Carbohydrates

Q: Why is sucrose a non-reducing sugar?

In sucrose, C-1 of α-D-glucose is bonded to C-2 of β-D-fructose, so both anomeric carbons are locked in the glycosidic link. No free hemiacetal or hemiketal group is left to open into a carbonyl group, so sucrose does not reduce Tollens' or Fehling's reagent and does not mutarotate.

Q: What is the difference between starch and cellulose?

Both are polymers of D-glucose. Starch uses α-glucose: amylose is a linear α(1→4) chain and amylopectin is branched through α(1→6) links; it is the food store of plants. Cellulose uses β(1→4) links, giving straight chains that hydrogen-bond into fibres. Humans digest starch but not cellulose.

ChemistryBiomoleculesFor JEE aspirants

Carbohydrates are optically active polyhydroxy aldehydes or ketones, or compounds that give them on hydrolysis: glucose, fructose, sucrose, starch and cellulose are the key examples. These notes on carbohydrates cover classification, the D/L system, the open-chain and ring structures of glucose and fructose, anomers, mutarotation, osazones, the glycosidic links in disaccharides and the structures of starch and cellulose, with every structure drawn. The topic follows the NCERT Biomolecules chapter and is asked every year in NEET and JEE Main.

On this page1Definition2Classes3D/L, epimers4Glucose5Ring, anomers6Fructose7Disaccharides8Polysaccharides9Flowchart10Examples

Key Formulas - Quick Reference

General formula $C_{x} (H_{2} O)_{y}$ . Exceptions: rhamnose $C_{6} H_{12} O_{5}$ and 2-deoxyribose $C_{5} H_{10} O_{4}$ are carbohydrates but do not fit it.
★ Must learn Maximum optical isomers $= 2^{n}$ ( $n$ = chiral carbons): aldohexose $2^{4} = 16$ , 2-ketohexose $2^{3} = 8$ , cyclic aldohexopyranose $2^{5} = 32$ .
★ Must learn D/L: OH on the chiral carbon farthest from C=O (C-5 in hexoses) on the right in the Fischer projection means D. The sign (+)/(−) is measured, not predicted.
Glucose: HI/red P $\to$ n-hexane; $B r_{2}$ water $\to$ gluconic acid; $HN O_{3}$ $\to$ saccharic acid; $(C H_{3} CO)_{2} O$ $\to$ pentaacetate.
★ Must learn Osazone: sugar + 3 $C_{6} H_{5} NHN H_{2}$ . Only C-1 and C-2 react, so glucose, mannose and fructose give the same osazone.
Epimers differ at one chiral carbon (glucose/mannose at C-2, glucose/galactose at C-4); anomers differ only at C-1 of an aldose or C-2 of a ketose.
★ Must learn Mutarotation: $α$ ( $+ 11 2^{\circ}$ ) $⇌$ open chain $⇌$ $β$ ( $+ 1 9^{\circ}$ ); equilibrium $+ 52. 7^{\circ}$ with about 36% $α$ and 64% $β$ .
★ Must learn Invert sugar: $[α]_{D} = \frac{( + 52. 7 ^{\circ} ) + ( - 92. 4 ^{\circ} )}{2} \approx - 19. 9^{\circ}$ ; sucrose itself is $+ 66. 5^{\circ}$ .
★ Must learn Links: sucrose $α 1 \to β 2$ (non-reducing); maltose $α (1 \to 4)$ and lactose $β (1 \to 4)$ (reducing); starch $α (1 \to 4)$ with $α (1 \to 6)$ branches; cellulose $β (1 \to 4)$ .
$HI O_{4}$ on open-chain glucose: 5 $HI O_{4}$ used, 5 $HCOOH$ + 1 $HCHO$ formed.

1. What Are Carbohydrates?

1.1 The old definition and why it failed

Carbohydrates got their name because early chemists found that most of them fit the formula $C_{x} (H_{2} O)_{y}$ , so they looked like hydrates of carbon. Glucose, $C_{6} H_{12} O_{6}$ , can be written as $C_{6} (H_{2} O)_{6}$ .

The definition did not last, for two reasons:

Some true carbohydrates do not fit the formula. Rhamnose, $C_{6} H_{12} O_{5}$ , and 2-deoxyribose, $C_{5} H_{10} O_{4}$ , behave as carbohydrates but are not hydrates of carbon.
Some compounds that fit the formula are not carbohydrates. Formaldehyde, $HCHO$ ( $C (H_{2} O)$ ), and acetic acid, $C H_{3} COOH$ ( $C_{2} (H_{2} O)_{2}$ ), do not behave like carbohydrates.

1.2 The modern definition

Carbohydrates are optically active polyhydroxy aldehydes or polyhydroxy ketones, or compounds that give these units on hydrolysis.

The aldehyde and ketone groups are usually not present as free groups. Each combines with one of the molecule's own OH groups to form a cyclic hemiacetal (in aldoses) or hemiketal (in ketoses), as Section 4.5 shows. Carbohydrates are also called saccharides (Latin saccharum, sugar). Dihydroxyacetone, the simplest ketose, is the one carbohydrate with no chiral carbon, so it is optically inactive.

1.3 Where carbohydrates come from

Carbohydrates are widespread. They make up to 80% of the dry weight of plants: cellulose gives plants their structure, starch is their energy store, and sugars such as sucrose and glucose occur in fruits and sap. Glucose is an essential constituent of blood in higher animals and is stored as glycogen in the liver and muscles. Sugar units are also part of adenosine triphosphate (ATP), which stores and transfers energy in cells, and of the nucleic acids that control the making of enzymes and pass on genetic information.

Green plants make carbohydrates by photosynthesis. Chlorophyll absorbs sunlight, and the energy converts carbon dioxide and water into a carbohydrate and oxygen:

x C O_{2} + y H_{2} O sunlight, chlorophyll carbohydrate C_{x} (H_{2} O)_{y} + x O_{2}

For glucose, $x = y = 6$ :

6C O_{2} + 6 H_{2} O sunlight, chlorophyll glucose C_{6} H_{12} O_{6} + 6 O_{2}

2. Classification of Carbohydrates

Carbohydrates are grouped by whether they can be hydrolysed and, if they can, by how many monosaccharide units they give (Figure 1).

2.1 Monosaccharides, oligosaccharides and polysaccharides

Monosaccharides are the simplest carbohydrates. They cannot be hydrolysed into smaller carbohydrates. Glucose and fructose, both $C_{6} H_{12} O_{6}$ , are examples.

glucose or fructose C_{6} H_{12} O_{6} + H_{2} O H^{+} no reaction

Oligosaccharides (Greek oligo, few) give a definite number of monosaccharide units, from two to ten, on hydrolysis:

Disaccharides give two units. Sucrose and maltose are both $C_{12} H_{22} O_{11}$ .
Trisaccharides give three units. Raffinose, $C_{18} H_{32} O_{16}$ , is an example.
Tetrasaccharides give four units, and so on.

sucrose C_{12} H_{22} O_{11} + H_{2} O H^{+} glucose C_{6} H_{12} O_{6} + fructose C_{6} H_{12} O_{6}

maltose C_{12} H_{22} O_{11} + H_{2} O H^{+} glucose 2 C_{6} H_{12} O_{6}

raffinose C_{18} H_{32} O_{16} + 2 H_{2} O H^{+} glucose C_{6} H_{12} O_{6} + fructose C_{6} H_{12} O_{6} + galactose C_{6} H_{12} O_{6}

Polysaccharides have high molecular masses and give a large number of monosaccharide units on hydrolysis. Starch and cellulose are both $(C_{6} H_{10} O_{5})_{n}$ .

starch (C_{6} H_{10} O_{5})_{n} + n H_{2} O H^{+} glucose n C_{6} H_{12} O_{6}

2.2 Sugars, non-sugars and reducing sugars

Monosaccharides and oligosaccharides are crystalline solids, soluble in water and sweet. Together they are called sugars. Polysaccharides are amorphous, insoluble in water and tasteless, so they are called non-sugars.

Carbohydrates that reduce Fehling's solution and Tollens' reagent are reducing sugars; the rest are non-reducing sugars. All monosaccharides are reducing sugars, and so are all the common disaccharides except sucrose. A sugar is reducing when it has a free aldehyde or ketone group, or a free hemiacetal or hemiketal carbon that can open to give one.

Figure 1: Carbohydrates are classified by what they give on hydrolysis. Monosaccharides and oligosaccharides are sugars; polysaccharides are non-sugars, and sucrose is the one common non-reducing disaccharide.

Key idea

Sugars (mono- and oligosaccharides) are sweet and soluble; polysaccharides are not. A sugar reduces Tollens' reagent only while an anomeric carbon is still free.

3. Monosaccharides: Aldoses, Ketoses and Configuration

3.1 Aldoses and ketoses

Monosaccharides are the base of carbohydrate chemistry, because every carbohydrate either is a monosaccharide or gives monosaccharides on hydrolysis. They fall into two classes:

Aldoses contain an aldehyde group, $- CHO$ . General formula $CHO - (CHOH)_{n} - C H_{2} OH$ with $n = 1$ to $4$ .
Ketoses contain a ketone group, $> C = O$ . General formula $C H_{2} OH - CO - (CHOH)_{n} - C H_{2} OH$ with $n = 0$ to $3$ .

They are further named by the number of carbon atoms: trioses (3 C), tetroses (4 C), pentoses (5 C), hexoses (6 C) and so on. A complete name gives both the carbonyl type and the carbon count, such as aldohexose or ketohexose. Glucose is an aldohexose and fructose a ketohexose. 2,3-Dihydroxypropanal (glyceraldehyde) is an aldotriose and 1,3-dihydroxypropanone (dihydroxyacetone) a ketotriose.

