student name:

Year 12 Chemistry Module 7

The purpose of this unit is to support student investigation into the many classes of organic compounds and their characteristic chemical reactions. By considering the primary, secondary and tertiary structures of organic materials, students are provided with opportunities to gain an understanding of the properties of materials – including strength, density and biodegradability – and relate these to proteins, carbohydrates and synthetic polymers.

Student Name: ………………………………………………………

HSC Chemistry Organic Chemistry P a g e | 2

Contents Introduction .................................................................................................................................... 5

Nomenclature ..................................................................................................................................... 6

Nomenclature of Hydrocarbons .................................................................................................. 8

Nomenclature of Organic compounds .................................................................................... 8

Question Set 1 ........................................................................................................................... 11

Nomenclature of Alkanols .......................................................................................................... 13

Question Set 2 ........................................................................................................................... 15

Nomenclature of Aldehydes and Ketones ................................................................................ 16

Alkanals..................................................................................................................................... 16

Alkanone ................................................................................................................................... 17

Question Set 3 ........................................................................................................................... 17

Nomenclature of Carboxylic Acids ............................................................................................ 18

Question Set 4 ........................................................................................................................... 19

Nomenclature of Amines and Amides ...................................................................................... 20

Amines ...................................................................................................................................... 20

Amides ...................................................................................................................................... 21

Question Set 5 ........................................................................................................................... 22

Nomenclature of halogenated organic compounds ................................................................ 23

Question Set 6 ........................................................................................................................... 24

Structural Isomerism ................................................................................................................... 26

Chain isomers ........................................................................................................................... 26

Position isomers ....................................................................................................................... 27

Functional group isomers ....................................................................................................... 27

Question Set 7 ........................................................................................................................... 28

Hydrocarbons ................................................................................................................................... 31

Reviewing the Hydrocarbons ..................................................................................................... 33

Question Set 8 ........................................................................................................................... 35

Properties of Alkanes .................................................................................................................. 37

Experiment 19: Properties of Hydrocarbons..................................................................... 38

Question Set 9 ........................................................................................................................... 39

Molecular shapes ......................................................................................................................... 41


Question Set 10 ......................................................................................................................... 44

Bonding within and between the hydrocarbons ...................................................................... 45

Question Set 11 ......................................................................................................................... 47

Safe Handling of Hydrocarbons................................................................................................. 49

Question Set 12 ......................................................................................................................... 51

Implications of Hydrocarbons .................................................................................................... 54

Question Set 13 ......................................................................................................................... 55

Products of Reactions Involving Hydrocarbons .......................................................................... 57

Addition Reactions ...................................................................................................................... 59

Experiment 20: Properties of Hydrocarbons..................................................................... 61

Question Set 14 ......................................................................................................................... 61

Substitution Reactions ................................................................................................................. 63

Question Set 15 ......................................................................................................................... 64

Alcohols ............................................................................................................................................ 65

Alkanols ........................................................................................................................................ 67

Question Set 16 ......................................................................................................................... 69

Properties of Alcohols ................................................................................................................. 70

Question Set 17 ......................................................................................................................... 71

Enthalpy of Combustion for Alcohols ....................................................................................... 73

Experiment 21: Enthalpy of Combustion for Alcohols .................................................... 74

Question Set 18 ......................................................................................................................... 75

Reactions involving Alcohols ..................................................................................................... 80

Question Set 19 ......................................................................................................................... 82

Production of Alcohols ................................................................................................................ 84

Question Set 20 ......................................................................................................................... 86

Production of Alcohols by Fermentation .................................................................................. 87

Experiment 22: Fermentation Reaction ............................................................................. 88

Question Set 21 ......................................................................................................................... 88

Oxidation of Alcohols .................................................................................................................. 91

Question Set 22 ......................................................................................................................... 92

Comparing Fuels .......................................................................................................................... 93

Question Set 23 ......................................................................................................................... 95


Reactions of Organic Acids and Bases........................................................................................... 98

Comparing Properties of different Organic Groups .............................................................. 100

Question Set 24 ....................................................................................................................... 102

Properties and Bonding ............................................................................................................. 105

Question Set 25 ....................................................................................................................... 108

Esterification ............................................................................................................................... 110

Refluxing ................................................................................................................................. 111

Experiment 23: Esterification ............................................................................................ 112

Question Set 26 ....................................................................................................................... 112

Organic Acids and Bases ........................................................................................................... 118

Question Set 27 ....................................................................................................................... 119

Saponification ............................................................................................................................. 121

Experiment 24: Saponification .............................................................................................. 124

Question Set 28 ....................................................................................................................... 125

Organic Reaction Pathways ...................................................................................................... 131

Question Set 29 ....................................................................................................................... 132

Polymers ......................................................................................................................................... 134

Some Common Polymers – Polyethylene (PE) ....................................................................... 136

Question Set 30 ....................................................................................................................... 138

Polyvinyl chloride (PVC) .......................................................................................................... 141

Question Set 31 ....................................................................................................................... 142

Polystyrene (PS) ......................................................................................................................... 143

Question Set 32 ....................................................................................................................... 144

Polytetrafluoroethylene (PTFE) ................................................................................................ 145

Question Set 33 ....................................................................................................................... 146

Condensation Polymers - Nylon .............................................................................................. 148

Question Set 34 ....................................................................................................................... 149

Condensation Polymers - Polyesters ....................................................................................... 152

Question Set 35 ....................................................................................................................... 153

References ....................................................................................................................................... 154


Introduction

Students focus on the principles and applications of chemical synthesis in the field of organic chemistry. Current and future applications of chemistry include techniques to synthesise new substances – including pharmaceuticals, fuels and polymers – to meet the needs of society.

Each class of organic compounds displays characteristic chemical properties and undergoes specific reactions based on the functional groups present. These reactions, including acid/base and oxidation reactions, are used to identify the class of an organic compound. In this module, students investigate the many classes of organic compounds and their characteristic chemical reactions. By considering the primary, secondary and tertiary structures of organic materials, students are provided with opportunities to gain an understanding of the properties of materials – including strength, density and biodegradability – and relate these to proteins, carbohydrates and synthetic polymers.

In this module, students focus on collecting, analysing and processing data and information to identify trends, patterns and relationships to solve problems and communicate scientific understanding of ideas about organic chemistry.

There are six main areas of content:

Nomenclature (N) Hydrocarbons (H) Products of Reactions involving Hydrocarbons (PRH) Alcohols (A) Reactions of Organic Acids and Bases (ROAB) Polymers (P)

Through these areas of content, you will have the opportunity to explore the following questions:

1) How do we systematically name organic chemical compounds? 2) How can hydrocarbons be classified based on their structure and reactivity? 3) What are the products of reactions of hydrocarbons and how do they react? 4) How can alcohols be produced and what are their properties? 5) What are the properties of organic acids and bases? 6) What are the properties and uses of polymers?


Nomenclature


Concept Map

Inquiry question 1•How do we systematically

name organic chemical compounds?


Nomenclature of Hydrocarbons

Watch OC#1 Naming Hydrocarbons

Nomenclature of Organic compounds

The system of nomenclature used for the naming of organic compounds follows the IUPAC system in the Blue Book (Favre & Powell, 2013). Organic Chemistry is an incredibly rich and diverse field of Chemistry and it is the only branch of Chemistry which is based on a single element – carbon. Organic chemistry is sometimes referred to as the chemistry of life and whilst it is true that many organic molecules are found in living things, the field of organic chemistry is even broader. Most organic compounds come from coal, oil, or plant and animal products. Organic compounds are carbon compounds.

A vast number of carbon compounds exist because:

The carbon atom is tetravalent which means that each carbon atom can share an electron pair with four other atoms.

The carbon atom itself is relatively small.

As a result, carbon atoms can form stable chains, networks and rings by means of carbon-carbon bonds. Even its elemental forms of diamond and graphite show this ability.

I can investigate the nomenclature of organic chemicals, up to C8, using IUPAC conventions, including simple methyl and ethyl branched chains, including alkanes, alkenes and alkynes.


Clarity in communication about chemical substances is of critical importance for a global subject like Chemistry, hence the rules exist which ensure all chemists know which compound or group of compounds is being discussed. In the first part of this module, we will look at some of the important rules of nomenclature as they apply to different groups of organic compounds.

A few simple rules will get us started:

Prefix: indicates the number of carbons in the chain, eg meth = 1, eth = 2, etc.

Suffix: indicates the functional group, eg ane (single bond), ene (double bond), yne (triple bond).

The first group for study is the hydrocarbons. Hydrocarbons are compounds containing only hydrogen and carbon. There are three important groups of hydrocarbons which we need to learn: the alkanes, alkenes and alkynes. NB The ‘Alk’ prefix is used to refer to the group.

