Chapter 10 · Biomolecules

10.1 Carbohydrates

Sugars, starch and cellulose: how carbohydrates are classified, how chemists worked out the open-chain and ring structures of glucose, and how glycosidic links build disaccharides and polysaccharides.

A spoonful of table sugar, a slice of bread, a cotton shirt and the paper of a notebook are made of almost the same thing. Sucrose, starch and cellulose are all built from the six-carbon sugar glucose. Yet sugar dissolves in tea, starch is digested within an hour of a meal, and cellulose passes through the human gut untouched and can last for centuries as wood. The difference lies in how the glucose units are joined: one small change in the direction of a single bond decides whether a polymer is food for people or food only for termites and cows.

Carbohydrates are the most abundant organic compounds on Earth. Green plants make them from carbon dioxide and water by photosynthesis, and almost every living cell burns glucose for energy. In this section you will classify carbohydrates, follow the chemical evidence that reveals the structure of glucose (and why a simple open chain is not the whole story), meet the ring forms called anomers, and see how glycosidic linkages join monosaccharides into sucrose, maltose, lactose, starch, cellulose and glycogen.

Classifying Carbohydrates

The name "carbohydrate" comes from the old observation that many of these compounds have the formula $\mathrm{C_{x}(H_{2}O)_{y}}$, as if they were hydrates of carbon: glucose is $\mathrm{C_{6}H_{12}O_{6}}$, or $\mathrm{C_{6}(H_{2}O)_{6}}$. The formula is not a reliable guide, though. Rhamnose, $\mathrm{C_{6}H_{12}O_{5}}$, is a carbohydrate that does not fit it, while ethanoic acid, $\mathrm{C_{2}H_{4}O_{2}}$, and methanal, $\mathrm{CH_{2}O}$, fit it but are not carbohydrates. Chemists therefore define carbohydrates by their functional groups.

📘 Definition 10.1 — Carbohydrates and their classification

Carbohydrates are optically active polyhydroxy aldehydes or polyhydroxy ketones, or compounds that give these on hydrolysis. By their behaviour on hydrolysis they are classified as:

  • Monosaccharides: cannot be hydrolysed into simpler polyhydroxy aldehydes or ketones. About 20 occur in nature; examples are glucose, fructose, galactose and ribose.
  • Oligosaccharides: give two to ten monosaccharide units on hydrolysis. A disaccharide gives two (identical or different) units: sucrose gives glucose and fructose; maltose gives two glucose units. Trisaccharides give three, tetrasaccharides four.
  • Polysaccharides: give a large number of monosaccharide units on hydrolysis: starch, cellulose, glycogen, gums.

Monosaccharides and oligosaccharides are crystalline, water-soluble and sweet; they are called sugars. Polysaccharides are not sweet and are called non-sugars.

📘 Definition 10.2 — Aldoses and ketoses; reducing and non-reducing sugars

A monosaccharide is an aldose if it contains an aldehyde group and a ketose if it contains a keto group. The number of carbon atoms is shown by the stem: triose (3), tetrose (4), pentose (5), hexose (6), heptose (7). Glucose is an aldohexose, fructose a ketohexose, ribose an aldopentose.

Carbohydrates that reduce Fehling's solution and Tollens' reagent are reducing sugars. All monosaccharides, aldoses and ketoses alike, are reducing sugars. A disaccharide is reducing if one of its units still has a free aldehyde or keto group (more exactly, a free ring carbon C1 or C2 that can open, see below), as in maltose and lactose. If the reducing groups of both units are used up in the bond that joins them, as in sucrose, it is a non-reducing sugar.

Carbon atomsGeneral termAldoseKetose
3triosealdotriose (glyceraldehyde)ketotriose (1,3-dihydroxypropanone)
4tetrosealdotetroseketotetrose
5pentosealdopentose (ribose)ketopentose
6hexosealdohexose (glucose, galactose)ketohexose (fructose)
7heptosealdoheptoseketoheptose
✏️ Example 1 — Classifying sugars
Classify each compound as a monosaccharide, disaccharide or polysaccharide; for each monosaccharide say whether it is an aldose or a ketose and name its class fully (e.g. aldohexose). (a) $\mathrm{HOCH_{2}(CHOH)_{3}CHO}$ (ribose); (b) $\mathrm{HOCH_{2}CO(CHOH)_{3}CH_{2}OH}$ (fructose); (c) $\mathrm{HOCH_{2}CHOHCHO}$; (d) maltose, which gives two molecules of glucose on hydrolysis; (e) cellulose.
(a) Five carbons ($\mathrm{CH_{2}OH}$ + three $\mathrm{CHOH}$ + $\mathrm{CHO}$) and an aldehyde group: monosaccharide, aldopentose.
(b) Six carbons with a keto group at C2: monosaccharide, ketohexose.
(c) Three carbons, aldehyde group: glyceraldehyde, an aldotriose.
(d) Two monosaccharide units: disaccharide (an oligosaccharide).
(e) Thousands of glucose units: polysaccharide (a non-sugar).
Check Count the carbons and find the $\mathrm{C{=}O}$: $\mathrm{{-}CHO}$ at the end of the chain means aldose; $\mathrm{{>}C{=}O}$ inside the chain means ketose.

