ChemistryCBSE Class 12 · Biomolecules

Glucose: Structure and Reactions

Updated 15 October 2026 · Fact-checked

Glucose, $C_6H_{12}O_6$, is an aldohexose. Its reactions show a straight chain of six carbons (HI gives n-hexane), an aldehyde group (oxime, cyanohydrin, gluconic acid), five –OH groups (pentaacetate) and a primary alcohol group (saccharic acid). Because it does not give Schiff's or bisulphite tests, glucose is shown to exist mainly as a six-membered cyclic hemiacetal with α and β anomers.

Each reagent probes one part of OHC–(CHOH)₄–CH₂OH: aldehyde end, five –OH groups, CH₂OH end, and HI shows the whole six-carbon straight chain.

Illustration drawn for QuestionBuddy.

Understand Glucose: Structure and Reactions

Preparation. (1) From sucrose: boiling sucrose with dilute HCl or $H_{2} S O_{4}$ in alcoholic solution gives glucose and fructose in equal amounts: $C_{12} H_{22} O_{11} + H_{2} O H^{+} C_{6} H_{12} O_{6} (glucose) + C_{6} H_{12} O_{6} (fructose)$ . (2) From starch: commercially, starch is hydrolysed by boiling with dilute $H_{2} S O_{4}$ at 393 K under 2–3 atm: $(C_{6} H_{10} O_{5})_{n} + n H_{2} O \to n C_{6} H_{12} O_{6}$ .

Evidence for the open-chain structure.
1. *Molecular formula* $C_{6} H_{12} O_{6}$ from analysis and molar mass.
2. *Straight chain:* prolonged heating with HI gives n-hexane, so the six carbons are in a straight chain.
3. *Carbonyl group:* glucose forms an oxime with hydroxylamine and a cyanohydrin with HCN.
4. *Aldehyde (not ketone):* mild oxidation with bromine water gives a six-carbon carboxylic acid, gluconic acid, so the carbonyl group is an aldehyde (–CHO at the end of the chain).
5. *Five –OH groups:* acetylation with acetic anhydride gives glucose pentaacetate, so there are five –OH groups. Because glucose is stable, they are on different carbons.
6. *Primary alcohol group:* oxidation with nitric acid gives saccharic acid (a dicarboxylic acid, glucaric acid) with six carbons; gluconic acid gives the same product, so the other end must be a $- C H_{2} O H$ group.

So the open-chain formula is $O H C - (C H O H)_{4} - C H_{2} O H$ . Its full name is D-(+)-glucose: D describes the configuration of the lowest asymmetric carbon (C5, OH on the right, related to D-glyceraldehyde), and (+) shows it is dextrorotatory. D and (+) are unrelated: D is configuration, (+) is measured rotation.

Facts the open chain cannot explain.
1. Despite having an aldehyde group, glucose does not give Schiff's test and does not form an addition product with $N a H S O_{3}$ .
2. Glucose pentaacetate does not react with hydroxylamine, which means no free –CHO is present.
3. Glucose exists in two crystalline forms: α-glucose (m.p. 419 K) is obtained by crystallising from a concentrated solution at 303 K, and β-glucose (m.p. 423 K) from a hot saturated solution at 371 K.

Cyclic structure. The –OH at C5 adds to the –CHO group at C1, forming a six-membered ring hemiacetal (five carbons and one oxygen). C1 becomes a new chiral centre, so there are two forms: α-D-glucose and β-D-glucose, which differ only in the configuration at C1. Such isomers are called anomers, and C1 is the anomeric carbon. The six-membered ring is called a pyranose ring by analogy with pyran, and is shown by the Haworth projection. In solution, a tiny amount of open-chain form is in equilibrium with the rings, which is why glucose still forms an oxime and reduces Tollens' and Fehling's reagents.

Fructose for comparison. Fructose is a ketohexose (C2 is a ketone) and forms a five-membered furanose ring by addition of the C5–OH to the C2 carbonyl.

