Organic & Biological Chem

BIOLOGICAL & ORGANIC CHEMISTRY

TOPIC 5: ORGANIC AND BIOLOGICAL CHEMISTRY

Most chemicals are compounds of carbon with other elements, mainly hydrogen, oxygen, and nitrogen, with many more being synthesised each year. The variety and importance of carbon compounds are so great that there is a specific branch of chemistry known as ‘organic chemistry’. In this topic students are introduced to the chemistry of the more common organic compounds.

Biological chemistry is one of the growing areas of research, including medical technology, genetic engineering, and the development of pharmaceuticals. In this topic students are introduced to the major groups of compounds of biological significance.

The reactions of the larger macromolecules can often be explained by referring to the reactions and properties of smaller molecules with the same functional groups.

Subtopic 5.1: Systematic Nomenclature

Key Ideas Intended Student Learning

The presence or absence of functional groups in an organic compound determines its physical and chemical properties.

Identify the functional groups in the structural formulae of alcohols, aldehydes, ketones, carboxylic acids, amines, esters, and amides.

Organic compounds are named systematically to provide unambiguous identification.

State, given its structural formula, the systematic name of an organic compound containing:

up to eight carbon atoms arranged as either a straight chain or a branched chain;

one or more of the same functional groups (with these limited to hydroxyl, aldehyde, ketone, carboxyl, or primary amino groups).

The structural formula of an organic compound can be deduced from its systematic name.

Draw, given its systematic name, the structural formula of an organic compound containing:

up to eight carbon atoms arranged as either a straight chain or a branched chain;

one or more of the same functional groups (with these limited to hydroxyl, aldehyde, ketone, carboxyl, or primary amino groups).

Esters are named as derivatives of a carboxylic acid. State the systematic names of methyl and ethyl esters of straight-chain acids containing up to eight carbon atoms.

The structural formula of an ester can be deduced from its systematic name.

Draw, given its systematic name, the structural formula of an organic methyl or ethyl ester of a straight-chain acid containing up to eight carbon atoms.

Introduction

Organic compounds are carbon-containing compounds. Carbon has the following unique properties:

It has a covalence of four

Carbon atoms can bond to each other to form straight chains, branched chains and rings.

Carbon atoms can form sing, double and triple covalent bonds with each other and other atoms.

As a consequence, there are thousands of different organic molecules containing carbon atoms.

1


Terminology

Term Meaning Example

Molecular formula represents the number of atoms in a molecule C3H8

Empirical formula the simplest ratio of atoms in a compound C4H8 CH2

Structural formula represents the arrangement of atoms in a molecule

CH3-CH=CH-CH3

Structural isomers molecules with the same molecular formula, but different structural formulae

CH3-CH=CH-CH3

CH2=CH-CH2-CH3

Homologous series a series of organic compounds which conform to a general formula, which have the same functional group, which behave similarly in chemical reactions but have slightly different physical properties

Alkanes

CnH2n+2

Saturated compounds have no double bond between C atoms alkanes

Unsaturated compounds

have at least one C=C or CC bond in a chain or ring

Alkenes and alkynes

Aromatic compounds a benzene ring forms part of the molecule

OR

C

C

C

C

C

C

H

H

H

H

H

H

Systematic Naming Of Organic Compounds

There is a set of rules established by IUPAC (the International Union of Pure and Applied Chemistry) to name this compounds. eg alkanes

1. Find the longest continuous chain of carbon atoms and assign the parent name by combining the prefix from the table below with the suffix –ane.

Number of C’s 1 2 3 4 5 6 7 8

Prefix meth- eth- prop- but- pent- hex- hept- oct-

The longest continuous chain in this molecule has 6 carbons so the parent name hexane:

CH3 CH CH2 CH2 CH2 CH3

CH3

2. Locate any side chains (alkyl groups) and name them according to the number of carbons in them:

-CH3 : methyl -C2H5 : ethyl etc


CH3 methyl group

2


Thus the molecule above is methylhexane.

3. The position of the branching alkyl group in the continuous chain is then indicated by a number that is the number of the carbon atom in the main chain to which the chain is attached. The chain is numbered from the end which gives the lower of the two possibilities.


CH3

1 62 43 5

Thus the molecule above is 2-methylhexane.

Note:

Numbers are only required when other isomers are possible.

Numbers in a name are separated from each other by commas and from the words by hyphens.

4. If a compound contains two or more identical branches, the following prefixes are used:

Number of branches 2 3 4

Prefix di- tri- tetra-

This molecule has 2 methyl groups and so is a dimethyl-, with the methyl groups attached to the second and third carbons this the name is 2,3-dimethylhexane.

CH3 CH CH CH2 CH2 CH3

CH3 CH3

5. If a molecule has two or more different branching alkyl groups, they are placed in alphabetical order in the name, hence the molecule below is 3-ethyl-2-methylhexane.

CH3 CH CH CH2 CH2 CH3

CH3 C2H5

6. Each homologous series has an associated functional group which is named as a suffix or prefix.

o For suffixes, e is removed and the suffix added eg –ol, oic acid. The number of the carbon can be

before the parent name or before the suffix eg 2-propanol or propan-2-ol.

o For prefixes it is added to the start of the parent name eg 2,2-dichloropropane.

7. For aldehydes, ketones, carboxylic acid and carboxylate ions, the longest carbon chain includes the carbon that is part of the functional group eg CH3-CH2-COOH is propanoic acid.

8. Esters are formed from the reaction between a carboxylic acid and an alcohol, the names of which are components of the name. The carboxylic acid gaines the –oate suffix instead of –oic acid and the alcohol gets the name of the alkyl group it contains. Example – propanoic acid and methanol make methyl propanoate.

CH3 CH2 C

O

O CH3

methyl frommethanol

propanoate frompropanoic acid

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Common Functional Groups

Homologous series Functional group General formula Naming

alkane C - C CnH2n+2 - ane

alkene C = C CnH2n - ene

alkyne C C CnH2n-2 - yne

alcohol R OH CnH2n+1OH - ol

aldehyde R C

O

H CnH2n+1CHO - al

ketone R C

O

R' CnH2nO - one

carboxylic acid CnH2n+1COOH - oic acid

carboxylate salt CnH2n+1COO- - oate ion

ester CnH2nO - oate

amine R - NH2 CnH2n+1NH2 - amine

Amindes R – CONH2 CnH2n+1CONH2 -amide

ASSIGNMENT 5.1 : SYSTEMATIC NOMENCLATURE

1. Write systematic names and give the molecular formulae (first 4only ) of the following molecules and ions:

Name Name

Molecular formula Molecular formula

Name Name

Molecular formula Molecular formula

4


5


2. Draw the structure of the following compounds

3-methylbutan-2-ol

3-methylpentan-3-ol

2,2-dimethylpentane- 1,3-diol

3 ,3-dimethylbutanal

3-methylhexane-1,6-dial

6


2,4-dimethylpentan-3-one

heptan-2,4-dione

3-methylbutanoic acid

octan-1,8-dioic acid

methylpropanamine

N-ethylbutanamine

heptane-1,7-diamine

Subtopic 5.2: Physical Properties


The melting-points and boiling-points of organic compounds that contain the same functional group increase with the length of carbon chain.

Predict and explain the melting-points and boiling-points of an organic compound in comparison with those of other compounds that contain the same functional group.

The boiling-points of organic compounds that display hydrogen bonding between molecules are higher than those of compounds of similar molar mass that do not display hydrogen bonding.

Predict and explain the boiling-points of alcohols in comparison with those of aldehydes and ketones of similar molar mass.

The boiling-points of esters are lower than those of isomeric acids because of the absence of hydrogen bonding between molecules of the ester.

Predict and explain the boiling-points of esters in comparison with those of isomeric acids.

Organic compounds are generally insoluble in water. Explain the insolubility in water of most organic compounds.

Hydrogen bonding between functional groups and water can explain the solubility in water of some smaller organic compounds.

Predict and explain the solubility in water of the smaller amino acids, carboxylic acids, alcohols, aldehydes, and ketones.

The solubility in water of an organic compound depends on its molar mass and the functional groups present.

Predict and explain the relative solubilities in water of two organic compounds, given their structural formulae.

Important physical properties of organic compounds

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Three important physical properties of organic compounds, which relate to their structure and bonding are melting point, boiling point and solubility in water. The size of the hydrocarbon chain and the polarity of the functional group(s) that make up a molecules structure influence each one of these properties.

Effect of carbon chain length on melting and boiling points

For compounds with identical functional groups, the boiling points increase as the length of the carbon chain increases. The secondary bonding forces operating between non-polar hydrocarbon chains are dispersion forces. The strength of the dispersion forces increases with molar mass (which is proportional to chain length). The trend for melting points is similar but not as clear cut, because melting also involves the complete separation of molecules.