Glucose: from an Open Chain to a Ring

Glucose occurs free in sweet fruits and honey, and combined in sucrose, starch, cellulose and glycogen. It is prepared

How do we know the structure of glucose? Each of the reactions below answers one question about the molecule. Write $\mathrm{R}$ for $\mathrm{HOCH_{2}(CHOH)_{4}{-}}$, so that the open-chain formula of glucose is $\mathrm{R{-}CHO}$.

📐 Rule 10.1 — Evidence for the open-chain structure of glucose
  1. Formula $\mathrm{C_{6}H_{12}O_{6}}$ from analysis and molar mass.
  2. Straight chain of six carbons. On prolonged heating with $\mathrm{HI}$ glucose gives hexane, $\mathrm{CH_{3}(CH_{2})_{4}CH_{3}}$, an unbranched chain.
  3. A carbonyl group. Glucose forms an oxime with hydroxylamine, $\mathrm{R{-}CH{=}NOH}$, and adds $\mathrm{HCN}$ to give a cyanohydrin, $\mathrm{R{-}CH(OH)CN}$.
  4. The carbonyl group is an aldehyde. The mild oxidising agent bromine water converts glucose to gluconic acid, $\mathrm{R{-}COOH}$, a six-carbon carboxylic acid with no loss of carbon.
  5. Five $\mathrm{{-}OH}$ groups on different carbons. Acetylation with ethanoic anhydride gives a stable glucose pentaacetate. (Two $\mathrm{OH}$ groups on one carbon would lose water and could not both be acetylated.)
  6. A primary alcohol group. Nitric acid oxidises glucose and gluconic acid alike to saccharic (glucaric) acid, $\mathrm{HOOC(CHOH)_{4}COOH}$: the $\mathrm{{-}CH_{2}OH}$ at the other end of the chain becomes a second $\mathrm{{-}COOH}$.

Conclusion: glucose is $\mathrm{OHC{-}(CHOH)_{4}{-}CH_{2}OH}$, a 2,3,4,5,6-pentahydroxyhexanal. Its four $\mathrm{CHOH}$ carbons are stereocentres; the arrangement found by Emil Fischer is drawn below.

📘 Definition 10.3 — D and L configuration of sugars

The prefixes D and L give the relative configuration of a sugar, compared with glyceraldehyde. In the Fischer projection (aldehyde or keto group at the top), look at the stereocentre farthest from C1, the lowest $\mathrm{CHOH}$ (C5 in glucose and fructose). If its $\mathrm{{-}OH}$ is on the right, as in (+)-glyceraldehyde, the sugar belongs to the D series; if it is on the left, as in (−)-glyceraldehyde, to the L series.

D and L have nothing to do with the sign of optical rotation, which is written separately as (+) for dextrorotatory and (−) for laevorotatory: natural glucose is D-(+)-glucose, natural fructose is D-(−)-fructose. (Mirror-image molecules and optical rotation are treated in Chapter 6, Haloalkanes and Haloarenes.)

⚠️ Where the open-chain structure fails

1. Missing aldehyde reactions. Although glucose has an aldehyde group, it does not give the 2,4-DNP test or Schiff's test, and it does not form an addition product with sodium hydrogensulphite, $\mathrm{NaHSO_{3}}$.

2. The pentaacetate is not an aldehyde. Glucose pentaacetate does not react with hydroxylamine, so it has no free $\mathrm{{-}CHO}$ group.

3. Two crystalline forms. Glucose crystallised from a concentrated solution at $303\ \mathrm{K}$ is the α form (m.p. $419\ \mathrm{K}$); crystallised from a hot saturated solution at $371\ \mathrm{K}$ it is the β form (m.p. $423\ \mathrm{K}$). One open-chain structure cannot explain two different compounds.

All three facts are explained if the $\mathrm{{-}OH}$ group on C5 adds to the aldehyde group of the same molecule. An alcohol adds to an aldehyde to give a hemiacetal; here the addition is internal, so it closes a ring of five carbon atoms and one oxygen atom. Only a tiny fraction (well under 1%) of glucose in water is in the open-chain form at any moment, which is why the slow aldehyde tests fail, while fast reactions that use up the open form (oxidation, oxime formation) pull the ring open and go to completion.

📘 Definition 10.4 — Cyclic glucose: pyranose ring and anomers

In its cyclic form glucose is a six-membered ring containing one oxygen atom, like the heterocycle pyran; such a ring sugar is called a pyranose. Ring closure turns C1 into a new stereocentre, with an $\mathrm{{-}OH}$ group (the hemiacetal $\mathrm{OH}$) and an $\mathrm{{-}H}$. C1 is the anomeric carbon.

The two possible configurations at C1 give two stereoisomers, α-D-glucopyranose and β-D-glucopyranose. Isomers of a cyclic sugar that differ only in the configuration at the anomeric carbon (C1 of an aldose, C2 of a ketose) are called anomers. In the Haworth projection of a D sugar, the anomeric $\mathrm{{-}OH}$ points down (opposite side to the $\mathrm{CH_{2}OH}$ group) in the α anomer and up (same side) in the β anomer.

The two anomers are interconverted through the open chain: in water, the ring of either anomer opens to the aldehyde and closes again, with the $\mathrm{OH}$ on C1 ending up on either side. (What this does to the optical rotation of a glucose solution is beyond the board syllabus; see the 🚀 box at the end of the section.)