Carbon atoms	Aldose (-CHO)	Ketose (C=O at C-2)
3 (triose)	aldotriose: glyceraldehyde	ketotriose: dihydroxyacetone
4 (tetrose)	aldotetroses: erythrose, threose	ketotetrose: erythrulose
5 (pentose)	aldopentoses: ribose, arabinose, xylose, lyxose	ketopentoses: ribulose, xylulose
6 (hexose)	aldohexoses: glucose, mannose, galactose	ketohexose: fructose

3.2 D and L configuration

The simplest carbohydrates are the trioses glyceraldehyde and dihydroxyacetone. Glyceraldehyde has one asymmetric (chiral) carbon, so it exists in two optically active forms, the D-form and the L-form. They are mirror images that cannot be superimposed, that is, enantiomers.

The two glyceraldehydes are the reference point for naming and drawing every other monosaccharide. In a Fischer projection the carbon chain is vertical, with the aldehyde or ketone group at the top. Horizontal bonds point towards the viewer and vertical bonds point away.

D/L rule (Rosanoff): if the OH on the chiral carbon farthest from the carbonyl group points to the right, as in (+)-glyceraldehyde, the sugar belongs to the D-family. If it points to the left, the sugar is in the L-family.

D and L describe configuration only. They have no fixed link with the sign of optical rotation, (+) or (−), which is measured with a polarimeter. Glucose is D-(+)-glucose, while natural fructose is D-(−)-fructose (Figure 2). Glucose is called dextrose because it occurs as the dextrorotatory isomer.

Figure 2: A sugar is D when the OH on its lowest chiral carbon points right, like D-glyceraldehyde. D-fructose is laevorotatory, so D/L and (+)/(

-

) are independent.

Exam Trick

Glucose code "R L R R": reading down from C-2 to C-5, the OH groups of D-glucose point Right, Left, Right, Right. Mannose flips the first one (L L R R), galactose flips the third (R L L R), and fructose keeps glucose's C-3 to C-5 pattern (L R R). Aldohexose names in order: "All altruists gladly make gum in gallon tanks" (allose, altrose, glucose, mannose, gulose, idose, galactose, talose).

3.3 How many stereoisomers?

The maximum number of optical isomers of a sugar depends on the number of chiral carbons, $n$ :

Maximum number of optical isomers = 2^{n}

Glyceraldehyde has one chiral carbon, so $2^{1} = 2$ , and there are indeed two glyceraldehydes.

Aldotetroses, $CHO - (CHOH)_{2} - C H_{2} OH$ , have two chiral carbons (C-2 and C-3), so $2^{2} = 4$ isomers exist, and all four have been made in the laboratory. D- and L-erythrose are mirror images (enantiomers): they rotate plane-polarised light by the same amount in opposite directions. An equal mixture of the two is a racemic mixture. It does not rotate plane-polarised light, but it can be separated into the dextrorotatory and laevorotatory isomers. The same holds for D- and L-threose. D-erythrose and L-threose, however, are not mirror images. Optical isomers that are not mirror images are diastereomers, and their rotations differ.

Aldotetrose	OH at C-2	OH at C-3	Relationship
D-(−)-erythrose	right	right	enantiomer of L-erythrose
L-(+)-erythrose	left	left	enantiomer of D-erythrose
D-(−)-threose	left	right	enantiomer of L-threose; C-2 epimer of D-erythrose
L-(+)-threose	right	left	enantiomer of D-threose

Aldopentoses have three chiral carbons, so $2^{3} = 8$ isomers: D-(−)- and L-(+)-ribose, D-(−)- and L-(+)-arabinose, D-(+)- and L-(−)-xylose, and D-(−)- and L-(+)-lyxose.

Aldohexoses have four chiral carbons, so $2^{4} = 16$ isomers: the D- and L-forms of allose, altrose, glucose, mannose, gulose, idose, galactose and talose. All sixteen have been synthesised, but only three occur in nature: D-glucose, D-mannose and D-galactose. None of the three is the mirror image of another, so they are diastereomers of each other.

Sugar type	Chiral C (open chain)	Isomers $2^{n}$	Members
Aldotriose	1	2	D- and L-glyceraldehyde
Aldotetrose	2	4	D/L-erythrose, D/L-threose
Aldopentose	3	8	D-(−)-ribose, D-(−)-arabinose, D-(+)-xylose, D-(−)-lyxose and their L-forms
Aldohexose	4	16	D and L forms of allose, altrose, glucose, mannose, gulose, idose, galactose, talose
2-Ketohexose	3	8	D-(−)-fructose and three others, with L-forms

3.4 Epimers

Epimers are a pair of diastereomers that differ in configuration at only one chiral carbon. The carbon is usually named; if it is not, it is taken to be C-2.

D-glucose and D-mannose differ only at C-2, the first chiral carbon, so they are C-2 epimers, or simply epimers. D-galactose is the C-4 epimer of D-glucose. Among the tetroses, D-threose is the C-2 epimer of D-erythrose (Figure 3).

Figure 3: Epimers differ at only one chiral carbon. Mannose is the C-2 epimer and galactose the C-4 epimer of glucose.

3.5 General characteristics of monosaccharides

All monosaccharides dissolve in water, because their many OH groups form hydrogen bonds with water molecules.
They taste sweet. On heating they char and give the smell of burnt sugar.
They are optically active because they contain chiral carbons (dihydroxyacetone is the exception).
Their chemistry comes from their OH groups and their carbonyl group, which is an aldehyde or a ketone.

Key idea

Read D or L from the chiral carbon farthest from C=O; measure (+) or (−). Epimers differ at exactly one chiral carbon.

Quick Recall: tap to check

Is D-fructose dextrorotatory?

No. It is laevorotatory, D-(−)-fructose. D describes the C-5 configuration only.

How many stereoisomers can an aldopentose have?

Three chiral carbons (C-2, C-3, C-4), so

2^{3} = 8

Glucose and galactose are epimers at which carbon?

C-4. Glucose and mannose are C-2 epimers.

4. Glucose

4.1 Occurrence and preparation

D-(+)-glucose, also called dextrose or grape sugar, is a monosaccharide with formula $C_{6} H_{12} O_{6}$ . It occurs free in ripe grapes, honey and many sweet fruits, and in human blood at about 0.1%. In the combined state it is present in many disaccharides and polysaccharides. It is optically active and dextrorotatory (+).

From sucrose. Sucrose is boiled with dilute HCl (or $H_{2} S O_{4}$ ) in alcoholic solution. Glucose is less soluble in ethanol, so it crystallises out when the mixture cools.

sucrose C_{12} H_{22} O_{11} + H_{2} O dil. HCl, ethanol glucose C_{6} H_{12} O_{6} + fructose C_{6} H_{12} O_{6}

From starch (commercial). Starch from cheap sources such as maize, potatoes and rice is boiled with dilute $H_{2} S O_{4}$ at 393 K under pressure.

starch (C_{6} H_{10} O_{5})_{n} + n H_{2} O H^{+}, 393 K 2-3 atm glucose n C_{6} H_{12} O_{6}

4.2 Open-chain structure: the evidence

Molecular formula. Elemental analysis and molar mass give $C_{6} H_{12} O_{6}$ .
A straight chain of six carbons. Heating with HI and red phosphorus gives 2-iodohexane on short heating and n-hexane on prolonged heating. Gluconic acid (below) reduced with excess HI gives n-hexanoic acid, $C H_{3} (C H_{2})_{4} COOH$ . So the six carbons form one straight chain.
$glucose OHC (CHOH)_{4} C H_{2} OH HI, red P prolonged heating n-hexane C H_{3} (C H_{2})_{4} C H_{3}$
A carbonyl group. Glucose forms a mono-oxime with hydroxylamine and adds one molecule of HCN to form a cyanohydrin.
$OHC (CHOH)_{4} C H_{2} OH + N H_{2} OH glucose oxime HON = CH (CHOH)_{4} C H_{2} OH + H_{2} O$
$OHC (CHOH)_{4} C H_{2} OH + HCN glucose cyanohydrin NCCH (OH) (CHOH)_{4} C H_{2} OH$
Five OH groups on five different carbons. Acetic anhydride (with pyridine) gives a pentaacetate (Ac = $C H_{3} CO$ ). Glucose is a stable compound, and a carbon carrying two OH groups would lose water, so the five OH groups sit on five different carbons.
$OHC (CHOH)_{4} C H_{2} OH + 5A c_{2} O pyridine glucose pentaacetate OHC (CHOAc)_{4} C H_{2} OAc + 5AcOH$
The carbonyl group is an aldehyde. Glucose reduces Tollens' reagent (silver mirror) and Fehling's solution (red $C u_{2} O$ ). Bromine water, a mild oxidant, converts it to gluconic acid with the same number of carbons. An aldehyde group is monovalent, so it must be at the end of the chain (C-1).
$OHC (CHOH)_{4} C H_{2} OH + [O] B r_{2} water gluconic acid HOOC (CHOH)_{4} C H_{2} OH$
A primary alcohol at C-6. Nitric acid oxidises both glucose and gluconic acid to the dicarboxylic acid saccharic (glucaric) acid. So the other end of the chain is a $- C H_{2} OH$ group.
$OHC (CHOH)_{4} C H_{2} OH + 3 [O] HN O_{3} saccharic acid HOOC (CHOH)_{4} COOH + H_{2} O$
Open-chain structure. Putting these facts together, and filling each carbon's remaining valencies with H atoms, gives the structure below, with four chiral carbons (*), C-2 to C-5.
$OHC - C * HOH - C * HOH - C * HOH - C * HOH - C H_{2} OH$

4.3 Configuration of D-(+)-glucose

With four chiral carbons, this structure can exist in $2^{4} = 16$ optically active forms, that is, eight pairs of enantiomers: the D- and L-forms of glucose, mannose, galactose, allose, altrose, gulose, idose and talose. All are known. Natural dextrorotatory glucose is only one of the sixteen. In it, the OH groups at C-2, C-4 and C-5 point right and the C-3 OH points left. Its C-5 OH is on the right, like (+)-glyceraldehyde, so it is D-(+)-glucose (Figure 2).