Name Formula Molecular mass Melting Point (ºC) Boiling Point (ºC)

Methane

Ethane

Propane

Butane

Pentane

Hexane

Heptane

Octane

Nonane

Decane

CH4

C2H6

C3H8

C4H10

C5H12

C6H14

C7H16

C8H18

C9H20

C10H22

16

30

44

58

72

86

100

114

128

142

-183

-172

-190

-135

-130

-94

-90

-57

-51

-30

-162

-89

-42

-1

36

69

98

126

151

174

Melting & Boiling Point vs Molecular Massfor the Alkanes

-250

-200

-150

-100

-50

0

50

100

150

200

0 20 40 60 80 100 120 140 160

Molecular Mass

Tem

pera

ture

melting point

boiling point

Effect of functional group on melting and boiling points

The polarity of the functional group affects the strength of the secondary bonding and hence the melting and boiling points. The most polar have the highest boiling point (keeping atomic mass about the same. The table below compares the boiling points of compounds containing different functional groups in molecules containing three carbons and about the same molecular mass.

Compound Molecular mass Boiling point (ºC)

Propanol 60 97.2

Propanal 58 47.9

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Propanone 58 56.1

Propanoic acid 74 140.8

Ethyl methanoate 74 54.5

propanamine 59 48.5

This suggests that the order of polarity amongst the functional groups is:

carboxylic acid > alcohol > aldehydes and ketones > amines

Solubility of organic compounds in water

Non-polar organic compounds, such as hydrocarbons, are not soluble in water because they cannot form hydrogen bonds with polar water molecules. Molecules with polar functional groups, but long carbon chains (of 6 or more carbon atoms) are also classified as non-polar because the non-polar carbon chain is the dominant structural feature of these molecules.

Polar groups with the capacity to form hydrogen bonds with water, can render an organic compound soluble in water, if the carbon chain is not too long. A molecule with more than one such polar functional group has increased solubility. The effect of chain length and polar functional groups is shown in the table below.

Solubility (g/100g water)

Carbon chain Alcohol Aldehyde Ketone Carboxylic acid

amine

methan- ∞ very - ∞ ∞

ethan- ∞ ∞ - ∞ ∞

propan- ∞ 16 ∞ ∞ ∞

butan- 8 7 26 ∞ ∞

pentan- slightly slightly 6 4 very

octan- insoluble insoluble insoluble insoluble insoluble

It can be seen from inspection of the data table above that the lower members of each homologous series are very soluble in water. As the non-polar carbon chain length increases, the molecules take on greater non-polar character overall, with a consequent decrease in solubility.

The diagram below shows the hydrogen bonding between:.

ethanoic acid and water methanal and water

ASSIGNMENT 5.2 : PHYSICAL PROPERTIES

1. The alkanes have the general formula CnH2n+2. The melting and boiling points for the first ten members of this homologous series are given in the table.

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Name Formula Melting Point (oC) Boiling Point (oC)

Methane

Ethane

Propane

Butane

Pentane

Hexane

Heptane

Octane

Nonane

Decane

CH4

C2H6

C3H8

C4H10

C5H12

C6H14

C7H16

C8H18

C9H20

C10H22

-183

-172

-190

-135

-130

-94

-90

-57

-51

-30

-162

-89

-42

-1

36

69

98

126

151

174

(a) Draw a graph with the boiling point temperatures on the vertical axis against the number of carbon atoms in each molecule.

(b) How is the trend explained?

(c) How would the graph for the first ten alcohols differ from the one drawn. Explain why.

2. You have just extracted a substance from a plant and you suspect that you may have been the first chemist to do so. After purifying it you examine its physical properties:

Your results: Tb = 78oC

Conductivity – non-conductor

Solubility – completely miscible with water

Solution conductivity - nil

pH of solution – 6.9

(a) What can you deduce from this information?

(b) How could you test your substance to show that it is not a mixture?

(c) What might be some of the analytical techniques you could use to find out more about your discovery?

3. (a) Name the following

(i) (ii) (iii) (iv) (v)

H C

H

H

C

OH

H

C

H

H

H H C

H

H

C

O

C

H

H

H H C

H

H

C

H

OH

H

H C

H

H

C

O

H

H C

H

H

C

O

O H

(b) Which substance has:

(i) the lowest boiling point?

(ii) the highest boiling point?

(c) Give reasons for your choices in part (b).

4. With esters you have to be careful to get the

C

O

O group the right way round.

For example, ethyl ethanoate is very different to its isomer methyl propanoate

(a) In what way would they differ?

(b) However, their Tb and lack of solubility in water are similar. Explain why.

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5. Draw diagrams to show hydrogen bonding between water molecules, between alcohol molecules and between water and alcohol.

6. This table shows the solubility of some alcohols in water:

Name Formula Solubility (g/100g H2O)

Methanol

Ethanol

Propan-1-ol

Butan-1-ol

Pentan-1-ol

Hexan-1-ol

CH3OH

CH3CH2OH

CH3CH2CH2OH

CH3CH2CH2CH2OH

CH3CH2CH2CH2CH2OH

CH3CH2CH2CH2CH2CH2OH

∞

∞

∞

8.0

2.7

0.6

(a) Explain the observed trend.

(b) Butan-1-ol has a boiling point of 117.7oC, whereas butane boils at –0.5oC. Explain this difference. (Diagrams must be used in your explanation.)

7. The table below shows information about some alcohols which form part of an homologous series.

Name Formula Boiling Point (°C) Molecular Mass

Methanol CH3OH 65 32

Ethanol C2H5OH 78 46

Propan-1-ol C3H7OH 97 60

Butan-1-ol C4H9OH 74

Pentan-1-ol C5H11OH 138

Hexan-1-ol C6H13OH 158 102

(a) (i) Identify the functional group common to all alcohols.

(ii) What is the general formula for these alcohols?

(b) What is the formula of the next alcohol in the series?

(c) Calculate the relative molecular mass of pentan-1-ol, using only the information given in the table. Explain your working.

(d) (i) Plot a graph of boiling point against number of carbon atoms in a molecule of the alcohol.

Use the graph to estimate the boiling points of:

(1) butan-1-ol

(2) C8H17OH

(e) State the connection between boiling point and the relative molecular mass of these alcohols.

8. Amino acids are the building blocks of proteins and contain both amine and carboxylic acid functional groups. Three amino acids are shown in the diagrams below:

CH C

O

O HH

N

H

H

CH C

O

O HCH3

N

H

HCH C

O

O HCH

N

H

H

CH3H3C

glycine alanine valine

(a) Which would have the highest boiling point? Why?

(b) Which would be most soluble in water? Why?

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SUBTOPIC 5.3: ALCOHOLS


Ethanol is produced by fermentation of glucose, which can be derived by hydrolysis of complex carbohydrates.

Describe the conditions, and write equations, for the hydrolysis of polysaccharides and the production of ethanol by fermentation of glucose.

Alcohols are classified as primary, secondary, or tertiary.

Identify a hydroxyl group in an alcohol as primary, secondary, or tertiary, given the structural formula.

Primary and secondary alcohols can be distinguished from tertiary alcohols by their reaction with acidified dichromate solution.

Describe how primary and secondary alcohols can be distinguished from tertiary alcohols by their reaction with acidified dichromate solution.

The type of product obtained by oxidising an alcohol depends on whether the alcohol is primary or secondary.

Predict the structural formula(e) of the product(s) of dichromate oxidation of a primary or secondary alcohol, given its structural formula.

Ethanol Production

On an industrial scale, ethanol is produced in two ways:

Synthetic ethanol – produced from ethene from crude oil or natural gas

Fermentation – produced from glucose (monosaccharide) in grape or sugar (disaccharide) in fruits or starch (polysaccharide) in vegetables and grains.

Fermentation

If grain is used for the production of ethanol, the starch has to be converted to glucose before fermentation. To do this they are firstly soaked in water, which causes the grains to germinate. Enzymes in the grain then hydrolyse the starch to glucose in two steps.

Step 1 Hydrolysis of the polysaccharides to disaccharides

(C6H10O5)n + n2 H2O

n2 C12H22O11

Step 2 Hydrolysis of the disaccharides to glucose

C12H22O11 + H2O 2C6H12O6

Some fruits contain disaacharides (like sucrose – cane sugar) so only the second step is needed.

Grape juice already contain glucose.

Step 3 The glucose is then fermented using yeast.

C6H12O6 2C2H5OH + 2CO2

Conditions needed

Temperature - 20C to 30C. Outside this range, the enzymes are less effective. The reaction is exothermic, so cooling is sometimes necessary to keep the temperature in the optimal range.

Exclusion of oxygen – fermentation is an anaerobic process and needs little oxygen. Too much oxygen causes oxidation of the alcohol to ethanoic acid (vinegar).