Interactive 1 — Closing the Ring: Fischer to Haworth

Follow D-glucose from its open chain to its ring. Step 1 is the Fischer projection; in step 2 the oxygen of the C5 $\mathrm{OH}$ (blue) swings round and attacks the aldehyde carbon C1; step 3 is the cyclic Fischer projection with the ring oxygen bridging C1 and C5; step 4 is the Haworth projection. The anomeric carbon C1 is ringed in amber: choose α or β to put its new $\mathrm{OH}$ on the other side. The graph on the right (🚀 beyond the board syllabus: mutarotation, see the box at the end of the section) follows the optical rotation of a fresh solution of the chosen anomer as it changes into the equilibrium mixture (time scale illustrative). Press Play to run through the steps for both anomers.

structure—
OH on C1—
specific rotation now—
at equilibrium—
✏️ Example 2 — Reading the evidence
Give the product of glucose with each reagent and state what each result tells us about its structure: (a) $\mathrm{HI}$, prolonged heating; (b) $\mathrm{NH_{2}OH}$; (c) bromine water; (d) ethanoic anhydride; (e) concentrated $\mathrm{HNO_{3}}$. (f) Which observation shows that glucose pentaacetate has no free aldehyde group?
(a) Hexane, $\mathrm{CH_{3}(CH_{2})_{4}CH_{3}}$: the six carbons form one unbranched chain.
(b) Glucose oxime, $\mathrm{R{-}CH{=}NOH}$: there is a carbonyl group.
(c) Gluconic acid, $\mathrm{R{-}COOH}$, still with six carbons: the carbonyl group is an aldehyde (a ketone is not oxidised by this mild reagent without breaking the chain).
(d) Glucose pentaacetate: five $\mathrm{OH}$ groups, on five different carbons.
(e) Saccharic acid, $\mathrm{HOOC(CHOH)_{4}COOH}$: besides $\mathrm{CHO}$ the chain ends in a primary alcohol group, $\mathrm{{-}CH_{2}OH}$.
(f) The pentaacetate does not react with hydroxylamine. The C1 $\mathrm{OH}$ of the ring is acetylated, which locks the ring shut.
✏️ Example 3 — Counting $\mathrm{OH}$ groups by acetylation
Each $\mathrm{{-}OH}$ that is acetylated becomes $\mathrm{{-}OCOCH_{3}}$, a gain of $\mathrm{C_{2}H_{2}O}$ ($42.04\ \mathrm{g\,mol^{-1}}$) per group. (a) Find the formula and molar mass of glucose pentaacetate and the mass obtained from $1.80\ \mathrm{g}$ of glucose. (b) $0.900\ \mathrm{g}$ of an aldohexose gives $1.95\ \mathrm{g}$ of a fully acetylated product. How many $\mathrm{OH}$ groups does the sugar have? Use $\mathrm{C}=12.011$, $\mathrm{H}=1.008$, $\mathrm{O}=15.999$.
Glucose $M(\mathrm{C_{6}H_{12}O_{6}})=6(12.011)+12(1.008)$ $+\,6(15.999)=180.16\ \mathrm{g\,mol^{-1}}$.
(a) $\mathrm{C_{6}H_{12}O_{6}}+5\,\mathrm{C_{2}H_{2}O}=\mathrm{C_{16}H_{22}O_{11}}$; $M=180.16+5(42.04)=390.34\ \mathrm{g\,mol^{-1}}$. From $1.80\ \mathrm{g}$: $\dfrac{1.80}{180.16}\times390.34=3.90\ \mathrm{g}$.
(b) $n=\dfrac{0.900}{180.16}=4.996\times10^{-3}\ \mathrm{mol}$. Gain $=1.95-0.900=1.05\ \mathrm{g}$, i.e. $\dfrac{1.05}{4.996\times10^{-3}}=210\ \mathrm{g\,mol^{-1}}$.
Answer $\dfrac{210}{42.04}=5.0$: five $\mathrm{OH}$ groups, as in glucose.

Fructose and the Disaccharides

Fructose, $\mathrm{C_{6}H_{12}O_{6}}$, is obtained together with glucose when sucrose is hydrolysed, and is the sweetest natural sugar (it is plentiful in fruits and honey). It is a ketohexose with the keto group at C2, and its C5 $\mathrm{OH}$ is on the right in the Fischer projection, so it is D-(−)-fructose. Like glucose it exists mainly as rings: the C5 $\mathrm{OH}$ adds to the C2 keto group to give a five-membered ring of four carbons and one oxygen, like the heterocycle furan. Such a ring sugar is a furanose, and fructose has α and β anomers that differ at C2.

📘 Definition 10.5 — Glycosidic linkage

Two monosaccharide units are joined when the anomeric $\mathrm{{-}OH}$ of one unit and an $\mathrm{{-}OH}$ of the other lose a molecule of water. The oxygen atom that links them, $\mathrm{C{-}O{-}C}$, is a glycosidic linkage. It is named by the anomer and the carbon numbers it joins: α(1→4) joins C1 of an α unit to C4 of the next. A disaccharide $\mathrm{C_{12}H_{22}O_{11}}$ is two hexoses minus one water: $2(180.16)-18.02=342.30\ \mathrm{g\,mol^{-1}}$. Hydrolysis by dilute acid or by an enzyme adds the water back and breaks the link.