Glucose: Structure and Reactions: formulas and key facts

Open-chain structure: $C H O - (C H O H)_{4} - C H_{2} O H$; Aldohexose; D-(+)-glucose.
Reactions proving the structure: $H I, Δ n -hexane; N H_{2} O H oxime; H C N cyanohydrin; B r_{2} w a t er gluconic acid; (C H_{3} C O)_{2} O pentaacetate; H N O_{3} saccharic acid$; Chain, C=O, CHO, five OH, $C H_{2} O H$ .
Bromine water oxidation: $C_{6} H_{12} O_{6} + B r_{2} + H_{2} O \to C_{6} H_{12} O_{7} (gluconic acid) + 2 H B r$; Only the –CHO is oxidised.
Acetylation: $C_{6} H_{12} O_{6} + 5 (C H_{3} C O)_{2} O \to C_{6} H_{7} O (O C O C H_{3})_{5} + 5 C H_{3} C O O H$; Pentaacetate: five –OH groups.
Preparation from sucrose and starch: $C_{12} H_{22} O_{11} + H_{2} O H^{+} C_{6} H_{12} O_{6} + C_{6} H_{12} O_{6}; (C_{6} H_{10} O_{5})_{n} + n H_{2} O H^{+}, 393 K n C_{6} H_{12} O_{6}$; Starch hydrolysis is the commercial method.
Anomers: $α -D-glucose (m.p. 419 K) ⇌ open chain ⇌ β -D-glucose (m.p. 423 K)$; Differ only at C1 (anomeric carbon); pyranose rings.

How to solve Glucose: Structure and Reactions questions

Glucose questions are mostly 'what does this reaction show' or 'which facts need the cyclic structure'. Link every reagent to the functional group it reveals.

1List the reaction and its product.
2Name the structural feature it proves: HI → straight chain; $N H_{2} O H$ /HCN → carbonyl; $B r_{2}$ water → aldehyde; acetic anhydride → five OH; $H N O_{3}$ → primary alcohol.
3Write the equation where asked (oxime, cyanohydrin, gluconic acid, pentaacetate, saccharic acid).
4For 'limitations of the open-chain structure', give the three NCERT facts: no Schiff's/ $N a H S O_{3}$ test, pentaacetate does not react with $N H_{2} O H$ , two forms with different melting points.
5For the cyclic form, say: C5–OH adds to C1 aldehyde, six-membered pyranose ring, hemiacetal, anomeric carbon C1, α and β anomers.
6For D and (+), explain configuration versus rotation separately.
7For preparation, write both the sucrose and the starch equations with conditions.

Glucose: Structure and Reactions in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Hydrolysis of sucrose gives glucose and fructose

Suppose a sweet shop makes a smooth syrup by boiling a solution of 34.2 g of table sugar (sucrose) with a little lemon juice. The acid hydrolyses the sucrose completely.

Write the equation and calculate the masses of glucose and fructose formed. (Sucrose 342 g mol⁻¹, glucose and fructose 180 g mol⁻¹ each)

Try it first, then show the solutionSolution

$C_{12} H_{22} O_{11} + H_{2} O H^{+} C_{6} H_{12} O_{6} (glucose) + C_{6} H_{12} O_{6} (fructose)$ .
Moles of sucrose $= 34.2/342 = 0.10$ mol.
Each mole gives 1 mol glucose and 1 mol fructose: 0.10 mol each.
Mass of each $= 0.10 \times 180 = 18.0$ g.

Answer: 18.0 g of glucose and 18.0 g of fructose.

Memory hook: One sucrose, split by acid and water, makes one glucose and one fructose.

Tricks to remember Glucose: Structure and Reactions

Short memory aids. Say each one aloud once, then use it on a question the same day.

