food chemistry

F FOOD CHEMISTRY 1CHEMISTRY FOR THE IB DIPLOMA © CAMBRIDGE UNIVERSITY PRESS 2011

F Food chemistry

F1 Food groups Learning objectives

• Distinguish between a food and a nutrient

• Describe the chemical composition of lipids, carbohydrates and proteins

The structure of carbohydrates is discussed in more detail in Option B, page 11 on the CD-ROM.

In solution, the cyclic form of a monosaccharide predominates, with only a small fraction existing as the linear structure.

Examiner’s tipThe words lipid and fat may be used interchangeably in many texts and exam questions.

Food – ‘Any substance, whether processed, semi-processed or raw, which is intended for human consumption, and includes drink, chewing gum and any substance which has been used in the manufacture, preparation or treatment of ‘food’, but does not include cosmetics or tobacco or substances used only as drugs’ – Codex Alimentarius (FAO/WHO) defi nition 2005.

Nutrients A nutrient is any substance obtained from food and used by the body to provide energy, to regulate growth or to maintain and repair the body’s tissues. Any item of food will contain at least one, and possibly many, diff erent types of nutrient. For example, bread is food and contains several nutrients including carbohydrates, proteins and lipids. Any substance that does not contain usable nutrients is not food.

There are seven principal classes of nutrient:• carbohydrate• protein• lipid (fats and oils)• minerals• vitamins• water• � bre is sometimes included because, although it is not absorbed

by the body, it is required for the maintenance of a properly functioning intestinal tract.

CarbohydrateCarbohydrates contain only the elements carbon, hydrogen and oxygen. The most basic carbohydrates, which are often referred to as simple sugars, are called monosaccharides. Examples of monosaccharides include glucose, fructose and galactose. They contain carbon, hydrogen and oxygen in the ratio 1 : 2 : 1 and therefore can be represented by the empirical formula CH2O. Monosaccharides can exist in two structural forms: linear or cyclic (Figure F1). In the cyclic form, the carbonyl group is broken to produce an intramolecular oxygen bridge (see Figure F1). The hydroxyl groups remain unchanged.

In the linear form, monosaccharides contain a carbonyl group (C=O) as part of either an aldehyde (e.g. glucose) or a ketone (e.g. fructose) and at least two hydroxyl (alcohol) –OH groups.

Monosaccharides can be combined to produce more complex carbohydrates such as disaccharides, e.g. sucrose and maltose (Figure F2)

2 CHEMISTRY FOR THE IB DIPLOMA © CAMBRIDGE UNIVERSITY PRESS 2011F FOOD CHEMISTRY

and polysaccharides (e.g. starch and cellulose). When monosaccharides are joined together, they release a molecule of water in a condensation reaction. The bond formed between the monomers is called a glycosidic bond and contains another oxygen atom that bridges the gap between two monosaccharide units.

Ha O

OHH C

HHO C

OHH C

OHH C

CH2OH

C

C

C

C

C O

HHO

OHH

OHH

CH2OH

CH2OH

b

HO

OH

OHH

O

OH

H

HH

H

CH2OH

C C

CC

C

Figure F1 (a) The chemical structures of glucose (left) and fructose (right) in Fischer projection; (b) a cyclic form of glucose.

The structures of polysaccharides are discussed in Option B, page 12 on the CD-ROM.

Fischer projection is a two-dimensional representation of a three-dimensional molecule. In the projected structure, horizontal bonds represent groups pointing towards the viewer and vertical bonds represent groups pointing away from the viewer.

Some proteins also contain the element sulfur.

The structure of proteins is discussed in more detail in Option B, page 2 on the CD-ROM.

Figure F2 A molecule of the disaccharide sugar maltose.

C

H

a b c

H

peptide bond

O

O

R

H

N C C

H

H

O

OH

H

N C

O

C

H

N

H

C

H

H

O

HO

H

C C N

H H HCH2

SHFigure F3 (a) The general structure of an amino acid; (b) the sulfur-containing amino acid methionine; (c) a dipeptide molecule containing a peptide bond.

HO O

OHH

O

OH

H

H

HH

CH2OH

H

OH

OHH

O

OH

H

HH

CH2OH

ProteinProtein always contains the elements carbon, hydrogen, oxygen and nitrogen. Proteins are polymers of smaller sub-units called amino acids (or, more precisely, 2-amino acids). The smallest protein (called a dipeptide) contains just 2 amino acids but many proteins contain thousands of amino acids.

Amino acids contain an amine group (NH2) at one end and a carboxylic acid group (COOH) at the other. The central carbon contains a bond to a variable group, R (Figure F3). Amino acids are sometimes referred to as 2-amino acids to indicate that the amine group is on the second carbon of the carboxylic acid (position 1) main chain.

When two amino acids join together in a condensation reaction (water eliminated) they form a peptide bond which contains the amide (CONH) functional group (Figure F3).

LipidsLipids contain only the elements carbon, hydrogen and oxygen. Lipids may be split into two classes: oils (usually from plant sources and liquids at room temperature) and fats (usually from animal sources and solids at room temperature).


Lipids are triester molecules, formed by reacting three long-chain fatty acid molecules with the three hydroxyl (–OH) groups of a glycerol (propane-1,2,3-triol) molecule. The resultant compounds are often referred to as triglycerides (Figure F4).

O

OC

H

H

H

H

H

C C

H

H

C

H

ester groups long hydrocarbon chain

H

C

H

H

C

H

H

C

H

H

C

H

H

C

H

H

C

H

H

C

H

H

C

H

H

C

H

H

C

H

H

C

H

H

C

H

H

C H

O

OCH

H

H

C C

H

H

C

H

H

C

H

H

C

H

H

C

H

H

C

H

H

C

H

H

C

H

H

C

H

H

C

H

H

C

H

H

C

H

H

C

H

H

C

H

H

C H

O

OCH

H

H

C C

H

H

C

H

H

C

H

H

C

H

H

C

H

H

C

H

H

C

H

H

C

H

H

C

H

H

C

H

H

C

H

H

C

H

H

C

H

H

C

H

H

C H

F2 Fats and oilsThe long-chain fatty acids that make up the bulk of the mass of a triglyceride molecule may be divided into two categories:(a) saturated(b) unsaturated, which can be further subdivided into: (i) mono-unsaturated (ii) poly-unsaturated

The structure of fatty acids

Saturated

Saturated fatty acids are those in which the carbon atoms are joined together only by C–C single bonds.

Saturated fatty acids therefore contain the maximum number of hydrogen atoms in the carbon chain. For example, stearic acid is a saturated fatty acid (Figure F5). It contains a carboxylic acid functional group and an alkane-like (saturated) hydrocarbon chain.

Figure F4 The structure of a triglyceride molecule.

Test yourself1 Give the molecular formula of the molecule,

X, shown and explain to which food group it belongs.

2 The molecule in Question 1 can react with another molecule in a condensation reaction to form the compound, Y, below. Work out the molecular formula of Y and give the structure of the molecule, Z, that must react with X to form Y.

HO OH

OHH

O

OH

H

H

HH

CH2OH

HO O

OH H

H

HOH

OHH

OO

OH

H

H

HH

HOCH2

CH2OH

CH2OH

Learning objectives

• Describe the structure of saturated and mono-unsaturated fatty acids and understand how structure aff ects melting point

• Deduce the stability of a lipid from its structure

• Describe the hydrogenation of unsaturated fats and its advantages and disadvantages

O H

H

C CH O

H

H

C

H

H

C

H

H

C

H

H

C

H

H

C

H

H

C

H

H

C

H

H

C

H

H

C

H

H

C

H

H

C

H

H

C

H

H

C

H

H

C C

H

H

H

H

C H

Figure F5 Stearic acid (C17H35COOH).


Unsaturated

Unsaturated fatty acids contain one (mono-unsaturated) or several (poly-unsaturated) C=C double bonds. They therefore do not contain the maximum number of hydrogen atoms in the carbon chain.

In a typical, naturally occurring lipid (oil or fat), there will be a mixture of saturated, mono-unsaturated and poly-unsaturated fatty acids (Figure F6). For example, butterfat (the main fat found in cow’s milk) contains predominantly saturated fatty acids. Table F1 compares the composition of butterfat to olive oil and sunfl ower oil, which are both derived from plants.

HO

Oa

b

c

HO

O

HO

O

Butterfat Olive oil Sunfl ower oil

% saturated 65 15 10

% mono-unsaturated 30 70 20

% poly-unsaturated 5 15 70

Table F1 Approximate fatty acid content of animal and plant lipids.

Fats and oils are categorised according to the predominant type of fatty acid present. Thus butterfat is classed as saturated, olive oil as mono-unsaturated and sunfl ower oil would be classed as poly-unsaturated.

Solid or liquid?Lipid-rich foods can have very diff erent physical properties. For example, beef dripping is a solid at room temperature but vegetable oil stays liquid even when placed in the fridge. The reason for this diff erence depends mainly on the structure of the fatty acids that make up the predominant triglyceride in the food.

Degree of crystallisationDegree of crystallisation is another way of describing the property of solidifi cation (i.e. how likely a substance is to be a solid) and is directly related to a fat’s melting point.

There are three main factors that aff ect the melting point of a fat: relative molecular mass (length of fatty acid chains), degree of unsaturation and nature of the double bonds (cis or trans).

Figure F6 (a) Elaidic acid; (b) linoleic acid; (c) arachidonic acid.

In general, saturated fatty acids will tend to be solid at room temperature, while mono- and poly-unsaturated ones will remain liquid.


Effect of relative molecular massAll other things being equal, increasing the relative molecular mass of a fatty acid will increase its melting point. Longer chain fatty acids tend to be solid at room temperature, e.g. stearic acid (C18H36O2 – Mr 284) melts at 69.6 °C whereas lauric acid (C12H24O2 – Mr 200) melts at 44.2 °C – both are saturated. The rise in melting point as the chain gets longer is due to an increase in the number and combined strength of the intermolecular van der Waals’ interactions along the length of the chains.

Effect of unsaturationUnsaturation (C=C bonds) in the hydrocarbon chain decreases the degree of crystallisation of lipids and they are more likely to be oils; this is why oils are a source of mono- and poly-unsaturated fatty acids. Some examples of saturated and unsaturated lipids are shown in Table F2.

Mono- and poly-unsaturated fatty acids generally have lower melting points than saturated fatty acids of similar relative molecular mass, because the double bond(s) causes many of them to adopt a ‘kinked’ or ‘curled’ structure (see Figure F7).

Effect of geometrical isomerismUnsaturated fatty acids can exist in two forms: cis and trans. These represent two diff erent arrangements of the hydrogens attached to the carbons in the double bond. In the cis form the hydrogens are both on the same side of the double bonds, whereas in the trans form they are positioned on opposite sides.

The greater the number of double bonds (i.e. the more unsaturated the molecule) the more distorted the chain will become, and hence the lower the melting point.