Sucrose (cane or beet sugar) is formed from α-D-glucose and β-D-fructose joined between C1 of glucose and C2 of fructose. Both anomeric carbons are tied up in the link, neither ring can open, and sucrose is a non-reducing sugar. Sucrose is dextrorotatory, $[\alpha]=+66.5^{\circ}$. Its hydrolysis gives equal amounts of D-(+)-glucose ($+52.5^{\circ}$) and D-(−)-fructose ($-92.4^{\circ}$); fructose wins, and the mixture is laevorotatory. Because the sign of rotation is inverted, the hydrolysis is called the inversion of cane sugar (a pseudo-first-order reaction, Section 3.2) and the product mixture is called invert sugar. The enzyme invertase in bees and yeast does the same job, which is why honey is largely invert sugar.

Maltose (malt sugar, from the partial hydrolysis of starch) is made of two α-D-glucose units joined C1 to C4, an α(1→4) link. The second unit keeps a free hemiacetal $\mathrm{OH}$ at C1, which can open to an aldehyde in solution, so maltose is a reducing sugar. Lactose (milk sugar) is made of β-D-galactose and D-glucose joined C1 of galactose to C4 of glucose, a β(1→4) link; its glucose C1 is free, so lactose is also reducing. (Galactose differs from glucose only in the configuration at C4.)

✏️ Example 4 — Following the inversion of cane sugar
A solution contains $10.0\ \mathrm{g}$ of sucrose in $100\ \mathrm{mL}$ and is placed in a polarimeter tube $2.00\ \mathrm{dm}$ long. The observed rotation is $\alpha=[\alpha]\,l\,c$, with $l$ in dm and $c$ in $\mathrm{g\,mL^{-1}}$. Use $[\alpha]$: sucrose $+66.5^{\circ}$, glucose $+52.5^{\circ}$, fructose $-92.4^{\circ}$; $M$: sucrose $342.30$, hexose $180.16\ \mathrm{g\,mol^{-1}}$. (a) Find the specific rotation of invert sugar. (b) Find the observed rotation at the start and after complete hydrolysis. (c) What fraction of the sucrose has been hydrolysed when the observed rotation is zero?
(a) Equal moles of glucose and fructose, and equal molar masses, so equal masses: $[\alpha]=\tfrac{1}{2}(52.5-92.4)=-19.95\approx-20.0^{\circ}$.
(b) start $\alpha_{0}=66.5\times2.00\times0.100=+13.3^{\circ}$.
(b) end $1\ \mathrm{mol}$ sucrose ($342.30\ \mathrm{g}$) gives $2\ \mathrm{mol}$ hexose ($360.32\ \mathrm{g}$), so $c=0.100\times\dfrac{360.32}{342.30}=0.1053\ \mathrm{g\,mL^{-1}}$ and $\alpha_{\infty}=-19.95\times2.00\times0.1053=-4.20^{\circ}$.
(c) With a fraction $f$ hydrolysed, $\alpha$ is the weighted sum: $13.3(1-f)-4.20f=0$, so $f=\dfrac{13.3}{13.3+4.20}=0.760$, i.e. $76.0\%$.
Check The rotation falls steadily from $+13.3^{\circ}$ through zero to $-4.20^{\circ}$; this change is how the rate of inversion is followed in the laboratory.
Interactive 2 — Sugar Test Lab: Who Reduces Tollens' and Fehling's?

Choose a carbohydrate and a test, then press Play (or change a choice) to warm the tube and watch. The right-hand panel shows the sugar's rings: an anomeric carbon ringed green is free (its ring can open to a reducing group); red means it is locked in a glycosidic link. Tick hydrolyse first to boil the sample with dilute acid (then neutralise) before testing: see what happens to sucrose and starch.

class—
free anomeric C—
reducing sugar?—
observation—
✏️ Example 5 — Telling sugars apart
Give one chemical test, with the observation, to distinguish each pair: (a) glucose and sucrose; (b) sucrose and maltose; (c) glucose and fructose; (d) starch and cellulose; (e) a sucrose solution before and after it has been boiled with dilute acid.
(a) Tollens' reagent, warm: glucose gives a silver mirror; sucrose (non-reducing) gives none.
(b) Fehling's solution, warm: maltose (free C1) gives a red precipitate of $\mathrm{Cu_{2}O}$; sucrose leaves the solution blue.
(c) Both reduce Tollens' and Fehling's reagents (in alkali fructose is converted partly into aldoses), so use bromine water, a mild oxidant that works in slightly acidic solution: glucose (an aldose) decolourises it, forming gluconic acid; fructose (a ketose) does not.
(d) Iodine solution: starch gives a deep blue-black colour; cellulose gives no blue colour.
(e) Fehling's solution: negative before boiling; positive (red $\mathrm{Cu_{2}O}$) after, because hydrolysis releases glucose and fructose, both reducing.

Polysaccharides and Why Carbohydrates Matter

Plants store glucose as starch, animals as glycogen, and plants build their cell walls from cellulose. All three are polymers of D-glucose; they differ in the anomer used and in the pattern of links.