Primary, secondary and tertiary alcohols

Alcohols are classified as primary, secondary or tertiary according to the position of the hydroxyl (alcohol) group in the molecular structure.

Primary alcohols – the OH group is on the end of a chain eg CH3-CH2-CH2-OH

Secondary alcohols – the OH group is bonded to a carbon inside a chain, that is bonded to 2 other carbons eg propan-2-ol

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H C

H

H

C

OH

H

C

H

H

H

Tertiary alcohols – the OH group is bonded to a carbon that is in turn bonded to 3 other carbons eg methyl propan-2-ol.

CH3 C

CH3

CH3

OH

Oxidation of alcohols

The type of product, if any, formed by the reaction of an alcohol with and oxidising agent such as acidified potassium dichromate depend on whether the alcohol is primary, secondary or tertiary,

A primary alcohol is oxidised first to the corresponding aldehyde, and, if there is continued contact with the oxidising agent, then the aldehyde is oxidised to the corresponding carboxylic acid. eg

H C

H

H

C

H

H

C

H

H

OH H C

H

H

C

H

H

C

H

O

H C

H

H

C

H

H

C

OH

O

propan-1-ol propanal propanoic acid

A secondary alcohol is oxidised to the corresponding ketone.

H C

H

H

C

OH

H

C

H

H

H H C

H

H

C

O

C

H

H

H

propan-2-ol propanone

A tertiary alcohol is not oxidised by acidified dichromate ions.

These tests can be use to distinguish between the alcohols. During the reaction, the orange dichromate is reduced to the green chromium (III) ion.

Note

The reaction is slow at room temperature and needs heating.

The alkyl group is not involved in the reaction, only the functional group.

ASSIGNMENT 5.3 : ALCOHOLS AND ALDEHYDES

1. Alcohols are organic compounds with an --OH group. It is called a hydroxyl group, not a hydroxide group. Why?

2. (a) Give systematic names for these alcohols:

(i) (ii) (iii)

CH3 CH2 CH2 CH2 OH

CH3 CH2 CH2 CH2 OH

CH3 CH2 CH

CH3

OH

CH3 CH2 CH

CH3

OH

CH3 CH2 CH

CH3

CH2 OH

CH3 CH2 CH

CH3

CH2 OH

Classify each as a primary, secondary or tertiary alcohol.

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3. Glucose solution was fermented under anaerobic conditions. The glucose solution was boiled and then cooled before the yeast was added.

(a) Why was the glucose solution boiled and cooled before the yeast was added?

(b) What does anaerobic mean?

(c) Explain the role of the S-shaped section of the tubing.

4. Ethanol is the starting material for a range of important organic chemicals. The reactions, which produce some of these, are summarised below. The reagents and conditions for the reactions have not been included.

State the reagents and conditions needed to bring about each of the reactions A – C above.

5. Propan-1-ol, C3H7OH, is refluxed with an acidified solution of potassium dichromate (VI) to produce propanoic acid. The acidified potassium dichromate (VI) acts as an oxidising agent.

(a) Explain what is meant by the term reflux.

(b) State what colour change takes place in the reaction mixture.

(c) Write a balanced equation for the oxidation of propan-1-ol to propanoic acid.

6. The production of body odour often begins with secretions from glands called apocrine glands which are most numerous in the armpits. Skin bacteria, which live in the armpits, use these secretions to produce energy and many different waste products. Scientists have isolated one of these products, compound E, which is shown below. Compound E can be oxidised to form compounds F and G.

C C

CH2OH

H

CH3

CH2CH2CH3

Compound F C C

COOH

H

CH3

CH2CH2CH3

Compound E Compound G

(a) Compound E contains two functional groups. Identify both functional groups and state how you could test for each.

(b) Suggest the structural formula of compound F.

7. Compound A is an alcohol, that has several structural isomers.

The molecular formula of A is C5H12O. Give and name a structural formula which is:

(a) a primary alcohol

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(b) a secondary alcohol that has a branched carbon chain

(c) a tertiary alcohol.

(d) Each of these alcohols reacts differently with a solution of potassium dichromate which has been acidified with sulfuric acid. For each example in (a), give the structure of the organic product from this treatment. If there is no product, say so.

8. Before the introduction of IR spectroscopy to determine blood alcohol content of drivers, gas liquid chromatography (GLC) was used.The chromatogram shown gives the results of a GLC analysis of the first five straight chained primary alcohols. The more volatile a compound the less its retention time.

(a) Record in table form, the name, the formula and retention time for each alcohol analysed.

(b) Estimate a retention time for hexan-1-ol.

In GLC analysis for BAC a doctor takes a blood sample and an exact amount of blood is mixed with a known amount of a standard solution of propan-1-ol. Here is a chromatogram from such an analysis.

(c) What is meant by the term, a standard solution?

(d) Why is a standard amount of propan-1-ol added to the blood?

(e) If the concentration of the propan-1-ol was 60 mg per 100 mL of solution, what was the approximate concentration of the ethanol?

9. (a) What is an aldehyde and what is a ketone?

(b) What is a carbonyl group?

10. (a) Aldehydes and ketones cannot form intermolecular hydrogen bonds. Explain why.

(b) What is a consequence of this?

11. Give the systematic name to each of the following:

(a) (b) (c) (d)

CH3CH2CH2CHO CH3CH2CH2CH2CHOH3C CH2 CH2 C CH2 CH3

O

CH3 C CH2 CH3

O

12. Compound A (butanal) and compound B (butanone) have the same molecular formula, C4H8O.

(a) What term is used to describe compounds with the same molecular formula?

15


(b) Name the functional group common to both A and B.

13. Three different compounds A, B, and C have the same molecular formula, C3H6O. Tests show that A is an unsaturated alcohol, B is an aldehyde and C a ketone.

(a) Write structures for A, B and C.

(b) When compound B was added to a solution of silver nitrate in aqueous ammonia (Tollen’s Reagent), metallic silver is formed.

(i) What is the name for this test?

(ii) Write an equation for the reaction.

(c) Another test used to distinguish between aldehydes and ketones uses acidified dichromate solution.

(i) What would you observe when this solution is added to B and then C?

(ii) Write the equation for the reaction with B.

14. Compound C4H10O, is oxidised in two steps to 2-methylpropanoic acid.CH3

CH

CH3

COOH

Write the structure of the original compound.

15. Write the full structural and molecular formula of each of the following molecules

(a)OH

OH

OH

HO

OHO

(b)OH

OH

OH

OH

O

HOH

Subtopic 5.4: Aldehydes and Ketones


Aldehydes and ketones are produced by the oxidation of the corresponding primary and secondary alcohols respectively. Aldehydes are readily oxidised, and so must be distilled off from the reaction mixture as they are formed.

Draw, given the structural formula of the aldehyde or ketone, the structural formula of the alcohol from which it could be produced by oxidation, and describe the necessary reaction conditions.

Aldehydes can be oxidised to form carboxylic acids or, in alkaline solutions, carboxylate ions.

Draw the structural formula of the oxidation product of a given aldehyde in either acidic or alkaline conditions.

Ketones cannot readily be oxidised. This difference in properties can be used to distinguish aldehydes from ketones.

Describe how acidified dichromate solution and Tollen’s reagent (ammoniacal silver nitrate solution) can be used to distinguish between aldehydes and ketones.

Aldehydes and ketone molecules contain the carbonyl group C O

. For aldehydes, the carbonyl group is in a terminal position, while it is in a non-terminal position for ketones.

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Aldehydes and ketones are widely used as solvents, flavourings, and fragrances.

Examples

C O

C3H7

H

C O

C3H7

C2H5

H3C CH2 C C

CH3

C

O

HH3C CH2 C CH CH2

O

butanal hexan-3-one 2-methylpent-2-enal (grassy) pent-1-en-3-one (mustard)

Preparation of aldehydes and ketones

Aldehydes are prepared by the controlled oxidation of primary alcohols.

A solution of the oxidising agent is dropped into the acid-alcohol mixture as it is heated. The aldehyde boils off immediately it forms, preventing the further oxidation to the carboxylic acid. The aldehyde is the lowest boiling point component of the mixture and consequently distils off at the lowest temperature.

R-CH2-OH R C

O

H

Ketones are prepared by the oxidation of secondary alcohols. Unlike aldehydes, they do not oxidise further. Thus the ketone product does not have to be distilled off as it forms. Heating of the mixture is required to increase the rate of the reaction.

C OH

R'

R

C O

R'

R

Apparatus used for the production of an aldehyde from a primary alcohol

Oxidation of aldehydes

When aldehydes are heated with acidified solution of dichromate. They undergo oxidation to carboxylic acids. The oxidising agent is the orange dichromate ion which is reduced to the green chromium (III) ion.