Name of food Predominant

type of fatty acid

State at room

temperature

Approximate

melting point / °C

butter saturated solid 35

lard saturated solid 30

shortening saturated solid 45

palm (oil) saturated solid 37

coconut (oil) saturated solid 26

olive oil mono-unsaturated liquid 0

peanut oil mono-unsaturated liquid −2

canola (rapeseed) oil mono-unsaturated liquid −10

saffl ower oil poly-unsaturated liquid −17

sunfl ower oil poly-unsaturated liquid −17

corn oil poly-unsaturated liquid −11

cod liver oil poly-unsaturated liquid −5

linoleic acid poly-unsaturated liquid −5

linolenic acid poly-unsaturated liquid −11

Table F2 Degree of saturation and approximate melting points of various lipids.

C

cis-but-2-ene trans-but-2-ene

H H

H3C CH3

C C

H

HH3C

CH3

C


In the trans isomer, the carbon chain remains straight and so contact points between chains are maximised; thus the strength of van der Waals’ interactions is greater. This means that trans fatty acids have a higher melting point than their cis isomers, which are kinked or bent and have a reduced number of potential contact points (Figure F7).

Rancidi� cation is the process by which fats, oils and other lipids undergo chemical decomposition.

a

b c

cis double bond

trans double bond

OH

H

C

C

CC

O CH

HOO

O

O CH

Figure F7 (a) Trans double bonds: chain is fairly straight; (b) cis double bond causes a kink in the chain; (c) cis double bonds cause the chains to be kinked and molecules cannot pack together as well.

Choice of fat for purposeFats and oils are chosen for cooking on the basis of their melting temperature. For example, cocoa butter melts at close to body temperature making it ideal for creating ‘melt-in-the-mouth’ chocolates; fats chosen for cake-making melt over a wide range of temperatures to provide diff erent baking properties and texture of the fi nal product.

StabilityIn general, fats that contain predominantly saturated fatty acids are more stable than those classifi ed as mono- or poly-unsaturated. This is due to the reactivity of the C=C double bond(s) present in the latter two. The absence of a reactive functional group in saturated fatty acids renders them fairly inert.

Saturated fats are more stable than unsaturated fats. They go rancid at a slower rate.

The C=C double bond in unsaturated fats may undergo a number of reactions: reaction with oxygen (auto-oxidation), reaction with hydrogen (hydrogenation), in the presence of light (photo-oxidation) and with enzymes/heat/water (hydrolysis).

Oxidation with atmospheric oxygen (auto-oxidation)Fats will go rancid even when kept at low temperatures; this is due to their reaction with oxygen in the air causing them to break down into a mixture of unpleasant-tasting compounds including hydrocarbons,


aldehydes and ketones. The mechanism involves the production of highly reactive free radical species. Generally, the more unsaturated fatty acids present in a fat, i.e. the greater the number of double bonds present, the faster the rate of oxidation.

HydrogenationIn this process, hydrogen gas is added across the double bond(s) present in mono- or, more usually, poly-unsaturated vegetable oils at a high temperature (140–225 °C) and pressure in the presence of a nickel, zinc or copper catalyst. This increases the saturated character of a fat (partial hydrogenation) and is used to turn liquid vegetable oils into semi-solid or solid fats; for example, margarine. Generally, the degree of saturation in the hydrogenated product is not 100% and, by carefully controlling the conditions, the percentage saturation can be varied to yield partially hydrogenated products with diff erent physical properties and uses. Hydrogenation is also used to produce other semi-solid fats used for baking.

Advantages of hydrogenationPartially hydrogenated vegetable oils are cheaper to produce than saturated fats from animals, and they also have an increased shelf life over their fully unsaturated precursors as the rate of oxidation decreases with increased saturation. Hydrogenation also increases hardness and plasticity (stiff ness) of the products.

Disadvantages of hydrogenationClinical evidence points to health benefi ts from a diet containing more mono- and poly-unsaturated fats than saturated. Saturated fats are more damaging to the heart and circulatory system. In addition, in the hydrogenation process, cis fatty acids can isomerise into the trans form. These have been implicated in a number of cardiovascular diseases. The body fi nds it diffi cult to metabolise trans fatty acids (lipase enzymes seem to recognise only cis forms) and so they tend to accumulate in the adipose (fatty) tissue rather than be excreted. Trans fatty acids also cause an increase in more harmful forms of circulating cholesterol such as low density lipoprotein (LDL). Finally, trans fats are a poorer quality source of energy compared to their cis counterparts.

Photo-oxidation (reaction in light)This is another free-radical-based reaction similar to auto-oxidation. In the presence of high light intensities, fats will oxidise and go rancid far more quickly than by auto-oxidation alone.

Hydrolysis (reaction with water, heat and enzymes)Hydrolysis involves splitting a molecule into smaller pieces using water. This chemical reaction occurs when the fat is heated with acid or alkali. It can also be catalysed by enzymes such as lipases. Water is able to cleave the ester bonds in a triglyceride between the fatty acids and the glycerol backbone. This will yield the triol (glycerol) and the original fatty acids (see Figure F8 on page 10). In the body this process is used in the digestion of fats and oils as fatty acids are more easily absorbed as separate units, rather than joined together in a triglyceride.

The release of fatty acids when fats undergo hydrolysis is one reason that fats can develop an unpleasant smell when they go rancid.

An enzyme is a biological catalyst that speeds up a specifi c biochemical reaction. Enzymes are proteins with a unique three-dimensional structure.


Test yourself3 The formulas of some fatty acids are shown below.

Deduce the number of C=C bonds in each and classify them as saturated, mono-unsaturated or poly-unsaturated.

C13H27COOH C15H23COOHC17H33COOH C19H35COOH

4 Arrange the following fatty acids in order of increasing melting point (lowest fi rst).

C17H35COOH C19H39COOHC17H31COOH

F3 Shelf life

Shelf life is the period of time for which a food retains the desired qualities expected by the consumer, such as � avour, texture and appearance. Once a food’s shelf life has been exceeded it is considered unsuitable for human consumption due to changes in the properties described above as well as the potential for microbial spoilage.

Learning objectives

• Understand the factors that aff ect the shelf life of a food

• Describe hydrolytic and oxidative rancidity of fats and how they can be reduced

• Describe traditional methods for extending shelf life

• Discuss natural and synthetic antioxidants including their structural features, sources, advantages and disadvantages of use

In most supermarkets, ‘best before’ dates are displayed clearly on packaging. These indicate the supermarket or manufacturer’s estimate of the food’s shelf life. Many items of food may still be edible beyond this date, but the food will not be in its optimum state. Shelf life is diff erent to the ‘expiration’ or ‘use by’ dates, after which a food product may become harmful.

Factors affecting shelf life and food qualitySeveral factors aff ect the shelf life and quality of food. These include water content, chemical changes (e.g. changes in pH), light, temperature and contact with air.

Water content Changes in the water content can alter the texture of food (e.g. if breakfast cereal absorbs water it becomes less crisp). There are several other reasons why changes in water content aff ect the shelf life of food.

The abundance of water increases the rate of hydrolysis and can lead to rancidity (due to hydrolysis of fats). Changes in water content can also result in degradation of nutrients such as vitamins, proteins and starch.

Changes in water content can also increase the rate of non-enzymatic and enzymatic browning of food (the process in which foods literally turn the colour brown), which is undesirable in foods such as fruits, vegetables and seafood. The rate of these processes is a maximum at an intermediate level of water content.

Bacteria fi nd it easier to multiply in foods containing higher quantities of water – drying or salting foods can increase their shelf life by reducing water content. These preserved foods are, however, particularly vulnerable to bacterial spoilage should they absorb moisture.


Chemical changes, e.g. pHIn a similar manner to increased water content, changes in the chemical composition of foods can lead to increased rate of hydrolysis. Both high and low pH levels can increase the speed at which proteins, starches and vitamins are degraded. Additionally, changes to the pH and chemical balance of a food will change its fl avour, colour and texture.

LightLight can cause rancidity in fats by the process of photo-oxidation (see page 7). Exposure to high light levels can also lead to chemical changes in many of the common vitamins, causing them to degrade more rapidly and rendering them less useful. Light is also able to bleach the natural dyes present in many foods, causing colours to fade. Appropriate packaging should be used to reduce exposure of the food to light.

TemperatureRaised temperatures generally act to increase the rate at which processes such as hydrolysis, photo-oxidation and microbial growth occur, thus reducing the shelf life of products. Keeping food in a cool, dry environment such as a pantry or fridge helps to alleviate the eff ect of raised temperature.

Contact with airAuto-oxidation of fats and other compounds such as alcohols occurs spontaneously in the presence of atmospheric oxygen. Fats turn rancid, while the ethanol in alcoholic beverages will be slowly converted to ethanoic acid by oxidation:

C2H5OH + O2 → CH3COOH + H2O

RancidityRancidity is a descriptive term relating to our perception of the fl avour of fats and oils: specifi cally, those fl avours that we perceive as unpleasant, indicating a food has gone off or is starting to spoil. Rancidity will be manifest as disagreeable changes in taste, smell, texture and appearance (Table F3).

Name of fatty acid Substance(s) detected in Characteristic smell/taste

butanoic acid milk, butter vomit

hexanoic acid milk goats, farmyard animals

octanoic acid milk musty, acrid

palmitic acid chocolate oil, soap

oleic acid chocolate wax, lard

stearic acid chocolate oil, fat

lauric acid palm/coconut oil, cocoa butter substitutes

soap

Table F3 Various ‘off-smelling’ naturally occurring fatty acids and their associated odours.


Hydrolytic rancidityThis is the process by which fats and oils are broken down into their constituent fatty acids and glycerol (Figure F8). The process can be speeded up by increased water content, the presence of enzymes (e.g. lipase), high (OH−) or low (H+) pH values and raised temperatures. During deep-fat frying, the extremely high temperatures and presence of water in foods can lead to rapid rancidifi cation.

H

H

glyceroltriglyceride fatty acid

H C OH

H OH +

H OH

C 3C17H35COOH

C17H35COC

OH

H

C17H35COC

O

H

C17H35COC

O

H

H

+ 3H2O

C

Figure F8 The hydrolysis of a triglyceride to yield one molecule of glycerol and three molecules of fatty acid.

Oily fi sh, like mackerel and herring, which contain poly-unsaturated fatty acids such as eicosapentaenoic acid (EPA) with fi ve C=C double bonds, are particularly susceptible to rancidifi cation.

Oxidative rancidityThe fatty acid chains liberated by hydrolysis can undergo oxidation via auto- and photo-oxidation in a free radical chain reaction; enzymes may also catalyse the mechanism. In this process, oxygen adds across C=C double bonds of unsaturated fatty acids and causes them to break down into unpleasant smelling and tasting volatile compounds such as ketones, aldehydes as well as some alcohols. The greater the degree of unsaturation present, the more predisposed a lipid will be to oxidative rancidifi cation.

Preventing rancidifi cation and increasing shelf lifeVarious techniques can be employed to reduce the rate at which lipids become rancid. These include processing the food, careful choice of packaging and adding additives.

ProcessingLow temperature storage/refrigeration reduces the rate of lipase-catalysed hydrolysis, particularly in dairy products. The lipase enzyme works best at its optimum temperature of around 37 °C.

Decreasing exposure to light by storing food in a dark environment (e.g. in a cupboard or pantry), or by using coloured glass for storage vessels will reduce the rate of adverse photochemical reactions.