HI-HCN-Br-Ac-HNO₃ ladder
HI: chain; HCN/NH₂OH: C=O; Br₂ water: CHO; Ac₂O: five OH; HNO₃: CH₂OH.
Use it when: Matching reactions with structural evidence.
Three things the chain can't explain
No Schiff's (and no bisulphite), Pentaacetate ignores NH₂OH, Two melting points: 'No Pink, No Partner, Two forms'.
Use it when: Questions on why the cyclic structure was proposed.
α is 419, β is 423
Alphabetical and numerical order match: α (lower letter) melts lower.
Use it when: Recalling anomer melting points.

Glucose: Structure and Reactions: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Glucose and HI · 1 mark · MCQBoard level

On prolonged heating with HI, glucose gives:
(a) gluconic acid (b) n-hexane (c) saccharic acid (d) sorbitol

Try it first, then show the solutionSolution

HI removes all oxygen and reduces the chain to a hydrocarbon with the same carbon skeleton.

Answer: (b) n-Hexane.

Example 2: Bromine water and nitric acid · 2 marks · Very short answerBoard level

Name the products of the oxidation of glucose by (a) bromine water and (b) nitric acid. What does each show about the structure of glucose?

Try it first, then show the solutionSolution

(a) Gluconic acid, $H O O C - (C H O H)_{4} - C H_{2} O H$ : the carbonyl group is an aldehyde (oxidised to –COOH without losing carbon) (1 mark).
(b) Saccharic acid, $H O O C - (C H O H)_{4} - C O O H$ : the other end has a primary alcohol group (–CH₂OH) (1 mark).

Answer: (a) Gluconic acid: –CHO present. (b) Saccharic acid: –CH₂OH present.

Example 3: Where the open chain fails · 3 marks · Short answerHarder board level

Give three observations that cannot be explained by the open-chain structure of glucose.

Try it first, then show the solutionSolution

Although glucose has an aldehyde group, it does not give Schiff's test and does not form a hydrogensulphite addition product with $N a H S O_{3}$ (1 mark).
Glucose pentaacetate does not react with hydroxylamine, showing that there is no free –CHO group in it (1 mark).
Glucose exists in two crystalline forms, α (m.p. 419 K) and β (m.p. 423 K), which the open chain cannot account for (1 mark).

Answer: No Schiff's/ $N a H S O_{3}$ reaction; pentaacetate does not react with $N H_{2} O H$ ; two forms (α and β).

Example 4: Anomers and the pyranose ring · 3 marks · Short answerHarder board level

(a) How is the cyclic structure of glucose formed? (b) What are anomers? Name the anomeric carbon. (c) Why is the cyclic form called a pyranose structure? (d) What do D and (+) mean in D-(+)-glucose?

Try it first, then show the solutionSolution

(a) The –OH at C5 adds to the C1 aldehyde group, forming a six-membered cyclic hemiacetal (½ mark).
(b) α- and β-D-glucose differ only in the configuration of the –OH at C1; such isomers are anomers, and C1 is the anomeric carbon (1 mark).
(c) The ring has five carbons and one oxygen, like pyran (½ mark).
(d) D: configuration at C5 (–OH on the right in the Fischer projection, related to D-glyceraldehyde). (+): dextrorotatory, measured experimentally (1 mark).

Answer: (a) C5–OH + C1 CHO → hemiacetal ring. (b) Isomers differing at C1. (c) Six-membered ring like pyran. (d) D = configuration, (+) = rotation.

Example 5: Building the structure from evidence · 5 marks · Long answerHOTS / competency

A hexose X ( $C_{6} H_{12} O_{6}$ ) gives: (i) n-hexane with HI; (ii) a pentaacetate with acetic anhydride; (iii) a monocarboxylic acid $C_{6} H_{12} O_{7}$ with bromine water; (iv) a dicarboxylic acid with nitric acid. (a) Deduce the open-chain structure of X, linking each observation to a feature. (b) Write the equation for (ii). (c) Calculate the mass of pentaacetate from 1.8 g of X. ( $C_{6} H_{12} O_{6}$ = 180, pentaacetate = 390 g mol⁻¹) (d) Why would the pentaacetate not react with hydroxylamine?