Alkanes

Methane

Ethane

Propane

Butane

Pentane

Hexane

Heptane

Octane

Alkenes

Ethene

Propene

But-1-ene

Pent-1-ene

Hex-1-ene

Hept-1-ene

C C C C C CH

H

H H

H

H

H

H

H

H

H

H

H

H

C C C

C

H

H

H H

H

H

H

H


Oct-1-ene

Alkynes

Ethyne

Propyne

But-1-yne

Pent-1-yne

Hex-1-yne

Hept-1-yne

Oct-1-yne

Naming Conventions

Organic compounds are classified as if they were derivatives of hydrocarbons. In order to agree on which hydrocarbon will be chosen, we count the longest chain of carbon atoms in the molecule without back-tracking. Having selected the chain, we number each carbon atom, starting from one end, and describe the other groups attached to the chain by naming them also and giving them a number to indicate the carbon atom to which they are attached. By convention we begin numbering the chain from the end which gives us the lowest numbers. The number should immediately precede the atom or group of atoms which it allocates. The names given to the parent hydrocarbon incorporate the following prefixes: 1 - meth, 2 - eth, 3 - prop, 4 - but, 5 - pent, 6 - hex, 7 - hept, 8 - oct.

A hydrocarbon with no multiple carbon - carbon bonds in the longest chain will have the ending -ane. Double bonds are indicated by the ending -ene. Triple bonds are described as -yne.

Organic chemical names are usually written as one word with total disregard of aesthetics or even grammar. Where a number of endings is necessary, they are simply tacked on with their appropriate numbers to the stem.

Alphabetical order is observed by convention.

Numbers are separated from the word by hyphens.

Multiples of side chains or functional groups are indicated by di (2), tri (3), tetra (4), etc

The hydrocarbon skeleton parts usually consist of an alkane less one hydrogen atom and they are named from the related alkanes. Eg. CH3 - methyl, C2H5 - ethyl, C3H7 - propyl, etc.

C C C H

H

H

H


Note: These are not whole molecules but only parts of molecules which always have other atoms combined with them. In general formulae they may be represented as "R".

Question Set 1

1. Draw the structural formula for methylpropane. (Odlum, 1999, p. 58) (1 mark)

2. Draw a structural formula for one possible compound with the molecular formula C4H6. Name the compound you have drawn. (2 marks)

3. What is the IUPAC name for the compound shown below? (Odlum, 1999, p. 52)

CH2 CCH3

CH2 CH

CH2 CH3

CH3

A. 2-methyl-4-ethyl-pent-1-ene B. 2,4-dimethyl-hex-1-ene C. 2,5-dimethyl-hex-5-ene D. 4-ethyl-2-methyl-pent-1-ene

4. (95, Q8)

The structural formula below represents a compound.

C

C

CC

CCC

C

HH

HH

H

H

HH

H

H

H

HHH


The IUPAC systematic name for this compound is

A. 2-methyl-4-ethyl-pent-1,3-diene

B. 2-ethyl-4-methyl-pent-2,4-diene

C. 2,4-dimethyl-hex-1,3-diene

D. 3,5-dimethyl-hex-3,5-diene

5. Two compounds which may be found in a school laboratory are cyclohexane (C6H12) and cyclohexene (C6H10). Draw diagrams to distinguish between these two compounds. (2 marks)

6. 3 Hydrocarbons – Alkanes (Schell & Hogan, 2018, pp. 5-6) 7. 4 Hydrocarbons – Alkenes (Schell & Hogan, 2018, p. 7) 8. 5 Hydrocarbons – Alkynes (Schell & Hogan, 2018, p. 8) 9. 6 Naming Hydrocarbons (Schell & Hogan, 2018, pp. 9-11) 10. Saturated and unsaturated hydrocarbons (de Vreeze & McMicking, 1998, pp. 32-33) 11. 17 Introduction to Carbon Chemistry 1 (Gribben & St Germain, 2005, pp. 37-38)


Nomenclature of Alkanols

Watch OC#2 Naming Alcohols

The functional group of an organic compound is the part that undergoes change in a reaction. Eg. -OH, -Br, -CHO, -COOH, -NH2. These give the molecules their characteristic properties.

Some homologous series related to the alkanes include:

• ALKANOLS (alcohols) ROH • ALKANALS RCHO • ALKANONES RCOR' • ALKANOIC ACIDS (carboxylic acids) RCOOH • HALOALKANES RX • ALKYL ALKANOATE (esters) RCOOR’

We shall look at the nomenclature for each of these groups, starting with the alkanols.

Alkanols (alcohols) are identified by the presence of a hydroxyl group (OH) attached to a carbon. The number of carbons to which that carbon is also bonded helps us classify the three different groups of alcohols.

Primary: The OH group is attached to an end (or terminating) carbon atom

I can investigate the nomenclature of organic chemicals, up to C8, using IUPAC conventions, including simple methyl and ethyl branched chains, including alcohols (primary, secondary and tertiary).


Secondary: The OH group is attached to a carbon which is also bonded to two other carbons.

Tertiary: The OH group is attached to a carbon which is also bonded to three other carbons.

The most common alcohol is ethanol - a colourless liquid with a faint, sharp odour. It is the key ingredient in alcoholic beverages. Ethanol is a primary alcohol and it looks like this:

C C OH

H

H

H

H

H

Figure 1 Ethanol (C2H5OH)

A common secondary alcohol is isopropanol or propan-2-ol (2-propanol). It looks like this:

C C C

OHH

H

H

H H

H

H

Figure 2 propan-2-ol (C3H7OH)

Notice the hydroxy group is now on a middle carbon which is attached to two other carbon atoms.

An example of a tertiary alcohol is methylpropan-2-ol. It looks like this:

C C C

C

OHH

H

HH

H

H

HHH

Figure 3 methylpropan-2-ol (C4H9OH)

Notice the hydroxy group is now on a middle carbon which is attached to three other carbon atoms.


Question Set 2

1. Draw the structural formula for the following compounds: a) Methanol

b) Propan-1-ol c) Butan-1,2-diol

2. (1995, Q16 a) 2 marks) Draw a structural formula for 2-methyl-2-pentanol and for 3-methyl-2-pentanol.

3. (1995, Q25, 2 marks)

a) Glycerol is the common name for 1,2,3-propan-triol. Draw the structural formula

for glycerol.

b) Name the functional group present in glycerol.

.......................................................................................................................................................

4. Ethylene glycol, also known as antifreeze, is a substance which is added to car radiators to change the freezing and boiling temperature of water to reduce the chances of damage. The IUPAC name for ethylene glycol (antifreeze) is 1,2-ethanediol. Draw the structural formula for this compound.

5. 7 Alcohols (Schell & Hogan, 2018, p. 12) 6. Primary, secondary and tertiary alkanols (de Vreeze & McMicking, 1998, pp. 39-40) 7. 7 Alkanols (Gribben, 2010, pp. 14-15)


Nomenclature of Aldehydes and Ketones

Watch OC#3 Naming aldehydes and ketones

Another important functional group is the double-bonded oxygen. There are two places where a double bonded oxygen can be located on a carbon chain; either at an end carbon or on a carbon atom attached to two other atoms. Its location affects its name and we have two homologous series:

Alkanals

These organic compounds are also known as aldehydes. They have a double-bonded oxygen attached to an end carbon. The simplest example is methanal (also known as formaldehyde).

C O

H

H

Figure 4 Methanal (CH2O)

I can investigate the nomenclature of organic chemicals, up to C8, using IUPAC conventions, including simple methyl and ethyl branched chains, including aldehydes and ketones.


Alkanone

These organic compounds are also known as ketones. They have a double-bonded oxygen attached to a central carbon which is bonded to two other carbon atoms. The simplest example is propanone (also known as acetone).

C O

C

CH

HH

H

H

H

Figure 5 propanone (C3H6O)

Question Set 3

1. Draw the structural formula for the following aldehydes:

a) Ethanal

b) Butanal

c) Heptanal

2. Draw the structural formula for the following ketones:

a) Butanone

b) Hexanone

c) Octanone

3. Alkanols, alkanals and alkanones (de Vreeze & McMicking, 1998, pp. 36-38)

4. 9 Aldehydes and Ketones (Schell & Hogan, 2018, p. 15)


Nomenclature of Carboxylic Acids

Watch OC#4 Naming carboxylic acids

In the previous module, we spent some time with organic acids even though we didn’t study their structure. Organic acids are usually weak acids, but many are also primary standards which makes them a great choice for titrations, especially for standardising sodium hydroxide.

Organic acids, more properly named alkanoic (or carboxylic) acids, have the COOH functional group. This involves an oxygen double bonded to a carbon with an additional hydroxy group off the same carbon. This only leaves one remaining bond for the carbon and hence this functional group must occur on an end (or terminal) carbon. The first few carboxylic acids are shown.

CO

OH

H

Figure 6 Methanoic acid (HCOOH)

I can investigate the nomenclature of organic chemicals, up to C8, using IUPAC conventions, including simple methyl and ethyl branched chains, including carboxylic acids.