Reducing moisture content by using methods such as salting, crystallising (adding sugar) and smoking decreases the eff ects of hydrolysis and microbial spoilage.

Packaging

Inert gasFilling packaging with an inert gas to expel atmospheric oxygen reduces the rate of auto- and photo-oxidation. Classic examples include fi lling packets of crisps (US: chips) and other snack foods with nitrogen.


Hermetic sealingUsing a fi lm packaging around the food, prevents it from coming into contact with oxygen in the air. Low-gas-permeability fi lm can be used as an alternative. Most fresh meat tends to be sold in hermetically sealed packaging.

Minimising the amount of air presentThis is used especially in bottled and canned foods. Reducing the quantity of air present at the top of the vessel will decrease the rate of oxidation.

Adding additivesSodium sul� te (Na2SO3), sodium hydrogensul� te (NaHSO3) and citric acid are added to foods as oxygen scavengers and help to prevent non-enzymatic browning. The sulfi tes are commonly added to fresh meat and dried fruit; citric acid is often used to preserve soft drinks.

Sodium and potassium nitrite and nitrate (NaNO2, NaNO3, KNO2, KNO3) are used to cure meat and fi sh, prevent the growth of bacteria and moulds and also as colour fi xatives.

Sodium benzoate (C6H5COONa) and benzoic acid (C6H5COOH) are used as preservatives preventing the growth of microbes in fruit juices, carbonated drinks and pickled foods (Figure F9).

Sorbic acid (C5H7COOH), propanoic acid (C2H5COOH), sodium and calcium propanoate (C2H5COONa and (C2H5COO)2Ca) can be added to bread and cheese to delay the growth of bacteria, yeasts and other moulds (Figure F10).

Ethanoic acid (CH3COOH) and benzoic acid prevent mould and bacterial growth in pickled meat and fi sh products. They also add to the fl avour. Vinegar is a solution of ethanoic acid.

The sulfi te, SO32−, ion is also

called the sulfate(IV) ion. The nitrite, NO2

−, ion is also called the nitrate(III) ion.

Osmosis is the passive process by which water moves from a region of low solute concentration (dilute solution) to a region of high solute concentration, through a partially permeable membrane (e.g. the cell membrane).

OH

O

Figure F9 Benzoic acid.

Oa

HC

CH

HC

CH

C

HC3 OH

Ob

CH3C

CH2 OH

Figure F10 (a) Sorbic acid; (b) propanoic acid.

Traditional methods for extending shelf lifeFermentation is used to increase shelf life of certain foods such as sauerkraut, dry sausages (salamis) and yogurt by using bacteria to break down the sugars present creating lactic acid which acts as the preservative.

Preserves (e.g. jams), are produced by boiling fruits to kill off microbes and then adding a gelling agent, such as pectin, as well as large amounts of sugar. The jelly formed prevents the growth of bacteria and moulds. The high sugar content also helps to reduce microbial growth by reducing available water. Other jellied preserves may use a gelling agent like aspic (e.g. jellied eels).

Pickling increases the shelf life of foods by keeping them in a liquid environment that is inherently antimicrobial. Brine and vinegar are classic pickling solutions as bacteria and mould fi nd it extremely diffi cult to multiply in the presence of these solutions. Examples include gherkins, pickled eggs and onions.

Salting is an ancient and highly successful preservation technique which involves reducing the water content of foods (by osmosis) and thereby decreasing the rates of hydrolysis, rancidifi cation and microbial growth. The Spanish delicacy bacalao is cod fi sh preserved by salting.

Figure F11 Lactic acid (2-hydroxy propanoic acid).

OH

HC

H3C

O

C

OH


HO

H3C

CH3

CH3

CH3

CH3 CH3 CH3

CH3O

Drying and smoking are other techniques with a long heritage. They also work by reducing the water content of foods. Smoked salmon and cured meats such as Parma ham and Westphalian ham are examples of foods preserved by this method.

AntioxidantsAntioxidants are substances that delay the onset or slow down the process of oxidation. They are commonly used as preservatives to extend the shelf life of foodstuff s. Naturally occurring antioxidants include vitamin C, vitamin E, β-carotene and selenium.

Vitamin C (ascorbic acid)Vitamin C (Figure F12) is found in high levels in citrus fruits, green-leaf vegetables, broccoli, green peppers, strawberries, raw cabbage and potatoes. It is broken down by cooking at high temperature and also dissolves in water during boiling.

Vitamin EVitamin E (Figure F13) is a fat-soluble antioxidant found in wheatgerm, nuts, seeds, whole grains, green-leaf vegetables, soya beans and vegetable oils such as canola. Vitamin E is part of a family of compounds called tocopherols. This family contains a number of compounds with vitamin E activity, based around a similar chemical structure to that in Figure F13, containing a methylated phenol group.

OHHO

HHO

HO O O

Figure F12 Ascorbic acid (vitamin C).

Figure F13 Vitamin E.

β-caroteneβ-carotene (Figure F14) is a precursor to vitamin A. It is highly fat-soluble and has a vivid red-orange colour. β-carotene is found in high quantities in carrots, squash, pumpkin, broccoli, sweet potato, tomato, kale cabbage, cantaloupe melon, peaches and apricots.

CH3 CH3

CH3

CH3

CH3

CH3 CH3 CH3

H3CH3C

Figure F14 β-carotene.

Selenium Selenium is found in fi sh, shellfi sh, red meat, eggs, grains, chicken and garlic.

Synthetic antioxidantsSynthetic antioxidants act as free-radical scavengers preventing oxidation via a free radical mechanism. The structures of some synthetic antioxidants are shown in Table F4.

The phenol group is common to all these compounds (Figure F15).

OH

Figure F15 Phenol.


Structure Name of antioxidant Arrangement

O

tertiary butylgroup

ethergroup

C

OH

CH3

CH3

CH3

CH32-tert-butyl-4-hydroxyanisole (2-BHA) 2- and 3-tert-butyl-4-hydroxyanisole (BHA):

differ by the position of the tertiary butyl group, (CH3)3C–, on the phenolic ring structure.3-BHA is the more effective antioxidant but commercially a mixture is used.The C–O–CH3 group is called an ether group. Specifi cally this is an aromatic ether.

O

C

OH

CH3

CH3

CH3

CH33-tert-butyl-4-hydroxyanisole (3-BHA)

OH

CCH3

CH3

CH3

CH3

CH3C

H3C

CH3 3,5-di-tert -butyl-4-hydroxytoluene (BHT) Contains a phenolic ring as well as two tertiary butyl side chains.Toluene is an alternative name for methylbenzene.

OH

OHHO

C O

O

H

H

C

H

H

C

H

H

C H

estergroup

propyl gallate (PG) contains 3 phenolic OH groups, has an ester functional group on a side chain

OH

OH

HO

C

O

H

H

C

H

H

C

H

H

C H

ketonegroup

2,4,5-trihydroxybutyrophenone (THBP) contains 3 phenolic OH groups, has a ketone functional group on a side chain

C

OH

OHCH3

CH3

CH3

tert-butylhydroquinone (TBHQ) contains a phenol group and a tertiary butyl group

Table F4 Structure of various synthetic antioxidants.


O

O

Figure F16 The fl avonoid core structure.

Use of antioxidants in food

Advantages

• Naturally occurring antioxidants have been shown to reduce the risk of heart disease and cancer by acting as free radical scavengers, preventing the damaging eff ects of these very reactive species.

• Vitamin C is essential for the synthesis of many hormones and the connective protein, collagen. It acts as a reducing agent in important enzymatic processes, transferring electrons to the enzymes when required.

• β-carotene can be added to margarine to provide colouring (yellow colour). It is also the principal precursor in the synthesis of vitamin A.

• Public consensus suggests that antioxidants are perceived as being benefi cial for healthy living and can be useful in the prevention of disease and ageing.

Disadvantages

• Synthetic antioxidants are generally perceived by the public to be less eff ective and possibly less safe because they are not naturally occurring.

• Natural antioxidants add colour and an aftertaste that is not always pleasant to foods. They also tend to be less eff ective at reducing the rate of oxidation when compared to their synthetic alternatives.

• There is a signifi cant amount of legislation, policy and scientifi c testing that must be in place before synthetic antioxidants can be used in food. This is needed to ensure that they are safe for human consumption.

• Across the globe and especially in developing countries, it is often very diffi cult to ensure the correct implementation and monitoring of the safe use and labelling of food additives in food.

Antioxidants found in traditional foodsMany traditional foods contain another class of antioxidant, the fl avonoids (Figure F16). The benefi cial eff ects of some of these foods are given in Table F5.

Traditional food Benefi cial effects

green tea reduced cholesterol, decreased risk of heart disease, lower blood pressure, reduced rate of cancer cell proliferation

turmeric antibacterial, anti-infl ammatory, potential benefi ts in the treatment of Alzheimer’s disease and cancer

oregano antimicrobial action

blueberries decrease cholesterol levels, possible anti-Alzheimer’s disease effect

cranberries possible benefi ts in preventing heart disease and cancer, boosting the immune system, prevent bacterial infections in the urinary tract

dark chocolate reduced blood pressure, prevention of heart disease

Table F5 Benefi cial effects of foods containing fl avonoid compounds.

Free radicals are chemical species that contain an unpaired electron. This causes them to be highly reactive. In a biological setting, free radicals such as the hydroxyl radical (OH•) and superoxide radical (O2

−•) can be generated from the metabolism of oxygen. Due to their reactive nature they can cause signifi cant cellular damage by reacting with proteins, lipids and DNA, leading to inactivation of enzymes, degradation of the cell membrane and mutation of the genetic code, among many other eff ects.


F4 Colour

A dye is a food-grade, synthetic, water-soluble colourant.A pigment is a naturally occurring colourant found in the cells of plants and animals.

Occurrence of colourNaturally occurring pigments possess colour due to their ability to absorb diff erent wavelengths of visible light and transmit or refl ect the portion of the spectrum that has not been absorbed. The transmitted or refl ected light is the complementary colour to the light absorbed. Thus a compound which absorbs the blue-green wavelengths of light will appear red-orange (Figure F17).

Test yourself5 The molecule shown here has been synthesised as

a potential antioxidant. Copy the molecule, then circle the tertiary butyl group and identify any other functional groups present.

OH

OH

C(CH3)3

O

C

H

H

C

H

H

CH3C O

Learning objectives

• Explain the occurrence of colour in naturally occurring pigments, including the range of colour and sources of carotenoids, anthocyanins, chlorophyll and heme

• Describe the factors that aff ect stability of colour in carotenoids, anthocyanins, chlorophyll and heme

• Discuss the issues associated with the use of synthetic colourants

• Understand the diff erence between caramelisation and the Maillard reaction in the browning of foodyellow orange-

yellow

orange

red-orange

violet-red

blue

green

yellow-green

blue-green

blue-violet

red

violet

Figure F17 A colour wheel.