17


H C

H

H

C

H

H

C

H

O

H C

H

H

C

H

H

C

OH

O

propanal propanoic acid

Under alkaline conditions, aldehydes also undergo oxidation to carboxylate ions for example when they are heated with ammoniacal silver nitrate (Tollen’s Reagent). The oxidising agent is the silver diamine ion which is reduced to metallic silver. Under favourable conditions, the silver can be made to deposit on the inside walls of the reaction vessel as a silver mirror. This is used as a test for aldehydes. Note that the carboxylate ion is formed rather than carboxylic acid, because of the alkaline conditions produced by the ammonia.

Distinguishing between an aldehyde and a ketone

Ketones do not undergo oxidation when heated with acidified dichromate solution or Tollen’s reagent.

Acidified dichromate Tollen’s Reagent

aldehydes Turns green Silver mirror

ketones No reaction No reaction

Subtopic 5.5: Carboxylic Acids


Carboxylic acids can be produced by the oxidation of aldehydes or primary alcohols.

Identify the aldehyde or primary alcohol from which a carboxylic acid could be produced by oxidation, given its structural formula.

Carboxylic acids are weak acids and ionise, to a small extent, in water.

Write an equation for the ionisation of a carboxylic acid in water.

Carboxylic acids react with bases to form ionic carboxylate salts.

Write equations for the reactions of carboxylic acids with hydroxides, carbonates, and hydrogen carbonates, and describe changes that accompany these reactions.

The salts of sodium and potassium carboxylates are soluble in water because of the ion–dipole attraction between the ions and water.

Explain why some drugs with carboxyl groups are usually taken in the form of their salts.

Carboxylic acid contain the carboxyl group –COOH

Preparation

Carboxylic acids can be prepared by heating primary alcohols or aldehydes with excess acidified dichromate solution.

R C

H

H

OHR C

H

O

R C

OH

O

primary alcohol aldehyde carboxylic acid

Ionisation of carboxylic acids in water

Carboxylic acids are weak acids that partially ionise in water to form hydronium and carboxylate ions.

RCOOH(s) or (l) + H2O H3 O(aq)+

+ RCOO(aq )−

18

C

O

O H


Neutralisation of carboxylic acids

Carboxylic acids are neutralised by hydroxide ions, carbonate ions and hydrocarbonate (bicarbonate) ions. These reactions are fast and exothermic.

Neutralisation by hydroxide ions

RCOOH(s) or (l) + OH(aq)−

RCOO(aq )−

+ H2 O(l )

Neutralisation by carbonate ions

2RCOOH(s) or (l) + CO3( aq)2−

2RCOO(aq )−

+ H2 O(l ) + CO2(g)

Neutralisation by hydrogencarbonate ions

RCOOH(s) or (l) + HCO3( aq )−

RCOO(aq )−

+ H2 O(l ) + CO2(g)

The carboxylate salts are ionic and soluble in water, especially if the cation is sodium or potassium.

The solubility of the carboxylate salts

The solubility of the carboxylate ion is due to the strong ion-dipole bond (note - new type of bond) that forms between the negatively charged carboxylate ion and water molecules. This bond is stronger than the hydrogen bonds that form between hydroxyl groups and water in carboxylic acids.

Drugs with carbonyl groups as part of their molecular structure

Two commonly used drugs are pain-relieving drugs are aspirin and ibuprofen. Both have carboxyl groups, but because of their size they are insoluble in water. They normally contain sodium hydrogencarbonate and when they are added to water, the carboxylic acid and the hydrogencarbonate react with each other to produce the soluble carboxylate ion. In this form, the drug is easier to administer and faster acting.

C

O

OHOC

O

CH3CH2 C

O

OH

CH2CH

CH3

CH3

Aspirin (acetyl salicyclic acid) Ibuprofen

(4-isobutyl--methylphenylacetic acid)

Once in the stomach, where it is acidic, it is converted back to the carboxylic acid

Subtopic 5.6: Amines


Owing to the presence of an unbonded electron pair, amines are able to act as bases and accept H+ ions.

Draw the protonated form of an amine, given its structural formula.

Amines are classified as primary, secondary, or tertiary.

Identify an amino group in an amine as primary, secondary, or tertiary, given the structural formula.

The salts of amines are soluble in water because of the Explain why some drugs with amine groups are

19



ion–dipole attraction between the ions and water. usually taken in the form of their salts.

Amines

Amines are derivatives of ammonia where one or more of the hydrogens have been substituted with and alkyl group. Depending on the number of replacements, they are classified as primary (for 1), secondary (for 2) and tertiary (for 3), although we mainly consider the primary amines.

ammonia primary amine secondary amine tertiary amine

All amines contain an unbonded pair of electrons on the nitrogen. Because of this, they can behave as bases and accept a proton and become protonated amines.

Solubility of amine salts

The protonated amine salts are soluble in water because of the ion-dipole attraction between the ions and water.

Many amine containing drugs are taken in the form of the salt, because they dissolve better and are therefore more easily absorbed.

procaine protonated form of procaine

Note that only the tertiary amine is protonated. The primary amine, because it is connected directly to the benzene ring, is difficult to protonate.

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Assignment 5.4, 5.5, 5.6 : Carboxylic Acids and Amines

1. (a) (i) What is a carboxylic acid?

(ii) Give the name and structure for the acid containing four carbon atoms

(b) Draw the carboxylic acid functional group, and show how the bonds are polarised in this group.

(c) Why are the carboxylic acids weak acids? Give an equation to illustrate that methanoic acid is a weak acid.

(d) How can carboxylic acids be prepared? Give an example with equations.

(e) Write the equation for the reaction of a carboxylic acid with a solution of sodium bicarbonate.

(f) Solid carboxylic acids do not react with solid carbonates. Why?

2. (a) Why would you expect the compounds CH3COOH and C3H7COOH to have similar chemical properties?

(b) Give, with equations, the reactions of both these compounds with:

(i) Sodium bicarbonate solution

(ii) Sodium hydroxide solution

(iii) Ethanol with concentrated sulfuric acid under reflux

(c) How would you expect the physical properties (ie. MPt and solubilities in water) to differ?

3. Ethanoic acid, CH3COOH, forms a weak acid when added to water.Some reactions of ethanoic acid are shown:

(a) Explain what is meant by a weak acid.

(b) Name the substance A that is added to ethanoic acid.

(c) Substance B is formed when ethanoic acid reacts with ethanol. What type of substance is B?

(d) Draw a structural formula for salt C.

(e) Write a balanced equation for the reaction between magnesium and ethanoic acid.

4. (a) A series of compounds called ‘prostaglandins’ are produced in the bodies of people who suffer from inflammatory diseases such as arthritis. An example of a prostaglandin is shown below:

O

COOH

OH

Draw the structure of the organic product obtained when samples of prostaglandin are treated in the following ways:

(i) Warmed with acidified potassium dichromate.

(ii) Refluxed with methanol and a small amount of concentrated sulfuric acid.

21


(b) A more recent mining technique is to react the required ore with sulfuric acid using underground pipes. The ‘dissolved’ ore is then pumped to the surface and treated. An environmental concern with this method is that excess sulfuric acid may leak into water supplies. To remove this acid the use of sulfate-reducing bacteria is being researched. One of the reactions that then occurs is shown by the following equation:

2CH3CHOHCOOH(aq) + SO42-

(aq) 2CH3COOH(aq) + 2CO2(g) + H2S(g) + 2OH-(aq)

(i) Give a reason for the resulting increase in pH.

(ii) The equation above suggests that these bacteria are anaerobic decomposers. Give a reason to support this statement.

5. The more complex the carboxylic acid the less soluble it is.

(a) Why is this?

However, many complex carboxylic acids are useful medicines. If they are not soluble they cannot be absorbed into the body.

(b) How can these acids be made into the more useful soluble compounds?

(c) Using a diagram, explain how the increase in solubility occurs.

(d) Many analgesics (pain-relieving medicines) contain aspirin.

C

OH

O

O C

O

CH3

Suggest a formula for ‘soluble aspirin’.

6. A plant known to the Chinese as Ma Huang provided them with a medicine which was described in 2760 BC as a valuable remedy for coughs and as a cardiac stimulant. The active ingredient of this plant is ephedrine. In 1933 during an attempt to find a synthetic analogue to ephedrine the compound amphetamine was found to have similar properties. Amphetamine was widely used as a nasal decongestant but later it was used, and now dangerously abused by some, as a stimulant.

C C N

OH

H

H

CH3

2-(methylamino)-1-phenylpropan-1-ol

Ephedrine

CH3

H

C C NH2

H

H

H

CH3

1-phenylpropan-2-amine

Amphetamine

(a) Both react with acids, eg. HCl(aq). Write equations to show the protonated form of these compounds.

(b) Comment on their solubility both before and after protonation.