📘 Definition 10.6 — Starch, cellulose and glycogen
  • Starch, the main storage carbohydrate of plants (cereals, potatoes, roots, tubers), is a polymer of α-D-glucose with two components. Amylose (about 15–20%) is water-soluble: a long unbranched chain of 200–1000 α-D-(+)-glucose units joined by C1–C4 glycosidic links. Amylopectin (about 80–85%) is insoluble in water: a branched polymer whose chains are joined C1–C4, with branches attached by C1–C6 links.
  • Cellulose, the main constituent of plant cell walls and the most abundant organic compound in the plant kingdom, is a straight-chain polymer of β-D-glucose joined C1–C4. Humans have no enzyme that breaks β(1→4) links, so we cannot digest cellulose; cattle and termites rely on microbes in their gut to do so.
  • Glycogen, the storage carbohydrate of animals ("animal starch"), has a structure like amylopectin but is even more highly branched. It is stored in the liver, muscles and brain and is broken down to glucose when the body needs energy; it also occurs in yeast and fungi.
✏️ Example 6 — How many glucose units?
(Illustrative data.) An amylose chain has a molar mass of $1.62\times10^{5}\ \mathrm{g\,mol^{-1}}$. (a) How many glucose units does it contain? (b) How many water molecules are needed to hydrolyse one chain completely? (c) What mass of glucose can be obtained from $1.00\ \mathrm{kg}$ of starch? Use $M$: glucose $180.16$, water $18.02\ \mathrm{g\,mol^{-1}}$.
Idea A chain of $n$ units is formed from $n$ glucose molecules with the loss of $n-1$ water molecules: $M=180.16n-18.02(n-1)$ $=162.14n+18.02$.
(a) $n=\dfrac{1.62\times10^{5}-18.02}{162.14}=999$ glucose units.
(b) $n-1=998$ molecules of water (one per glycosidic link).
(c) For long chains each unit is $\mathrm{C_{6}H_{10}O_{5}}$ ($162.14\ \mathrm{g\,mol^{-1}}$) and becomes one glucose: $1.00\times\dfrac{180.16}{162.14}=1.11\ \mathrm{kg}$. The mass increases because water is added.

Why carbohydrates matter. Carbohydrates are the main fuel of living things: glucose is oxidised in cells to carbon dioxide and water, releasing energy, and honey has long been used as an instant source of energy in traditional systems of medicine such as Ayurveda. They are the energy stores of plants (starch) and animals (glycogen), and the structural material of plants (cellulose in wood and cotton). Carbohydrates are raw materials for the textile, paper, lacquer and brewing industries, and two sugars, D-ribose and 2-deoxy-D-ribose, are part of the nucleic acids (Section 10.3).

🧭 Problem-solving strategy — Is this carbohydrate a reducing sugar?
  1. Monosaccharide? Then yes, aldose or ketose.
  2. Find every anomeric carbon: C1 of each aldose unit, C2 of each ketose unit.
  3. Is any of them free, carrying a ring $\mathrm{OH}$ rather than a glycosidic $\mathrm{O}$? If yes, that ring can open to $\mathrm{{-}CHO}$ (or $\mathrm{{>}C{=}O}$): reducing (maltose, lactose). If all are locked: non-reducing (sucrose).
  4. Polysaccharides have at most one free end per huge chain and count as non-reducing.
  5. After hydrolysis every product is a monosaccharide, so the test turns positive.
⚠️ Common mistakes with carbohydrates

1. D/L is not (+)/(−). D-fructose is laevorotatory. D/L comes from a drawing; (+)/(−) from a polarimeter.

2. α/β is not D/L. α and β differ only at the anomeric carbon; D and L describe the lowest stereocentre.

3. Sucrose is non-reducing although both its units are reducing sugars: the link uses up both anomeric carbons.

4. Starch and cellulose are both glucose polymers. They differ in the anomer (α vs β), not in the monomer.

5. Furanose vs pyranose. Glucose rings are six-membered (pyranose); fructose rings in sucrose are five-membered (furanose), closed from C5 to C2.

🚀 Beyond the board syllabus (JEE Main · NEET · JEE Advanced) — mutarotation

In JEE Main, NEET and JEE Advanced 2026 (glucose; JEE Advanced names anomers); not required by the CBSE syllabus.

A fresh solution of pure α-D-glucose has $[\alpha]=+112^{\circ}$; one of pure β-D-glucose has $+18.7^{\circ}$. In each the rotation drifts to the same final value, the $+52.5^{\circ}$ quoted for glucose, as the rings open and close (Interactive 1). This change of rotation is called mutarotation. If $x$ is the fraction of α at equilibrium, $112x+18.7(1-x)=52.5$ gives $x=0.362$: about $36\%$ α and $64\%$ β. The β anomer wins because, in the real puckered (chair) shape of the ring, all its bulky groups, including the C1 $\mathrm{OH}$, can sit pointing outwards from the ring, where they are least crowded.

