Chemistry · Biomolecules

Monosaccharides — Glucose & Fructose (Structure)

Glucose and fructose are the two monosaccharides every NEET aspirant must be able to draw, justify and contrast. Following NCERT Class XII §10.1.2 and NIOS §29.1.2, this note builds the open-chain Fischer structure of glucose from experimental evidence, exposes the reactions that the open chain cannot explain, and arrives at the cyclic pyranose and furanose forms. Structural questions from this subtopic recur almost every year, so the reasoning behind each structure matters as much as the structure itself.

What Is a Monosaccharide

Carbohydrates are defined chemically as optically active polyhydroxy aldehydes or ketones, or compounds that yield such units on hydrolysis. A monosaccharide is the simplest member: a carbohydrate that cannot be hydrolysed further into a simpler polyhydroxy aldehyde or ketone. About twenty monosaccharides occur in nature; glucose, fructose and ribose are the common examples named in NCERT.

Monosaccharides are classified two ways at once — by the number of carbon atoms and by the functional group present. A unit carrying an aldehyde group is an aldose; one carrying a keto group is a ketose. The carbon count is folded into the name, so a six-carbon aldose is an aldohexose and a six-carbon ketose is a ketohexose. Glucose is the prototype aldohexose, fructose the prototype ketohexose; both share the molecular formula $\ce{C6H12O6}$.

Carbon atomsGeneral termAldose exampleKetose example
3TrioseGlyceraldehyde (aldotriose)Dihydroxyacetone (ketotriose)
5PentoseRibose (aldopentose)Ketopentose
6HexoseGlucose (aldohexose)Fructose (ketohexose)

All monosaccharides — whether aldose or ketose — are reducing sugars: they reduce Fehling's solution and Tollens' reagent because a free or potentially free carbonyl group is available. This single fact ties together much of the chemistry that follows.

Preparation of Glucose

Glucose occurs freely in sweet fruits, honey and ripe grapes, and in combined form in starch and cellulose. Two laboratory and commercial routes are prescribed by NCERT, both hydrolyses of larger carbohydrates.

1. From sucrose (cane sugar)

Boiling sucrose with dilute $\ce{HCl}$ or $\ce{H2SO4}$ in alcoholic solution cleaves the glycosidic linkage, giving glucose and fructose in equal amounts:

$$\ce{\underset{Sucrose}{C12H22O11} + H2O ->[H+] \underset{Glucose}{C6H12O6} + \underset{Fructose}{C6H12O6}}$$

2. From starch

Commercially, glucose is obtained by hydrolysing starch with dilute $\ce{H2SO4}$ at 393 K under a pressure of 2–3 atm:

$$\ce{\underset{Starch}{(C6H10O5)_n} + nH2O ->[H+][393\,K] n\,\underset{Glucose}{C6H12O6}}$$

Glucose is an aldohexose, also known as dextrose, and is the monomer of starch and cellulose. It is probably the most abundant organic compound on earth.

Open-Chain Structure of Glucose: The Evidence

The straight-chain structure of glucose was not assumed — it was deduced, step by step, from chemical behaviour. NCERT lists six pieces of evidence; each one fixes a structural feature.

ObservationReagent / conditionConclusion
Molecular formulaAnalysis$\ce{C6H12O6}$
Forms n-hexaneProlonged heating with $\ce{HI}$Six carbons in an unbranched straight chain
Forms oxime; adds HCN to give cyanohydrin$\ce{NH2OH}$; $\ce{HCN}$A carbonyl group $\ce{(>C=O)}$ is present
Oxidised to gluconic acid (a six-carbon monocarboxylic acid)Mild oxidant, $\ce{Br2}$ waterThe carbonyl is specifically an aldehyde $\ce{(-CHO)}$
Forms a stable pentaacetateAcetic anhydrideFive $\ce{-OH}$ groups, each on a different carbon
Oxidised to saccharic (glucaric) acid, a dicarboxylic acidStrong oxidant, $\ce{HNO3}$A terminal primary alcohol $\ce{-CH2OH}$ is present

Read together, these tell a complete story. The $\ce{HI}$ reduction proves a continuous six-carbon backbone. The mild oxidation to gluconic acid converts only the carbonyl carbon, confirming it sits at the end of the chain as an aldehyde:

$$\ce{\underset{Glucose}{CHO-(CHOH)4-CH2OH} ->[Br2/H2O] \underset{Gluconic\ acid}{COOH-(CHOH)4-CH2OH}}$$

Strong oxidation with $\ce{HNO3}$ goes further, attacking the terminal $\ce{-CH2OH}$ as well, so both ends become $\ce{-COOH}$. Since gluconic acid too gives the same dicarboxylic acid, the primary alcohol must lie at the opposite end of the chain from the aldehyde:

$$\ce{\underset{Gluconic\ acid}{COOH-(CHOH)4-CH2OH} ->[HNO3] \underset{Saccharic\ (glucaric)\ acid}{COOH-(CHOH)4-COOH}}$$

NEET Trap

Mild vs strong oxidation give different acids

Examiners routinely test the oxidation products. Bromine water (mild) touches only the aldehyde, giving the monocarboxylic gluconic acid. Nitric acid (strong) oxidises both the $\ce{-CHO}$ and the terminal $\ce{-CH2OH}$, giving the dicarboxylic saccharic acid (also called glucaric acid).