7. The relative basicities of some nitrogen-containing compounds are as follows:

NH3>C2H5NH2>C6H5NH2

(a) Explain why these compounds behave as bases.

(b) Explain the variation in basicity.

8. Adrenalin is a hormone which raises blood pressure, increases the depth of breathing and delays fatigue in muscles, thus allowing people to show great strength under stress.Benzedrine is a pharmaceutical that stimulates the central nervous system in a similar manner to adrenalin.

22


C C NH2

H

H

H

CH3

1-phenylpropan-2-amine

Benzedrine

C C N

H

OH

H

H

4-(1-hydroxy-2-(methylamino)ethyl)benzene-1,2-diol

AdrenalinHO

HO

H

CH3

(a) Suggest why adrenalin is more soluble in water than is Benzedrine.

(b) Give the structural formulae of the organic products obtained when Benzedrine reacts with an aqueous acid such as dilute hydrochloric acid.

9. Consider the chemistry of love. The chemicals providing a romantic high are phenylethylamine (lots found in chocolate), dopamine and norepinephrine. The release of these is triggered by genetics, psychological experiences and even smells. These chemicals are related to epinephrine (also called adrenalin) which is secreted by the adrenal gland in times of stress.Circle and name all of the functional groups in these molecules.

C C N

H

H

H

H

4-(2-aminoethyl)benzene-1,2-diol

DopamineHO

HO

H

H

C C N

H

H

H

H

2-phenylethanamine

Phenylethylamine

H

H

C C N

OH

H

H

H

4-(2-amino-1-hydroxyethyl)benzene-1,2-diol

NorepinephrineHO

HO

H

H

C C N

H

OH

H

H

4-(1-hydroxy-2-(methylamino)ethyl)benzene-1,2-diol

AdrenalinHO

HO

H

CH3

Subtopic 5.7: Esters


An ester can be produced by a condensation reaction between an alcohol and a carboxylic acid.

Draw the structural formula of the ester that could be produced by the condensation reaction between an alcohol and a carboxylic acid, given their structural formulae, and write an equation for the reaction.

The production of an ester from the reaction of an alcohol and a carboxylic acid is slow at 25°C.

Explain the use of heating under reflux and the presence of a trace of concentrated sulfuric acid in the laboratory production of esters.

Esters may be hydrolysed under acidic or alkaline conditions.

Identify the products of hydrolysis of an ester, given its structural formula.

Esters

Esters contain the ester functional group

Commonly occurring esters include fragrances of flowers and fruits (with short carbon chains) and fats and oils (with long carbon chains).

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Preparation of esters

An ester is prepared by reacting, under reflux, an alcohol with a carboxylic acid in the presence of concentrated sulfuric acid as a catalyst. These reactions are called esterifications. Water is the other product and so it is also called a condensation reaction.

H C C

H

H

H

H

C

OH

O

+ HO C

H

H

C

H

H

C

H

H

H H C C

H

H

H

H

C

O

O

C3H7

+ H2O

Propanoic acid propan-1-ol propylpropanoate

These reactions are slow and so an extended period of heating is required to achieve a satisfactory yield. Reflux is a process by which a mixture of reactants is boiled for a prescribed period of time. The reactant and product vapours pass into a condenser where they are condensed to liquid drops that then fall back into the reaction mixture. The reflux process allows extended heating (to speed up the reaction) of the reaction mixture without loss of volatile reactants and products by evaporation.

The reactions are slow and reversible and so an equilibrium is established. It is pushed to the reactant side, by using excess alcohol.

Hydrolysis of esters

Esters undergo hydrolysis when refluxed with aqueous acid or base. Hydrolysis can be considered to be the reverse of esterification – water is consumed as a reactant during hydrolysis.

In acidic conditions, the products of the hydrolysis are a carboxylic acid and an alcohol. The acid is acting as a catalyst. Because the reaction is slow, the mixture is heated under reflux.

24


R C

O

O R'+ H2O R C

O

O H+ R’- OH

In basic conditions, such as with sodium hydroxide solution, the products of hydrolysis are a carboxylate salt and an alcohol. Refluxing is necessary to bring about a reaction at a reasonable rate.

R C

O

O R'+ OH- R C

O

O-+ R’- OH

Note the formation of water soluble carboxylate salts under alkaline conditions.To convert this salt to a carboxylic acid a solution of a strong acid such as hydrochloric acid is needed.

SUBTOPIC 5.8: AMIDES


An amide can be produced by a condensation reaction between an amine and a carboxylic acid.

Draw the structural formula of the amide that could be produced by the condensation reaction between an amine and a carboxylic acid, given their structural formulae.

Amides may be hydrolysed under acidic or alkaline conditions.

Identify the products of hydrolysis of an amide, given its structural formula.

Amides

Amides have the following functional group

While small amides are not very common in nature, larger ones in polymetric forms are in proteins.

Some simple amides of commercial importance:

Propenamide (polymer manufacture) Glutamine (infusion solution)

Caprolactum (polymer manufacture) Tick repellant

Preparation of amides

Theoretcially amides can be prepared from a condensation reaction between ammonia or an amine with a carboxylic acid:

25


In practice, when a carboxylic acid is reacted with ammonia or an amine, an acid base reaction occurs forming carboxylate and ammonium ions.

It usually requires refluxing to produce an amide.

Hydrolysis of amides

Amides are much more difficult than esters to hydrolyse and require extended refluxing.

Alkaline hydrolysis

Alkaline hydrolysis, with concentrated sodium hydroxide solution produces ammonia (or and amine) and a carboxylate salt.

Acid hydrolysis

Acid hydroysis, with concentrated hydrochloric acid, produces an ammonium (or protonated amine) ion and a carboxylic acid.

Assignment 5.7, 5.8 : Esters and Amides

1. Here are some ester fragrances:

ethyl methanoate - raspberries

3-methylbutyl ethanoate – pears

ethyl 2-methylbutanoate - apples

(a) Write the names of the alcohols and acids from which they are derived.

(b) Draw structures for these alcohols, acids and esters.

(c) How are 3-methylbutyl ethanoate and ethyl 2-methylbutanoate related?

2. Ethanol and acetic acid were heated with a little concentrated sulfuric acid in the flask of the apparatus shown:

(a) Name the organic product expected from the reaction.

(b) Write the equation for the reaction.

(c) State the function of the sulfuric acid.

(d) Explain why the condenser is used as shown in the diagram.

(e) Explain why the reactants must be heated for some time to obtain a satisfactory yield of product.

3. (a) Name the following compounds

26


(i)

H C

O

O CH2CH3

(ii)

H3C(H2C)3 C

O

O CH3

(a) Describe the expected odour of ethyl butanoate.

(b) Write an equation for its preparation.

(c) What is the catalyst used?

(d) Compare the water solubility of the reactants with that of ethyl butanoate and hence explain which components of the reaction mixture are mainly in the ‘organic’layer and which are mainly in the aqueous layer, when isolating the ester from the reactions in the separating funnel.

4. Consider the instructions below (1 – 9) and then answer the questions.

AIM : To hydrolyse the ester, ethyl benzoate, under alkaline conditions, and test product qualitatively.

INSTRUCTIONS.

1.Place about 5 mL of ethyl benzoate, 1rg of pumice (or boiling chips) and 25 mL of 10% sodium hydroxide solution (bench strength) in a flask and reflux for 30 – 40 minutes.

2.Set the condensor for distillation and collect 5 – 10 mL of distillate. Allow the residue to cool.

3.Add fused CaCl2 to the distillate – enough to cover the bottom of the container. Leave for 5 minutes and then

4.Test some of the liquid with acidified dichromate solution (warm). Note colour changes.

5.Test some for flammability. (A drop on a spatula in the Bunsen flame should ignite)

6.If time permits, determine the boiling point.

7.Acidify the cooled residue with concentrated sulfuric acid (CARE)

8.If time permits, filter, wash and dry the acidified residue.

9.Determine its melting point.

(a) Write a balanced equation for the reaction.

(b) Explain what reflux is and why it is necessary.

(c) Explain the purpose of the distillation.

(d) Explain the purpose of the fused calcium chloride.

(e) Explain the colour changes with the help of an annotated equation.

(f) Write an equation for the combustion.

(g) Write an equation for the acidification step. Explain why the residue does not distill.

(h) Explain why filtration of the acidified product is possible.

5. Ethyl ethanoate is an example of an ester.

(a) State three important commercial uses of esters.

(b) Fats and vegetable oils contain naturally occurring esters. Give the general formula of these esters and explain why they are important commercially.

(c) Describe, giving outline practical details, reaction conditions and equations, how you could produce a sample of ethyl ethanoate using ethanol as the only organic starting material.