Try it first, then show the solutionSolution

(a) (i) unbranched six-carbon chain; (ii) five –OH groups on different carbons; (iii) an aldehyde group oxidised to –COOH; (iv) a $- C H_{2} O H$ at the other end. So X is $O H C - (C H O H)_{4} - C H_{2} O H$ , an aldohexose like glucose (2 marks).
(b) $C_{6} H_{12} O_{6} + 5 (C H_{3} C O)_{2} O \to C_{6} H_{7} O (O C O C H_{3})_{5} + 5 C H_{3} C O O H$ (1 mark).
(c) Moles of X $= 1.8/180 = 0.010$ ; mass of pentaacetate $= 0.010 \times 390 = 3.9$ g (1 mark).
(d) In the pentaacetate the anomeric –OH at C1 is acetylated, so the ring cannot open to give a free –CHO group (1 mark).

Answer: (a) $O H C (C H O H)_{4} C H_{2} O H$ . (b) As shown. (c) 3.9 g. (d) The ring is locked; no free –CHO.

Where students lose marks in Glucose: Structure and Reactions

The slips examiners see most often, and what to write instead.

Mistake: Linking bromine water with the primary alcohol
Do this: Bromine water (mild) oxidises only –CHO → gluconic acid; nitric acid oxidises both ends → saccharic acid.
Usually costs: 1 mark.
Mistake: Saying α and β glucose differ at C5
Do this: They differ at C1, the anomeric carbon.
Usually costs: 1 mark.
Mistake: Confusing D with dextrorotatory
Do this: D is configuration; (+) is rotation. D-fructose is laevorotatory, for example.
Usually costs: ½ to 1 mark.
Mistake: Calling the glucose ring a furanose
Do this: Glucose: six-membered pyranose. Fructose: five-membered furanose.
Usually costs: ½ mark.

Exam tips

Learn the six structure-proving reactions as a table: reagent, product, inference.
When asked for 'reactions not explained by the open-chain structure', give exactly the three NCERT points.
Mention melting points (419 K and 423 K) and crystallisation conditions for α and β forms.
In equations, write glucose as $C_{6} H_{12} O_{6}$ or as the open-chain structure; both are accepted if clear.
Keep D/L and (+)/(−) separate in explanations.