Chlorophylls are the green pigments that give most plant leaves their characteristic colour. Chlorophylls have a complex structure containing multiple C=C double bonds that allow the compounds to absorb specifi c wavelengths of light in the visible spectrum. Chlorophyll a (Figure F18) absorbs visible light in the blue (400–500 nm) and red (620–700 nm) regions of the visible spectrum. It does not absorb light in the green


part of the spectrum (500–600 nm) and so these wavelengths of light are transmitted. Chlorophylls therefore appear green as do the leaves in which these compounds are found in high concentrations.

Foods may have colour due to the presence of naturally occurring pigments or the addition of artifi cial dyes. Dyes appeared coloured for exactly the same reasons as described above.

Naturally occurring pigmentsNaturally occurring pigments include anthocyanins, carotenoids, chlorophyll and heme.

AnthocyaninsAnthocyanins are the most widely occurring pigments in plants. They all have a characteristic structure based on the same core unit. Anthocyanins are the principal pigments responsible for the pink, red, blue and purple colours of many fruits and vegetables including red cabbage, blackcurrants, strawberries, cranberries, blueberries, raspberries and grapes (Figure F19).

a b

N

N

H

O

O

O

O

O

N–N– Mg2+

Rela

tive

abso

rban

ce

300ultraviolet violet blue green yellow red infrared

400 500 600

Wavelength / nm

700 800

Figure F18 (a) The structure of chlorophyll a; (b) its visible light spectrum.

OCH3

OCH3

OH

OH

OHOH HOHO

HO

a

O

O O

+

b

OH

OH

OH

OH

OHOH HOHO

HO O

O O

+

Figure F19 (a) Oenin found in purple grape skin; (b) myrtillin found in blackcurrants and blueberries.


CarotenoidsCarotenoids are the most widespread pigment found in nature; primarily due to their abundance in algae. Many of the carotenoids are precursors of vitamin A. The structural similarities between vitamin A and β-carotene can be seen in Figure F20.

Carotenoids generally transmit or refl ect longer wavelengths of the visible spectrum (absorb in the blue-violet region) and so have colours in the yellow-orange-red region. Carotenoids are present in bananas, carrots, tomatoes, watermelon, sweet peppers and saff ron (Figure F21).

The red carotenoid astaxanthin (Figure F22), when it is complexed with protein, gives live lobsters and crabs their blue-green hue. When the shellfi sh is cooked, the astaxanthin becomes dissociated from the protein and the shell turns red. Astaxanthin is also responsible for the deep pink colour of wild salmon. Farmed salmon are fed supplements that include astaxanthin so that their fl esh matches the colour of wild salmon.

H3C

OH

CH3

CH3 CH3 CH3 H3C

H3C

CH3 CH3 CH3CH3

CH3

CH3 CH3

CH3

a

b

OH CH3 CH3

CH3 CH3

O

O

O O

O

OH

HO

HO

HO

HO

OH

OH

OH

OH

O

O O

O

O

OH

HO

OH

OH

a

b

CH3 CH3 CH3 CH3

CH3 CH3 CH3 CH3

CH3

CH3

OH CH3 CH3

CH3 CH3

O

O

O O

O

OH

HO

HO

HO

HO

OH

OH

OH

OH

O

O O

O

O

OH

HO

OH

OH

a

b

CH3 CH3 CH3 CH3

CH3 CH3 CH3 CH3

CH3

CH3

Figure F20 (a) Retinol (vitamin A); (b) β-carotene.

Figure F21 (a) Lycopene found in tomatoes and watermelons; (b) crocin found in saffron.


ChlorophyllsThese are the major pigments in green plants responsible for harvesting light energy – the fi rst stage in the process of photosynthesis.

HemeHeme is the red pigment found in the haemoglobin of red blood cells and the myoglobin of muscle tissue (Figure F23). The heme in myoglobin is what gives meat its purple-red colour.

Stability of pigment coloursNumerous factors aff ect the stability of pigments. Variation in these factors may result in the loss or change of pigment colour.

Major factors aff ecting pigment colour are: temperature, pH, oxidation, presence of metal ions and the oxidation state of these metal ions. These factors contribute to changes in the structure of the pigment and/or the way in which it absorbs certain wavelengths of light.

Effects on anthocyanins(A) (AH+) (B) (C)

In aqueous solution, anthocyanins exist in a complex equilibrium between four diff erent structural forms (Table F6). The position of the equilibrium depends on the pH and temperature of the solution. At low pH (1–3) the red fl avylium form predominates. When the pH is increased, the blue quinoid (pH 6–7) and colourless carbinol (pH 4–5) forms may be produced. In basic conditions, the chalcone is formed.

O

OOH

HO

H3C

H3C

H3C

CH3

CH3CH3 CH3

CH3 CH3CH3

Fe2+

C

N –N

NN–

H

CH

CH

CH2

H3C

H3C

HOOC COOH

HC

CH3

CH2CH3

Figure F22 Astaxanthin.

Figure F23 The structure of the heme prosthetic group.

Anthocyanin Structural

form

Colour

(A) quinoid blue

(AH+) fl avylium red

(B) carbinol base colourless

(C) chalcone colourless

Table F6 Structural forms and colour of anthocyanins.

O+

OH

OH

HO

flavylium cation (red)

pH 1–2 pH 4–5

carbinol

OH

O

OH

OH

OHHO

OH

O

OH

OH

O

quinoidal base (blue)

pH 6–7 pH > 7

chalcone

OH

OH

OH

HO OHO

OH


At low temperatures, the coloured forms are abundant, but as the temperature rises, the equilibrium shifts to the right causing the pigments to lose their colour. This also causes them to lose their potent antioxidant ability, which leads to browning of the food.

Anthocyanins are able to form vivid, deep-coloured complexes with Al3+ and Fe3+ ions. These metal ions can come from the packaging in canned fruit and often causes discoloration in these food products.

Effects on carotenoidsDue to the presence of a conjugated, multiple C=C double bond system, carotenoids are not only coloured but highly susceptible to photo-oxidation and oxidation by metals and organic hydroperoxides (R–O–O–H). On oxidation, the conjugation is destroyed and the pigment is bleached (decolorised). The oxidised species are also unable to be converted to vitamin A and often have an unpleasant odour.

Carotenoids are thermally stable up to 50 °C and structurally stable in acidic conditions (pH range 2–7). They tend, therefore, to withstand most forms of food processing without being degraded. When heated above 50 °C, however, the abundant naturally occurring all-trans form rearranges to yield a variety of cis isomers.

Effects on chlorophyllsChlorophylls can be destabilised by high temperatures depending on the pH of the environment. In an alkaline solution (pH 9), chlorophyll remains stable, but in an acidic solution (pH 3) it is highly unstable. When plant material is heated, i.e. cooked, the plant cell membranes break down and cause acid to be released, thus decreasing the pH of the surrounding solution. The magnesium ion located at the centre of the porphin ring (see Figure F18a on page 16) is normally stable and diffi cult to remove, but at low pH, the Mg2+ ion is displaced by two H+ ions, resulting in formation of an olive-brown pheophytin complex.

Generally, brighter-green vegetables are viewed as more appealing than darker coloured ones; by reducing cooking time and boiling vegetables with the pan lid off to allow the escape of volatile acids, the production of pheophytin and discoloration can be minimised. Adding a small amount of sodium hydrogencarbonate (bicarbonate of soda, NaHCO3) to the water during boiling may also help to keep the vegetables green as this raises the pH.

A conjugated system is a system of alternating C=C double bonds and C–C single bonds.

The heat-induced cellular degradation can also predispose the chlorophyll pigments to photo-degradation in which they chemically decompose in the presence of light.

HemeFreshly cut meat has a purple-red appearance due to the predominance of the pigment myoglobin in the tissue. If freshly cut meat is exposed to air it changes colour within an hour. When oxygen in the air binds to the purple-red myoglobin (Mb) it forms oxymyoglobin (MbO2) which is red – the colour most people associate with meat being fresh. In both these complexes, the iron ion at the centre of the heme porphyrin ring


(see Figure F23 on page 18) is in the +2 oxidation state (Fe2+). However, if meat is left exposed to air for several days, the myoglobin and oxymyoglobin undergo auto-oxidation in which Fe2+ is oxidised to Fe3+ which results in the formation of metmyoglobin (MMb) which has an undesirable brown-red colour and indicates that the meat is no longer fresh.

oxymyoglobin (MbO2)deoxygenation

oxygenationred purple-red (Fe2+) brown-red (Fe3+)

oxidation

myoglobin (Mb) metmyoglobin (MMb)oxidation

reduction

This interconversion between the three forms of myoglobin happens readily. Formation of metmyoglobin in meat products can be reduced by packaging the food in low-gas permeability fi lm or hermetically sealing them in oxygen-free conditions. The storage gas used in the packaging of many meat products is 100% CO2, which prevents oxidation from occurring and the meat from discolouring.

Safety issues associated with dyes and synthetic colourantsOver time, myriad artifi cial additives have been put in food to make it more fl avoursome and colourful – in general, to make it more appealing and saleable. Many of the products you have in your fridge and pantry will not actually contain sources of the food they purport to be. So-called ‘crab’ sticks do not actually contain any crab meat, but rather are made of white fi sh (surimi) containing an artifi cial fl avour and cochineal to make them taste and look like crab meat.

Little was known about the safety of food additive compounds and the associated health implications of their distribution into the populous during their early development. Since then many problems have been encountered, some of them extremely serious.

One commonly used food colourant, Sunset Yellow (E110), has been linked to severe allergic reactions and also to hyperactivity in children. There have been repeated calls for its withdrawal from the marketplace. Another colourant, Amaranth (E123), has been banned in the USA for over 30 years as it is suspected by the Food and Drugs Administration (FDA) of being a carcinogen. It is still legal in the UK, however, and is used to give glacé cherries their distinctive red colour. Other synthetic dyes with alleged harmful eff ects include Brilliant Blue (E133) and Erythrosine (E127).

The main problem facing society concerning food colourants is the current lack of uni� ed global legislation governing their use.

Each country may have its own specifi c set of guidelines and policies for the use of additives in foodstuff s. A suspected carcinogen banned in one country may be a commonly used additive in another. Due to the international trade in food, this lack of uniformity is of major concern as imported foods may contain potentially harmful chemicals.

Non-enzymatic browningThere are two principal processes by which non-enzymatic browning of foods occurs: the Maillard reaction and caramelisation.

Food additives, including synthetic colourants, are given ‘E numbers’ in Europe, a code known elsewhere in the world as the International Numbering System (INS).


The Maillard reactionThis form of non-enzymatic browning occurs in foods containing both protein and carbohydrate. The process involves a reaction between the amino group of an amino acid and a reducing sugar (e.g. glucose, maltose, lactose) to yield a mixture of products and water as it is a condensation reaction. Milk contains both protein and lactose sugar and so browns by the Maillard reaction.

Examples of the Maillard reaction include the formation of brown caramel, fudge and toff ee by heating sugar (carbohydrate) with cream (protein), the production of milk chocolate and toast browning.

The Maillard reaction produces desirable outcomes such as a beautiful golden-brown colour (though undesirable colours can also be produced) and a change in smell and/or taste to give a caramel or nutty aroma.