(d) Esters may be hydrolysed by warming them with dilute aqueous acid or base. There are two possible bonds that could be broken in the ester as shown in the diagram below.

R C

O

O R'

Y

X

27


To identify which bond, (X) or (Y), is broken in the reaction, water enriched with oxygen-18, H2

18O, is used. After hydrolysis the oxygen-18 appears in the ethanoic acid and not in the ethanol.

C CH3

O

OC2H5

+ H218O OHC2H5 +

C CH3

O

18OH

(i) Use this information to identify which ester bond is broken during hydrolysis of the ester.

(ii) Suggest why it is this bond that is broken in preference to any other bond.

6. (a) Write a structural formula for the amide functional group.

(b) Complete the following equation:

(i)

R N

H

H+

HO C

O

R'

(ii) Why is this called a condensation reaction?

(c) Write the structural formula for ethanamide and identify the type of amide.

7. The C-N bonds in amides can be broken by hydrolysis.

(a) What is hydrolysis?

(b) This reaction requires reflux of the amide with either aqueous acid or alkali.

Complete these reactions with the appropriate structural formulae.

(i)

H3C C

O

NH2

(ii)C2H5 C

O

NHCH3

(iii)H3C C

O

NHC2H5

8. Nutrasweet, or aspartame, was discovered by accident in 1965. The patent expired in 1987 and sales world wide are now worth millions. It is 200 times sweeter than sucrose (cane sugar) so can be used in small quantities. It is hydrolysed in water or in the body.

Using structural formulae, complete this equation:

CH2

CH

CNH

OCH3

O

C

O

CH

H2N

CH2

C

HO

O

+ H2O

Aspartame

Subtopic 5.9: Proteins


Amino acids contain a carboxyl group and an amino group.

Determine whether or not a compound is an amino acid, given its structural formula.

Amino acids can self-ionise to produce an ion. Draw the structural formula of the product formed when an amino acid self-ionises.

Proteins are large molecules in which amide groups Identify the amide group and the repeating unit

28



link monomer units. and deduce the structural formula(e) of the monomer(s), given the structural formula of a section of a polyamide.

Proteins are polyamides consisting of covalently bonded long chains of amino acid units.

Write the general formula of amino acids and recognise their structural formulae.

In proteins the amide group is called a ‘peptide link’ or ‘peptide bond’.

Identify the amide groups in a protein.

Proteins have sites that allow hydrogen bonding between sections of chains and between the chain and water.

Identify where hydrogen bonding can occur between protein chains or between the chain and water, given the structural formula of a section of the chain.

The biological function of a protein is a consequence of its unique spatial arrangement.

Explain why the biological function of a protein (e.g. an enzyme) is altered if its spatial arrangement is altered.

Changes in pH and temperature disrupt the secondary interactions, and hence the spatial arrangements, of a protein chain.

Explain why proteins are sensitive to changes in pH and temperature.

Amino acids

Amino acids are the building blocks of proteins. They consist of small molecules with at least one amine functional group and one carboxyl functional group, a ‘R’ group. They have a central carbon that is bonded to:

One hydrogen atom;

One amine functional group;

One carboxyl functional group;

One other atom or group of atoms as R.

Some examples of R groups

R group

amino acid glycine valine aspartic acid tyrosine

Zwitterions

In a neutral environment, amino acids undergo self-ionisation whereby a proton is transferred from the carboxyl group to the amine group in the same molecule. The product of this self ionisation is a dipolar ion, called a zwitterion. Note it has no net charge.

N C

H

H

H

R

C

O

O H N+ C

H

H

H

R

C

O

O-H

Interesting point - The word zwitter in German means hermaphrodite.

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Proteins

While condensation reactions involving carboxylic acids and amines is difficult to achieve in the laboratory, condensation reactions between the carboxyl group from one amino acid can undergo a condensation reaction with an amine group from another amino acid with the aid of specific enzymes. The product of such a condensation is a dipeptide. The amide group that forms is called a peptide link. In the example below, there are two possibilities. (note water is also formed)

The continuation of this type of condensation process involving more amino acid molecules leads to the formation of along chain molecule called a polypeptide with many peptide links. These polypeptides are proteins that have the following general structural formula:

Hydrogen bonding within the protein chains

The peptide links within protein chains are polar. Hydrogen bonding between these links can occur both within a protein chain and between protein chains as shown in the diagrams below.

30


The polar peptide links in protein chains can also from hydrogen bonds with water molecules.

This hydrogen bonding can explain the amount of water in proteins like muscle tissue (eg moisture of fresh meat) and the ability of natural protein fibres like wood and silk to absorb water.

The structure and function of proteins

The structure of a protein and its biological function are inextricably linked. A protein enzyme has a structure that is compatible with the molecular structure of the substance that is undergoing an enzyme-catalysed conversion. Thus specific proteins can only do certain conversions.

The structure of proteins is described at four levels.

Primary structure is the sequence in which the amino acids are linked together by peptide bonds to form the polypeptide chain. eg DNA

Secondary structure refers to the shape that that the polypeptide chain folds or twists into. One example is the - helix shown below. The chains are held in these configurations by hydrogen bonds between peptides groups. In this case the amino groups are pointing up and the carbonyls down and the hydrogen bonding between them holds them together.

31


Tertiary structure refers to the folding of the structure onto itself. This structure is stabilised by different types of bonding between different R groups along the coils.

Quaternary structure Some protein consist of a number of polypeptide chains packed together. Quaternary structure refers to the 3-D arrangement in which chains pack together. It is mainly dispersion forces that hold these together.

Denaturation of proteins

If the normal secondary, tertiary or quaternary structure of a protein is altered, it loses its capacity to perform biological functions. The protein is said to be denatured.

Structural changes are commonly cause by changes in the temperature or pH of the environment of the protein. These changes can break the secondary bonds holding these structures in their normal shapes. eg setting of egg white or curdling of milk

Effect of pH

Changes in pH have the greatest effect in ionic bonding between –NH3+ and –COO- side groups. Increasing

the pH (with concentrated NaOH) converts the –NH3+ to molecular –NH2 and decreasing the pH (with

concentrated H2SO4) concerts the –COO- to the molecular –COOH. Thus ionic bonds cannot form and the structure is destabilised.

Effect of heat

Raising the temperature of a protein to above about 50C is sufficient to break secondary bonds - dispersion and hydrogen bonds. The protein structure then unravels as in the diagram below.

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ASSIGNMENT 5.9: Proteins

1. The general structure of an -amino acid is:

H2N CH COOH

R

Write equations for the following reactions:

(a) alanine (R = CH3) with hydrochloric acid.

(b) serine (R = CH2OH) with sodium hydroxide solution.

(c) lysine (R = (CH2)4NH2) with hydrochloric acid

(d) aspartic acid (R = CH2COOH) with sodium hydroxide solution.

(e) when alanine self ionises.

2. Amino acids are examples of bifunctional compounds, compounds with two functional groups. In alanine

C COOHH2N

CH3

H

(a) Circle and name the two functional groups present.

(b) Which group is the proton donor?

(c) Which group is the proton acceptor?

(d) Draw the zwitterion that forms from alanine.

3. Supply the missing words in this paragraph.

When amino acids combine to form a protein like insulin the __________ group from one amino acid joins into the _________ group on the next, and a molecule of __________ is lost from between them. This process is called ______ and the –CONH- group which links the amino acid residues in the protein is called the __________ link.

4. Two amino acids joined by the –CONH – group make a dipeptide.

Use these structures to show the formulation of a dipeptide.

R

H2C C

OH

O

H

N

H

CH

R'

C

OH

O

5. Look at the structure of the polymer shown below:

(CH2)5 C N

H

(CH2)5 C N (CH2)5 C

O O H O

(a) Redraw it to show the repeating unit.

(b) Draw the structures of the monomers used to make the polymers above.

6. Proteins are polymers made up of amino acids.

(a) Write the general formula of an amino acid.

(b) The structure of a protein molecule is shown below.

33


NH CH

CH2

OH

C

O

NH CH

CH2

C NH CH

CH2

OH OH

O

(i) Draw the structure of the peptide group.

(ii) Describe the type of bonding that is likely to occur between two of these protein molecules.

(iii) Draw a sketch to show how they are attracted to each other.

7. Amino acids are components of proteins. Draw diagrams to show the two dipeptides that may be formed between the amino acid glycine, H2NCH2CO2H, and alanine, H2NCH(CH3)CO2H.

8. Phenylketonuria (PKU) is a potentially lethal inherited disease. Babies suffering from PKU lack the ability to break down the surplus phenylalanine that is present in their diet. Excess phenylalanine is toxic and causes mental retardation. Its structural formula is given below:

C

H

H

C

H

NH2

COOH

(a) (i) Draw the structural formula of the zwitterion present in an aqueous solution of phenylalanine.