Exercises

I
Classification and tests.
  • 1 Define carbohydrates. Why is the name "hydrates of carbon" misleading? Give one compound that fits $\mathrm{C_{x}(H_{2}O)_{y}}$ but is not a carbohydrate, and one carbohydrate that does not fit it.
  • 2 Classify as monosaccharide, oligosaccharide or polysaccharide: glucose, sucrose, starch, maltose, raffinose (which gives three monosaccharide units on hydrolysis), glycogen.
  • 3 Name the class (e.g. aldohexose) of: glyceraldehyde, 1,3-dihydroxypropanone, ribose, fructose, galactose.
  • 4 What is a reducing sugar? Give two disaccharides that are reducing and one that is not.
  • 5 Why are polysaccharides called non-sugars?
  • 6 Which of these give a silver mirror with Tollens' reagent: glucose, fructose, sucrose, maltose, lactose, starch?
  • 7 A water-soluble, sweet carbohydrate X does not reduce Fehling's solution, but after it is boiled with dilute $\mathrm{HCl}$ and neutralised it does. What is X likely to be? Explain.
  • 8 Give a chemical test to distinguish glucose from fructose and one to distinguish starch from glucose.
1 Optically active polyhydroxy aldehydes or ketones, or compounds that give them on hydrolysis. The formula is a coincidence: ethanoic acid $\mathrm{C_{2}H_{4}O_{2}}$ (or methanal) fits but is not a carbohydrate; rhamnose $\mathrm{C_{6}H_{12}O_{5}}$ is a carbohydrate but does not fit.
2 Mono: glucose. Oligo: sucrose, maltose (disaccharides), raffinose (trisaccharide). Poly: starch, glycogen.
3 Aldotriose; ketotriose; aldopentose; ketohexose; aldohexose.
4 A carbohydrate that reduces Fehling's solution and Tollens' reagent (it has a free aldehyde or keto group, or a free anomeric carbon). Reducing: maltose, lactose. Non-reducing: sucrose.
5 They are not sweet (and are mostly insoluble, non-crystalline polymers).
6 Glucose, fructose, maltose, lactose. Not sucrose (both anomeric carbons locked) or starch.
7 Sucrose: it is non-reducing, but acid hydrolysis gives glucose and fructose, which reduce Fehling's solution.
8 Bromine water: decolourised by glucose (aldose), not by fructose. Iodine solution: blue-black with starch, no blue with glucose.
II
The structure of glucose. Use $\mathrm{R}=\mathrm{HOCH_{2}(CHOH)_{4}{-}}$ where convenient.
  • 1 Write the product of glucose with $\mathrm{HI}$ (prolonged heating). What does it prove?
  • 2 Write the product of glucose with $\mathrm{HCN}$ and with hydroxylamine. What do they prove?
  • 3 Glucose is oxidised by bromine water. Name and write the product. What does the reaction prove?
  • 4 What is formed when glucose is treated with ethanoic anhydride? What does it prove?
  • 5 What is formed when glucose is oxidised by nitric acid? What does it show?
  • 6 State three facts that the open-chain structure of glucose cannot explain.
  • 7 What are anomers? Draw the Haworth projection of β-D-glucopyranose and mark the anomeric carbon.
  • 8 Explain the meaning of "D" and "(+)" in D-(+)-glucose. Is D-fructose dextrorotatory?
  • 9 Glucose has an aldehyde group but does not react with sodium hydrogensulphite. How does the cyclic structure explain this?
1 Hexane, $\mathrm{CH_{3}(CH_{2})_{4}CH_{3}}$: all six carbons are in one straight chain.
2 Cyanohydrin $\mathrm{R{-}CH(OH)CN}$; oxime $\mathrm{R{-}CH{=}NOH}$: a carbonyl group is present.
3 Gluconic acid, $\mathrm{R{-}COOH}$ ($\mathrm{C_{6}H_{12}O_{7}}$): the carbonyl group is an aldehyde, $\mathrm{{-}CHO}$.
4 Glucose pentaacetate: five $\mathrm{{-}OH}$ groups, on different carbon atoms.
5 Saccharic acid, $\mathrm{HOOC(CHOH)_{4}COOH}$: both ends are oxidised, so besides $\mathrm{{-}CHO}$ there is a primary alcohol group $\mathrm{{-}CH_{2}OH}$.
6 No 2,4-DNP or Schiff's test and no $\mathrm{NaHSO_{3}}$ adduct; the pentaacetate does not react with $\mathrm{NH_{2}OH}$; two crystalline forms, α (m.p. $419\ \mathrm{K}$) and β (m.p. $423\ \mathrm{K}$).
7 Cyclic forms that differ only in configuration at the anomeric carbon (C1 of glucose). Haworth: as in the figure, with C1 $\mathrm{OH}$ up (same side as $\mathrm{CH_{2}OH}$), C2 down, C3 up, C4 down; C1 is the anomeric carbon.
8 D: the $\mathrm{OH}$ on C5 (the lowest stereocentre) is on the right in the Fischer projection, as in D-glyceraldehyde. (+): the compound rotates plane-polarised light to the right. D-fructose is laevorotatory, D-(−)-fructose.
9 Almost all glucose is in ring (hemiacetal) form, with no free $\mathrm{CHO}$; the open-chain form is present only in a tiny amount, too little for this slow, reversible addition.