4.4 Objections to the open-chain structure

The open-chain structure explains most reactions of (+)-glucose, but not these facts:

Glucose does not restore the colour of Schiff's reagent and does not give the 2,4-DNP test, although it has an aldehyde group.
It does not form a bisulphite (hydrogensulphite, $NaHS O_{3}$ ) addition product or an aldehyde-ammonia compound.
It forms two isomeric pentaacetates, and neither reacts with hydroxylamine or other carbonyl reagents.
Two isomeric forms of glucose exist, and the specific rotation of each changes with time in solution (mutarotation). An open chain cannot explain this.
With methanol and dry HCl gas, glucose forms two isomeric methyl glucosides, each with only one $OC H_{3}$ group.

4.5 Cyclic structure and anomers

These facts show that the carbonyl group of glucose is not free. Aldehydes and ketones react with alcohols, first forming hemiacetals (and hemiketals), then acetals (and ketals):

RCHO + R^{'} OH ⇌ RCH (OH) O R^{'}

RCH (OH) O R^{'} + R^{'} OH dry HCl RCH (O R^{'})_{2} + H_{2} O

R_{2} C = O + R^{'} OH ⇌ R_{2} C (OH) O R^{'}

R_{2} C (OH) O R^{'} + R^{'} OH dry HCl R_{2} C (O R^{'})_{2} + H_{2} O

A monosaccharide has both groups in one molecule, so one of its own OH groups (usually C-4 or C-5 in aldohexoses, C-5 or C-6 in ketohexoses) adds to the carbonyl group, forming an intramolecular hemiacetal or hemiketal. The real structures are therefore rings of five or six atoms, one of them oxygen. A six-membered ring is called a pyranose (after pyran) and a five-membered ring a furanose (after furan). Free pentoses and hexoses exist mainly in the pyranose form, while in the combined state some sugars, such as ribose, 2-deoxyribose and fructose, exist as furanoses.

In D-glucose, the C-5 OH adds to the C-1 aldehyde. C-1 becomes a new chiral centre, so the ring can close in two ways, giving $α$ -D-glucose and $β$ -D-glucose (Figure 4). In the ring formula drawn in Fischer style, the C-1 OH is on the right in the $α$ -form and on the left in the $β$ -form of the D-series (and the reverse in the L-series).

Anomers are a pair of stereoisomers that differ in configuration only at C-1 of an aldose (or C-2 of a ketose). That carbon is called the anomeric carbon (or glycosidic carbon).

Epimersdiffer at one chiral carbon (any one)
glucose and mannose: C-2
glucose and galactose: C-4
separate sugars; do not interconvert

Anomersdiffer only at the anomeric carbon

α

- and

β

-D-glucose: C-1
exist only as rings
interconvert in water (mutarotation)

Haworth projection. Ring structures are drawn more clearly as Haworth projections:

Draw a hexagon with the ring oxygen at the upper right corner (at the back). The thick lower edge is nearest to you.
Groups on the right in the Fischer projection go below the ring; groups on the left go above.
For D-sugars the terminal $C H_{2} OH$ group (C-6) is placed above the ring.
In $α$ -D-glucopyranose the C-1 OH is below the ring (opposite to $C H_{2} OH$ ); in the $β$ -form it is above (same side).

Figure 4: The C-5 OH adds to the C-1 aldehyde, so C-1 becomes a new chiral centre (anomeric carbon). In water both anomers pass through the open chain and settle at

+ 52. 7^{\circ}

: mutarotation.

Exam Trick

"Right goes down." Any OH on the right in the Fischer projection is below the ring in the Haworth projection. For D-sugars, $β$ = "both up": the anomeric OH and the $C H_{2} OH$ are on the same side; in $α$ the anomeric OH points down, away from it.

History. In 1883 Tollens proposed a five-membered ring ( $γ$ -oxide ring) between C-1 and C-4, based on the stability of the $γ$ -lactones of aldonic acids. In 1926, Haworth, Hirst and co-workers showed that glucose has a six-membered ring (the amylene oxide or $δ$ -oxide ring) between C-1 and C-5: the pyranose ring used today.

4.6 How the ring explains the objections

No free aldehyde group. Weak carbonyl reagents (Schiff's reagent, $NaHS O_{3}$ , ammonia) cannot open the ring, so they do not react. Strong reagents such as HCN, $N H_{2} OH$ and $C_{6} H_{5} NHN H_{2}$ react with the small amount of open chain in equilibrium and pull the ring open, so the cyanohydrin, oxime and phenylhydrazone still form.
Two forms of everything. C-1 is a new chiral centre, so glucose, methyl glucoside and glucose pentaacetate each exist as an $α$ and a $β$ form. In the pentaacetates the C-1 OH is acetylated, the ring is locked, and no free CHO can form, so hydroxylamine does not react.
Mutarotation is the ring opening and closing again (Section 4.7).

4.7 Mutarotation

$α$ -D-Glucose and $β$ -D-glucose are separate crystalline compounds with different melting points and specific rotations:

Property	$α$ -D-glucose	$β$ -D-glucose
C-1 OH (Haworth)	below the ring	above the ring
Melting point	419 K (146 °C)	423 K (150 °C)
$[α]_{D}$ of fresh solution	$+ 11 2^{\circ}$	$+ 1 9^{\circ}$
Share at equilibrium	about 36%	about 64%

When either form dissolves in water and stands, the ring opens to a trace of the open-chain form and closes again, either the same way or the other way. The mixture settles at about 36% $α$ , 64% $β$ and a trace of open chain. The specific rotation of a fresh $α$ -D-glucose solution falls from $+ 11 2^{\circ}$ to $+ 52. 7^{\circ}$ , and that of $β$ -D-glucose rises from $+ 1 9^{\circ}$ to $+ 52. 7^{\circ}$ . Older texts describe the intermediate as the aldehydrol (the hydrate of the aldehyde, $- CH (OH)_{2}$ ); either way, the ring must open for the two forms to interconvert.

36%, [α]_{D} = + 11 2^{\circ} α -D-glucose ⇌ trace open-chain form ⇌ 64%, [α]_{D} = + 1 9^{\circ} β -D-glucose

Mutarotation is the change in the specific rotation of an optically active compound in solution with time, until it reaches a fixed equilibrium value. All reducing sugars mutarotate in water: monosaccharides and disaccharides such as maltose and lactose. Acids and bases speed it up.

Plotted against time, the two rotations close in on each other exponentially and meet at the same equilibrium value, which proves that both solutions end up as the same mixture.

Figure 5: Both anomers relax exponentially to the same value,

[α]_{D} = 52.7 + (112 - 52.7) e^{- t / τ}

for

α

and

52.7 - 33.7 e^{- t / τ}

for

β

. After

5 τ

the gap is under 1%, and the mixture is about 36%

α

and 64%

β

4.8 Reactions of glucose

Figure 6 collects the main reactions. Reduction, oxidation, the oxime, the cyanohydrin and the pentaacetate are written out in Section 4.2; the rest follow.

Figure 6: Each reaction of glucose proves one structural fact: a straight six-carbon chain, a C=O that is an aldehyde, five OH groups and a terminal

- C H_{2} OH

(a) Reduction. HI and red phosphorus give n-hexane. Sodium amalgam and water reduce the CHO group to $C H_{2} OH$ , giving the hexahydric alcohol sorbitol.

OHC (CHOH)_{4} C H_{2} OH + 2 [H] Na-Hg, H_{2} O sorbitol HOC H_{2} (CHOH)_{4} C H_{2} OH

(b) With water. Glucose dissolves to give a neutral solution.

(c) With Tollens' reagent and Fehling's solution. Glucose gives a silver mirror and a red precipitate of $C u_{2} O$ , and is itself oxidised to gluconate ( $R$ = $HOC H_{2} (CHOH)_{4} -$ ):

RCHO + 2 [Ag (N H_{3})_{2}]^{+} + 3O H^{-} warm RCO O^{-} + 2Ag ↓ + 4N H_{3} + 2 H_{2} O

RCHO + 2C u^{2 +} + 5O H^{-} warm RCO O^{-} + C u_{2} O ↓ + 3 H_{2} O

(d) With concentrated HCl. Glucose is degraded to laevulinic acid (4-oxopentanoic acid) and formic acid.

glucose C_{6} H_{12} O_{6} conc. HCl, Δ laevulinic acid C H_{3} COC H_{2} C H_{2} COOH + HCOOH + H_{2} O

(e) Glycoside formation. When a little dry HCl gas is passed into a solution of D-(+)-glucose in methanol, the anomeric OH is replaced by $OC H_{3}$ , giving a mixture of methyl $α$ -D-glucopyranoside and methyl $β$ -D-glucopyranoside (anomeric methyl acetals). Carbohydrate acetals are called glycosides, and an acetal of glucose is a glucoside. Only one $OC H_{3}$ group enters, which shows that the free CHO group has already become a ring $- CHOH -$ (hemiacetal) group. Glycosides do not reduce Tollens' reagent and do not mutarotate.

glucose C_{6} H_{12} O_{6} + C H_{3} OH dry HCl methyl glucosides C_{6} H_{11} O_{5} - OC H_{3} + H_{2} O

(f) With periodic acid. $HI O_{4}$ cleaves every C-C bond between two carbons that carry OH or C=O groups. Open-chain glucose uses 5 $HI O_{4}$ : C-1 to C-5 each become $HCOOH$ and C-6 becomes $HCHO$ .