The rate at which the Maillard reaction takes place can be accelerated by increasing the temperature (frying, roasting), low moisture levels, a higher content of monosaccharide sugars and increased levels of specifi c amino acids such as lysine. Lysine-containing foods (e.g. milk) brown very nicely, but cysteine-containing foods develop a minimal brown colour.

The Maillard reaction occurs spontaneously at room temperature, but the eff ects of the reaction are not noticeable because the rate is slow. The browning eff ect of the reaction becomes appreciable only at the temperatures involved in cooking.

CaramelisationCaramelisation is a form of non-enzymatic browning that occurs in foods with a high content of carbohydrates, especially sucrose and reducing sugars. Nitrogen-containing compounds must be absent otherwise the Maillard reaction will occur. Caramelisation involves the process of pyrolysis, in which a high temperature causes the sugars to oxidise and decompose to form volatile compounds that may then go on to polymerise. The products of decomposition give food its resultant nutty, caramel-like aroma, while the products of polymerisation make the food light-brown. Baked-egg dishes go brown on top as a result of caramelisation.

The rate of caramelisation can be increased by catalysis with a base (pH > 9) or an acid (pH < 3), as well as by using high temperatures in excess of 120 °C, which occur when baking and roasting foods with a high sugar content.

When foods are boiled, neither caramelisation nor the Maillard reaction occur appreciably as the temperature is not suffi ciently high. Many recipes therefore suggest that meat is browned and vegetables such as onions are caramelised to develop fl avour and colour before they are cooked using a method requiring a temperature below 100 °C (e.g. stewing).

The brown colour of chocolate does not come from the Maillard reaction as cocoa is naturally brown. However, the Maillard reaction is partly responsible for the fl avour of milk chocolate.

Cysteine and lysine are amino acids.

Examiner’s tipDespite the obvious name, caramel is not formed by caramelisation but rather by the Maillard reaction.

Test yourself6 Use the colour wheel on page 15 to suggest the

colour of compounds that absorb the following colours of visible light:

a green b orange c red

7 Name two classes of naturally occurring pigments.


F5 Genetically modifi ed foodsA genetically modifi ed (GM) food is defi ned as a food derived or produced from a genetically modifi ed organism. The food can be substantially diff erent from or essentially the same as the conventional food, in terms of composition, nutrition, taste, smell, texture and functional characteristics.

GM foods have been available in the marketplace since the 1990s. Most GM food is plant-based and includes products such as vegetable oils, fruits, vegetables and rice. Some animal-derived GM foods have been developed including (in 2006) a transgenic pig genetically modifi ed to be rich in ‘good’ omega-3 fatty acids.

The process of genetic modifi cation involves the insertion or deletion of specifi c genes in the GM organism to produce the desired characteristics. In the production of transgenic species, the inserted genes come from a completely unrelated organism and this is partly where the controversy lies.

Benefi ts of GM foods 1 Increased yield and nutritional content of crops – of particular benefi t

to developing countries with exploding populations and those prone to famine or endemic malnutrition. So-called ‘Golden Rice’ has been genetically modifi ed by inserting genes from daff odil and a bacterium, so that it is enriched with vitamin A. Increasing yield also permits diversifi cation in the uses of some crops (e.g. use of vegetable oils as biofuel – a renewable alternative to gasoline).

2 Reduced maturation time for crops – increases annual yield and permits more frequent harvests. Ideal for feeding a growing population or meeting increased consumer demand for a product.

3 Ability of crops to grow in adverse conditions – for example, the ability to grow in a hot climate with limited water and mineral availability. It might be possible to grow GM plants that are able to remove pollutants (e.g. heavy metals), from barren soil effi ciently, allowing it to be farmed again.

4 Improved crop resistance to disease, drought, pests and herbicides – inserted genes could confer these benefi cial characteristics on crops to prevent damage from microbial infection or attack by damaging pests such as beetles and locusts.

5 Enhanced taste, texture, quality and shelf life – improves upon currently available foods and makes the GM products more desirable and saleable.

6 Production of ‘healthy’ crops – which can produce higher levels of useful unsaturated fatty acids and fewer saturated ones. It may also be possible to modify crops so that they contain ‘medicinal’ compounds such as antioxidants or anticancer compounds.

7 Improved productivity and feed effi ciency in GM animals – means that livestock would require less food per kilogram gain in body mass.

8 Improved animal health – GM animals could also be developed to be more resistant to disease.

9 Enhanced yield of dairy products – such as milk and eggs. 10 Replacement of chemical insecticides and herbicides – with

‘environmentally friendly’ bioherbicides and bioinsecticides to eradicate damaging weeds and pests.

Learning objectives

• Understand what is meant by a genetically modifi ed organism

• Discuss the advantages and disadvantages of using genetically modifi ed foods

In the transgenic pig, the inserted gene designed to make the pig produce more omega-3 oils came from the microscopic nematode worm, Caenorhabditis elegans.


11 Improved conservation of water, soil and energy – GM crops can be engineered to require less space, less water and less management. Better natural waste management is an additional benefi t.

Concerns about GM foodsThere is signifi cant lobbying and opposition to the use and globalisation of GM foods. Some of the arguments against genetic modifi cation include:1 Not enough is known about the long-term eff ects – as GM foods are

a fairly modern invention, not enough is known about the long-term eff ects of their introduction. Eff ects on the environment and human health may take many more years to be apparent. There is still much to learn about the way in which genes operate and interact.

2 Escape of transgenic material into the ‘wild’ – possible contamination of indigenous, natural species with unknown eff ects. Imagine the eff ects of a disease- and herbicide-resistant gene becoming fully integrated into several species of pervasive weeds. A gene conferring antibiotic resistance could possibly be transferred to a species of bacteria making it more diffi cult to treat infection.

3 Alleged increase in allergies – especially in those working directly with GM organisms.

4 Damage to the environment and unbalancing of ecosystems – may lead to the endangerment of multiple species and upset the current balance of ecosystems and food chains worldwide.

5 Alteration of the composition of a balanced diet – by changing the nutritional content and quality of foods.

6 Exploitation and monopolisation – by countries and companies with the intellectual property rights to the technology. Developing countries may become dependent on GM foods.

7 Ethical considerations – include tampering with nature, general disgust at consuming modifi ed animal products and animal welfare.

F6 TextureThe texture of foods can be described using many diff erent adjectives including soft, gummy, crispy, viscous and rough. Many individual foods have the same textural qualities throughout – there is no change of texture within the food. This uniformity is described as homogeneity. Many foods appear homogeneous although they actually contain components that are immiscible with each other. This is because they contain very small particles (too small to be seen with the naked eye) of one phase dispersed in another. Imagine smoke rising from a fi re – the smoke appears as a uniform and continuous grey blanket, when it is actually billions of microscopic solid particles fl oating in air. The two phases (solid smoke particles and gaseous air) are immiscible – they remain separate and discrete and yet the overall impression is one of homogeneity (though the solid particles could be seen easily under a microscope). This type of system is called a dispersed system or a colloid.

Phases are diff erent substances separated by a boundary (e.g. in mayonnaise the oil and vinegar are diff erent phases). The particles which are fi nely dispersed make up the dispersed phase. The phase they are dispersed in is called the continuous phase.

A dispersed system can be de� ned as a kinetically stable mixture of one phase (dispersed phase) within another, largely immiscible, phase (continuous phase).

Learning objectives

• Understand what is meant by a dispersed system and their relationship to the texture of food

• Distinguish between emulsions, suspensions and foams in food

• Describe the action of emulsifi ers


There are various dispersed systems possible (many are found in foods) depending on the phases involved. In most foods the continuous phase is a liquid and the state of the dispersed phase varies. Examples are shown in Table F7.

Type of dispersed system Continuous phase Dispersed phase Example

sol(can also be called a suspension)

liquid solid melted chocolate (continuous phase – molten cocoa butter, dispersed phase – cocoa solids), many natural food colourings

emulsion liquid liquid mayonnaise (continuous phase – oil, dispersed phase – vinegar), cream, salad dressing

foam liquid gas whipped cream (continuous phase – liquid cream, dispersed phase – air), beer froth, raw meringue

gel solid liquid jelly

solid foam solid gas cooked meringue, souffl é

Table F7 Dispersed systems.

Emulsifi ersEmulsifi ers are substances that increase the kinetic stability of a dispersed system. They aid the formation of emulsions or foams. Emulsifi ers act as an interface between the liquid continuous phase and the solid or gas dispersed phase holding them together.

Generally, emulsi� ers have both hydrophilic (water-liking) and lipophilic (fat-liking) properties and can therefore dissolve in both the continuous and dispersed phases helping bind them together.

Emulsifi ers are added to emulsions and foams to prevent the individual phases from separating from one another and to maintain homogeneity. Surface tension causes liquids to try to have minimum surface area. When two droplets collide and combine to form a single, larger droplet the surface area is less, therefore it is favourable for the colloid to separate out. Emulsifi ers lower the surface tension between two liquids and stabilise the emulsion.

Typical emulsifi ers used in food production include mustard and egg yolk. Mustard contains mucilage which consists of long-chain polysaccharides and polypeptides. This allows it to bind to both oil and water and to prevent the small droplets of each from separating. This is useful for making vinaigrettes. Egg yolk contains a substance called lecithin. Lecithin is used in chocolate bars to help the cocoa solids and butter mix and in mayonnaise to help the oil and vinegar form an emulsion.

Lecithin contains phospholipid and glycolipid molecules. These have fatty acid chains which will dissolve in the oily phase of the emulsion (lipophilic) and polar phosphate/sugar heads which will dissolve in the aqueous phase (hydrophilic), thus keeping the two phases bonded together (Figure F24).

To make an emulsion, an oil, water and an emulsifi er need to be beaten, mixed or whisked together using mechanical energy. This breaks the intermolecular forces between the molecules in each of the phases and causes them to form minute droplets that take some time to coalesce back into large globules; the rate of coalescence is signifi cantly reduced

The colloidal particles of the dispersed phase scatter light so you cannot see through a colloid. Compare a cup of black coff ee (solution – see-through) to a cup of white coff ee (dispersed system – not see-through).


in the presence of stabilisers, such as pectin or agar, and emulsifi ers. While emulsifi ers help the two phases of a dispersed system to form an emulsion, a stabiliser works to prevent the phases from separating.

hydrophilic head

lipophilic tail

emulsifier

OIL

water molecules

Figure F24 Emulsifi er structure.

Test yourself8 For each of the following dispersed systems

(colloids) give the state of the continuous phase and dispersed phase:

emulsion foam

9 State what types of dispersed system (colloid) are present in each of the following foods:

mayonnaise froth on cappuccino beaten egg white hot chocolate

F7 Oxidative rancidity (auto-oxidation)

Oxidative rancidity (auto-oxidation)Fats turn rancid spontaneously in the presence of atmospheric oxygen by the process of auto-oxidation. This yields a mixture of volatile and unpleasant ‘off ’ smelling compounds including aldehydes, ketones and alcohols. The principal mechanism by which auto-oxidation is thought to occur is a free-radical chain reaction. The key intermediates in the process are organic hydroperoxides (ROOH). As with other free-radical chain reactions, the process can be split into initiation, propagation and termination steps.