(ii) Explain why the addition of small quantities of acid or alkali to an aqueous solution of phenylalanine causes little change to the pH of the solution.

(b) Another amino acid, alanine, has the following structural formula:

H2N CH COOH

CH3

(i) Draw structural formulae for the two different dipeptides which can be formed when phenylalanine and alanine join to each other.

(ii) On your formulae, draw a circle around the peptide link.

(iii) What reagent and conditions are usually used in the lab to break the peptide link?

9. Explain why the biological function of a protein is altered if its special arrangement is altered.

10. Why do proteins often not work properly if there is a change in:

(a) Temperature

(b) pH

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11. Proteins are made up of amino acid units joined together. The structural formula of a fragment of a plant protein made up of three different amino acid units is shown below:

N

H

CH

H2C

C

O

N

H

CH2

C

O

N

H

CH

CH3

C

O

N

H

CH

CH2

C

O

N

S

H

CH2C

O

S

N

H

CH

H2C

C

O

N

H

CH2

C

O

N

H

CH

CH3

C

O

N

H

CH

CH2

HC

O

N

S

H

CH2

C

O

S

P

Q

(a) Circle and label an amide group in the protein fragment shown above.

(b) The unit P is derived from the amino acid alanine. Draw the structural formula of alanine.

(c) The two protein strands shown are cross linked by two different types of bonds in two different ways: by a disulfide link and by a hydrogen bond. On the structure above:

(i) Indicate the polarity of the hydrogen bond by using delta notation.

(ii) Indicate another hydrogen bond by using the same notation.

(iii) Circle the disulfide link and name the type of bond between the sulfur atoms.

(d) The amino acids represented by the units P and Q could bond together to from a polymer.

(e) Draw the structural formula of the resulting polymer, using two units each of P and Q alternating along the chain.

(f) Write the formula for the small molecule that will be released when the amino acids condense to form the polymer.

Subtopic 5.10: Triglycerides


Edible oils and fats are esters of propan-1,2,3-triol (glycerol) and various carboxylic acids. The carboxylic acids are unbranched and usually contain an even number of carbon atoms between twelve and twenty.

Draw the structural formula of an edible oil or fat, given the structural formula(e) of the carboxylic acid(s) from which it is derived.

Triglycerides can be hydrolysed to produce propan-1,2,3-triol and various carboxylic acids.

Identify the alcohol and the acid(s) from which a triglyceride is derived, given its structural formula.

Edible oils are liquids at 25°C and are commonly obtained from plants and fish. Edible fats are solids at 25°C and are commonly obtained from land animals.

Identify the most likely source of a triglyceride, given its state at 25°C.

Most liquid triglycerides contain a larger proportion of unsaturated carbon chains than solid triglycerides contain.

Describe the use of a solution of bromine or iodine to determine the degree of unsaturation of a triglyceride.

Liquid triglycerides can be converted into triglycerides Explain the role of pressure, temperature, and a

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of higher melting-point by a process that involves the addition of hydrogen, under pressure, at increased temperature, and in the presence of a catalyst.

catalyst in the hydrogenation of liquid triglycerides.

Triglycerides

Triglycerides are edible fats and oils derived from plants and animals. They are esters of propan-1,2,3-triol (glycerol) and various carboxylic acids. The general structural formula of a trigylceride is as follows:

These carboxylic acids are often referred to as fatty acids. Almost without exception they contain an even number of carbon atoms (commonly from 12 to 22) In the carbon chain, including the carbon atom of the carboxyl group. The chains are both saturated and unsaturated. If more than one C = C bond is present in the carbon chain, it is described as polyunsaturated.

The structural formulae of some of the more common fatty acids are given in the table below. It should be noted that these structural formulae are usually written in condensed form as in the following examples:

CH3CHCH2CH2CH2CH2CH2CH2CH2CH2CH2COOH is written as CH3(CH2)10COOH

CH3CH2CH2CH2CH2CH2CH = CHCH2CH2CH2CH2CH2CH2CH2COOH

is written as CH3(CH2)5CH = CH(CH2)7COOH

Fatty acid

common name

(…… acid)

systematic name

(…….. acid)structural formula

lauric dodecanoic CH3(CH2)10COOH

palmitic hexadecanoic CH3(CH2)14COOH

stearic octadecanoic CH3(CH2)16COOH

palmitoleic hexadec-9-enoic CH3(CH2)5CH = CH(CH2)7COOH

oleic octadec-9-enoic CH3(CH2)7CH = CH(CH2)7COOH

linoleic octadec-9,12-dienoic CH3(CH2)4CH = CHCH2CH = CH(CH2)7COOH

Naturally occurring triesters of propane-1,2,3-triol nearly always have three different fatty acids forming the ester linkages as shown In the following example in which linoleic, stearic and oleic acids have formed ester links with the propane-1,2,3-triol:

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Hydrolysis of triglycerides

Fats and oils are highly concentrated stores of energy. The first stage in the use of fats as an energy source is the hydrolysis of the triglyceride molecules catalysed by enzymes called lipases. The products of the hydrolysis of each fat molecule are one molecule of glycerol and three fatty acid molecules as shown by the following general equation:

Without the aid of these lipases, extreme conditions are needed to perform this hydrolysis. This usually involves extended heating with concentrated acids or alkalis.

Edible fats and oils

Edible oils are liquids at 25°C and are commonly obtained from plants and fish. Edible fats are solids at 25°C and are commonly obtained from land animals. Some examples are shown below:

fat or oil mol % of saturatedfatty acids

mol % of unsaturatedfatty acids

beef fat 55 45

pig fat 45 55

corn (maize) oil 15 85

olive oil 15 85

cod liver oil 20 80

coconut oil 95 5

Note - melting point:

Increases with chain length

Decrease as the degree of unsaturation increases. This is due to the less ordered structure of the unsaturated chains which cannot then pack together as closely, weakening the dispersion forces.

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The ordered and compact structure of a saturated triglyceride

The disordered and loose structure of an unsaturated trigyceride

Determining the degree of unsaturation of a triglyceride

Alkanes which contain the C=C group can undergo addition reactions with diatomic molecules like the halogens and hydrogen. The general reaction can be represented as:

Fats and oils which contain unsaturated groups will also undergo these addition reactions if mixed with a solution containing bromine or iodine in a saturated hydrocarbon solvent such as cyclohexane. The colour of the solution disappears as the products of the addition reaction are colourless.

The degree of saturation can be be determined by measuring the volume of bromine solution required to react with a solution of a fat or oil in cyclohexane eg by titration.

Hydrogenation of fats

Liquid oils can be converted to solid fats by a process called hydrogenation – the process of adding hydrogen across a C=C bond. To do this a vegetable oil is heated with hydrogen under pressure in the presence of a nickel catalyst. Sufficient hydrogen is added to form a product that is a solid at room temperature, but still retain a degree of unsaturation.

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Subtopic 5.11: Carbohydrates


Carbohydrates are naturally occurring sugars and their polymers. They usually have the general formula CxH2yOy. They are defined more precisely as either polyhydroxy aldehydes or polyhydroxy ketones, or their polymers.

Determine, given its structural formula, the molecular formula of an organic compound, and whether or not it is a carbohydrate.

Carbohydrates can be classified as monosaccharides, disaccharides, or polysaccharides.

Write molecular formulae for glucose, and disaccharides and polysaccharides based on glucose monomers.

Polysaccharides are produced by the condensation of many monosaccharide units linked in chains by covalent bonds.

Identify the repeating unit and draw the structural formula of the monomer, given the structural formula of a section of a polysaccharide derived from one monomer.

Glucose molecules can occur in either a chain form or a ring form. There is equilibrium between the two structures. In the chain form an aldehyde group is present.

Explain the ability of glucose to react as an aldehyde when in chain form but not when in ring form.

Many simple carbohydrates are soluble in water whereas polysaccharides are insoluble in water.

Explain the differences in solubility in water of simple carbohydrates and polysaccharides in terms of the size of the molecules and the number of hydroxyl groups.

Carbohydrates

Carbohydrates are naturally occurring sugars and their polymers. They usually have the general formula Cx(H2O)y where x is often equal to y, and can be considered hydrates of carbon. They are defined more precisely as either polyhydroxy aldehydes or polyhydroxy ketones, or their polymers. They are also classified as monosaccharides, disaccharides and polysaccharides depending on the number of simple sugar units that comprise the molecular structure.

Monosaccharides

These are the monomers from which all disaccharide and polysaccharide molecules are constructed. The following generalisations can be applied to monosaccharides:

their general formula is CXH2XOX with x having values from 3 to 8;

they are water-soluble compounds (they have a low ratio of carbon to hydroxyl groups - almost 1:1);

they have the name ending …ose. For example, glucose and fructose.

they are sweet to taste and are often called simple sugars;

structurally the molecules of six carbon polyhydroxyketones and five and six carbon polyhydroxyaldehydes can exist in chain or cyclic (ring) form. The two forms exist in equilibrium with each other. In aqueous solution the equilibrium lies very much in favour of the cyclic form.

they are solids at room temperature (extensive hydrogen bonding).