Mild → gluconic (mono-acid). Strong → saccharic / glucaric (di-acid). Do not swap them.

The pentaacetate result is the subtle one: glucose has five hydroxyl groups, and because the acetate is stable, those five $\ce{-OH}$ groups must be on five different carbons — never two on the same carbon (a gem-diol would be unstable). With one aldehyde, five hydroxyls and a six-carbon straight chain, only the arrangement $\ce{CHO-(CHOH)4-CH2OH}$ fits.

Fischer Projection & D-Configuration

The skeleton is now fixed; the spatial arrangement of the four interior $\ce{-OH}$ groups was settled by Emil Fischer. The molecule is drawn as a Fischer projection with the most oxidised carbon (the $\ce{-CHO}$) at the top and the $\ce{-CH2OH}$ at the bottom. Horizontal bonds project toward the viewer; vertical bonds recede.

Figure 1
CHO C1 H OH C2 HO H C3 H OH C4 H OH C5 CH₂OH C6
Fischer projection of open-chain D-(+)-glucose. The OH on the lowest asymmetric carbon, C5 (red), lies on the right — this fixes the D-configuration. OH pattern down the chain: right, left, right, right.

The letters D and L describe relative configuration, not optical rotation. They are assigned by comparing the lowest asymmetric carbon (C5 in glucose) with the reference compound glyceraldehyde. In D-(+)-glyceraldehyde the $\ce{-OH}$ sits on the right; because glucose's C5 hydroxyl is also on the right, glucose is assigned the D-configuration. The accompanying $(+)$ records that glucose is experimentally dextrorotatory. The full, correct name is therefore D-(+)-glucose.

NEET Trap

D/L is not the same as (+)/(–)

The capital D/L describes configuration (which side the C5 –OH lies on, relative to glyceraldehyde). The sign (+)/(–) is the measured direction of optical rotation. They are independent: D-(+)-glucose is dextrorotatory, yet D-(–)-fructose is laevorotatory despite both being D-sugars.

D ≠ (+). The label fixes geometry; the sign is measured in a polarimeter.

Build the bigger picture

New to aldose/ketose and the mono–di–poly hierarchy? Start with Classification of Carbohydrates before locking in these structures.

Anomalies the Open Chain Cannot Explain

The open-chain Fischer structure accounts for most of glucose's chemistry, but three stubborn observations refuse to fit. These are the heart of the subtopic and a perennial NEET favourite.

AnomalyWhy the open chain fails
Glucose does not give Schiff's test and does not form the bisulphite adduct with $\ce{NaHSO3}$ — yet it supposedly has a $\ce{-CHO}$.A free aldehyde would respond to both. Its silence implies the $\ce{-CHO}$ is not actually free.
The pentaacetate of glucose does not react with $\ce{NH2OH}$ (no oxime).Confirms there is no free carbonyl group available for the hydroxylamine.
Glucose exists in two crystalline forms, α (m.p. 419 K) and β (m.p. 423 K), and shows mutarotation.A single open-chain structure permits only one form and one fixed rotation.

The 2,4-DNP and Schiff behaviours signal the same thing: at any instant essentially none of the glucose carries a genuine free aldehyde. Yet the molecule still adds $\ce{HCN}$ and reduces Tollens' reagent — because a tiny equilibrium concentration of open-chain aldehyde regenerates whenever the carbonyl-demanding reagent consumes it. The structure must therefore be one in which the aldehyde is normally masked but reversibly available.

Cyclic Pyranose Structure & Anomers

The resolution: one of glucose's own hydroxyl groups adds intramolecularly across the $\ce{-CHO}$ to form a cyclic hemiacetal. Specifically, the $\ce{-OH}$ on C5 attacks the C1 aldehyde, closing a six-membered ring containing one oxygen — a pyranose ring, named in analogy with pyran.

Ring closure makes C1 a new stereocentre. The two possible orientations of the newly formed C1 hydroxyl give two diastereomers, the α- and β-anomers; C1 is the anomeric carbon. In the Haworth projection the α-anomer has its C1 $\ce{-OH}$ pointing down (trans to the $\ce{-CH2OH}$), the β-anomer up (cis to the $\ce{-CH2OH}$). These two cyclic forms exist in dynamic equilibrium with a trace of the open chain.