III
Disaccharides and polysaccharides.
  • 1 Name the products of complete hydrolysis of sucrose, maltose and lactose.
  • 2 Describe the glycosidic linkage in each of sucrose, maltose and lactose (which carbons, which anomers).
  • 3 Why is sucrose a non-reducing sugar while maltose is a reducing sugar?
  • 4 What is invert sugar, and why is it so called?
  • 5 Give three differences between amylose and amylopectin.
  • 6 Cows can live on grass, but humans cannot digest cellulose. Explain.
  • 7 Compare glycogen with amylopectin. Where is glycogen stored in the body?
  • 8 Starch and cellulose both give only D-glucose on complete hydrolysis. Why are their properties so different?
1 Sucrose: D-glucose + D-fructose. Maltose: two D-glucose. Lactose: D-galactose + D-glucose.
2 Sucrose: C1 of α-D-glucose to C2 of β-D-fructose. Maltose: C1 of α-D-glucose to C4 of the second glucose, α(1→4). Lactose: C1 of β-D-galactose to C4 of glucose, β(1→4).
3 In sucrose both anomeric carbons (glucose C1, fructose C2) are used in the link, so no ring can open. In maltose C1 of the second glucose is free and opens to $\mathrm{{-}CHO}$.
4 The equimolar mixture of D-glucose and D-fructose from hydrolysis of sucrose. Sucrose is dextrorotatory ($+66.5^{\circ}$) but the mixture is laevorotatory (fructose $-92.4^{\circ}$ outweighs glucose $+52.5^{\circ}$): the sign is inverted.
5 Amylose: 15–20% of starch, water-soluble, unbranched, C1–C4 links only. Amylopectin: 80–85%, insoluble, branched, C1–C4 chains with C1–C6 branch points.
6 Cellulose has β(1→4) links. Humans lack an enzyme that hydrolyses them; cows carry gut microbes that make such enzymes and release glucose from cellulose.
7 Both are branched α-glucose polymers with C1–C4 chains and C1–C6 branches, but glycogen is more highly branched. It is stored in the liver, muscles and brain.
8 The links differ: α(1→4) in starch gives coiled chains that enzymes in our gut can hydrolyse; β(1→4) in cellulose gives straight chains that pack side by side, held by hydrogen bonds, into strong insoluble fibres.
IV
Numerical problems. Use $\mathrm{C}=12.011$, $\mathrm{H}=1.008$, $\mathrm{O}=15.999\ \mathrm{g\,mol^{-1}}$; $[\alpha]$: sucrose $+66.5^{\circ}$, glucose $+52.5^{\circ}$, fructose $-92.4^{\circ}$; observed rotation $\alpha=[\alpha]\,l\,c$ ($l$ in dm, $c$ in $\mathrm{g\,mL^{-1}}$).
  • 1 Calculate the molar masses of glucose and of sucrose.
  • 2 What masses of glucose and of fructose are obtained by complete hydrolysis of $68.46\ \mathrm{g}$ of sucrose?
  • 3 A trisaccharide is formed from three hexose units. Find its molecular formula and molar mass.
  • 4 What mass of glucose pentaacetate is formed by complete acetylation of $2.70\ \mathrm{g}$ of glucose?
  • 5 (Illustrative.) A glycogen molecule contains $30\,000$ glucose units. Estimate its molar mass.
  • 6 An amylose chain of $600$ glucose units is formed from glucose. How many glycosidic links does it contain, and how many water molecules are released?
  • 7 $34.23\ \mathrm{g}$ of sucrose is dissolved to make $100\ \mathrm{mL}$ of solution in a $1.00\ \mathrm{dm}$ tube. Find the observed rotation at the start and after complete inversion.
  • 8 Food science. A cola contains $10.6\ \mathrm{g}$ of sugar per $100\ \mathrm{mL}$ (taken as sucrose, illustrative). How many moles of sucrose are in a $330\ \mathrm{mL}$ can, and what mass of glucose would complete hydrolysis give?
1 Glucose $\mathrm{C_{6}H_{12}O_{6}}$: $180.16\ \mathrm{g\,mol^{-1}}$. Sucrose $\mathrm{C_{12}H_{22}O_{11}}$: $2(180.16)-18.02=342.30\ \mathrm{g\,mol^{-1}}$.
2 $n=\dfrac{68.46}{342.30}=0.2000\ \mathrm{mol}$; each hexose $0.2000\times180.16=36.03\ \mathrm{g}$ (glucose and fructose).
3 $3\,\mathrm{C_{6}H_{12}O_{6}}-2\,\mathrm{H_{2}O}=\mathrm{C_{18}H_{32}O_{16}}$; $M=3(180.16)-2(18.02)=504.44\ \mathrm{g\,mol^{-1}}$.
4 $\dfrac{2.70}{180.16}\times390.34=5.85\ \mathrm{g}$.
5 $M=162.14\times30\,000+18.02$ $\approx4.86\times10^{6}\ \mathrm{g\,mol^{-1}}$.
6 $599$ links; $599$ water molecules released (one per link).
7 Start: $c=0.3423\ \mathrm{g\,mL^{-1}}$, $\alpha=66.5\times1.00\times0.3423=+22.8^{\circ}$. After: $c=0.3423\times\dfrac{360.32}{342.30}=0.3603\ \mathrm{g\,mL^{-1}}$ of invert sugar ($[\alpha]=-19.95^{\circ}$): $\alpha=-7.19^{\circ}$.
8 Mass $=10.6\times3.30=35.0\ \mathrm{g}$; $n=\dfrac{35.0}{342.30}=0.102\ \mathrm{mol}$; glucose $0.102\times180.16=18.4\ \mathrm{g}$ (and the same of fructose).
🎯 Exam corner BoardJEE MainNEET