C C

O

OH

H

H

H

Figure 7 ethanoic acid (CH3COOH)

C C C

O

OH

H

H

H

H

H

Figure 8 propanoic acid (C2H5COOH)

Question Set 4

1. Draw the structural formula for the following carboxylic acids:

a) Butanoic acid (butyric acid)

b) Hexanoic acid

c) Octanoic acid

2. Draw the structural formula for the following branched carboxylic acids: a) Methylpropanoic acid

b) 2,3-dimethylpentanoic acid

3. 8 Carboxylic Acids (Schell & Hogan, 2018, pp. 13-14) 4. 18 Introduction to Carbon Chemistry 2 (Gribben & St Germain, 2005, p. 39) 5. 8 Organic Acids (Gribben, 2010, p. 16)


Nomenclature of Amines and Amides

Watch OC#5 Naming amines and amides

Our previous experience with the ‘am’ prefix has been with the cation ammonium and the weak base ammonia. Both are combinations of N and H. Likewise when we are naming organic compounds with N in the structure, we also use the ‘am’ prefix, although now it is a functional group and hence is used as a suffix! There are two important homologous series you need to know which contain N; the amines and the amides.

Amines

Amines have an NH2 functional group with the nitrogen containing a lone pair of unbonded electrons. Just like alcohols, the two hydrogen atoms can be replaced by carbon atoms to produce a secondary or tertiary amine. Amines are named similarly to alcohols, with the chain attached to the amine numbered and the appropriate prefix selected prior to the amine suffix.

The sequence for naming an amine is:

1. Find the longest continuous carbon chain which contains the functional group. 2. Name any branched groups (substituents) with the standard prefix along with the yl

suffix.

I can investigate the nomenclature of organic chemicals, up to C8, using IUPAC conventions, including simple methyl and ethyl branched chains, including amines and amides.


3. Number the longest chain sequentially (with the carbon attached to the amine having the lowest number.

4. Observe alphabetic rules for side groups (with di, tri, etc for multiple side groups). 5. Include numerical or N prefix depending on whether the side chain is attached to the

main chain or the N atom.

C

C

C

N

HH

H

H

H

HH

H

H

Figure 9 propan-2-amine (or isopropanamine) (C3H7NH2)

CC

C

C

NC

C

HH

H

H

H

H

HH

H

H

H

H

H

H

H

Figure 10 N-ethyl-2-butanamine (C4H9NHC2H5)

Amides

Amides are a little more tricky. The easiest way to think of them is as derivatives of a carboxylic acid, where the OH group has been replaced by an NH2 group. Like the amines, amides can be primary, secondary or tertiary and again may include both a number and/or an N to indicate the position of R groups.

CC

O

N

H

HH

H

H

Figure 11 ethanamide (CH3CONH2)


CC

O

H

HH

NC

H

H

H

H

Figure 12 N-methylethanamide (CH3CONHCH3)

Whilst it is unclear, it is probable that secondary and tertiary amides are beyond the scope of this course. However, it is worth seeing and naming one or two, just in case.

Question Set 5

1. Draw the structural formula for the following amines:

a) Methanamine

b) Ethanamine

c) Butanamine

2. Draw the structural formula for the following amides:

a) Methanamide

b) Ethanamide

c) Butanamide

3. Give the systematic name of:

a) CH3(CH2)3CHNH2(CH2)2CH3

4. 10 Amines and amides (Schell & Hogan, 2018, p. 16)


Nomenclature of halogenated organic compounds

Watch OC#6 Naming halogenated organic compounds

The IUPAC naming system for haloalkanes involves the following steps: 1. The longest unbranched hydrocarbon chain determines the ending of the molecule’s

name and the name corresponds to the alkane with that number of carbon atoms, eg.

butane where the longest chain is 4 carbons long.

2. The name(s) of the halogens involved are prefixes (eg. fluoro, chloro, bromo or iodo)

and are placed in the name in alphabetical order.

3. The position of the halogens along the longest unbranched carbon chain are each

indicated by a number. They should be numbered to give the lowest possible numbers

to the halogens.

4. The prefixes di-, tri-, tetra- etc are used when multiple halogens are attached to the

hydrocarbon chain e.g. 1,2-dibromo-1,1,3-trichloropropane.

5. If the above rules give more than one possible name, then the correct name is the one

that gives the lowest numbers to the alphabetically ordered halogen.

[NB: We will follow the nomenclature guidelines produced by RACI which are based on

I can investigate the nomenclature of organic chemicals, up to C8, using IUPAC conventions, including simple methyl and ethyl branched chains, including halogenated organic compounds.


the IUPAC system.] Some examples are shown in the table below:

Name Structural Formula Trichloromethane

1,1-dibromo-2-chloroethane

1,1,2-trichloro-1,2,2-trifluoroethane

(from (Schiller, 2013)) Note correction in nomenclature of final compound in table as per IUPAC guide (International Union of Pure and Applied Chemistry, 2017).

Question Set 6

1. (2009, Q12)

What is the IUPAC name of the following compound?

………………………………………………………………………………………

2. (2005, Q13, 1 mark)

What is the name of the following compound?

http://www.easychem.com.au/monitoring-and-management/the-atmosphere/haloalkane-isomers/Trichloromethane.png?attredirects=0

http://www.easychem.com.au/monitoring-and-management/the-atmosphere/haloalkane-isomers/1,1-dibromo-2-chloroethane.png?attredirects=0

http://www.easychem.com.au/monitoring-and-management/the-atmosphere/haloalkane-isomers/1,2,2-trichloro-1,1,2-trifluoroethane.png?attredirects=0


……………………………………………………………………………………….

3. (2003, Q9, 1 mark)

What is the name of the compound shown?

……………………………………………………………………………………………..

4. (2002, Q25, 6 marks)

What is the systematic name of the CFC in the diagram? (1 mark)

...............................................................................................................................

5. 11 Halogenated organic compounds (Schell & Hogan, 2018, pp. 17-18) 6. 2 Nomenclature (Gribben, 2010, pp. 6-8)


Structural Isomerism

Watch OC#7 Structural Isomers

An isomer is one of a series of compounds that have the same molecular formula, but different structural formulae. It is also possible for isomers to exist with the same general structural formula but a slightly different spatial arrangement. We will concentrate on the three structural isomers; namely chain isomers, position isomers and functional group isomers.

Chain isomers

Chain isomers are isomers where the structural differences are in the carbon chain. Isomers may have different chain lengths and include one or more side branches. One example could be butane and methylpropane.

I can explore and distinguish the different types of structural isomers, including saturated and unsaturated hydrocarbons, including chain isomers, position isomers and functional group isomers.


C C C CH

H

H

H

H

H

H

H

H

H

C

H

H

H C

C

H

HH

H

C

H

H

H

Figure 13 butane (C4H10) Figure 14 methylpropane (C4H10)

Position isomers

Position isomers are isomers where the structural differences are in the location of the functional group. This could be as simple as changing the location of a double bond in an alkene. One possible example could be but-1-ene and but-2-ene.

C

C

C

C

H

H

H

H

H

H

H

H

H

C

C

C

CH

HH

H

H

H

H

Figure 15 but-1-ene (C4H8) Figure 16 but-2-ene (C4H8)

Functional group isomers

Functional group isomers are isomers where the structural differences are in the functional group. One simple example may be propanal (an alkanal or aldehyde) and propanone (an alkanone or ketone).

C C C

OH

H

H

H

H

H

C

C

C

O

H

HH

H

HH

Figure 17 propanal (C3H6O) Figure 18 propanone (C3H6O)


Figure 19 https://www.compoundchem.com/2014/05/22/typesofisomerism/

The graphic above reviews the three types of structural isomers required by the syllabus. It also provides an idea of the spatial types of isomers we call stereoisomers. These are interesting and of great interest in evolutionary studies but are not specifically required so are included for interest.

ATAR Notes Key Point: • Chain isomers vary based on the carbon backbone. • Position isomers vary based on the location of specific functional groups. • Functional group isomers vary based on changes to functional groups. (Silove,

2018, p. 54)

Question Set 7

1. (2014, Q9)

Four compounds W, X, Y and Z are represented below.


Which of these compounds represent a pair of isomers?

………………………………………………………………………………………

2. (2011, Q14)

How many isomers are there for C3H6BrCl? A) 3 B) 4 C) 5 D) 6

3. (2010, Q4, 1 mark)

The diagram shows the structural formula of a gas.

How many isomers does this compound have? ………………………….

4. (2006, Q5, 1 mark)

How many isomers are there for C3H6Cl2? ………………………………..

5. (04, Q4, 1 mark)

Which term describes the relationship between the compounds shown below?


……………………………………………………………………………………………

6. Isomers and chemical formulas (de Vreeze & McMicking, 1998, pp. 44-45) 7. 12 Structural isomers (Schell & Hogan, 2018, pp. 19-20) 8. 3 Isomerism (Gribben, 2010, p. 9)


Hydrocarbons


Concept Map

Inquiry question 2• How can hydrocarbons be

classified based on their structure and reactivity?