Figure 2
O CH₂OH OH C1 α α-D-glucopyranose open-chain O CH₂OH OH C1 β β-D-glucopyranose
Haworth projections of the two glucose anomers. The anomeric C1 hydroxyl (red) points down in α-D-glucopyranose and up in β-D-glucopyranose; the two interconvert through the open-chain aldehyde, which is the basis of mutarotation.

This cyclic hemiacetal structure dissolves every anomaly at once. There is no permanently free $\ce{-CHO}$, so Schiff's test and the $\ce{NaHSO3}$ adduct both fail and the pentaacetate gives no oxime. Two distinct ring closures explain the two crystalline forms. And mutarotation — the slow drift of specific rotation of a freshly dissolved pure anomer to a fixed equilibrium value — follows because α and β interconvert through the open-chain aldehyde until equilibrium is reached.

NEET Trap

"No free CHO" does not mean "non-reducing"

Glucose has no free aldehyde at any instant, yet it is still a reducing sugar and reduces Tollens' and Fehling's reagents. The cyclic hemiacetal is in equilibrium with a small amount of open-chain aldehyde, which is continuously regenerated to react. Loss of reducing power happens only when the anomeric –OH is locked in a glycosidic bond, as in sucrose.

Hemiacetal at C1 ⇒ reducing. Glycoside at C1 ⇒ non-reducing.

Fructose: The Ketohexose

Fructose is the most important ketohexose. It is obtained, along with glucose, by hydrolysis of sucrose, and occurs naturally in fruits, honey and vegetables; in pure form it is used as a sweetener. Like glucose it has the molecular formula $\ce{C6H12O6}$, but its chemistry places the carbonyl differently.

Reaction studies show fructose contains a ketonic group at C2 and six carbons in a straight chain. It belongs to the D-series but is laevorotatory, so its correct name is D-(–)-fructose. The open-chain structure carries $\ce{-CH2OH}$ at C1, a keto group $\ce{(>C=O)}$ at C2, three $\ce{-CHOH}$ centres, and $\ce{-CH2OH}$ at C6.

Figure 3
CH₂OH C1 O C C2 (keto) HO H C3 H OH C4 H OH C5 CH₂OH C6
Open-chain Fischer projection of D-(–)-fructose, a ketohexose: the carbonyl is a keto group at C2 (red), not an aldehyde. The lowest asymmetric carbon C5 carries its OH on the right, placing fructose in the D-series.

Fructose also exists in cyclic form. Here the C5 $\ce{-OH}$ adds across the C2 keto group, closing a five-membered ring — a furanose, named in analogy with furan (one oxygen, four carbons). As with glucose, ring closure creates an anomeric centre at C2, giving α- and β-fructofuranose, represented by Haworth structures. In sucrose it is specifically β-D-fructofuranose that is bonded to α-D-glucose.

The take-home structural contrast is sharp: glucose closes to a six-membered pyranose through C1–C5, fructose to a five-membered furanose through C2–C5.

Glucose vs Fructose at a Glance

Most NEET questions on this subtopic reduce to a clean side-by-side comparison. The table below collects every contrast worth memorising.

FeatureGlucoseFructose
Molecular formula$\ce{C6H12O6}$$\ce{C6H12O6}$
ClassAldohexoseKetohexose
Carbonyl position$\ce{-CHO}$ at C1$\ce{>C=O}$ (keto) at C2
Configuration / rotationD-(+) — dextrorotatoryD-(–) — laevorotatory
Cyclic ringSix-membered pyranose (C1–C5)Five-membered furanose (C2–C5)
Anomeric carbonC1C2
Reducing sugar?YesYes
Common sourceGrapes, honey, starch hydrolysisFruits, honey, sucrose hydrolysis
Quick Recap

Lock these in before the exam

  • Glucose is prepared by acid hydrolysis of sucrose (with fructose) or of starch (393 K, 2–3 atm).
  • Open-chain evidence: $\ce{HI}$ → n-hexane (straight chain); $\ce{Br2}$ water → gluconic acid (aldehyde); $\ce{HNO3}$ → saccharic acid (primary $\ce{-OH}$); pentaacetate → five $\ce{-OH}$ on different carbons.
  • D-(+) refers to C5 configuration (right) and dextrorotation separately; D ≠ (+).
  • Open chain fails to explain: no Schiff's test, no $\ce{NaHSO3}$ adduct, no oxime from pentaacetate, two crystalline forms, mutarotation.
  • Cyclic hemiacetal (C5–OH onto C1–CHO) gives the pyranose ring; α/β anomers differ only at the anomeric C1.
  • Fructose: ketohexose, keto at C2, D-(–)-laevorotatory, five-membered furanose ring via C2–C5.