OHC (CHOH)_{4} C H_{2} OH + 5HI O_{4} 5HCOOH + HCHO + 5HI O_{3}

Glucose or fructose? Tollens' reagent and Fehling's solution cannot tell them apart, because both sugars reduce them. Bromine water can: it oxidises aldoses (glucose) to aldonic acids but leaves ketoses (fructose) unchanged.

4.9 Osazone formation

Like ordinary aldehydes, glucose reacts with one molecule of phenylhydrazine to form a phenylhydrazone. Unlike ordinary aldehydes, it reacts with excess phenylhydrazine (three molecules) to form glucosazone. According to Fischer, this happens in three steps (Ph = $C_{6} H_{5}$ ; R = $(CHOH)_{3} C H_{2} OH$ , the part from C-3 to C-6, which does not change):

The first molecule condenses with the aldehyde group (C-1) to give glucose phenylhydrazone.
$glucose OHC - CHOH - R + PhNHN H_{2} phenylhydrazone PhNHN = CH - CHOH - R + H_{2} O$
The second molecule oxidises the neighbouring secondary alcohol group (C-2, $- CHOH -$ ) to a keto group and is itself reduced to aniline and ammonia.
$PhNHN = CH - CHOH - R + PhNHN H_{2} keto derivative PhNHN = CH - CO - R + aniline PhN H_{2} + N H_{3}$
The new keto group condenses with the third molecule to give glucosazone.
$PhNHN = CH - CO - R + PhNHN H_{2} glucosazone PhNHN = CH - C (= NNHPh) - R + H_{2} O$

Overall:

C_{6} H_{12} O_{6} + 3 C_{6} H_{5} NHN H_{2} glucosazone C_{18} H_{22} N_{4} O_{4} + C_{6} H_{5} N H_{2} + N H_{3} + 2 H_{2} O

Glucosazone is a yellow crystalline compound, sparingly soluble in water, with a sharp melting point, so osazones are used to identify sugars.

Osazone formation involves only C-1 and C-2. Glucose and fructose give the same osazone, so they must have the same configuration at C-3, C-4 and C-5. (+)-Mannose, another aldohexose, also gives the same osazone, so it too matches glucose at C-3, C-4 and C-5 and differs only at C-2. That makes glucose and mannose epimers (Figure 7).

Figure 7: Osazone formation destroys the difference at C-1 and C-2, so glucose, mannose and fructose (same C-3 to C-6) give the same osazone.

JEE Advanced

Kiliani-Fischer synthesis (lengthening the chain). HCN adds to the aldehyde group of an aldose. C-1 becomes a new stereocentre, so two epimeric cyanohydrins form; as diastereomers they are easy to separate. Each is hydrolysed to an aldonic acid, which is turned into its lactone and reduced with Na-Hg in acid solution ( $H_{2} S O_{4}$ ) to an aldose with one more carbon. D-Glyceraldehyde gives D-erythrose and D-threose. Both are D-sugars because the original stereocentre of D-glyceraldehyde is not touched.

Ruff degradation (shortening the chain). The reverse change, in two steps: bromine water oxidises the aldose to its aldonic acid, then the calcium salt of the acid is treated with $H_{2} O_{2}$ and a ferric salt (Fenton's reagent, e.g. $F e_{2} (S O_{4})_{3}$ ). C-1 leaves as $C O_{2}$ and C-2 becomes the new aldehyde group. D-Ribose gives D-erythrose, and D-glucose and D-mannose both give D-arabinose.

Key idea

Glucose is a pentahydroxy aldehyde that lives as a six-membered ring; C-1 is the anomeric carbon, and

α

and

β

interconvert only through the open chain.

5. Fructose

5.1 Structure

D-(−)-fructose (fruit sugar, laevulose), $C_{6} H_{12} O_{6}$ , occurs free in fruits and honey and, joined to glucose, in sucrose. It forms together with glucose when sucrose is hydrolysed. Fructose is a ketohexose: its keto group is at C-2, and C-3, C-4 and C-5 have the same configuration as in glucose. It belongs to the D-series because its C-5 OH is on the right, yet it is laevorotatory (Figure 2).

The evidence parallels glucose. Heating with HI and red phosphorus gives n-hexane (a straight six-carbon chain); acetic anhydride gives a pentaacetate (five OH groups); fructose forms an oxime and a cyanohydrin (a C=O group). Bromine water does not oxidise it, so the C=O is a ketone, not an aldehyde.

Glucose (aldohexose)CHO at C-1; D-(+),

+ 52. 7^{\circ}

pyranose ring: C-5 OH + C-1
decolourises bromine water

Fructose (ketohexose)C=O at C-2; D-(−),

- 92. 4^{\circ}

furanose in sucrose: C-5 OH + C-2
no reaction with bromine water

5.2 Cyclic structure

Like glucose, fructose has a cyclic structure. Because it contains a keto group, it forms an intramolecular hemiketal. When the C-5 OH adds to the C-2 keto group, C-2 becomes chiral and the $C H_{2} OH$ (C-1) and OH groups can take two arrangements around it: $α$ - and $β$ -D-fructofuranose. This five-membered ring (C-2, C-3, C-4, C-5 and O) is the furanose form found in sucrose and other combined sugars, and it is the form NCERT draws (Figure 8). When the C-6 OH closes the ring instead, the six-membered $α$ - and $β$ -D-fructopyranoses form; crystalline free fructose is $β$ -D-fructopyranose.

Figure 8: The C-5 OH of fructose adds to the C-2 keto group, giving a five-membered hemiketal ring (furanose). In the

β

form the C-2 OH is on the same side as the C-6

C H_{2} OH

Why a ketose reduces Tollens' reagent. Fructose is a ketone, yet it reduces Tollens' reagent and Fehling's solution. Both reagents are alkaline. In dilute alkali the $α$ -hydroxy ketone unit at C-1 and C-2 tautomerises through a 1,2-enediol, $- C (OH) = CH (OH)$ , to the aldoses glucose and mannose (the Lobry de Bruyn-van Ekenstein rearrangement), and these aldoses reduce the reagent. The same enediol interconverts glucose, mannose and fructose in base, which is why all three are linked so closely.

Quick Recall: tap to check

Which carbon of glucose is anomeric?

C-1, the former aldehyde carbon (C-2 in fructose).

Why does glucose not react with

NaHS O_{3}

Almost all of it is the ring (hemiacetal); a weak reagent cannot open it, so there is no free CHO.

How many molecules of phenylhydrazine make one osazone?

Three: one condenses at C-1, one oxidises C-2, one condenses at the new C=O.

6. Disaccharides

6.1 The glycosidic linkage

Disaccharides give two molecules of the same or different monosaccharides when hydrolysed by an acid or an enzyme. Their general formula is $C_{12} H_{22} O_{11}$ . The three most important are sucrose, maltose and lactose:

sucrose C_{12} H_{22} O_{11} + H_{2} O H^{+} or invertase D-glucose C_{6} H_{12} O_{6} + D-fructose C_{6} H_{12} O_{6}

maltose C_{12} H_{22} O_{11} + H_{2} O H^{+} or maltase D-glucose 2 C_{6} H_{12} O_{6}

lactose C_{12} H_{22} O_{11} + H_{2} O H^{+} or lactase D-glucose C_{6} H_{12} O_{6} + D-galactose C_{6} H_{12} O_{6}

A disaccharide can be seen as two monosaccharides condensed with the loss of one water molecule. One sugar acts as the hemiacetal and the other as the alcohol, and together they form an acetal. The C-O-C bridge between the two units is the glycosidic linkage.

6.2 Sucrose

Sucrose (cane sugar, table sugar) is abundant in sugar cane and sugar beet, which contain 14-20% of it by weight. It is formed from one molecule of $α$ -D-glucose and one of $β$ -D-fructose, joined through C-1 of glucose and C-2 of fructose: $α$ -D-glucopyranosyl- $(1 \to 2)$ - $β$ -D-fructofuranoside. Both reducing centres (the anomeric carbons) are used in the link, so, unlike maltose and lactose, sucrose is a non-reducing sugar and does not mutarotate (Figure 10).

Hydrolysis: invert sugar. Hot dilute acid or the enzyme invertase hydrolyses sucrose into D-glucose and D-fructose. Sucrose is dextrorotatory ( $+ 66. 5^{\circ}$ ). Glucose is also dextrorotatory ( $+ 52. 7^{\circ}$ ), but fructose has a larger negative rotation ( $- 92. 4^{\circ}$ ), so the equimolar product mixture is laevorotatory:

[α]_{D} (invert sugar) = \frac{( + 52. 7 ^{\circ} ) + ( - 92. 4 ^{\circ} )}{2} \approx - 19. 9^{\circ}

The sign of rotation changes from dextro (+) to laevo (−), so this hydrolysis is called the inversion of sucrose, and the equimolar mixture of glucose and fructose is invert sugar (invertose).

Figure 9: Hydrolysis moves the rotation from

+ 66. 5^{\circ}

(sucrose) to

\frac{52.7 - 92.4}{2} = - 19. 9^{\circ}

(invert sugar). Fructose's larger rotation wins, so the sign flips: the inversion of cane sugar.

Exam Trick

Average, do not subtract. Invert sugar is a 1 : 1 mixture, so its rotation is the average of $+ 52. 7^{\circ}$ and $- 92. 4^{\circ}$ . Fructose's bigger number wins, and the sign flips to minus.