Reviewing the Hydrocarbons

Watch OC#8 Reviewing the hydrocarbons

Homologous series are groups of compounds which are characterised by

a common general formula a common functional group similar structures and chemical properties gradations in their physical properties (as MM increases)

Alkanes

These contain only single carbon-carbon bonds (functional group) and are thus called saturated hydrocarbons. Their general formula is:

CnH2n+2 or R1(CH2)nR2 where R1 and R2 are either H or an alkyl group

The alkanes are the principle compounds in natural gas and petroleum.

The C — C single bond is extremely stable and so saturated hydrocarbons are relatively inert.

I can construct models, identify the functional group, and write structural and molecular formulae for homologous series of organic chemical compounds, up to C8 for the alkanes, alkenes and alkynes.


One important reaction however is their combustion reactions where they are important fuels. They all burn in sufficient air or oxygen to produce carbon dioxide and water.

Figure 20 ethane (C2H6)

Alkenes

These contain a double carbon = carbon bond (functional group) at least once in the main carbon chain of the molecule and single carbon-carbon bonds elsewhere. Their general formula is:

CnH2n or R1CH = CHR2 where R1 and R2 are either H or an alkyl group

Figure 21 ethene (C2H4)

Alkynes

These contain a triple carbon ≡ carbon bond (functional group) at one end (1-alkyne) of the molecule and single carbon-carbon bonds elsewhere. Their general formula is:

CnH2n-2 or R1C ≡ CR2 where R1 and R2 are either H or an alkyl group


Figure 22 ethyne (C2H2)

Both alkenes and alkynes are called unsaturated hydrocarbons as they can react by addition reactions at the double or triple bond to combine with more atoms as single bonds are produced.

All hydrocarbons can exist as long chains or rings. If a hydrocarbon has a number of carbons joined together to form an enclosed ring, we use the prefix ‘cyclo’ to indicate this circle or ring structure, eg cyclobutane is shown below.

NB Cyclic forms will not conform to the general formula for their homologous series.

C C

CC

H

H H

H

H

HH

H

Figure 23 cyclobutane (C4H8)

Question Set 8

1. Write the molecular formula and draw the structural formula for the following alkanes: a) Methylbutane

b) 2,3-dimethylhexane


c) Cyclopentane

2. Write the molecular formula and draw the structural formula for the following alkenes: a) Methylbut-1-ene

b) 2,3-dimethylhex-3-ene c) Cyclohepta-1,2-diene

3. Write the molecular formula and draw the structural formula for the following alkynes: a) Methylbut-1-yne

b) 2,3-dimethylhex-1-yne c) Cyclooctyne

4. 13 Revision of Nomenclature (Schell & Hogan, 2018, pp. 21-22)


Properties of Alkanes

Watch OC#9 Properties of Alkanes

When studying the properties of the hydrocarbons, there are two groups of properties which are important; physical properties and chemical properties.

Chemical properties

Chemical properties relate to the functional group. As the carbon and hydrogen atoms have similar electronegativities, there is no polarity in the bonds, nor the molecules. Both alkenes and alkynes are chemically more reactive than alkanes due to the presence of a double or triple bond. Chemical properties will be examined later in the module.

Physical properties

In terms of physical properties, the presence of non-polar bonds and non-polar molecules affects the physical properties of the compound. As these are discrete molecules, they exist in the solid phase as covalent molecular networks and are held together by the weak dispersion forces. However as the chain length increases, so the number of dispersion forces between molecules increase and hence larger molecules have higher melting and boiling points. This is why small alkanes exist as a gas at SLC, butane, octane and several others are

I can conduct an investigation to compare the properties of organic chemical compounds within a homologous series and explain these differences in terms of bonding.


liquids, whilst the longer chains, such as the paraffin waxes (20+ carbon atoms) are solids at SLC.

The non-polar nature of the hydrocarbon molecules also makes them insoluble in a polar substance such as water. However as our rule is like dissolves like, they will dissolve in other non-polar solvents.

The lack of free moving electrons also affects electrical conductivity and the hydrocarbons are poor conductors.

Experiment 19: Properties of Hydrocarbons

Introduction

Hydrocarbons are compounds which contain only carbon and hydrogen. They can be classified into several types, depending on their structure.

Aliphatic hydrocarbons are divided into three classes:

• Alkanes which have only single bonds and are said to be saturated. • Alkenes which have at least one carbon-carbon double bonds and are said to be

unsaturated. • Alkynes which have at least one carbon-carbon triple bonds and are said to be

unsaturated.

Aromatic Hydrocarbons are cyclic compounds whose structure is related to that of benzene. We will not focus much on this group of compounds in this course.

In this experiment you will perform experiments to illustrate some of the properties of saturated and unsaturated hydrocarbons.

Method

Solubility and Density of Hydrocarbons 1. Collect a 5mL sample of each of the following hydrocarbons; hexane (H), hex-1-ene

(He), cyclohexane (CH), cyclohexene (CHe). 2. Collect four test tubes and place them in a test tube rack. Label the test tubes H, He,

CH and CHe. 3. Add 5 mL of water to each test tube. 4. Test the solubility of the four hydrocarbons in water by adding 1mL of one

hydrocarbon with a dropper to the 5mL of water in the test tube marked with its identifier (ie, add 1mL of hexane to the test tube of water marked H).

5. Repeat step 4 for each hydrocarbon.


6. Look for the formation of any separate layers and determine which is the water layer in each case.

7. Collect another four test tubes and place them in a separate test tube rack. Label the test tubes H, He, CH and CHe as before.

8. Add 5 mL of dichloromethane to each test tube. 9. Test the solubility of the four hydrocarbons in dichloromethane by adding 1mL of

one hydrocarbon with a dropper to the 5mL of water in the test tube marked with its identifier (ie, add 1mL of hexane to the test tube of water marked H).

10. Repeat step 8 for each hydrocarbon. 11. Test the solubility of the four hydrocarbons in dichloromethane. If separate layers

are formed, determine which is dichloromethane in each case. 12. Record your results in the table provided.

Question Set 9

1. In general, the alkene homologous series with increasing number of carbons show little variation in what? (Shenfield & Silove, 2018, p. 72) A) Freezing point B) Strength of intermolecular forces C) Density at room temperature D) Flammability

2. Alkanes are generally unreactive species. This can be explained by which of the following? (Shenfield & Silove, 2018, p. 72) A) Consist only of highly stable carbon-carbon bonds B) Lack of functional groups C) Highly non-polar nature D) Relatively low boiling points

3. 15 Revision – Intramolecular Forces (Schell & Hogan, 2018, pp. 24-25) 4. 16 Revision – Intermolecular Forces (Schell & Hogan, 2018, pp. 26-27) 5. 1 Introduction to Organic Chemistry (Gribben, 2010, pp. 4-5) 6. Complete the structure strip on the next page (with thanks to Leah Anderson-

Griffiths via facebook NSW Chem teachers group).


Mod 7 – Question 8b (4 marks) Primary, unbranched alcohols and alkanes of the same carbon length have quite different boiling points. Explain the difference in boiling point of these organic compounds, showing all intermolecular forces. Support your answer with diagrams.

Command Verb “Explain” Explain means to show cause and Effect.

Explain what an alkane is in terms of: Bonding of C &

C and C & H Polarity

………………………………………………………………………………

………………………………………………………………………………

………………………………………………………………………………

Explain what the intermolecular forces are, and how this relates to BP

………………………………………………………………………………

………………………………………………………………………………

………………………………………………………………………………

Draw a diagram of the intermolecular forces in alkanes, labelling all parts.

Explain how alkenes are different to alkanes in terms of: Functional

group Polarity.

………………………………………………………………………………

………………………………………………………………………………

………………………………………………………………………………


Molecular shapes

Watch OC#10 Molecular Shapes

Hybridisation

A quick word about orbitals and hybridisation. According to Thickett (Thickett, 2018) hybridisation is “a procedure used in the valence bond theory in which atomic orbitals are combined to produce molecular orbitals” (p. 125). This process will have an effect on the molecular geometry.

We need to recognise the shape of the distribution of atoms around any carbon atom. Usually this will fit one of three types:

Tetrahedral Trigonal planar Linear

C

H

H H

H

C C

H

H

H

H

C CH H

I can analyse the shape of molecules formed between carbon atoms when a single, double or triple bond is formed between them.


In simple terms, if one of the electrons in the 2s orbital jumps to a 2p orbital, the excited carbon atom can hybridise this state to form 4 identical hybrid orbitals which are equivalent in energy, size and shape. This is shown below:

Figure 24 sp3 orbitals https://www.chemistrysteps.com/sp3-sp2-and-sp-hybridization-organic-chemistry/

The formation of these hybrid orbitals slightly lowers the energy conferring some stability. The orbitals are arranged in an overlapping organisation which accounts for the tetrahedral arrangement of hydrogen atoms around a central carbon atom.