Name and molecular formula

Chain structure Cyclic structure

Glyceraldehyde

C3H6O3

N/A

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Erythose

C4H8O4

N/A

Ribose

C5H10O5

Ribulose

C5H10O5

N/A

Glucose

C6H12O6

Fructose

C6H12O6

Disaccharides

These are water-soluble carbohydrates, with two monosaccharide units per molecule. A disaccharide molecule is formed by a condensation reaction of two monosaccharide molecules with the elimination of a water molecule. Hydrolysis of disaccharides reverses the formation process to produce monosaccharides. In nature, these condensation and hydrolysis reactions are catalysed by enzymes.

There are several disaccharides comprised of glucose monomers. They differ in the positions in which the glucose units are linked and in the relative positions of H atoms and OH groups in the ring structures. Most of them only occur as hydrolysis products of polysaccharides. Maltose is an example of a disaccharide.

Polysaccharides

These are insoluble in water (due to their very large molecular size) but do absorb water and are virtually tasteless. They are formed by successive condensation reactions of monosaccharides. Partial hydrolysis produces di- and tri-saccharides. Complete hydrolysis of a polysaccharide produces monosaccharides such as glucose. Four of the most common and biologically most important polysaccharides are all composed of glucose units. They are:

cellulose - the main structural material of plants;

glycogen - the main storage polysaccharide in animals;

amylose - one of the two polysaccharides that make up starch;

amylopectin - the other polysaccharide that makes up starch.

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Cellulose and amylose are straight chain polysaccharides whereas glycogen and amylopectin are branched chain polymers.

Cellulose

On average, each cellulose molecule is made up of about 3,000 glucose units linked as shown below.

Glucose as a reducing agent

Glucose exists in both chain and cyclic forms, which in aqueous solution are in equilibrium. The aldehyde functional group that is present in the chain form can be oxidised by Tollen's Reagent. The silver ion in the Tollen's Reagent is reduced to silver metal in the form of a mirror, while at the same time the aldehyde group is oxidised to a carboxylate group. There is no aldehyde group present in the cyclic form and consequently this form does not react with Tollen's Reagent. However, consumption of the chain form by Tollen's Reagent upsets the chain - cyclic equilibrium driving it towards the formation of more of the chain form (Le Chatelier's Principle).

The interaction of carbohydrates with water

Carbohydrate molecules can undergo extensive hydrogen bonding with polar water molecules because of the presence of hydroxyl groups as part of their molecular structure. However, only the mono- and disaccharides have molecules small enough in size to actually mix with the water molecules to form aqueous solutions.

Assignment 5.10 & 5.11 : Triglycerides and Carbohydrates

1. (a) What structural feature of a molecule identifies a compound as an ester?

(b) Write down the full structural formula of the triester made from glycerol and ethanoic acid.

(c) Give an equation to show what happens when this ester is heated with sodium hydroxide solution.

2. Below is the structure of a triester present in a naturally occurring substance.

H2C

HC

H2C O

O

O C

C

C

O

O

O

(CH2)7CH

(CH2)7CH

(CH2)7CH

CH(CH2)7CH3

CHCH2CH CH(CH2)4CH3

CH(CH2)7CH3

(a) What is the name given to this tiester and similar triesters?

(b) Is this triester likely to be a fat or an oil? Give reasons for your answer.

(c) The triester was hydrolysed by heating with sodium hydroxide solution, and the resulting solution was neutralised by addition of hydrochloric acid.

(i) Draw structural formulae for the hydrolysis products formed. Use short form as above.

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(ii) Name the alcohol formed.

(d) How many moles of each product would be obtained from one mole of the triester?

3. Explain the difference between the following terms, which often appear on food labels.

(a) Saturated fat

(b) Monounsaturated fat

(c) Polyunsaturated fat

4. Animal fat is made up almost entirely of saturated fats. One compound present in sheep fat produces only glycerol and stearic acid, CH3(CH2)16COOH, on hydrolysis. Draw out the structural formula of this compound. (use the short form as in question 2)

5. A food company buys a vegetable oil and converts it to margarine by treatment with hydrogen in the presence of a metal catalyst. During the process some of the double bonds in the oil are hydrogenated.

(a) Name a suitable catalyst and give conditions under which the hydrogenation could be carried out.

(b) Assume that the oil contains only the triester shown in question 2. 1 tonne (1,000 kg) of the oil requires 4.90 kg of hydrogen to make margarine. What percentage of the double bonds in the oil have been hydrogenated?

6. The ester spermaceti is a soft wax that in the past was obtained from the head of the sperm whale. It is used in skin lotions and has the following formula:

C C10H22C15H31

O

O

Write the formula of the carboxylic acid and the alcohol from which spermaceti can be prepared.

7. Oleic acid is an unsaturated carboxylic acid obtained from the hydrolysis of soya bean oil. The structure of oleic acid is shown below:

CH3(CH2)7CH=CH(CH2)2COOH

(a) Write the structural formula of a triglyceride that would give oleic acid on hydrolysis.

(b) When heated, liquid oils may be converted into solid fats, as shown below:

OIL + A FAT

Identify the required reagent A, the substance B and the condition C necessary for the conversion to occur.

8. Trehalose, aspartame, and cyclamate are three substances used as sweeteners.

Trehalose Aspartame cyclomate

OC C

CC

C

HOH2C

HO

OH

OH

O

H HH

HH

OCC

CC

C

CH2OH

OH

OH

HO

HHH

HH

CH2

CH

CNH

OCH3

O

C

O

CH

H2N

CH2

C

HO

O

NH

SO3H

(a) Which one is the carbohydrate and why?

(b) Write down the molecular formula of this carbohydrate.

(c) State the conditions needed to convert the carbohydrate to ethanol.

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(d) Glucose burns in air above 500C. Explain how it is possible for living organisms to carry out this reaction swiftly and efficiently at 37C.

(e) Write the equation for the conversion of the carbohydrate, glucose, to ethanol.

9. Sucrose is a natural sweetener derived from sugar cane. The structure of sucrose is shown below:

OC C

CC

C

HOH2C

HO

OH

OH

O

H H

H

H

H

OCC

C C

CH2OH

OHHO

H

H H

CH2OH

Sugars may be classified as monosaccharides, disaccharides, or polysaccharides.

(a) State the type of saccharide that sucrose belongs to.

(b) A crushed sample of sucrose is warmed with ammoniacal silver nitrate solution. State the observation you would expect. Give a reason for your answer.

(c) Sucrose can be hydrolysed to produce a mixture of fructose (F) and Glucose (G). The structure of F and G are shown in open-chain forms below:

F G

(d) State the saccharide classification to which both fructose and glucose belong.

(e) The mixture of F and G is warmed with ammoniacal silver nitrate solution. State the observation you would expect and name the functional group responsible for it.

10. Sucrose, C12H22O11, is so soluble in water that a saturated solution at room temperature is only one third solvent by weight. The solubility of sucrose in several solvents at 20oC is given.

Solvent Solubility of sucrose

Water

Methanol

Ethanol

Petrol

1 g/0.5 mL water

1 g/100 mL methanol

1 g/170 mL ethanol

insoluble

Explain these differences in solubility.

11. Glucose exists in equilibrium between a cyclic form and an open chain form shown below:

(a) Name the functional groups present in open form.

(b) (i) What would you observe if a sample of the above mixture was shaken and warmed with some Tollen’s reagent?

(ii) What type of reaction has the glucose undergone?

(iii) Describe the changes to the above equilibrium resulting from the addition of Tollen’s reagent.

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12. The structural formulae of glucose and fructose are shown below:

(a) (i) Circle and name the functional groups in the two structures above.

(ii) Write down the molecular formula for both these sugars.

(iii) Explain the high solubility of glucose in water.

(b) Glucose units can chemically bond together to form long chain molecules. Give two examples.

(c) Glucose, which is found in fruit and honey, is the most abundant sugar found in nature. Why do athletes consume glucose, rather than other sugars, as a source of ‘instant’ energy? Support your answer with a chemical equation.

(d) What would you observe if glucose was treated with:

(i) acidified dichromate ions?

(ii) ammoniacal silver nitrate solution?

(e) Can fructose undergo the two reactions in d) above? Explain.

(f) (i) Draw the structure of the compound formed when fructose is treated with some acetic acid, in the presence of a few drops of concentrated sulfuric acid.

(ii) Name the new functional group produced.

(iii) What name is given to the reaction?

(iv) What is the function of the concentrated sulfuric acid?

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