Practice questions on Glucose: Structure and Reactions

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
The formation of a pentaacetate by glucose shows the presence of:
(a) an aldehyde group (b) five –OH groups (c) a ketone group (d) a straight chain
Hint
Acetylation of –OH groups.
Answer
(b).
Q2 · 2 marks · Very short answerBoard level
Write equations for the preparation of glucose from (a) sucrose and (b) starch.
Hint
Acid hydrolysis.
Answer
(a) $C_{12} H_{22} O_{11} + H_{2} O H^{+} C_{6} H_{12} O_{6} (glucose) + C_{6} H_{12} O_{6} (fructose)$ . (b) $(C_{6} H_{10} O_{5})_{n} + n H_{2} O H^{+}, 393 K, 2 - 3 atm n C_{6} H_{12} O_{6}$ .
Q3 · 2 marks · Very short answerHarder board level
Write the products of glucose with (a) hydroxylamine and (b) hydrogen cyanide, and state what these reactions show.
Hint
Both react with C=O.
Answer
(a) Glucose oxime, $H O C H_{2} (C H O H)_{4} C H = N O H$ . (b) Glucose cyanohydrin, $H O C H_{2} (C H O H)_{4} C H (O H) C N$ . Both show a carbonyl group.
Q4 · 1 mark · Assertion-reasonHarder board level
Assertion (A): Glucose does not give Schiff's test although it contains an aldehyde group.
Reason (R): In solution glucose exists mainly in a cyclic hemiacetal form with very little free aldehyde.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Cyclic structure.
Answer
(a).
Q5 · 2 marks · Very short answerHarder board level
Explain what 'D' and '(+)' mean in D-(+)-glucose. Are they related?
Hint
Configuration vs observed rotation.
Answer
D means the –OH on the lowest asymmetric carbon (C5) is on the right in the Fischer projection, matching D-glyceraldehyde. (+) means the compound rotates plane-polarised light to the right (dextrorotatory). They are unrelated: one is configuration, the other an experimental measurement.
Q6 · 3 marks · Short answerHarder board level
Write the structures of gluconic acid and saccharic acid, and explain why the formation of saccharic acid shows that glucose contains a primary alcohol group.
Hint
Saccharic acid is a six-carbon dicarboxylic acid.
Answer
Gluconic acid: $H O O C - (C H O H)_{4} - C H_{2} O H$ . Saccharic acid: $H O O C - (C H O H)_{4} - C O O H$ . Nitric acid oxidises both ends to –COOH without losing a carbon. Only a primary alcohol ( $- C H_{2} O H$ ) can be oxidised to –COOH without C–C cleavage, so C6 must carry a $C H_{2} O H$ group.
Q7 · 3 marks · Short answerHOTS / competency
Glucose reacts with hydroxylamine, but glucose pentaacetate does not. Explain using the cyclic structure, and state what this tells us about the –OH group at C1.
Hint
Ring opening needs a free C1–OH.
Answer
Glucose exists as a cyclic hemiacetal in equilibrium with a small amount of open-chain form; hydroxylamine reacts with that free –CHO, pulling the equilibrium. In the pentaacetate, the C1 hydroxyl (the hemiacetal –OH) is acetylated, so the ring cannot open and no –CHO is available. So C1 carries an –OH in cyclic glucose: it is the anomeric hemiacetal carbon.
Q8 · 4 marks · Case-basedHOTS / competency
Read and answer.
Suppose a student crystallises glucose in two ways: (P) from a concentrated solution at 303 K, and (Q) from a hot saturated solution at 371 K. The two samples melt at 419 K and 423 K respectively.
(a) Name the forms obtained in P and Q. (1 mark)
(b) In what way do they differ structurally? (1 mark)
(c) Why can the open-chain formula not explain two forms? (1 mark)
(d) Draw (in words) the ring present in both forms and name it. (1 mark)
Hint
Anomers.
Answer
(a) P: α-D-glucose (419 K). Q: β-D-glucose (423 K). (b) They differ only in the configuration of the –OH at C1 (the anomeric carbon). (c) In the open chain, C1 is a planar CHO carbon and not chiral, so no such pair of isomers is possible. (d) A six-membered ring of five carbon atoms and one oxygen atom (from C5–O bonding to C1): the pyranose ring.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Product of glucose with bromine water?Show answer
Gluconic acid.
2.Product of glucose with HNO₃?Show answer
Saccharic acid.
3.What is the anomeric carbon of glucose?Show answer
C1.
4.Melting point of β-glucose?Show answer
423 K.
5.Ring size of glucose?Show answer
Six-membered (pyranose).
6.What does pentaacetate formation show?Show answer
Five –OH groups.

Glucose: Structure and Reactions: frequently asked questions

What reactions prove the open-chain structure of glucose in Class 12?

HI gives n-hexane (straight chain), hydroxylamine and HCN react (carbonyl group), bromine water gives gluconic acid (aldehyde), acetic anhydride gives a pentaacetate (five OH) and nitric acid gives saccharic acid (primary alcohol).

Why does glucose not give Schiff's test?

In solution glucose exists almost entirely as a cyclic hemiacetal, so there is too little free aldehyde to react with Schiff's reagent.

What are anomers?

Cyclic forms of a sugar that differ only in the configuration at the anomeric carbon (C1 in glucose), such as α- and β-D-glucose.

What is the difference between pyranose and furanose?

A pyranose is a six-membered ring (five carbons and one oxygen) as in glucose; a furanose is a five-membered ring as in fructose.

How is glucose prepared commercially?

By hydrolysing starch with dilute sulphuric acid at 393 K under 2–3 atm pressure.