The single bonds representing the 4 sp3 hybrid orbitals are called sigma (σ) bonds. The sigma bond forms when the heads of 2 atomic or molecular orbitals overlap. As a result “you can have a sigma bond between an sp3 orbital on a carbon atom and a 1s orbital on a neighbouring hydrogen atom as well as from the overlap of 2 sp3 orbitals between neighbouring C atoms”. (Walker, 2019)

Figure 25 sigma bonds in ethane (Hardinger, 2019)

Put more simply: “A sigma bond is a covalent bond formed by overlap of atomic orbitals and/or hybrid orbitals along the bond axis (i.e., along a line connected the two bonded atoms)”. (Walker, 2019)

https://www.chemistrysteps.com/sp3-sp2-and-sp-hybridization-organic-chemistry/


Figure 26 sp2 orbitals https://www.chemistrysteps.com/sp3-sp2-and-sp-hybridization-organic-chemistry/

However excited carbon atoms do not just stabilise through the formation of 4 hybridised orbitals, they can also do so with three or two. When this happens, other interesting things happen too.

Figure 27 sp orbitals https://www.chemistrysteps.com/sp3-sp2-and-sp-hybridization-organic-chemistry/

When these types of hybrids are formed, the geometry about the central atom changes and the additional p orbitals will sit at right angles to the sp hybrids. The latter are known as pi (π) bonds. They are weaker than sigma bonds and usually manifest as the double or triple bonds in alkenes and alkynes.

One final note. It is tricky to show 3D structures on paper, so there are some conventions which you might see.

Methane (CH4) could look like this:

C H

H

H

H

This is simple to draw but does not give us any idea of the 3D nature of the molecule.




Or methane could look like this:

C

H

H HH

This tries to hint at the tetrahedral structure by showing a dotted bond which suggests the bond is disappearing back into the page and a solid wedge which suggests a bond coming out of the page.

Question Set 10

1. Select the true statement concerning bonding in hydrocarbon molecules: (Thickett, 2018, p. 133) A) Carbon-carbon single bonds are formed when 4 electrons

are shared B) Double covalent bonds between carbon atoms consist of one electron pair C) Triple covalent bonds between carbon atoms in alkynes consist of three electron

pairs D) C-H bonds are formed when sp3 hybrid orbitals from each atom interact.

2. The structure of hydrocarbons is heavily dependent on the bonding within and between the carbon backbone. a) Identify the functional group of the linear hydrocarbon with the chemical

formula C4H6. Include a structural formula in your answer (2 marks)

b) Describe the different geometries around each carbon from the structural formula in part a). (Shenfield & Silove, 2018, p. 73)

3. 19 Shapes of Hydrocarbon Molecules (Schell & Hogan, 2018, p. 32)


Bonding within and between the hydrocarbons

Watch OC#11 Bonding in the Hydrocarbons

In very broad terms, the chemical properties of a compound or a series of compounds are associated with the intramolecular bonding. Alkanes are characterised by single carbon-carbon and carbon-hydrogen bonds. The paired electrons are arranged in a tetrahedral arrangement and this confers a level of stability to the compounds, especially in view of the fact that the electronegativity values for both carbon and hydrogen are very similar and hence there is no polarity in the covalent bonds.

Table 1 Molar Mass and Boiling Point for the first 6 alkanes

Alkane Molecular Formula Molar Mass (g/mol) Boiling Point (oC) Mthane CH4 16.042 -162 Ethane C2H6 30.068 -89 Propane C3H8 44.094 -45 Butane C4H10 58.12 -0.5 Pentane C5H12 72.146 36 Hexane C6H14 86.172 69

Conversely, the physical properties of the alkanes are related to the intermolecular bonds between molecules. As a result the physical properties are not just based on the non-polarity of the covalent bonds within the molecules, but also on their size and shape. The dominant intermolecular bonds are dispersion forces. Table 1 is taken (and modified) from the Pearson

I can explain the properties within and between the homologous series of alkanes with reference to the intermolecular and intramolecular bonding present.


Textbook (Chan, et al., 2019, p. 297).

Looking at Table 1, there is a clear trend of increasing boiling point. It is even easier to see if we graph it.

The trend line is almost perfectly linear. It suggests a strong relationship between boiling point and molar mass for alkanes. The only difference between each member of the alkanes is the addition of a CH2 group. This additional group increases the size of the molecule and also increases the strength of the dispersion forces between the molecule. These are the forces which must be overcome for the molecules in a liquid to be free as gas molecules.

One other factor which can affect the physical properties is the shape of the molecule.

Pentane, methyl butane and dimethyl propane all have the same molecular formula and hence the same molar mass (72.146 g/mol) however, the boiling points are as follows:

Pentane 36oC

Methyl butane (isopentane) 27.8oC

Dimethyl propane (neopentane) 10oC

This suggests that as the compound becomes more branched, the dispersion forces holding the molecules together become slightly weaker and hence the highest melting point for the most linear compound (pentane).

-200

-150

-100

-50

0

50

100

0 20 40 60 80 100

Boiling Point (oC) vs Molar Mass for Alkanes


Question Set 11

1. Draw the structural formulae for pentane, methyl butane and dimethyl propane and use these structures to account for the differences in boiling points between these compounds.

2. Alkanes are insoluble in water. Use a diagram to explain this property.

3. The density of a number of liquid alkanes is shown below, but data for undecane is

missing.

Alkane Formula Density (g.mL-1)

Pentane C5H12 0.63

Hexane C6H14 0.66

Heptane C7H16 0.68

Octane C8H18 0.70

Nonane C9H20 0.72

Decane C10H22 0.73

Undecane 0.74


a) Use the table to predict the molecular formula of undecane.

b) Draw a graph to allow you to estimate the density of C12H26.

c) Describe and explain the relationship between chain length and density for this

homologous series.

d) Extrapolate from your graph to estimate the density of C12H26.

4. 18 Properties and Bonding of Hydrocarbons (Schell & Hogan, 2018, pp. 29-31)


Safe Handling of Hydrocarbons

Watch OC#12 Safe handling of hydrocarbons

The most critical aspect to the safe handling of hydrocarbons is their flammability. Most hydrocarbons readily combust and hence must be kept well away from a naked flame.

The alkanes from methane to octane are gases or volatile liquids at SLC and have low flash points. Petrol is a mixture of hydrocarbons and other components which lower its flash point and make it potentially dangerous to store and use. There are warnings in petrol stations regarding smoking and the use of mobile phones to ensure there is no opportunity to ignite the volatile liquid.

Weak dispersion forces between the smaller alkanes lowers their boiling points and increases their volatility. They have low flash points and can be readily ignited even during the lower temperatures in winter.

Volatility is a term used to describe substances which vapourise at room temperature to produce a high vapour concentration above the solid or liquid phase. This high concentration is known as the ‘equilibrium vapour pressure’.

Some hydrocarbons vapourise readily at standard laboratory conditions. In general, the lower the molecular weight, the greater the rate of evaporation and the higher the volatility, due to weaker dispersion forces between molecules.

I can describe the procedures required to safely handle and dispose of organic substances.


Boiling point is a good indicator of volatility. Volatility increases with increasing temperature.

Flash Point is a term used to describe the minimum temperature at which the vapour pressure of a hydrocarbon is sufficient to combust in air. Combustible fuel-air mixtures can be dangerous as a spark or flame can ignite them. Flash points can vary, however in general the higher the boiling point of a hydrocarbon, the higher its flash point. Solid waxes have higher boiling points and flash points than liquid hydrocarbons. Stronger dispersion forces makes them less volatile. This lowers the vapour pressure and raises the flash point. Liquids like octane have weaker dispersion forces between the molecules and are more volatile with lower flash points. This makes them potentially dangerous to store and use.

One common safety precaution in cars is to locate the fuel tank as far from the engine as possible. This ensures any volatile fuel/air mixture has little chance of coming in contact with a spark which might ignite it.

Gases are stored in air tight cylinders which have been carefully sealed and should be checked regularly for leaks.

Liquids should be stored in clearly labelled metal containers with a narrow opening and close-fitting lids.

Fuels should be stored in cool, well ventilated spaces to prevent the build-up of a volatile fuel/air mixture.

Liquid transfers should be done outside in a well-ventilated area to prevent any build-up of combustible mixtures.

Flammable liquids have specific labels (HazChem Codes) which should always be displayed to warn of the dangers.

Keep a regularly serviced fire extinguisher near any storage area used for volatile compounds.

In a laboratory, laws regarding the safe handling of hydrocarbons apply. Riskassess produces safety information (Safety Data Sheet, SDS https://www.safeworkaustralia.gov.au/sds) for all chemicals used in a laboratory. The information on the following page relates to the alkane hexane (Crisp, 2019).

https://www.safeworkaustralia.gov.au/sds


Figure 28 RiskAssess information for hexane

If we wanted to experiment with a hydrocarbon, a safer choice might be cyclohexane (Crisp, 2019).