6.3 Maltose

Maltose (malt sugar) contains two $α$ -D-glucose units. C-1 of one unit is linked to C-4 of the other: an $α (1 \to 4)$ glycosidic link. C-1 of the second unit is still a free hemiacetal, which opens to a free aldehyde in solution, so maltose is a reducing sugar and mutarotates. It forms when the enzyme diastase (amylase) breaks down starch, and maltase hydrolyses it to two glucose molecules.

6.4 Lactose and cellobiose

Lactose (milk sugar) is found in milk. It is made of $β$ -D-galactose and $β$ -D-glucose, linked from C-1 of galactose to C-4 of glucose: a $β (1 \to 4)$ link. The C-1 of the glucose unit is free, so lactose is a reducing sugar. The enzyme lactase hydrolyses it. Cellobiose, obtained by partial hydrolysis of cellulose, is two $β$ -D-glucose units joined by a $β (1 \to 4)$ link; it is also reducing.

Figure 10: A disaccharide is reducing only if one anomeric carbon stays free. Sucrose joins C-1 of glucose to C-2 of fructose, so no free hemiacetal remains.

Disaccharide	Units	Glycosidic link	Reducing?	Enzyme	Source
Sucrose	$α$ -D-glucose + $β$ -D-fructose	C-1 to C-2 ( $α 1 \to β 2$ )	No	invertase	cane, beet
Maltose	2 $α$ -D-glucose	$α (1 \to 4)$	Yes	maltase	malt, starch digestion
Lactose	$β$ -D-galactose + $β$ -D-glucose	$β (1 \to 4)$	Yes	lactase	milk
Cellobiose	2 $β$ -D-glucose	$β (1 \to 4)$	Yes	emulsin	partial hydrolysis of cellulose

Key idea

A disaccharide is reducing only if one anomeric carbon stays free: maltose and lactose yes, sucrose no.

7. Polysaccharides

Polysaccharides form when a large number (hundreds to thousands) of monosaccharide units join with the loss of water molecules. They are condensation polymers whose units are held together by glycosidic links. Natural polysaccharides usually contain about 100-3000 units. Important ones are cellulose, starch, glycogen, gums and pectins. The three most abundant, cellulose, starch and glycogen, are all built from the same monomer, glucose.

7.1 Starch

Starch, $(C_{6} H_{10} O_{5})_{n}$ , is the chief food reserve (storage polysaccharide) of plants. It is found mainly in seeds, roots and tubers; wheat, rice, potatoes, corn and bananas are rich sources. The value of $n$ varies from source to source. Starch is not a single compound but a mixture of two polymers of $α$ -D-glucose:

Amylose (15-20% of starch) is water-soluble. It is a long unbranched chain of 200-1000 $α$ -D-glucose units joined by $α (1 \to 4)$ links: C-1 of one unit to C-4 of the next.
Amylopectin (80-85%) is insoluble in water and highly branched. It is made of many short chains of 20-25 glucose units joined by $α (1 \to 4)$ links. The C-1 of the end unit of each chain is joined to C-6 of a unit in the next chain by an $α (1 \to 6)$ link, which creates the branches.

Hydrolysis. Hot dilute acids or enzymes break starch into dextrins of decreasing size, then maltose, and finally D-glucose. Starch does not reduce Tollens' reagent or Fehling's solution. With iodine it gives a deep blue colour, the standard test for starch.

Uses. Starch is a major food, eaten daily as potatoes, bread, cakes and rice. It is used to coat and size paper to improve its writing quality, and to treat textile fibres so that they can be woven into cloth without breaking. It is a raw material for dextrins, glucose and ethyl alcohol, and for starch nitrate, which is used as an explosive.

7.2 Glycogen

Glycogen is the storage carbohydrate of animals, often called animal starch. Its structure is like amylopectin but more highly branched. It is stored in the liver, muscles and brain, and enzymes break it down to glucose when the body needs energy. It is also found in yeast and fungi.

7.3 Cellulose

Cellulose is the chief component of wood and plant fibres; cotton is nearly pure cellulose. It occurs only in plants and is the most abundant organic compound in the plant kingdom. It is insoluble in water, tasteless and non-reducing, largely because of its very high molecular mass.

Structure. Complete acid hydrolysis of cellulose, $(C_{6} H_{10} O_{5})_{n}$ , gives D-(+)-glucose as the only monosaccharide. Hydrolysis of fully methylated cellulose gives mainly 2,3,6-tri-O-methyl-D-glucose, so each unit is linked through C-1 and C-4 (C-5 is used in the ring). Treatment with acetic anhydride and sulfuric acid gives octa-O-acetylcellobiose, which shows that every glycosidic link is $β$ , as in (+)-cellobiose. Cellulose is therefore a straight chain of $β$ -D-glucose units joined by $β (1 \to 4)$ links (Figure 11).

Size and fibres. Physical methods give molecular masses from 250 000 to 1 000 000 or more, so each molecule has at least 1500 glucose units; end-group analysis (by methylation and by periodic acid oxidation) gives chains of 1000 or more units. X-ray analysis and electron microscopy show the long chains lying side by side in bundles, held by hydrogen bonds between the many neighbouring OH groups. The bundles twist into rope-like strands, which group into the fibres we can see. In wood these cellulose "ropes" are embedded in lignin, a structure often compared to reinforced concrete. Humans cannot digest cellulose because they lack the enzyme that breaks $β (1 \to 4)$ links; cattle and other ruminants digest it with the help of microbes in their gut.

Figure 11: Starch is built from

α

-glucose (amylose linear, amylopectin branched), while cellulose uses

β (1 \to 4)

links that give straight, H-bonded fibres. Glycogen resembles amylopectin but is more branched.

Property	Amylose	Amylopectin	Glycogen	Cellulose
Monomer	$α$ -D-glucose	$α$ -D-glucose	$α$ -D-glucose	$β$ -D-glucose
Links	$α (1 \to 4)$	$α (1 \to 4)$ + $α (1 \to 6)$	$α (1 \to 4)$ + $α (1 \to 6)$	$β (1 \to 4)$
Shape	unbranched chain	branched (every 20-25 units)	more highly branched	straight chains in H-bonded fibres
In water	soluble	insoluble	disperses	insoluble
Role	plant food store	plant food store	animal food store (liver, muscles, brain)	plant cell walls

Starch

α

-D-glucose units

α (1 \to 4)

, branches

α (1 \to 6)

coils; blue-black with iodine
digested by amylase

Cellulose

β

-D-glucose units

β (1 \to 4)

only, unbranched
straight H-bonded fibres
not digested by humans

7.4 Cellulose derivatives

Acid breaks the glycosidic links of cellulose, so each molecule gives many molecules of D-(+)-glucose. Other reactions leave the chain almost intact and change the properties of this cheap, ready-made polymer, and they are of great industrial importance. Each glucose unit carries three free OH groups (at C-2, C-3 and C-6); these are the reaction sites.

Esters: nitrates. Like any alcohol, cellulose forms esters. A mixture of nitric and sulfuric acids converts it to cellulose nitrate, whose properties depend on how far nitration goes. Guncotton, used in smokeless powder, is almost completely nitrated and is often called cellulose trinitrate (three nitrate groups per glucose unit). Pyroxylin is less nitrated (two to three nitrate groups per unit) and is used for celluloid, collodion, photographic film and lacquers. Its drawbacks are that it burns easily and gives highly toxic nitrogen oxides.

one glucose unit C_{6} H_{7} O_{2} (OH)_{3} + 3HN O_{3} conc. H_{2} S O_{4} cellulose trinitrate unit C_{6} H_{7} O_{2} (ON O_{2})_{3} + 3 H_{2} O

Esters: acetate. With acetic anhydride, acetic acid and a little sulfuric acid, cellulose becomes the triacetate. Partial hydrolysis removes some acetate groups and breaks the chains into fragments of 200-300 units, giving commercial cellulose acetate (roughly a diacetate). It is less flammable than cellulose nitrate and has replaced it in many uses, such as safety photographic film. A solution of cellulose acetate in acetone forced through the fine holes of a spinneret loses its solvent and leaves solid filaments, whose threads make acetate rayon.

Xanthates: viscose rayon and cellophane. An alcohol treated with carbon disulfide and aqueous sodium hydroxide gives a xanthate; acid regenerates the starting materials:

RONa + C S_{2} a xanthate RO - C (= S) - SNa

RO - C (= S) - SNa + H^{+} ROH + C S_{2} + N a^{+}

Cellulose reacts in the same way to form cellulose xanthate, which dissolves in the alkali as a viscous colloidal dispersion called viscose. Forced through a spinneret into an acid bath, viscose regenerates cellulose as fine filaments that make rayon (viscose rayon); the viscose process is still the main route to rayon. Forced through a narrow slit, it gives thin sheets which, softened with glycerol, are used as protective film (cellophane).

Ethers. Industrially, cellulose is converted to ethers with alkyl chlorides (cheaper than alkyl sulfates) in the presence of alkali. Some breakdown of the long chains cannot be avoided. Methyl, ethyl and benzyl ethers of cellulose are used to make textiles, films and plastic objects.

Derivative	How it is made	Uses
Guncotton (cellulose trinitrate)	nearly complete nitration: about three nitrate groups per glucose unit	smokeless powder
Pyroxylin	two to three nitrate groups per unit	celluloid, collodion, photographic film, lacquers (flammable)
Cellulose acetate	$(C H_{3} CO)_{2} O$ + $C H_{3} COOH$ + a little $H_{2} S O_{4}$ gives the triacetate; partial hydrolysis gives roughly the diacetate	safety photographic film, acetate rayon
Viscose rayon and cellophane	$C S_{2}$ + NaOH give the xanthate (viscose); an acid bath regenerates cellulose	rayon fibres, transparent film
Cellulose ethers	alkyl chlorides + alkali	textiles, films, plastic objects

7.5 Gums and pectins

Plant gums and pectins are other natural polysaccharides. Pectins are the gelling agent in jams and jellies.