NEET PYQ Snapshot — Monosaccharides: Glucose & Fructose

Real NEET questions from the official papers, focused on monosaccharide structure and reactions.

NEET 2024 · Q.76

The reagents with which glucose does not react to give the corresponding tests/products are: A. Tollen's reagent · B. Schiff's reagent · C. HCN · D. $\ce{NH2OH}$ · E. $\ce{NaHSO3}$

  • (1) B and C
  • (2) A and D
  • (3) B and E
  • (4) E and D
Answer: (3) B and E

Because glucose exists as the cyclic hemiacetal with no free $\ce{-CHO}$, it fails the Schiff's test (B) and forms no hydrogensulphite adduct with $\ce{NaHSO3}$ (E). It still reduces Tollens' (A), adds HCN (C) and forms an oxime with $\ce{NH2OH}$ (D) via the small equilibrium amount of open-chain aldehyde.

NEET 2025 · Q.78

Sugar 'X': A. is found in honey · B. is a keto sugar · C. exists in α- and β-anomeric forms · D. is laevorotatory. 'X' is:

  • (1) Sucrose
  • (2) D-Glucose
  • (3) D-Fructose
  • (4) Maltose
Answer: (3) D-Fructose

All four clues point to fructose: it occurs in honey, is a ketohexose, cyclises to α/β furanose anomers, and is laevorotatory — D-(–)-fructose.

NEET 2020 · Q.165

Sucrose on hydrolysis gives:

  • (1) α-D-Glucose + β-D-Glucose
  • (2) α-D-Glucose + β-D-Fructose
  • (3) β-D-Fructose + β-D-Fructose
  • (4) β-D-Glucose + β-D-Fructose
Answer: (2) α-D-Glucose + β-D-Fructose

Acid (or enzymic) hydrolysis of sucrose cleaves the glycosidic linkage between α-D-glucose and β-D-fructose, releasing the two monosaccharides — the same preparation route used to obtain glucose and fructose.

FAQs — Monosaccharides: Glucose & Fructose

The structure questions students lose marks on most often.

Why does glucose not give the Schiff's test or form a hydrogensulphite adduct with NaHSO3 despite having an aldehyde group?

In the solid state and at equilibrium in solution, glucose exists overwhelmingly as the cyclic pyranose hemiacetal, in which the C1 aldehyde is locked up as a ring carbon bonded to two oxygens. Because there is no appreciable free –CHO group at any instant, glucose fails to give the Schiff's test and does not form the NaHSO3 hydrogensulphite addition product. These two negative tests are the classic evidence that the open-chain structure alone is incomplete.

What is mutarotation in glucose?

Mutarotation is the gradual change in specific rotation of a freshly prepared solution of a pure anomer of glucose until it reaches a fixed equilibrium value. Pure alpha-D-glucose (m.p. 419 K) and pure beta-D-glucose (m.p. 423 K) each dissolve to give an initial rotation that drifts to a common value, because the two cyclic anomers interconvert through the small amount of open-chain aldehyde form present at equilibrium.

What are anomers and which carbon is the anomeric carbon in glucose?

Anomers are the two cyclic hemiacetal forms (alpha and beta) of a monosaccharide that differ only in the configuration of the hydroxyl group at the anomeric carbon. In glucose the anomeric carbon is C1, the original aldehyde carbon, which becomes a new stereocentre on ring closure. In the alpha-anomer the C1 –OH points down (opposite to CH2OH) in the Haworth projection; in the beta-anomer it points up.

Why is glucose called D-(+)-glucose?

The 'D' denotes the relative configuration: the –OH on the lowest asymmetric carbon (C5) lies on the right in the Fischer projection, correlating with D-(+)-glyceraldehyde. The '(+)' is the experimentally observed dextrorotatory optical rotation. The D/L label describes configuration only and has no fixed relationship with the sign of optical rotation.

How does fructose differ structurally from glucose?

Both have the molecular formula C6H12O6. Glucose is an aldohexose with a –CHO group at C1, whereas fructose is a ketohexose with a keto (>C=O) group at C2. Both belong to the D-series, but glucose is dextrorotatory, D-(+)-glucose, while fructose is laevorotatory, D-(–)-fructose. On cyclisation glucose forms a six-membered pyranose ring whereas fructose forms a five-membered furanose ring.

What is the difference between pyranose and furanose forms?

A pyranose is a six-membered cyclic sugar (five carbons and one oxygen), named in analogy with pyran; glucose adopts this form using the C5 –OH. A furanose is a five-membered cyclic sugar (four carbons and one oxygen), named in analogy with furan; fructose adopts this form using the C5 –OH adding across the C2 keto group.