Figure 29 RiskAssess for cyclohexane

Question Set 12

1. Explain the terms ‘flash point’, ‘ignition temperature’ and ‘vapour pressure’ and why they are relevant when discussing safety in relation to handling of hydrocarbons. (3 marks)


2. Look at the two RiskAssess labels above and on the previous page. Evaluate the choice of cyclohexane as a safer option over hexane for experiments involving hydrocarbons. (3 marks)

3. What is the safest way to dispose of hydrocarbons once you have concluded your

experiments? Justify your response (2 marks)

4. Explain why each of the following safety precautions is taken when handling or

storing hydrocarbon fuels:

a) LPG gas bottles should be inspected and tested regularly

b) Lawn mower fuel should be poured into the mower’s fuel tank out in the open.

c) The use of mobile phones is not permitted around petrol stations.

d) Large quantities of petrol should not be stored at home.


e) Petrol tankers are grounded through attachment to steel chains.

f) The petrol tank in a car is usually at the opposite end to the engine.

5. 20 Safety with Organic Compounds (Schell & Hogan, 2018, p. 33)


Implications of Hydrocarbons

Watch OC#13 Implications of hydrocarbons

Hydrocarbons are commonly found in high concentrations in fossil fuels. Two of the simplest hydrocarbons are methane and ethane, the primary components of natural gas.

Petroleum is a critical commodity in the modern world. What sorts of compounds can be obtained from a barrel of crude oil? How are they obtained?

Crude petroleum is very viscous due to the presence of straight, branched and cyclic hydrocarbons.

Fractional distillation is the process used to separate a mixture of substances with different boiling points. Crude oil is heated in a column with the lighter fractions boiling off first (lower bp). This includes methane,

I can examine the environmental, economic and sociocultural implications of obtaining and using hydrocarbons from the Earth.

Figure 30 Fractional Distillation Column


ethane, propane, butane, etc, some of which are liquefied and distilled to separate these fractions. The larger fractions are separated in a column or tower, based on differences in boiling points.

The crude mixture is heated to about 370oC. Steam and vapours rise up the column and cool, condensing out to be collected at different levels in the column.

Heavy fractions can be cracked into smaller molecules with the use of catalysts. Industrially, one of the most important substances obtained from crude oil is ethene. Ethene is generally obtained by the catalytic cracking of some of the fractions obtained in the refining of raw petroleum. It is a critical raw material for the polymer industry.

Along with the expansion of the petrochemical industry from energy generation to polymer production and the associated economic benefits of both industries, environmental considerations have become more critical, especially as scientists link climate change to our reliance on the burning of fossil fuels. This has spawned new industries devoted to greener energy production and investigations into alternate fuel sources which might reduce the impact on the environment.

Question Set 13

1. (2006, Q1) Which is the main industrial source of ethylene? (‘06, Q1)

A) Ethanol B) Glucose C) Petroleum D) Polyethylene 2. The process of fractional distillation is used to separate crude oil into different

fractions. One of the compounds obtained from fractional distillation is C10H22. This compound undergoes catalytic cracking as follows: (08, Q16, 5 marks) (modified). C10H22 → C8H18 + X a) Identify X.

b) What is the functional group for X? c) To which homologous series does C8H18 belong?


3. Carry out some research to complete the table below. environmental, economic and sociocultural implications of obtaining and using hydrocarbons from the Earth

Obtaining Hydrocarbons Using Hydrocarbons

Description

Environmental Implications

Economic Implications

Sociocultural Implications

4. Worksheet 7.2 Families of hydrocarbons (Commons, 2018, pp. 120-122) 5. 21 Revision – Fossil Fuels (Schell & Hogan, 2018, pp. 34-36) 6. 24 Revision – The Greenhouse Effect (Schell & Hogan, 2018, pp. 41-42)


Products of Reactions Involving

Hydrocarbons


Concept Map

Inquiry question 3• What are the products of

reactions of hydrocarbons and how do they react?


Addition Reactions

Watch OC#14 Addition Reactions

Alkenes are more chemically reactive than alkanes. The double bond is a site of high electron density and will readily combine with species with high electronegativities.

Ethene (ethylene) is a small molecule with a highly reactive double bond. This means it can be readily transformed into many useful products.

1. Hydrogenation – Production of ethane

Hydrogenation is the process of adding hydrogen atoms across a double bond in an alkene (or across a triple bond in an alkyne). Ethene can be hydrogenated to produce ethane. A metal catalyst, eg Ni or Pt, is often used.

CH2=CH2 + H2 CH3-CH3

I can investigate, write equations and construct models to represent the reactions of unsaturated hydrocarbons when added to a range of chemicals, including but not limited to hydrogen (H2), halogens (X2), hydrogen halides (HX) and water (H2O).


2. Halogenation - Production of dichloroethane Halogenation is the process of adding halogen atoms (F, Cl, Br, I) across a double bond in an alkene (or across a triple bond in an alkyne). Ethene can be halogenated with chlorine to produce 1,2-dichloroethane. No catalyst is needed. CH2=CH2 + Cl2 CH2Cl-CH2Cl

3. Hydrohalogenation - Production of chloroethane

Hydrohalogenation is the process of adding one hydrogen and one halogen atom (F, Cl, Br, I) across a double bond in an alkene (or across a triple bond in an alkyne). Ethene can be hydrohalogenated with hydrogen chloride to produce chloroethane. CH2=CH2 + HCl CH3-CH2Cl

NB To work out which carbon will take the hydrogen and which will take the halogen,

we use Markovnikov’s Rule. This rule states that the hydrogen atom from the hydrogen halide will bond with the carbon which had the greater number of hydrogens prior to the addition reaction.

4. Hydration - Production of Industrial Alcohol (ethanol)

Hydration is the process of adding water molecules, or the equivalent of water molecules, to a substance. Ethylene can be hydrated to produce ethanol when heated with a dilute sulfuric acid (H2SO4) catalyst.

5. Production of 1,2-ethanediol (ethylene glycol) CH2=CH2 CH2OH-CH2OH requires O2/catalyst and H2O


ATAR Notes Key Point: Unsaturated hydrocarbons undergo addition reactions by ‘opening’ double bonds and incorporating the additional molecule. (Silove, 2018, p. 61)

Experiment 20: Properties of Hydrocarbons

Introduction

There is a distinct difference between the chemistry of alkenes and that of alkanes. Alkenes are unsaturated hydrocarbons that react readily by the process of addition. Alkanes are saturated hydrocarbons that react slowly by the process of substitution. This difference in reactivity is used as the chemical basis for this experiment. Compare the reactivities of alkenes with alkanes in bromine water. Write chemical equations for all reactions observed.

Materials

1x bromine water (dropper bottle) 1x safety glasses test tubes + rack + labels hydrocarbon test samples (e.g. cyclohexene, cyclohexane/hexane*)

waste container for the safe disposal of test samples Construct a table to show the name of each compound tested, whether it is an alkane or alkene and what you observed after the addition of bromine water. Write a conclusion based on your observations.

Question Set 14

1. (10, Q24, 4 marks) In the margarine industry, alkenes are often hydrogenated to convert unsaturated oils into solid fats that have a greater proportion of saturated molecules. a) Using ethene as an example, write an equation for this reaction and state the type of

reaction this represents. (2 marks)

…………………………………………………………………………………………………

…………………………………………………………………………………………………

…………………………………………………………………………………………………


…………………………………………………………………………………………………

b) Describe a test that could be used to confirm that all the ethene has been converted. (2 marks)

…………………………………………………………………………………………………

…………………………………………………………………………………………………

…………………………………………………………………………………………………

…………………………………………………………………………………………………

2. (2011, Q11) Which compound can form when bromine water reacts with propene? (A) 1-bromopropane (B) 2-bromopropane (C) 1,1-dibromopropane (D) 1,2-dibromopropane

3. (2009, Q6)

Bromine, Br2, dissolves in unsaturated hydrocarbons and reacts immediately. Which of the following is the best description of this process? (A) Bromine is polar and reacts by adding bromine atoms across the double bond. (B) Bromine is polar and reacts by substituting hydrogen atoms with bromine

atoms. (C) Bromine is non-polar and reacts by substituting hydrogen atoms with bromine

atoms. (D) Bromine is non-polar and reacts by adding bromine atoms across the double

bond.

4. (2011, Q1) Which of the following industrial processes is used to produce ethanol from ethylene? (A) Hydration (B) Dehydration (C) Addition polymerisation (D) Condensation polymerisation

5. Use an example to demonstrate your understanding of Markovnikov’s rule.

6. 27 Addition Reactions (Schell & Hogan, 2018, pp. 47-48)


Substitution Reactions

Watch OC#15 Substitution Reactions

Substitution Reaction – The production of chloroalkane.

The low reactivity of alkanes means they do not readily react with halogens unless in the presence of ultraviolet light. Even so, the reaction is not an addition reaction, but a substitution reaction. One of the halogen atoms substitutes for (or replaces) a hydrogen atom and a haloalkane is formed as a result. A second product will also form from the combination of the substituted hydrogen with the remaining halogen atom.