Quick Recall: tap to check

Which glycosidic link does cellulose have?

β (1 \to 4)

between

β

-D-glucose units.

Why does amylopectin not dissolve in water, while amylose does?

Amylopectin is a very large, highly branched molecule; amylose is a smaller unbranched chain.

Which polysaccharide is the animal food store?

Glycogen, stored in the liver, muscles and brain.

8. Importance of Carbohydrates

Carbohydrates are essential for life in plants and animals and form a major part of our food. Honey has long been used as an instant source of energy in Ayurveda.
They store energy: starch in plants and glycogen in animals.
Cellulose forms plant cell walls, and bacterial cell walls are also built from polysaccharides.
Wood (furniture), cotton fibre (cloth) and cellulose products are used every day.
They are raw materials for industries such as textiles, paper, lacquers and breweries.
Ribose and 2-deoxyribose are part of the nucleic acids RNA and DNA.

9. Identify a Carbohydrate: Flowchart and Mind Map

Most identification questions use the same four tests in a fixed order. The flowchart below is that order; the mind map after it puts every idea on this page on one screen.

Figure 12: Flowchart: four tests in a fixed order sort the common carbohydrates. Iodine first (starch), then Fehling's (only sucrose fails), then hydrolysis, and bromine water last, because it is the only test that separates an aldose from a ketose.

Figure 13: Mind map: the whole page on one screen. Two ideas run through every branch: which carbon is anomeric, and whether it is still free.

10. Solved Examples

Solved Example 1

The reaction of glucose with acetic anhydride and with Tollens' reagent suggests that glucose is
(A) a pentahydroxy aldehyde
(B) a hydrate of carbon
(C) a polyhydroxy ketone
(D) an alcohol

Solution:

Answer: (A). Acetic anhydride converts glucose into a pentaacetate (and 5 $C H_{3} COOH$ ), so glucose has five OH groups on five different carbons. Tollens' reagent gives a silver mirror, so the carbonyl group is an aldehyde. Together these make glucose a pentahydroxy aldehyde.

Solved Example 2

Glucose and fructose give the same osazone. One may therefore conclude that
(A) glucose and fructose have identical structures
(B) glucose and fructose are anomers
(C) the structures of glucose and fructose are mirror images
(D) the structures of glucose and fructose differ only at the carbon atoms that take part in osazone formation

Solution:

Answer: (D). Osazone formation involves only C-1 and C-2. Glucose (CHO at C-1, CHOH at C-2) and fructose ( $C H_{2} OH$ at C-1, C=O at C-2) differ only there; C-3 to C-6 are identical (Figure 7). They are functional isomers, not anomers or mirror images.

Solved Example 3

Open-chain glucose on oxidation with

HI O_{4}

gives
(A) 5

HCOOH

HCHO

(B) 4

HCOOH

+ 2

HCHO

HCOOH

+ 3

HCHO

(D) 2

HCOOH

+ 4

HCHO

Solution:

Answer: (A). Periodic acid cuts every C-C bond between carbons carrying OH or C=O. The CHO carbon and the four $CHOH$ carbons each become $HCOOH$ ; the terminal $C H_{2} OH$ becomes $HCHO$ . Five C-C bonds break, so 5 $HI O_{4}$ are used.

OHC (CHOH)_{4} C H_{2} OH + 5HI O_{4} 5HCOOH + HCHO + 5HI O_{3}

Solved Example 4

Which description matches the correct Haworth projection of sucrose?
(A)

α

-D-glucopyranose C-1 joined to C-4 of

β

-D-fructofuranose
(B)

α

-D-glucopyranose C-1 joined to C-2 of

β

-D-fructofuranose
(C)

β

-D-glucopyranose C-1 joined to C-2 of

α

-D-fructofuranose
(D)

α

-D-glucopyranose C-1 joined to C-2 of

β

-D-fructopyranose

Solution:

Answer: (B). Sucrose is $α$ -D-glucopyranosyl- $(1 \to 2)$ - $β$ -D-fructofuranoside: the $α$ -glucose unit is a six-membered ring, the $β$ -fructose unit a five-membered ring, and the link joins the two anomeric carbons (Figure 10).

Solved Example 5

When sucrose is heated with concentrated

HN O_{3}

, the product is
(A) sucrose nitrate
(B) oxalic acid
(C) formic acid
(D) citric acid

Solution:

Answer: (B). Hot concentrated nitric acid is a strong oxidant. It breaks the sugar chains down to oxalic acid, $(COOH)_{2}$ ; this was an old industrial route to oxalic acid from sugar.

Solved Example 6

The glycosidic linkages present in amylopectin are
(A)

α

-1,6 only
(B)

α

-1,4 only
(C)

β

-1,4
(D) both

α

-1,4 and

α

-1,6

Solution:

Answer: (D). Amylopectin is made of $α$ -D-glucose chains joined by $α (1 \to 4)$ links, and the chains are joined to each other by $α (1 \to 6)$ links at the branch points (Figure 11).

Solved Example 7

D-Allose (OH groups at C-2, C-3, C-4 and C-5 all on the right in the Fischer projection) is treated with

B r_{2}

water, and the calcium salt of the product with

H_{2} O_{2}

F e_{2} (S O_{4})_{3}

. In the aldopentose A formed, the OH groups at C-2, C-3 and C-4 point
(A) right, left, left
(B) right, right, right
(C) right, right, left
(D) left, left, left

Solution:

Answer: (B). This is a Ruff degradation. Bromine water turns CHO into COOH, and Fenton's reagent removes that carbon as $C O_{2}$ . Old C-2 becomes the new aldehyde, and the old C-3, C-4 and C-5 (all OH right) become the new C-2, C-3 and C-4. A is D-ribose, with all three OH groups on the right.

Solved Example 8

Write the hemiacetal formation for glucose.

Solution:

The C-1 aldehyde group reacts with one of glucose's own OH groups:

The C-5 OH gives a six-membered ring: D-glucopyranose, the more stable form and the one present in solution.
The C-4 OH gives a five-membered ring: D-glucofuranose, which is less stable.

In both cases C-1 becomes a new stereocentre, so each ring exists as an $α$ and a $β$ anomer (Figure 4).

Solved Example 9

Convert D-fructose,

HOC H_{2} - CO

- (CHOH)_{3} - C H_{2} OH

, into the

γ

-lactone of the corresponding hexonic acid (ring through O from C-1 to C-4).

Solution:

Dilute NaOH: fructose isomerises through the enediol to the aldoses glucose and mannose (Lobry de Bruyn-van Ekenstein rearrangement).
Bromine water: oxidises only the CHO group, giving the hexonic acids $HOOC (CHOH)_{4} C H_{2} OH$ (gluconic and mannonic acids). Nitric acid cannot be used here, because it also oxidises the C-6 $C H_{2} OH$ and gives saccharic acid.
Warm, losing water: the C-4 OH esterifies the C-1 COOH, closing the five-membered $γ$ -lactone.

hexonic acid HOOC (CHOH)_{4} C H_{2} OH warm lactone C_{6} H_{10} O_{6} + H_{2} O

Solved Example 10

Compound A,

C_{5} H_{10} O_{5}

, gives a tetraacetate with

(C H_{3} CO)_{2} O

. Oxidation of A with

B r_{2}

water gives an acid,

C_{5} H_{10} O_{6}

. Reduction of A with HI and red phosphorus gives 2-methylbutane. What is the structure of A?

Solution:

A tetraacetate means four OH groups. Oxidation by bromine water to an acid with one more O means a CHO group. Reduction to 2-methylbutane (not n-pentane) means the carbon skeleton is branched: one carbon hangs off the chain.

A is 2,3,4-trihydroxy-3-(hydroxymethyl)butanal (the branched sugar apiose):

OHC - CH (OH) - C (OH) (C H_{2} OH) - C H_{2} OH

Check: it has four OH groups; HI/P removes every O and turns CHO into $C H_{3}$ , giving $C H_{3} - C H_{2}$ $- CH (C H_{3}) - C H_{3}$ .

Solved Example 11

Compound A,

C_{5} H_{10} O_{4}

, is oxidised by

B r_{2}

water to an acid,

C_{5} H_{10} O_{5}

. A forms a triacetate and is reduced by HI to n-pentane. Oxidation of A with

HI O_{4}

gives, among other products, one molecule of

HCHO

and one of

HCOOH

. What are the possible structures of A, and how can you tell them apart?

Solution:

A is an aldehyde (bromine water) with three OH groups (triacetate) on a straight five-carbon chain (n-pentane). $C_{5} H_{10} O_{4}$ has one O fewer than an aldopentose, so A is a deoxy sugar. Moving the $C H_{2}$ group down the chain gives four candidates:

Structure	$HI O_{4}$ products	Fits?
I: $OHC - C H_{2} - CHOH$ $- CHOH - C H_{2} OH$	$OHC - C H_{2} - CHO$ + $HCOOH$ + $HCHO$	Yes
II: $OHC - CHOH - C H_{2}$ $- CHOH - C H_{2} OH$	$HCOOH$ + $OHC - C H_{2} - CHO$ + $HCHO$	Yes
III: $OHC - CHOH - CHOH$ $- C H_{2} - C H_{2} OH$	2 $HCOOH$ + $OHC - C H_{2} - C H_{2} OH$	No
IV: $OHC - CHOH - CHOH - CHOH - C H_{3}$	3 $HCOOH$ + $C H_{3} CHO$	No

Only I and II give one $HCHO$ and one $HCOOH$ . They are told apart with phenylhydrazine: I (C-2 is $C H_{2}$ ) forms only a phenylhydrazone, while II (C-2 carries OH) forms an osazone.