InitiationIn the fi rst step, a C–H bond undergoes homolytic � ssion to produce two free radicals.

RH → R• + H•

The homolytic fi ssion of the C–H bond requires a signifi cant amount of energy input due to the strength of the C–H bond (412 kJ mol−1).

HL Learning objectives

• Describe the free radical mechanism of oxidative rancidity in fatty acids

• Understand the role of lipid peroxides as key intermediates during rancidifi cation

RH represents any unsaturated fatty acid with hydrogen atoms that can be lost to form free radicals (e.g. linoleic acid).


HL PropagationThe alkyl radical (R•) produced in the initiation step now reacts with oxygen to form a peroxyl radical (ROO•). This peroxyl radical is able to remove a hydrogen from an intact fatty acid molecule to form a hydroperoxide and regenerate the alkyl radical, which can subsequently react with oxygen and continue the chain reaction.

R• + O2 → ROO•

ROO• + RH → ROOH + R•

This mechanism shows how a large number of hydroperoxide molecules can be formed from a small number of free radicals.

TerminationIn the termination steps, two free radicals combine to form a non-radical species. This stage produces a mixture of compounds.

R• + R• → RR

ROO• + R• → ROOR

ROO• + ROO• → ROOR + O2

Further reactions of the organic hydroperoxidesThe organic hydroperoxides (ROOH) undergo further oxidation and decomposition to produce more free radicals and a wide variety of products. The decomposition of the hydroperoxide is dependent on photolysis (split by light) of the O–O bond, or on catalysis with transition metals.

When the hydroperoxide decomposes, a hydroxyl radical is also formed. These new free radicals can react with any of the other free radicals or non-radical species present to produce the aldehydes, ketones and alcohols responsible for the disagreeable odour of rancid fat (Figure F25).

Examiner’s tipRemember, the partial hydrogenation of unsaturated fats uses metals such as nickel as catalysts, so there are trace amounts of these metals in products such as margarine. Iron is a commonly occurring transition metal in many foods and also promotes rancidifi cation.

Examiner’s tipThe equations for the decomposition reactions of lipid hydroperoxides are not required for the exam, but an understanding that decomposition occurs to yield ketones, aldehydes and other products, as well as additional free radicals, is expected.

Hultraviolet

lipid peroxide

aldehyde

radiation

ketone

CR1 R2

H

H

CR1 R2 OH+

HUV

CR1 R2 R RH+ +CR1 R2

+CR1 R2

OO

O

O O

O

H

Figure F25 The formation of volatile molecules from the decomposition of a lipid peroxide.

Splitting of the O–O bond requires less energy input than the initial C–H homolysis due to a signifi cantly lower bond enthalpy (O–O 146 kJ mol−1).

Extension

The exact mechanism of initiation in auto-oxidation is poorly understood, it is believed that the removal of a hydrogen atom from the fatty acid chain actually requires an initiator molecule which may be oxygen itself or a free radical present in the substance.

In• + RH → In–H + R•

Test yourself 10 Draw out a reaction sequence to show how pentan-3-one

could be formed from a suitable organic hydroperoxide.


F8 AntioxidantsThere are three main types of antioxidants: free-radical scavengers (free-radical quenchers), chelating agents and reducing agents.

Free-radical scavengers (quenchers)Antioxidants (AH), such as vitamin E (tocopherols), BHA, BHT and TBHQ (see pages 12 and 13) act as free radical scavengers. That is, they prevent the formation of free radicals during the initiation step, or they ‘mop up’ free radicals during the propagation steps of lipid auto-oxidation and thus prevent a chain reaction.

The antioxidant reacts with free radicals so that they are not available to react with lipid molecules, oxygen or other free radicals. In doing so, an antioxidant radical (A•) is produced, but these are less reactive than the other free radicals present because of delocalisation of the free electron (Figure F26).

Eventually, the antioxidant radical is removed in a termination step forming a non-radical species.

During initiation:

RH → R• + H•

AH + R• → RH + A•

where AH is the antioxidant molecule.During propagation:

R• + O2 → ROO•

ROO• + AH → ROOH + A•

ROOH → RO• + HO•

RO• + AH → ROH + A•

HO• + AH → H2O + A•

HL Learning objectives

• Describe the mechanism of action of free radical scavengers

• Understand the role of chelating agents as antioxidants

• Explain the function of reducing agents as antioxidants

HO

OH3C

CH3

CH3

CH3 CH3 CH3

CH3

CH3

a

b

O

OH3C

CH3

CH3

CH3 CH3 CH3

CH3

CH3

Figure F26 (a) Vitamin E (as AH); (b) vitamin E radical (A•).


During termination:

R• + A• → RA

RO• + A• → ROA

ROO• + A• → ROOA

HO• + A• → AOH

The antioxidants mentioned here (BHA, BHT, TBHQ and tocopherols) all contain a phenolic ring structure which is excellent at sequestering free radicals. The hydrogen on the phenolic –OH group is easily removed.

Chelating agentsThe decomposition of organic hydroperoxides may occur in the presence of metal ions and so removing these ions from solution acts to reduce the rate of hydroperoxide decomposition and rancidifi cation of fats. One method of removing ‘active’ metal ions from solution is to use a chelating agent (from the Greek meaning ‘claw’) which essentially wraps the metal ion up inside a molecular cage, like clutching a coin in a closed fi st. This prevents the metal ion from coming into contact with other species such as organic hydroperoxides and thereby prevents the auto-oxidation reaction.

One well-known and commonly used synthetic chelating agent is ethylenediaminetetraacetic acid or EDTA for short (Figure F27).

HL

C

O O

O O

C C

CC

C

CH2 CH2

CH2 CH2

HO OH

OHHO

NNH2 H2

Figure F27 Ethylenediaminetetraacetic acid (EDTA) in its uncharged form.

OM3+

–

–

–

N

O

O

N

O

O

O

O

O

–

Figure F28 The [EDTA]4− ion bound to a central metal ion in an octahedral coordination complex.

EDTA is called a hexadentate ligand, literally meaning six teeth.

The four acidic protons on EDTA partially dissociate in aqueous solution to produce an ion with a 4− charge, [EDTA]4−, with available lone pairs of electrons on the four O− atoms and the two nitrogens. These six atoms are able to form an octahedral coordination complex with a central metal ion (Figure F28).

In doing so, the metal ion is almost entirely enclosed in the structure of the [EDTA]4− ion. In many food and cosmetics products, EDTA is added as di- or tetrasodium EDTA in which the [EDTA]2−/4− ion is already present.

By preventing metal-catalysed oxidation, [EDTA]4− is eff ective at: reducing non-enzymatic browning of potato products, fi sh and shellfi sh; preventing fl avour changes in milk due to oxidative rancidifi cation; preventing oxidation of meat products; and preventing discoloration of canned fruits and vegetables by chelating Al3+ and Fe3+ ions.

Natural chelating agents are found in many plant extracts such as rosemary, tea and ground mustard.


Reducing agentsAntioxidants falling into this class include the carotenoids and ascorbic acid (vitamin C). They work by acting as electron and hydrogen donors which causes molecular oxygen to be reduced and hence removed (or at least lowered in concentration) from the environment.

Ascorbic acid reduces oxygen to water and, in the process, is converted to the oxidised form, dehydroascorbic acid (Figure F29).

F9 Stereochemistry in food

Naming enantiomeric stereoisomersSeveral diff erent systems are employed for naming the enantiomers present in food.

The D and L systemThe D and L convention for naming enantiomeric forms of a molecule tends to be used for carbohydrates and amino acids.

The D and L system is based on the spatial con� guration of groups around a chiral carbon atom.

When applying the D and L system, the molecule glyceraldehyde (2,3-dihydroxypropanal) is used as a reference. The chiral centre is on the central carbon in glyceraldehyde (Figure F30).

When applying the D and L system, it is the chiral carbon furthest away from the carbonyl group that is important. When shown in the Fischer projection (see page 2), if the –OH group on this important carbon is on the right, the molecule is given the D notation; if the –OH group is on the left, it is given the L notation.

Looking at two forms of glucose in the Fischer projection, we can compare them to glyceraldehyde to determine which is the D enantiomer and which is the L form (Figure F31).

If the molecule is not shown in Fischer projection it must be considered whether the molecule could be converted by a series of chemical reactions into D- or L-glyceraldehyde. If it could be converted into D-glyceraldehyde then it is assigned the label D.

Amino acidsWhen applying the D and L system to amino acids, the ‘CORN’ rule is used. This rule looks at the positions of the COOH, R and NH2 groups around the chiral carbon.

HL

HO

HO

O

a b

OH

OHHO

HO

HO

O OH

OO

Figure F29 (a) The reduced form of ascorbic acid; (b) the oxidised form of ascorbic acid, dehydroascorbic acid.

Learning objectives

• Apply the D/L, R/S and (+)/(−) systems for naming enantiomers

• Discuss the diff erent occurrence of enantiomers in food

• Distinguish the properties of diff erent enantiomers found in food

Examiner’s tipRemind yourself about enantiomers by looking at Chapter 10, page 495 in the Coursebook.

OHH C

CH2OH

a

D-glyceraldehyde L-glyceraldehyde

CHO

HHO C

CH2OH

CHO

b


CHO CHO

C CHO

H H OH

CH2OHHOH2C

Figure F30 The reference compound glyceraldehyde in (a) Fischer projection and (b) tetrahedral projection.


The molecule is drawn in a classic tetrahedral projection with the chiral carbon at the centre and the single hydrogen atom pointing away from the viewer. If looking from the COOH to R to NH2 you go in a clockwise direction, the structure is the D form; if you must go anticlockwise it is the L form (Figure F32).

The R and S systemUsing this system we assign the absolute confi guration of a chiral centre based on the spatial arrangement of groups.

The groups around each chiral centre are assigned a priority value according to a set of rules called the Cahn–Ingold–Prelog priority rules (CIP). These rules give higher priority to atoms attached to the chiral carbon that have a higher atomic number. Once the order of priority has been determined, the molecule is drawn in a tetrahedral projection with the lowest priority group (priority 4) pointing away from the viewer. If you look at the remaining three groups in priority order (1 → 2 → 3) and you go in a clockwise direction from 1 → 2 → 3, the enantiomer is the R form (standing for rectus, Latin for ‘straight); if going from 1 → 2 → 3 you go in an anticlockwise direction, the enantiomer is the S form (standing for sinister, Latin for ‘on the left’) (Figure F33).

HL

OHH C

CH2OH


D-glucosesame arrangementof H and OH

molecules arrangedwith CHO at top

L-glucose

CHO

HHO C

CH2OH

CHO

HHO

OHH

OHH

OHH

CH2OH

CHO

HO

HO

HO

OH

H

H

H

H

CH2OH

CHO

Figure F31 Fischer projections of the two stereoisomers of glucose and their relationship to glyceraldehyde.

Figure F32 D and L stereoisomers of the amino acid alanine.