Ethane + chlorine → chloroethane CH3—CH3 + Cl2 → CH3—CH2Cl + HCl

In the presence of excess halogen (chlorine in the example given), the substitution product (in this case chloroethane) may undergo additional substitution reactions increasing the

I can investigate, write equations and construct models to represent the reactions of saturated hydrocarbons when substituted with halogens.

Figure 31 chloroethane (C2H5Cl) a hydrogen atom has been substituted with a chlorine atom


number of halogen atoms on the original hydrocarbon. CH3—CH2Cl + Cl2 → CH2Cl—CH2Cl + HCl

Question Set 15

1. Excess fluorine is added to methane in the presence of uv light. What would be the expected products? Justify your response. (3 marks)

2. When reacting a hydrogen halide with an unsaturated hydrocarbon, the halogen atom tends to attach to the more substituted carbon. Which of the following best explains this phenomenon? (Shenfield & Silove, 2018, p. 77)

A) The hydrogen is more attracted to the less substituted carbon due to the presence of other hydrogens.

B) The electronegativity of the halogen atom is better stabilised by surrounding groups.

C) The less substituted carbon forms a more unstable intermediate. D) The hydrogen attaches to the molecule first and is more easily attached to a less

crowded carbon. 3. Intense conditions are required to undergo halogenation of an alkane. Which of the

following do not contribute to these conditions? (Shenfield & Silove, 2018, p. 78) A) High temperatures (400oC) B) UV light C) High halogen concentration D) Concentrated acid catalyst

4. Explain why the chlorination of methane, if not carefully controlled, can produce tetrachloromethane. In your answer, include relevant reactions. (3 marks) (Shenfield & Silove, 2018, p. 78)

5. Practical Activity 7.2 Modelling hydrocarbons (Commons, 2018, pp. 143-146) 6. 26 Substitution Reactions (Schell & Hogan, 2018, pp. 45-46)


Alcohols


Concept Map

Inquiry question 4•How can alcohols be

produced and what are their properties?


Alkanols

Watch OC#16 Alkanols

Recall that a functional group is either a type of bond, atom, or group of atoms which link the members of a homologous series and are important to the chemical properties of each member of the series. For alcohols, the functional group is the hydroxy (OH) group.

Primary alcohols eg propan-1-ol (or 1-propanol) Note in a primary alcohol, the hydroxy group is attached to an end carbon. This is denoted by the green arrow in the photo of the model. This carbon is only attached to one other carbon atom, hence it is a primary alkanol.

The presence of the hydroxy group changes some of the physical and chemical properties of the compound, primarily due to the polarity of the C-O and O-H bonds.


I can investigate the structural formulae, properties and functional group including primary, secondary and tertiary alcohols.


Secondary alcohols eg propan-2-ol (or 2-propanol)

Note in a secondary alcohol, the hydroxy group is attached to a non-terminal (middle) carbon. This is denoted by the green arrow in the photo of the model. This carbon is attached to two other carbon atoms hence it is a secondary alkanol.


Tertiary alcohols eg methylpropan-2-ol (or methyl-2-propanol)

Note in a tertiary alcohol, the hydroxy group is attached to a non-terminal (middle) carbon which is also attached to an additional carbon off the main chain. This is denoted by the green arrow in the photo of the model. This carbon is attached to three other carbon atoms hence it is a tertiary alkanol.

The hydroxy functional group, -OH, in alkanols provides their characteristic properties, such as their polar nature at one end. Many alcohols make good solvents as they often have a polar region and a non-polar region making them at least partially soluble in a range of solvents. They also have the ability to act as solvents.

The presence of the polar covalent bond and subsequent hydrogen bonding between molecules also means they have a higher BP than their corresponding alkane.

Figure 34 methylpropan-2-ol (C4H9OH)


Question Set 16

1. (2007, Q4) What is the IUPAC name for the following compound?

A) Hexan-3-ol B) Hexan-4-ol C) Heptan-3-ol D) Heptan-5-ol

2. Draw the structural formula, and classify, two isomers of methyl propan-2-ol.

3. Comparing ethane with ethanol, and using examples, explain the effect of the presence of the hydroxy group on the properties of ethanol.

4. 29 Alcohols (Schell & Hogan, 2018, pp. 50-51)


Properties of Alcohols

Watch OC#17 Properties of Alcohols

Ethanol is an excellent solvent due to its highly polar nature. One end of the molecule has non-polar C-H bonds, while the other end has an hydroxyl (OH) group. The high electronegativity of oxygen allows hydrogen bonding to take place with other molecules.

As a result, the hydroxyl end of the ethanol molecule attracts polar and ionic substances. The ethyl (C2H5) group in ethanol is non-polar, so this end attracts non-polar substances.

As a result, ethanol can dissolve both polar and non-polar substances.

In industrial and consumer products, this makes ethanol the second most important solvent after water.

Ethanol is the least toxic of the alcohols (though it is poisonous in large amounts), which also makes it more suitable for use in industry and consumer products. As a result ethanol is a common solvent in:

Cosmetics (eg. perfumes). Food colourings and flavourings (eg. vanilla essence). Medicinal preparations (eg. antiseptics and rubbing alcohol). Some cleaning agents.

Industry

I can explain the properties within and between the homologous series of alcohols with reference to the intermolecular and intramolecular bonding present.


Nature of intramolecular bonds in alkanols: Intramolecular bonds in alkanols are covalent bonds. C-C and C-H are non-polar bonds, but C-O and O-H bonds are polar bonds. Differences in electronegativity between O and both C and H account for these differences and cumulatively they create polarity in the molecule.

Nature of intermolecular bonds in alkanols: The alkanols have a dual nature. For one part of their structure there are only C-C or C-H

bonds. These are non-polar and hence can only form dispersion forces between adjacent molecules. However, the other part of their structure is polar due to the polarity of the C-O and O-H bond. As a result, alkanols can link together via hydrogen bonds which are much stronger than dispersion forces. This affects properties such as solubility in water and melting and boiling point.

Figure 35 Hydrogen bonding in liquid ethanol (Davis, Disney, & Smith, 2018, p. 269)

Some important properties trends in the alcohols:

BP of alcohols increase as chain length, and molar mass, increases due to increase in dispersion forces between molecules

Secondary and tertiary alcohols with identical molar masses to a primary alcohol will have a lower BP due to reduction in strength of hydrogen bonding between molecules

Small primary alcohols, eg methanol and ethanol, are readily soluble in water, but as chain length, and molar mass, increases, solubility decreases.

Question Set 17

1. (2011, Q21, 4 marks) What features of the molecular structure of ethanol account for its extensive use as a solvent? Include a diagram in your answer. …………………………………………………………………………………………………

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2. (2009, Q17, 4 marks) Water and ethanol are both used as solvents. Explain the differences and similarities in their solvent behaviour in terms of their molecular structures. Include a diagram in your answer. …………………………………………………………………………………………………

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3. M-4 Fixed points of Organic Compounds (Gribben, 1999, p. 21)


Enthalpy of Combustion for Alcohols

Watch OC#18 Enthalpy of combustion for alcohols

Molar Heat of Combustion refers to the heat released when one mole of a certain compound undergoes complete combustion with oxygen at a constant pressure of exactly one atmosphere (100 kPa) and at 25°C with the final products being carbon dioxide gas and liquid water.

Heats of combustion are quoted as positive numbers while the enthalpy changes of combustion reactions (ΔH) are quoted as negative numbers. This is because combustion reactions are always exothermic.

Heats of combustion are expressed in kilojoules per mole (kJ/mol. or kJ mol.-1).

The accepted value for the molar heat of combustion of ethanol is 1360 kJ mol.-1.

To process the results from an experiment designed to calculate the enthalpy of combustion of an alcohol, substitute the experimental results into the formula below to determine the enthalpy change: ΔH = -q, where q = mCΔT

where: ΔH = enthalpy change in joules, m = mass of water, C = thermal capacity (4.2 for water), ◦ΔT = change in temperature in degrees Celsius

Determine the number of moles of ethanol combusted (number of moles = mass/MM), and divide the enthalpy change in kilojoules by this number to determine the experimental value

I can conduct a practical investigation to measure and reliably compare the enthalpy of combustion for a range of alcohols.


of the molar heat of combustion of ethanol in kilojoules per mole.

Note: Due to heat loss to surroundings, the experimental value of the molar heat of combustion of ethanol will often be significantly lower than the accepted value.

If we conduct the same experiment for another alkanol, such as methanol, and the difference between the experimental value and accepted value was found, this difference could be used to calibrate the experimental results for ethanol and produce a more accurate experimental result.

The process described above can be applied to any alkanol and can be modified slightly in order to find the heat of combustion in kilojoules per gram instead of kilojoules per mole.

Experiment 21: Enthalpy of Combustion for Alcohols

Aim The aim of this experiment is to determine and compare the heat of combustion values for methanol, ethanol and propanol by measuring and processing calorimetry data.