Solved Example 12

Starch is a polymer of
(A) fructose
(B) glucose
(C) lactose
(D) none of these

Solution:

Answer: (B). Starch is a homopolysaccharide of $α$ -D-glucose: its two components, amylose and amylopectin, both give only glucose on hydrolysis.

Solved Example 13

The commonest disaccharide has the molecular formula
(A)

C_{10} H_{18} O_{9}

(B)

C_{10} H_{20} O_{10}

(C)

C_{11} H_{22} O_{11}

(D)

C_{12} H_{22} O_{11}

Solution:

Answer: (D). The commonest disaccharide is sucrose, $C_{12} H_{22} O_{11}$ : two hexoses ( $2 \times$ $C_{6} H_{12} O_{6}$ ) minus one $H_{2} O$ .

Solved Example 14

Sugars are identified by their osazones. Which pair gives identical osazones?
(A) glucose and lactose
(B) glucose and fructose
(C) glucose and arabinose
(D) glucose and maltose

Solution:

Answer: (B). Glucose and fructose differ only at C-1 and C-2, the two carbons that react with phenylhydrazine, so both give glucosazone.

Solved Example 15

Cane sugar on hydrolysis yields
(A) glucose and maltose
(B) glucose and lactose
(C) glucose and fructose
(D) only glucose

Solution:

Answer: (C). Cane sugar is sucrose, which gives equal amounts of D-glucose and D-fructose. The process is the inversion of cane sugar.

Solved Example 16

Sucrose on hydrolysis yields a mixture that is
(A) optically inactive
(B) dextrorotatory
(C) laevorotatory
(D) racemic

Solution:

Answer: (C). The mixture is equimolar D-(+)-glucose ( $+ 52. 7^{\circ}$ ) and D-(−)-fructose ( $- 92. 4^{\circ}$ ). Fructose has the larger rotation, so the mixture is laevorotatory, about $- 19. 9^{\circ}$ . It is not racemic, because the two sugars are not enantiomers.

Solved Example 17

Fresh solutions of

α

-D-glucose (

+ 11 2^{\circ}

) and

β

-D-glucose (

+ 1 9^{\circ}

) both reach

+ 52. 7^{\circ}

. Calculate the percentage of the

α

-form at equilibrium, ignoring the open chain.

Solution:

Let the fraction of $α$ be $x$ . The rotation of the mixture is the weighted average:

112 x + 19 (1 - x) = 52.7 \Rightarrow 93 x = 33.7 \Rightarrow x = 0.362

Answer: about 36% $α$ and 64% $β$ . The $β$ -form dominates because all its large groups, including the C-1 OH, are equatorial in the chair form.

Solved Example 18

How many stereoisomers are possible for (i) an aldopentose, (ii) a 2-ketohexose and (iii) the pyranose (ring) form of an aldohexose?

Solution:

(i) Aldopentose: C-2, C-3, C-4 are chiral, so $2^{3} = 8$ .
(ii) 2-Ketohexose: C-3, C-4, C-5 are chiral (C-2 is the C=O), so $2^{3} = 8$ .
(iii) Aldohexopyranose: ring closure makes C-1 chiral too, so C-1 to C-5 give $2^{5} = 32$ .

Solved Example 19

Which of these are reducing sugars: glucose, fructose, sucrose, maltose, lactose, starch? Give the reason.

Solution:

Reducing: glucose, fructose, maltose and lactose. Each has a free aldehyde or ketone group, or a free hemiacetal carbon that opens in solution.

Non-reducing: sucrose, whose two anomeric carbons are used in the glycosidic link, and starch, a huge molecule whose few free chain ends give no practical test.

Solved Example 20

How many moles of

HI O_{4}

are used, and which one-carbon products form, when methyl

α

-D-glucopyranoside and methyl

α

-D-glucofuranoside are oxidised? How does this prove the ring size of glucose?

Solution:

Pyranoside (ring O between C-1 and C-5): free OH at C-2, C-3, C-4 and C-6. Vicinal pairs C-2/C-3 and C-3/C-4 are cut, so 2 $HI O_{4}$ are used and the middle carbon, C-3, leaves as 1 $HCOOH$ . No $HCHO$ forms.

Furanoside (ring O between C-1 and C-4): free OH at C-2, C-3, C-5 and C-6. The pairs C-2/C-3 and C-5/C-6 are cut, so again 2 $HI O_{4}$ , but C-6 leaves as 1 $HCHO$ and no $HCOOH$ forms.

Methyl glucoside actually gives formic acid and no formaldehyde, which is how the six-membered (pyranose) ring of glucose was proved.

Solved Example 21

A carbohydrate X is not hydrolysed by dilute acid and reduces Tollens' reagent, but it does not decolourise bromine water. X is
(A) glucose
(B) fructose
(C) sucrose
(D) maltose

Solution:

Answer: (B). Not hydrolysed, so X is a monosaccharide. It reduces Tollens' reagent (fructose does, through the enediol in alkali), but bromine water oxidises only aldoses, so X is the ketose fructose. Glucose would decolourise bromine water; sucrose and maltose are hydrolysed.

Solved Example 22

A fresh solution of

α

-D-glucose reads

+ 11 2^{\circ}

. Some time later it reads

+ 82.3 5^{\circ}

. What fraction of the change to equilibrium is complete, and how many relaxation times

τ

have passed?

Solution:

The total change is $112 - 52.7 = 59. 3^{\circ}$ . The gap still left is $82.35 - 52.7 = 29.6 5^{\circ}$ , exactly half of 59.3. So the change is 50% complete.

The gap falls as $e^{- t / τ}$ (Figure 5), so $e^{- t / τ} = 0.5$ and $t = τ ln 2 \approx 0.69 τ$ .

Solved Example 23

Which of the following is non-reducing and gives only one monosaccharide on complete hydrolysis?
(A) maltose
(B) lactose
(C) starch
(D) sucrose

Solution:

Answer: (C). Starch gives only D-glucose and is non-reducing (a huge molecule with almost no free chain ends). Sucrose is non-reducing too, but gives two sugars; maltose gives only glucose but is reducing; lactose is reducing and gives glucose and galactose.