E1. Board · 2 marks What happens when D-glucose is treated with (a) $\mathrm{HI}$ on prolonged heating, (b) bromine water? Write the reactions and say what each shows.

(a) $\mathrm{R{-}CHO\xrightarrow{HI,\ \Delta}CH_{3}(CH_{2})_{4}CH_{3}}$ (hexane): six carbons in a straight chain. (1 mark)
(b) $\mathrm{R{-}CHO\xrightarrow{Br_{2}\ water}R{-}COOH}$ (gluconic acid): the carbonyl group is an aldehyde. (1 mark)

E2. NEET Which of the following is a non-reducing sugar? (A) maltose  (B) lactose  (C) sucrose  (D) fructose

Idea Look for a free anomeric carbon. Fructose is a monosaccharide (reducing); maltose and lactose each have a free glucose C1.
Answer (C): in sucrose glucose C1 and fructose C2 are both in the glycosidic link.

E3. JEE Main Amylose and cellulose are both linear polymers of D-glucose. They differ in that (A) amylose has C1–C6 links (B) cellulose has α(1→4) links and amylose β(1→4) (C) amylose has α(1→4) links and cellulose β(1→4) (D) cellulose is branched and amylose is not.

Idea Both are unbranched C1–C4 chains (C1–C6 branches belong to amylopectin and glycogen). The difference is the anomer.
Answer (C).

E4. JEE Main Numerical value. $0.0100\ \mathrm{mol}$ of sucrose is completely acetylated with ethanoic anhydride. By how many grams does the mass increase? (Each acetylated $\mathrm{OH}$ adds $\mathrm{C_{2}H_{2}O}=42.04\ \mathrm{g\,mol^{-1}}$.)

Count OH Glucose and fructose have five $\mathrm{OH}$ each; forming the glycosidic link uses two (one from each, with loss of water): $10-2=8$ free $\mathrm{OH}$ groups.
Mass $0.0100\times8\times42.04=3.36\ \mathrm{g}$.
Answer $\mathbf{3.36}$ (sucrose octaacetate, $\mathrm{C_{28}H_{38}O_{19}}$).

E5. Board · 3 marks Give three observations which show that glucose cannot be fully described by an open-chain structure, and state how the cyclic structure accounts for them.

1 No Schiff's test, no 2,4-DNP derivative, no $\mathrm{NaHSO_{3}}$ adduct: in the ring (hemiacetal) form there is no free $\mathrm{CHO}$. (1 mark)
2 The pentaacetate does not react with $\mathrm{NH_{2}OH}$: acetylating the C1 $\mathrm{OH}$ locks the ring, so no $\mathrm{CHO}$ can form. (1 mark)
3 Two forms, α (m.p. $419\ \mathrm{K}$) and β (m.p. $423\ \mathrm{K}$): ring closure makes C1 a stereocentre, giving two anomers. (1 mark)

E6. Board · Case study Many adults produce little lactase, the intestinal enzyme that hydrolyses milk sugar. Undigested lactose passes to the large intestine, where bacteria ferment it, causing discomfort. "Lactose-free" milk is made by adding lactase to milk in the dairy; it tastes noticeably sweeter than ordinary milk. (a) Name the products of hydrolysis of lactose and the linkage broken. (2 marks) (b) Is lactose a reducing sugar? Explain. (1 mark) (c) Suggest why lactose-free milk tastes sweeter. (1 mark)

(a) β-D-galactose and D-glucose; the β(1→4) glycosidic link between C1 of galactose and C4 of glucose.
(b) Yes: C1 of the glucose unit is free, so its ring can open to an aldehyde group.
(c) One molecule of lactose becomes two molecules of monosaccharide, and glucose and galactose taste sweeter than lactose.
🔬 Research–Teaching NexusResearch-tutored · optional, about 30 minutes

Research question. Could fuel for cars be made from straw and wood chips instead of from food crops, and what limits the yield?

Why it matters. Brazil, the USA and India blend ethanol into petrol, mostly made from sugar cane, molasses or maize starch, which competes with food. Cellulose from crop waste is far more plentiful, but its β(1→4) links are hard to break. Second-generation ("cellulosic") ethanol plants, including ones being built for rice and wheat straw in India, depend on enzymes that hydrolyse cellulose to glucose.

What to do.

  1. Explain, with a sketch of two linked rings, why the enzymes that digest starch cannot digest cellulose.
  2. (Illustrative, constructed for this exercise.) $1.00$ tonne of pure cellulose is hydrolysed completely. Calculate the mass of glucose formed (each $\mathrm{C_{6}H_{10}O_{5}}$ unit gives one $\mathrm{C_{6}H_{12}O_{6}}$).
  3. Yeast converts glucose to ethanol: $\mathrm{C_{6}H_{12}O_{6}\rightarrow2C_{2}H_{5}OH+2CO_{2}}$. Find the maximum mass of ethanol and its volume (density $0.789\ \mathrm{g\,mL^{-1}}$).
  4. Real straw is only about 35–40% cellulose, and only part of it is hydrolysed. Estimate the ethanol from one tonne of straw if 38% is cellulose and 70% of that is converted (illustrative figures).

Tools: A calculator; molar masses from this section; news or agency reports on cellulosic ethanol plants.

Hand in: Your sketch, the calculations, and a paragraph weighing straw-based against grain-based ethanol.

Going further. Find out which microbes make cellulase enzymes, and why burning crop residue in fields is a problem that cellulosic ethanol might help to solve.

Instructor note (all yields illustrative, constructed for this exercise): glucose $=1.00\times\dfrac{180.16}{162.14}=1.11\ \mathrm{t}$; ethanol $=1.111\times\dfrac{2(46.07)}{180.16}=0.568\ \mathrm{t}$, i.e. $568\ \mathrm{kg}$ or about $720\ \mathrm{L}$. For straw: $0.38\times0.70\times568\approx151\ \mathrm{kg}$ of ethanol, about $192\ \mathrm{L}$ per tonne.

🧪 Self-check quiz

0/5 correct

Q1. α-D-glucopyranose and β-D-glucopyranose differ in the configuration at

Q2. Which carbohydrate gives a silver mirror with Tollens' reagent?

Q3. Invert sugar is

Q4. Cellulose is a polymer of

Q5. Which reaction shows that the six carbon atoms of glucose are in a straight chain?