

'D' and 'L' are not related to '(+)' and '(–)' (or to lowercase 'd' and 'l'). A compound with D-configuration can be dextrorotatory or laevorotatory; the two labels are independent and must be determined separately.

The D/L configurational label is assigned relative to glyceraldehyde, the simplest sugar with one asymmetric carbon, which exists as two enantiomers:

The two enantiomeric Fischer projections of glyceraldehyde, with the –OH on the right in the (+) isomer and on the left in the (–) isomer.
The two enantiomeric Fischer projections of glyceraldehyde, with the –OH on the right in the (+) isomer and on the left in the (–) isomer.

Drawn by us to help you understand the concept clearly, and verified to make sure it's accurate. For exams, practice from your textbook's own diagram.

Redrawn from the NCERT page with the structures, printed labels (CHO, OH, CH2OH, HO) and reagent placement exactly as the textbook prints them. Every element of this display was checked against the printed page during the sweep's blind-judge verification pass, so wh …

By convention, D-(+)-glyceraldehyde has its −OH-\text{OH} on the right when drawn with −CHO-\text{CHO} at the top. Any compound that can be chemically correlated to this D-(+)-glyceraldehyde is assigned D-configuration; any compound correlated instead to L-(–)-glyceraldehyde (OH on the left) is assigned L-configuration.

For a monosaccharide with several asymmetric carbons, only one of them is used for this comparison: the lowest asymmetric carbon — the one nearest the −CH2OH-\text{CH}_2\text{OH} end, farthest from the most-oxidised carbon (which is conventionally drawn at the top of the structure). In D-(+)-glucose, that lowest asymmetric carbon is C5, and its −OH-\text{OH} points to the right, exactly matching D-(+)-glyceraldehyde — so glucose is assigned D-configuration. (The configuration at the other asymmetric carbons, C2–C4, plays no role in this particular comparison.)

Fischer projections of D-(+)-glyceraldehyde and D-(+)-glucose with dashed boxes around the lowest asymmetric carbon and CH2OH end, the fragment compared to assign the D-configuration.
Fischer projections of D-(+)-glyceraldehyde and D-(+)-glucose with dashed boxes around the lowest asymmetric carbon and CH2OH end, the fragment compared to assign the D-configuration.

Drawn by us to help you understand the concept clearly, and verified to make sure it's accurate. For exams, practice from your textbook's own diagram.

Redrawn from the NCERT page with the structures, printed labels (CHO, OH, CH2OH, HO) and reagent placement exactly as the textbook prints them. Every element of this display was checked against the printed page during the sweep's blind-judge verification pass, so wh …

Why the Open-Chain Structure Was Not the Whole Story

The open-chain structure explains most of glucose's reactions, but three observations resisted it completely:

Note

What the open-chain (aldehyde) structure could not explain:

  1. Glucose fails Schiff's test, and it does not form the expected hydrogensulphite (NaHSO3\text{NaHSO}_3) addition product — both surprising for a molecule assumed to carry a free aldehyde group.
  2. Glucose pentaacetate does not react with hydroxylamine, implying there is no free −CHO-\text{CHO} group left in that form.
  3. Glucose is found to exist in two distinct crystalline forms, labelled α\alpha and β\beta: the α\alpha-form (m.p. 419 K) crystallises from a concentrated glucose solution at 303 K, while the β\beta-form (m.p. 423 K) crystallises from a hot, saturated aqueous solution at 371 K.

None of this fits a molecule that is simply an open chain ending in a free aldehyde. The resolution: one of the chain's own −OH-\text{OH} groups can add across the −CHO-\text{CHO} group intramolecularly, forming a cyclic hemiacetal. Specifically, the −OH-\text{OH} on C5 attacks the carbonyl carbon (C1), closing a six-membered ring. This single step explains both puzzles at once: the aldehyde carbon is now tied up in a hemiacetal linkage (so no free −CHO-\text{CHO} remains, resolving points 1 and 2), and the new hemiacetal carbon, C1, becomes a fresh stereocentre that can form in two different spatial arrangements — giving exactly the two crystalline forms of point 3. …

The equilibrium between alpha-D-(+)-glucose, the open-chain aldehyde form with the C5 –OH adding to the –CHO group, and beta-D-(+)-glucose, showing how the two cyclic hemiacetal anomers form.
The equilibrium between alpha-D-(+)-glucose, the open-chain aldehyde form with the C5 –OH adding to the –CHO group, and beta-D-(+)-glucose, showing how the two cyclic hemiacetal anomers form.

Drawn by us to help you understand the concept clearly, and verified to make sure it's accurate. For exams, practice from your textbook's own diagram.

Redrawn from the NCERT page with the structures, printed labels (OH, HO, CH2OH) and reagent placement exactly as the textbook prints them. Every element of this display was checked against the printed page during the sweep's blind-judge verification pass, so wha …

The pyran ring beside the Haworth structures of alpha-D-(+)-glucopyranose and beta-D-(+)-glucopyranose, which differ only in the orientation of the –OH at the anomeric carbon C1.
The pyran ring beside the Haworth structures of alpha-D-(+)-glucopyranose and beta-D-(+)-glucopyranose, which differ only in the orientation of the –OH at the anomeric carbon C1.

Drawn by us to help you understand the concept clearly, and verified to make sure it's accurate. For exams, practice from your textbook's own diagram.

Redrawn from the NCERT page with the structures and reagent placement exactly as the textbook prints them. Every element of this display was checked against the printed page during the sweep's blind-judge verification pass, so what you …