Practice Questions

A disaccharide does not reduce Fehling's reagent and does not mutarotate. It is hydrolysed by maltase (an $α$ -glucosidase) and by invertase to D-glucose and D-fructose. Methylation and hydrolysis give 2,3,4,6-tetra-O-methyl-D-glucopyranose and a tetramethyl-D-fructose. Give its structure.Answer: Sucrose, $α$ -D-glucopyranosyl- $(1 \to 2)$ - $β$ -D-fructofuranoside (Figure 10).
Glucose forms an oxime but glucose pentaacetate does not. Explain.Answer: In the pentaacetate the C-1 OH is acetylated, so the ring cannot open and no free CHO is available to react with $N H_{2} OH$ .
Fructose reduces Tollens' reagent because of: (A) its ketonic group (B) the alkali ( $N H_{4} OH$ ) in Tollens' reagent (C) its rearrangement into a mixture of glucose and mannose (D) both B and CAnswer: (D). The alkali isomerises fructose (through an enediol) into glucose and mannose, which reduce the reagent.
$C_{4} H_{8} C l_{2}$ (A) on hydrolysis forms $C_{4} H_{8} O$ (B), which forms an oxime, does not reduce Fehling's solution and gives the iodoform test. Identify A and B.Answer: A is 2,2-dichlorobutane, $C H_{3} CC l_{2} C H_{2} C H_{3}$ ; its gem-diol loses water to give B, butan-2-one, $C H_{3} COC H_{2} C H_{3}$ (a methyl ketone).
Explain the terms anomers and epimers with examples from carbohydrate chemistry.Answer: Anomers differ only at the anomeric carbon ( $α$ - and $β$ -D-glucose); epimers differ at one chiral carbon (D-glucose and D-mannose at C-2, D-glucose and D-galactose at C-4).
Deduce the molecular formula of glucose: C = 40%, H = 6.7%, O = 53.3%; a solution of 9.0 g in 100 g of water freezes at $- 0.93^{\circ} C$ ( $K_{f}$ = 1.86 K kg mol $^{- 1}$ ).Answer: Empirical formula $C H_{2} O$ (30). Molality $= 0.93/1.86 = 0.50$ mol kg $^{- 1}$ , so $M = 90/0.50 = 180$ g mol $^{- 1}$ and the formula is $C_{6} H_{12} O_{6}$ .
Prove by one reaction that glucose contains hydroxyl groups.Answer: Acetic anhydride converts it to a pentaacetate, showing five OH groups.
Glucose, mannose and fructose give identical osazones. Explain.Answer: Only C-1 and C-2 react with phenylhydrazine; C-3 to C-6 are identical in all three.
Sucrose on hydrolysis gives D-(+)-glucose ( $+ 52. 7^{\circ}$ ) and D-(−)-fructose ( $- 92. 4^{\circ}$ ). Calculate the rotation of invert sugar.Answer: $\frac{52.7 - 92.4}{2} \approx - 19. 9^{\circ}$ .
Why is cellulose insoluble in water although it has so many OH groups?Answer: Its OH groups are already used in hydrogen bonds between neighbouring chains in crystalline bundles, and its molecular mass is very high, so water cannot pull the chains apart.
$α$ - and $β$ -D-glucose have different specific rotations, but in water each changes until the same value is reached. What is this change called?Answer: Mutarotation; both reach $+ 52. 7^{\circ}$ through the open-chain form.
Why is the acid hydrolysis of sucrose called an inversion reaction?Answer: The sign of rotation changes from + (sucrose, $+ 66. 5^{\circ}$ ) to − (invert sugar, $- 19. 9^{\circ}$ ).
Glucose and fructose are both reducing sugars. Why is sucrose non-reducing?Answer: Their anomeric carbons (C-1 of glucose, C-2 of fructose) form the glycosidic link, so no free hemiacetal or hemiketal is left.
D-(−)-fructose gives the same osazone as D-(+)-glucose and D-(+)-mannose. How is its configuration related to theirs?Answer: It has the same configuration at C-3, C-4 and C-5; it differs only at C-1 and C-2.
(A), $C_{6} H_{12} O_{6}$ , is oxidised by bromine water to a monobasic acid and reduces Tollens' and Fehling's reagents. With HCN, then $H_{2} O$ / $H^{+}$ , then HI/P, it gives n-heptanoic acid; with excess phenylhydrazine it gives D-glucosazone. Identify (A).Answer: D-glucose (a straight-chain aldohexose); D-mannose also fits these data.
Cellulose ( $β$ -D-glucose units) has a stronger, more compact structure than starch ( $α$ -D-glucose units). Explain.Answer: $β (1 \to 4)$ links give straight chains with all bulky groups equatorial; they pack side by side and hydrogen-bond into fibres. $α (1 \to 4)$ links make the chain coil into a helix, and amylopectin also branches.
Identify A, B and C: D-glucose $P h_{3} CCl, pyridine$ A $A c_{2} O, pyridine$ B $dry HBr, AcOH$ C.Answer: A = 6-O-trityl-D-glucose (the bulky trityl group reacts at the primary C-6 OH); B = 1,2,3,4-tetra-O-acetyl-6-O-trityl-D-glucose; C = 1,2,3,4-tetra-O-acetyl-D-glucose (HBr removes the trityl group as $P h_{3} CBr$ , freeing the C-6 OH).
An optically active compound does not reduce Tollens' or Fehling's reagent. Dilute acid gives one aldose and one ketose, which both reduce these reagents. It forms an octaacetate and an octamethyl derivative, is not oxidised by bromine water, and conc. $HN O_{3}$ oxidises it to oxalic acid. What is it?Answer: Sucrose.
D-glucose and D-fructose have different structures. Why do they give the same product with excess phenylhydrazine?Answer: Both give glucosazone: the reaction changes only C-1 and C-2, and their C-3 to C-6 are identical.
(a) Draw L-glucose. (b) Give the reaction of L-glucose with Tollens' reagent.Answer: (a) L-glucose is the mirror image of D-glucose: OH at C-2 left, C-3 right, C-4 left, C-5 left. (Inverting only C-5 would give L-idose.) (b) It is oxidised to L-gluconate and deposits a silver mirror.
The general formula of carbohydrates is (A) $C_{n} H_{2 n + 1} O$ (B) $C_{n} H_{2 n} O$ (C) $C_{n} (H_{2} O)_{n}$ or $C_{x} (H_{2} O)_{y}$ (D) $C_{n} (H_{2} O)_{2 n}$ Answer: (C).
Which of the following is a disaccharide? (A) sucrose (B) glucose (C) fructose (D) starchAnswer: (A).
Anomers have different (A) properties (B) melting points (C) specific rotations (D) all of theseAnswer: (D). Anomers are diastereomers, so they differ in all physical properties ( $α$ 419 K, $+ 11 2^{\circ}$ ; $β$ 423 K, $+ 1 9^{\circ}$ ).
Glucose and fructose are (A) tautomers (B) chain isomers (C) functional isomers (D) geometrical isomersAnswer: (C): an aldehyde and a ketone of the same formula.
Glucose is (A) an aldopentose (B) an aldohexose (C) a ketopentose (D) a ketohexoseAnswer: (B).
The reagent used in Ruff's degradation is (A) Baeyer's reagent (B) Tollens' reagent (C) Fenton's reagent (D) Benedict's reagentAnswer: (C): $H_{2} O_{2}$ with a ferric salt.
Carbohydrates that differ in configuration at the glycosidic carbon (C-1 in aldoses, C-2 in ketoses) are called (A) anomers (B) epimers (C) diastereomers (D) enantiomersAnswer: (A).
The hydrolysis of sucrose is called (A) saponification (B) inversion (C) esterification (D) hydrationAnswer: (B).
Glucose gives a silver mirror with ammoniacal silver nitrate because it contains (A) an aldehyde group (B) an ester group (C) a ketone group (D) an amide groupAnswer: (A).
Oligosaccharides contain (A) 2 to 10 (B) 4 to 8 (C) 6 to 12 (D) 6 to 10 simple sugar unitsAnswer: (A).
A pair of diastereomers that differ only in the configuration at a single carbon atom are called (A) anomers (B) epimers (C) conformers (D) enantiomersAnswer: (B).
Cellulose is a linear polymer of (A) $α$ -glucose (B) $β$ -glucose (C) $α$ -fructose (D) none of theseAnswer: (B).
One glucose molecule reacts with X molecules of phenylhydrazine to give an osazone. X is (A) three (B) two (C) one (D) fourAnswer: (A).

Common Mistakes to Avoid

Watch out

Reading D/L as the sign of rotation. D-fructose is laevorotatory: D/L comes from the drawn structure, (+)/(−) from a polarimeter.
Assigning D/L from C-2. Use the chiral carbon farthest from the C=O group: C-5 in hexoses, C-4 in pentoses.
Calling glucose and fructose epimers. They are functional isomers (aldose and ketose); epimers such as glucose and mannose differ at only one chiral carbon.
Thinking $α$ / $β$ means D/L. $α$ - and $β$ -D-glucose are both D; they differ only at the anomeric carbon.
Assuming sucrose is reducing because glucose and fructose are. Its two anomeric carbons are locked in the glycosidic link.
Subtracting rotations for invert sugar. Take the average: $\frac{52.7 + ( - 92.4 )}{2} = - 19. 9^{\circ}$ , not $- 3 9^{\circ}$ or $+ 14 5^{\circ}$ .
Writing the ring formed by the C-5 OH of fructose as a pyranose. C-5 OH + C-2 gives the five-membered furanose; the pyranose needs the C-6 OH.
Using Tollens' or Fehling's reagent to tell glucose from fructose. Both give a positive test; use bromine water, which oxidises only the aldose.
Mixing up the starch and cellulose links. Starch: $α (1 \to 4)$ with $α (1 \to 6)$ branches; cellulose: $β (1 \to 4)$ , which humans cannot digest.

Frequently Asked Questions

What are carbohydrates and how are they classified?

Carbohydrates are optically active polyhydroxy aldehydes or ketones, or compounds that give them on hydrolysis. They are classified as monosaccharides (glucose, fructose; not hydrolysed), oligosaccharides (two to ten units, such as sucrose and maltose) and polysaccharides (many units, such as starch, cellulose and glycogen).

Why is sucrose a non-reducing sugar?

In sucrose, C-1 of $α$ -D-glucose is bonded to C-2 of $β$ -D-fructose, so both anomeric carbons are locked in the glycosidic link. No free hemiacetal or hemiketal group is left to open into a carbonyl group, so sucrose does not reduce Tollens' or Fehling's reagent and does not mutarotate.

What is the difference between anomers and epimers?

Epimers are diastereomers that differ in configuration at only one chiral carbon, like D-glucose and D-mannose (C-2) or D-glucose and D-galactose (C-4). Anomers are cyclic forms that differ only at the anomeric carbon, C-1 of an aldose or C-2 of a ketose, such as $α$ - and $β$ -D-glucose.

What is mutarotation of glucose?

Mutarotation is the slow change in specific rotation of a fresh sugar solution to a fixed value. $α$ -D-glucose ( $+ 11 2^{\circ}$ ) and $β$ -D-glucose ( $+ 1 9^{\circ}$ ) both open to the chain form and close again until the solution holds about 36% $α$ and 64% $β$ , with a rotation of $+ 52. 7^{\circ}$ . All reducing sugars mutarotate.

Why do glucose, fructose and mannose form the same osazone?

Osazone formation uses three molecules of phenylhydrazine and changes only C-1 and C-2 into C=N-NHC $_{6}$ H $_{5}$ groups. Glucose, mannose and fructose differ only at C-1 and C-2, while C-3, C-4 and C-5 have the same configuration. Once C-1 and C-2 react, the three products are identical: glucosazone.

What is the difference between starch and cellulose?

Both are polymers of D-glucose. Starch uses $α$ -glucose: amylose is a linear $α (1 \to 4)$ chain and amylopectin is branched through $α (1 \to 6)$ links; it is the food store of plants. Cellulose uses $β (1 \to 4)$ links, giving straight chains that hydrogen-bond into fibres. Humans digest starch but not cellulose.

Which carbohydrate topics are most important for NEET?

NEET questions follow NCERT lines closely: classification and reducing sugars, the reactions that prove glucose's structure (HI, bromine water, nitric acid, acetylation), objections to the open-chain structure, $α$ and $β$ anomers, the links in sucrose, maltose and lactose, invert sugar, and the $α$ versus $β$ links of starch and cellulose.

How are carbohydrates tested in JEE Main and JEE Advanced?

JEE Main asks one-step facts: D/L assignment, the number of stereoisomers ( $2^{n}$ ), osazones, reducing sugars, epimers and anomers. JEE Advanced adds structure puzzles: periodic acid counts, Ruff and Kiliani-Fischer steps, identifying a sugar from its reactions, and deciding whether a drawn Haworth structure is reducing.

Previous year questions on Carbohydrates

28 questions from past papers, each with a step-by-step solution.

Show all 28 questions

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