D-alanineL-alanine

anticlockwise therefore L clockwise therefore D

H pointing away from you

mirrorplane

CH3 CH3

C CN

R

H COOH HOOCCOH2N NH2

H


Consider the enantiomer of lactic acid (2-hydroxypropanoic acid) shown in Figure F34. The OH group has the highest priority as O has a higher atomic number than C and H. The COOH group is assigned a higher priority than CH3 as, when two atoms attached to the central C have the same atomic number, we must look at the atoms joined to those atoms. In the COOH group the C has an O attached (higher atomic number), whereas the CH3 only has Hs (lower atomic number) attached.

Priority of other groupsWith larger groups which have the same atom attached to the central C, we keep going along the chain until there is a diff erence between the groups. Thus –CH2CH3 has a higher priority than –CH3, and in the groups shown here the group on the left has higher priority as the C with O (higher atomic number) is closer to the chiral centre.

anticlockwise, therefore S

3 3

2 2

1 1

4 4

clockwise, therefore R

lowest priority grouppointing away from you

S-enantiomer


R-enantiomermirrorplane

Br Br

C C

F FI ICI CI

clockwise,therefore R

3

2

1

4


HO

COOH

CH3

CH

Figure F33 R and S enantiomers of CFClBrI.

Figure F34 An enantiomer of lactic acid.

H >H C

HHO C

H C H

H

HH C

HH C

H C OH

H

> >H

HHO

decreasing priority

CCO OH

CO H

HL

Double bonded atoms count as two groups attached to an atom, thus CHO has higher priority than CH2OH as it is counted as if there are two O atoms attached to the C.

The (+) and (−) systemThese isomers are also known as (d) and (l) forms. This system is not based on the spatial positioning of the groups around a chiral centre, but rather the direction in which each enantiomer rotates the plane of polarised light. If the plane of plane polarised light is rotated clockwise (when viewed with the light approaching the viewer) then the enantiomer is designated (+) or (d); if the plane of plane polarised light is rotated anticlockwise, the enantiomer is the (−) or (l) form. (d) and (l) stand for dextrorotatory (from the Latin dexter meaning ‘right’) and levorotatory (from the Latin laevus meaning ‘left’), referring to the direction of rotation of the light.

The labels (d/l) and (+/−) can only be assigned using a polarimeter. R and S can be assigned from the structure of the molecule obtained using a technique such as X-ray crystallography.

Examiner’s tipDo not confuse the (d) and (l) system with D and L system as they are based on diff erent things. The three systems do not always match up with each other; for example, D-glyceraldehyde is the (+) form, but D-alanine is the (−) form.


Stereoisomers in foodKnowing about the enantiomeric forms of certain compounds present in the food we eat is extremely important because diff erent optical isomers can possess diff erent tastes, odours and toxicities.

HL

A classic example of varying toxicity shown by enantiomeric forms of the same compound is the diff erence in eff ect between the R and S isomers of the drug thalidomide. The R form is highly eff ective in treating morning sickness, while the S form is implicated in causing severe birth defects. Unfortunately, thousands of pregnant women were dosed with a mixture of both enantiomers (racemate) and many of them gave birth to children with limb abnormalities.

In food science, being able to distinguish between two enantiomeric forms of a compound based on smell and taste can be very useful in determining the authenticity of the ingredients and the level of processing to which the food has been subjected. For example, S(+)-carvone tastes of caraways seeds and dill, while the R(−)-carvone enantiomer tastes of spearmint (Figure F35).

Most naturally occurring amino acids are in the L form, which means they have little-to-no taste; the D forms of amino acids taste sweet. Conversely, most naturally occurring carbohydrates are of the D form and taste sweet. This is due to the way in which the enantiomers interact with taste receptors.

Other examples of taste being aff ected by the enantiomeric form of a molecule present include:1 R(+)-limonene, which smells of oranges. S(−)-limonene, which is described as having a piney or turpentine-like

odour (Figure F36).

H H

O O

R(–) S(+)

Figure F35 The R(−) and S(+) enantiomers of the fragrant molecule carvone.

R(+)-limonene S(–)-limonene

Figure F36 R(+) and S(−) enantiomers of limonene.

The diff erence in smell arises because of the way that protein receptors (olfactory receptors) in our nose detect the molecules.

Examiner’s tipThe IB syllabus describes the fragrance of S(−)-limonene as that of lemons, but there is signifi cant evidence in available published literature to support the belief that it does, in fact, smell of pine.

2 R(+)-alpha-ionone gives raspberries their natural fl avour. S(−)-alpha-ionone has less raspberry fl avour, but cedar wood overtones. See Figure F37.Many other synthetic food fl avourings contain a racemic mixture of both enantiomers, where only one enantiomer provides the natural fl avour. Measuring optical activity could, therefore, allow us to distinguish between natural food components and components that have been produced synthetically.

Synthetic raspberry fl avourings contain a mixture of both enantiomers which is why they never taste quite like the real thing!

Figure F37 R(+) and S(−) enantiomers of alpha-ionone.

O

O


F10 Chemical structure and colour

AnthocyaninsAnthocyanins are based around a fl avonoid/fl avylium core structure with a characteristic C6C3C6 skeleton and conjugated double bond system (Figure F38).

HLTest yourself 11 Draw the structure of the 2-amino acid that does not show

optical activity.

12 Classify each of the following amino acids as D or L.

13 Classify the chiral centre in each of the following as R or S.

a b c

CH2OH

H C

COOHH2N

H

CH3

C

COOH

H2N

CH2C6H5

H

C

COOH

H2N

a b c

CH2OH

CH2CH

3

H C

CH3

HO

CH3

H C

CIH2N

OH

OHC

CH3C

Learning objectives

• Describe the similarities and diff erences in the structures of natural pigments, including anthocyanins, carotenoids, chlorophyll and heme

• Explain why natural pigments are coloured

• Deduce the solubility rules for carotenoids and anthocyanins

O+

C6

C3

C6

Figure F38 The fl avylium ion core structure found in anthocyanin pigments.

Individual anthocyanins diff er in:

• the number of hydroxyl (–OH) and methoxy (–OCH3) side groups present on the rings

• the identity, number and position of attached sugar moieties

• the types and number of aliphatic or aromatic acids attached to the sugars.

Some examples are shown in Figure F39.

CarotenoidsMost carotenoids are derived from a 40-carbon polyene (multiple C=C double bonds) chain. The ends of the chain may terminate in a cyclic (ring) group which may or may not have oxygen-containing functional groups attached. Some examples are shown in Figure F40.

Examiner’s tipThe molecule quercetin is not technically an anthocyanin despite being included on the IB syllabus as an example. It does not contain the positively charged fl avylium ion core, nor does it have sugar groups attached. Quercetin is a fl avonol.


OH

OH

OH O

HO O

OH

sugar

a b

OCH3

OCH3

OH

OH

OHOH HOHO

HO O

O O

+

c

CH3

OH

OH

O

O

OHOH HOHO

HO O

O O

OHOH

OH+

d OH

HOO

CH3

O

OH

OH

OH

OH

OH

OH HO

OH

OH

OH

HO

HO

HO

HO O+

O

O

O

O O

OO

O

O

O

O

HO

HO

OH

OH

OH

OH

OH

OH

HO

OO

O

O

O

O

Figure F39 (a) Quercetin (no sugars); (b) oenin (one sugar); (c) antirrhinin (two sugars); (d) Heavenly Blue anthocyanin (from Ipomoea tricolor Mexican Morning Glory) which contains caffeic acid groups (in blue) between the sugar molecules.

HL

Those carotenoids that contain solely carbon and hydrogen are called carotenes, while those containing oxygen atoms are called xanthophylls.


Porphin rings (heme and chlorophyll)The principal structure in both heme and chlorophyll is a complex, planar heterocyclic unit called a porphin ring, which contains a system of conjugated C=C double bonds.

A porphin ring with side group attached at position 1–8 is called a porphyrin (see structure of heme, Figure F23 on page 18) Chlorophylls contain a porphyrin unit complexed to a central Mg2+ ion. It exists in two main forms, which are shown in Figure F41.

Heme is a complex between a porphyrin unit and an Fe2+ ion. Heme acts as a prosthetic group in both myoglobin (the pigment in muscles) and haemoglobin (the pigment in red blood cells). In both these pigments, a single heme group is bound in a complex with a single molecule of the protein globin.

Colour and conjugationAnthocyanins, carotenoids, chlorophyll and heme are all pigments – they are all coloured. They are coloured because they are able to absorb specifi c wavelengths of light in the visible region of the electromagnetic spectrum and therefore transmit coloured light that is missing one or more parts of the full visible spectrum (the complementary colour to the light absorbed). If only electromagnetic radiation in the ultraviolet (UV) or infrared (IR) region of the spectrum is absorbed, a compound will be colourless.

OH

HO

H

a

b

c

d OH

Figure F40 (a) Lycopene (C40H56) – no cyclic groups; (b) α-carotene (C40H56) – cyclic groups; (c) retinol/vitamin A (C20H30O) – half a carotene unit with cyclic groups; (d) lutein (C40H56O2) – xanthophyll with oxygen-containing groups.

HL

Heterocyclic refers to the presence of a ring structure (cyclic) containing carbon and at least one other element (hetero), typically sulfur or nitrogen in chlorophyll.

Myoglobin exists as a monomer, whereas haemoglobin exists as a tetramer comprising four globin molecules and four heme units.


The part of a molecule that absorbs light and is therefore responsible for the property of colour is called a chromophore. In the classes of molecule here, the chromophore is a conjugated system alternating single and double bonds that absorbs energy from photons to promote electrons from a lower energy level to a higher one. Given that the energy required to do so is quantised (i.e. takes discrete values), only certain parts of the spectrum are absorbed depending on the molecule.

The longer the conjugated (delocalised) system of alternating single and double bonds, the longer the wavelength of light absorbed (Figure F42). As the size and arrangement of the π-bonds are diff erent in the various anthocyanins, carotenoids, chlorophyll and heme they all absorb light in slightly diff erent parts of the visible spectrum and so have diff erent colours.

An organic molecule will absorb visible light and appear coloured if the conjugated system is su� ciently long (usually more than eight conjugated double bonds). A benzene ring counts as three conjugated bonds.

N–

N

O1

2

3

a b c

4

5

6

78

O O

O

O

H

C20H39

R=CH3

N

N

N

HNNH

CH3

chlorophyll a

double bond

porphin structure

double bond missing

N–Mg2+ N–

N

O

O O

O

O

H

R=CHO

N

H

O

chlorophyll b

N–Mg2+

Figure F41 (a) The ring structure of a porphin unit. In both (b) chlorophyll a and (c) chlorophyll b, a C=C is missing so that the carbon atoms at positions 7 and 8 are sp3 hybridised and tetrahedral, where previously all carbons in the porphyrin unit were planar and sp2 hybridised. At position 3, the substituent R group is a methyl (–CH3) in chlorophyll a and an aldehyde (–CHO) in chlorophyll b.

HLExaminer’s tipConjugation and chromophores are discussed in detail in Option A, pages 38 and 40–41 on the CD-ROM.


Test yourself

SolubilityAnthocyanins are water-soluble because they contain numerous hydroxyl (–OH) groups which can hydrogen bond with water molecules. Many of them possess an overall positive charge in the fl avylium ring structure permitting electrostatic forces to form between the molecule and water.

Anthocyanins are not fat-soluble due to the extensive hydrogen bonding between molecules due to the –OH groups.

Carotenoids are not water soluble because most of the molecule consists of a non-polar hydrocarbon chain. The primary type of intermolecular force is van der Waals’ forces as most of the molecule is non-polar. They contain no/few –OH groups so cannot form any/many hydrogen bonds to water molecules.

Carotenoids are fat soluble. The forces between fat molecules are mainly van der Waals’ forces. Carotenoids are able to form signifi cant van der Waals’ forces with fatty acids, hydrocarbons and other non-polar molecules.

a bH

C

H

H

C

H

H

C CH3

H

C

H

C

H

C

H

C

H

C

H

H

C

H

H

C

H

H

C

H

H

C

H

C

H

C

H

C

Figure 42 (a) Octa-1,3,5-triene has a conjugated system of three double bonds; (b) octa-1,4,7-triene – the double bonds are not conjugated and this will absorb a shorter wavelength of light than octa-1,3,5-triene.

HL

14 Explain which of the molecules shown here will absorb longer wavelength electromagnetic radiation.

15 Would you expect the molecules shown in Question 14 to be water soluble or fat soluble?

CH3

O

CH3

CH3

H3C

H3C

CH3

O

CH3 CH3 CH3

CH3

H3C

a

b

H3C


Exam-style questions

1 The study of the chemistry of food is an essential part of ensuring food quality and providing consumers with better fl avours, smells, textures and appealing colours in the foods they buy.

a Describe what is meant by a food. [2]

The composition of foods can be broken down into a number of discrete nutrient groups each with a diff erent chemistry and all need to be represented in a balanced diet.

b Give details on the structure and chemical composition of: i a named amino acid [3] ii a named monosaccharide [3]

c In which nutrient groups could the following features be found: i a peptide bond [1] ii a glycosidic bond [1] iii an ester linkage [1]

2 Oleic acid is a mono-unsaturated fatty acid, eicosapentaenoic acid is a poly-unsaturated fatty acid containing fi ve C=C double bonds. Olive oil has a melting point around 0 °C while that of fi sh oil is < −5 °C.

a State and explain, in terms of structure and bonding, which fatty acid is most likely to be present in each of the oils. [4]

Vaccenic acid, C18H34O2, is a naturally occurring mono-unsaturated fatty acid found in dairy products. It exists as both trans and cis stereoisomers.

b i State and explain, in terms of structure and bonding, the relative melting points of the two isomers. [3]

One of the stereoisomers of vaccenic acid is believed to be more harmful to human health than the other.

ii State which isomer is more harmful and give two possible reasons for your choice. [3]

Lipids are prone to rancidifi cation, during which they react with oxygen to form a number of unpleasant smelling and tasting breakdown products.

c i Place the following fatty acids in order (fastest � rst) of the rate at which they undergo auto-oxidation: stearic acid oleic acid linoleic acid [1] ii Give one reason for your choice. [1]

Margarine and butter are two similar products used in cooking that diff er chemically. Butter tends to be produced from saturated animal fats while margarine is manufactured from unsaturated vegetable oils.

d Suggest two reasons why it is benefi cial to partially hydrogenate the unsaturated fatty acids during the production of margarine. [2]

3 We are all familiar with the ‘best before’ dates seen on food products at the supermarket. This date is an indication of the shelf life of the product. Shelf life is an important concept in the selling and marketing of foods.

a State two properties of a food that might be adversely aff ected when it exceeds its shelf life. [2]

b Quantitatively describe the eff ects of three named factors on the shelf life of a food. For one of these factors, explain the eff ect described. [4]

Fatty foods have a shelf life generally determined by the rate at which the lipids decompose to form off -smelling byproducts. This process is called rancidifi cation and is a form of oxidation.

c i By what sort of mechanism do fats undergo rancidifi cation? [1] ii Name a functional group present in the products of rancidifi cation. [1]


The shelf life of products can be increased by a number of methods. There is a signifi cant amount of chemistry behind these processes.

d i Explain how fi lm packaging helps to increase the shelf life of food. [3]

The rate at which food becomes unsafe to eat can be reduced by combining foods with additives such as antioxidants. There are a number of naturally occurring antioxidants but, increasingly, synthetic antioxidants are added to perishable items to extend their shelf life.

ii For a named synthetic antioxidant, explain how its chemical structure gives it antioxidant properties. [2] iii State and explain two ways in which traditional preservation techniques increase the shelf life of food. [4]

4 Coloration in foods is due to the presence of naturally occurring pigments within the substance. One function of colour in foods is to make them more appealing to consumers, including animals in the natural world.

a i Explain why pigments and dyes are coloured. [2] ii Name two naturally occurring pigments and give one source for each. [4]

b i Name one factor that aff ects the colour of a pigment. [1] ii Describe and explain the eff ect of varying this factor on the colour of an anthocyanin. [3]

To make bland foods appear more nutritious and tasty, synthetic colourants are added by manufacturers.

c Suggest two problems associated with the worldwide use of synthetic food colourants. [2]

The Maillard reaction and caramelisation are two mechanisms by which foods develop fl avours and a brown coloration.

d i Describe the main ways in which these two processes are diff erent. [3] ii Name one food in which the Maillard reaction predominates and one in which caramelisation is the

predominant process of browning. [2] iii How might the rate of the Maillard reaction be increased? [1]

5 In a world with a rapidly increasingly population, food and fuel production struggle to keep up. There is a very real danger that many countries will simply become unable to feed their people. Genetic modifi cation of plants and animals is one possible partial solution to this potential disaster.

a Describe the principles behind genetic modifi cation of an organism. [2]

b Give two arguments for and two arguments against the global use of genetically modifi ed organisms in the food industry. [4]

6 Besides fl avour, odour and colour, the texture of a food is an extremely important factor to humans in determining the ‘likeability’ of a product. The texture of a food depends to a large extent on the phases of matter that exist within a food.

a Explain what is meant by a dispersed system. [2]

b For each of the following foods, state which type of dispersed system is present: i melted chocolate [1] ii whipped cream [1] iii jelly [1] iv Name a further type of dispersed system found in foods not included in parts i–iii and describe

the phases present. [2]

The texture and appearance of some foods can be manipulated by the addition of substances with special properties. A separated mixture of oil and vinegar would be unpalatable to most, but shake it with an emulsifying agent to prevent the separation process and a vinaigrette is produced.

c Explain how structure and bonding make lecithin useful as an emulsifi er. [4]


7 During rancidifi cation of fat-containing foods, fatty acids molecules undergo oxidation. The key intermediate in the auto-oxidation of fatty acids is an organic hydroperoxide.

a i Draw the structure of a hydroperoxide, use R to represent the side group. [1] ii Why are hydroperoxide molecules prone to photolysis? [2]

b Describe, using equations, the process of fatty acid auto-oxidation. Include the correct terminology for each step. [5]

c Write equations for the decomposition of an organic hydroperoxyl radical into: i an aldehyde [1] ii a ketone [1]

d Suggest two reasons why margarine might undergo rancidifi cation more rapidly than butter. [2]

8 Antioxidants, synthetic and natural, are important in maintaining the chemical environment in which other molecules exist. They are able to slow down the natural process of oxidation, during which foods undergo changes in texture, fl avour, odour and colour.

a Explain, including appropriate equations, how some antioxidants are able to prevent the auto-oxidation of fatty acids. [4]

To be an eff ective antioxidant, ethylenediaminetetraacetic acid (EDTA) needs to exist as a hexadentate ligand.

b i State what is meant by a hexadentate ligand and the shape of the EDTA–metal ion coordination complex. [2]

ii By considering the answer to part i, why might EDTA be expected to lose antioxidant potency in solutions with a very low pH? [3]

iii Why is the EDTA useful in preventing the discoloration of canned fruit, such as peaches? [2]

c Many fruits go brown when their fl esh is exposed to air or when they are chopped and stored in water. This process occurs as the colourless molecule catechol (1,2-dihydroxybenzene) is converted to 1,2-benzoquinone which is red.

i What name is given to the type of reaction which converts catechol to 1,2-benzoquinone? [1] ii Suggest a chemical species which might bring about this conversion when fruit is either exposed

to air or stored in water. [1] iii Why does adding lemon juice to the water used to store cut fruit reduce the rate at which it

goes brown? [3]

9 Compounds with identical chemical formulae may exist in various diff erent forms, or isomers. One type of isomerism is called stereoisomerism. There are several diff erent naming conventions for distinguishing the enantiomers of a molecule.

a Name the compound used as the basis for the D and L system for naming stereoisomers. [1]

The D and L system for naming enantiomers of amino acids can be applied by using the CORN rule.

b Explain how the CORN rule can be used to determine whether an amino acid is the D or L stereoisomer. [4]

OH

OH

O

catechol 1,2-benzoquinone

O


c Determine whether the following compounds are D or L enantiomers: i alanine ii galactose iii fructose [3]

d Describe the taste of each of the specifi c compounds shown in part c. [3]

e Use the R and S system to determine which enantiomer of the amino acid alanine is shown in part ci. [1]

f Explain why it is possible to have pairs of enantiomers designated as S(−) and R(+) as well as S(+) and R(−). [3]

g Suggest one property shared by both (+) and (−) enantiomers of a compound, and one property that might diff er. [2]

Menthol exists as a pair of enantiomers. Most naturally occurring menthol is (−)-menthol.

h Suggest why manufacturers of menthol have spent signifi cant time and money in developing a synthetic route that yields the (−) enantiomer and not a mixture of the (−) and (+) enantiomers. [2]

10 The absorption spectra of the anthocyanins oenin and antirrhinin are very similar, with an absorption maximum of around 520 nm. They appear close in colour to the naked eye but, under close inspection, there is a small variation of a few nanometres in their absorption maxima.

a By comparing the structures of oenin and antirrhinin, suggest one reason they appear the same colour, and two reasons for the small diff erence in the wavelength of the maximal absorption. [3]

Another type of naturally occurring pigment is the carotenoid family, members of which are found in carrots, tomatoes and the shell of some shellfi sh.

b Identify the subcategory of carotenoid to which each of the following molecules belongs:

i [1]

H OH

CHO

CH2OH

H2N

CH3

C

HO HC

HOOH

O

alanine galactose fructose

HHC

H OHC

O

CH2OH

CH2OH

C

HO H

H OH

H OH

OH

O

O

OH

ii [1]

iii Give a reason for your choices. [1]


The structure and bonding present in a molecule is an important factor in determining whether a molecule will be coloured or not.

c i Describe and explain what would be expected to happen to the position of the absorption peak for lycopene located in the visible region of the light spectrum, when bromine water is added to a solution of lycopene in hexane. [3]

ii In terms of structure and bonding, explain why the solvent used in part ci is hexane rather than water. [2]

iii State and explain whether the naturally occurring molecule betanin could be used as a food colouring. [2]

OH

OH

betaninOH

HO

HO

HO

HO

O

O O

O

O

O–

N+

HN

food chemistry

Documents

treatment of food

item of food

f food chemistryf1 food

cyclic form of glucose

linear form

examples of monosaccharides

cyclic figure f1

complex carbohydrates