Method 1. Warning: Wear safety glasses during this experiment! 2. Warning: This experiment involves flammable fuels. Your teacher will review fire

accident procedures before you start. 3. Warning: Ensure that your equipment is placed safely and is not a danger to

neighbouring students. 4. Warning: Methanol is toxic. Avoid skin contact. 5. Before starting this experiment, you need to identify which data to collect in order to

calculate the heat of combustion values. For guidance, study the Data processing section before you start. Transfer exactly 100.0 mL (m = 100.0 g) of water into a 100 mL conical flask. Immerse the thermometer. Stop and think what to do next.

6. Obtain a spirit burner filled with methanol. Stop and think what to do before you light the burner.

7. Clamp the neck of the flask over the burner and adjust the height a few centimetres above the wick. Ignite the wick. Heat the flask until the water temperature rises by about 20ºC.

8. Blow out the flame. Gently stir the flask with the thermometer. Stop and think about what temperature reading needs to be recorded.

9. Remove the spirit burner. Stop and think about what data need to be measured and recorded.

10. Empty the flask and shake dry. 11. Repeat steps 1–10 for spirit burners filled with ethanol and propanol (or whichever

alcohols are available).


Question Set 18

1. (2011, Q17) The molar heat of combustion of pentan-1-ol is 2800 kJ mol−1. A quantity of pentan-1-ol was combusted, generating 108 kJ of heat. What mass of pentan-1-ol was combusted? A) 2.29 g B) 2.86 g C) 3.32 g D) 3.40 g

2. (2010, Q17) A student completed an experiment to determine the amount of energy absorbed by a volume of water. The following data were recorded.

Mass of beaker 215.6 g Mass of beaker plus water 336.1 g Final temperature of water 71.0°C Energy absorbed 21.2 kJ

What was the initial temperature of the water? A) 15°C B) 25°C C) 29°C D) 42°C

3. (2008, Q6) What is the specific heat of a compound? (A) The quantity of heat required to boil 1 g of the compound (B) The quantity of heat required to melt 1 g of the compound (C) The quantity of heat required to increase the temperature of the compound by 1°C (D) The quantity of heat required to increase the temperature of 1 g of the compound by 1°C

4. (2008, Q10) The molar heat of combustion of ethanol is 1367 kJ mol–1. What quantity of ethanol must be combusted to raise the temperature of 1.0 kg water from 50°C to boiling point at sea level (assuming no loss of heat to the surroundings)? A) 6.5 g B) 7.0 g C) 209 g D) 300 g

5. (2009, Q5) The apparatus shown is used in a first-hand investigation to determine and compare the heat of combustion of three different liquid alkanols.


Which is the independent variable? (A) Type of alkanol used (B) Amount of water used (C) Amount of alkanol used (D) Temperature change in the water

6. A student used the apparatus shown to determine the molar heat of combustion of ethanol.

The following results were obtained.

Initial mass of burner 133.20 g Final mass of burner 132.05 g Initial temperature of water 25.0 °C Final temperature of water 45.5 °C

What is the molar heat of combustion calculated from this data? A) 22.4 kJ mol–1 B) 25.7 kJ mol–1 C) 1030 kJ mol–1 D) 1180 kJ mol–1

7. (2009, Q20 4 marks) a) Calculate the mass of ethanol that must be burnt to increase the temperature of

210g of water by 65°C, if exactly half of the heat released by this combustion is lost to the surroundings. The heat of combustion of ethanol is 1367 kJ.mol−1. (3 marks)

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b) What are TWO ways to limit heat loss from the apparatus when performing a first-hand investigation to determine and compare heat of combustion of different liquid alkanols? (1 mark)

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8. (2010, Q23 a) 1 mark) Write a balanced chemical equation for the complete combustion of 1-butanol. …………………………………………………………………………………………………

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9. (2010, Q23 b) 2 marks) A student measured the heat of combustion of three different fuels. The results are shown in the table.

Fuel Heat of Combustion (kJ g−1)

A B C

−48 −38 −28

The published value for the heat of combustion of 1-butanol is 2676 kJ mol−1. Which fuel from the table is likely to be 1-butanol? Justify your answer.

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10. (2007, Q24) The heats of combustion (–ΔHc) of three alkanols were determined. The results are shown in the table.

Alkanol Heat of combustion

(kJ mol-1) methanol ethanol butan-1-ol

480 920 1800

(a) Plot a graph of the heat of combustion versus the molecular weight for the three

alkanols. (3 marks)


(b) (i) Use the graph to estimate the heat of combustion of propan-1-ol. (1 mark)

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(ii) The theoretical value for the heat of combustion of propan-1-ol is more than 2000 kJ mol–1. Suggest a chemical reason, other than heat loss, for the difference between this value and the estimated value from part (b) (i). (1 mark)

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11. (2005, Q17, 3 marks) The heat of combustion of ethanol is 1367 kJ mol–1. In a first-hand investigation to

determine the heat of combustion of ethanol, the experimental value determined

differed from the theoretical value.

(a) Identify a reason for this difference. (1 mark)

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(b) Calculate the theoretical mass of ethanol required to heat 200 mL of water from

21.0°C to 45.0°C. (2 marks)

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12. 30 Enthalpy of Combustion of Alcohols (Schell & Hogan, 2018, pp. 52-53)


Reactions involving Alcohols

Watch OC#19 Reactions involving Alcohols

There are a number of important reactions involving alcohols which we need to investigate. These include combustion, dehydration, substitution with a hydrogen halide and oxidation.

Combustion reactions involving alcohols In the previous experiment we looked at the combustion reaction for alcohol. This reaction is dependent on the amount of oxygen which is present. If there is insufficient oxygen, which is often the case for the spirit burners, the fuel won’t combust completely and carbon will be a byproduct. As chain length increases so the tendency towards incomplete combustion increases.

Incomplete combustion of ethanol:

Complete combustion of ethanol:

I can write equations, state conditions and predict products to represent the reactions of alcohols, including but not limited to combustion, dehydration, substitution with HX and oxidation.


Dehydration of alcohols Dehydration is the process of removing water molecules, or the equivalent of water molecules, from a substance. Ethanol can be dehydrated to produce ethylene and water when heated with a concentrated sulfuric acid (H2SO4) or phosphoric acid (H3PO4) catalyst. Tertiary alcohols react quickly, secondary alcohols slowly and primary alcohols only with very strong heating.

Write a chemical equation to represent the dehydration of propan-2-ol.

Substitution Reactions involving alcohols Alcohols can react with hydrogen halides to form an alkyl halide and water. This happens when the halogen substitutes for the hydroxyl group, hence they are called substitution reactions.

Write a chemical equation to represent the substitution reaction involving hydrogen iodide and propan-2-ol.

Tertiary alcohols are the most reactive when involved in substitution reactions, then secondary and primary. There is also a sequence for the halogens, with HI the most reactive and HF the least.

“The close relationship among tert-butyl alcohol, isobutylene, and tert-butyl chloride is summarized in Fig. 5-10. In the formation of all three, the t-butyl cation plays a central role. A positive charge can be placed with relative ease on a tertiary carbon compared to a secondary or primary carbon, since alkyl groups have more electrons than do hydrogen atoms with which to stabilize a positive charge. It is therefore easier to form t-butyl chloride or isobutylene from t-butyl alcohol by way of the t-butyl cation than it is, for example, to form 2-chlorobutane or butenes from 2-butanol by way of the sec-butyl cation. The varying reactivities of alcohols are thus seen to have the same underlying explanation as Markovnikov's rule.”

Figure 36 Why are tertiary alcohols more reactive than secondary and primary alcohols? (University of Illinois, 2019)


Oxidation Reactions Involving Alcohols Primary and secondary alcohols are readily attacked by oxidising agents. Tertiary alcohols do not react. Usual reagents for the oxidation of alcohols are acidified solutions of KMnO4 or K2Cr2O7. Primary alkanols are oxidised initially to alkanals and then to alkanoic acids. Secondary alkanols are oxidised to alkanones.

Primary alcohols may be converted directly into alkanoic acids when strongly heated with acidified KMnO4 solution. Under cold conditions, or without excess KMnO4, the alkanal will be produced as an intermediate product.

CH3CH2OH + hot acid. KMnO4 → CH3COOH

CH3CH2OH + cold KMnO4 → CH3CHO

As these are true oxidation reactions, we can write them as redox couples. We shall investigate this further later in this module.

Methanol can be oxidised to CO2 and H2O in strong oxidising conditions.

Question Set 19

1. (2010, Q23, 3 marks) a) Write a balanced chemical equation for the complete

combustion of 1-butanol. (1 mark)

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b) A student measured the heat of combustion of three different fuels. The results are shown in the table.

Fuel Heat of Combustion (kJ.g-1)

A B C

-48 -38 -28

The published value for the heat of combustion of 1-butanol is 2676 kJ mol−1. Which fuel from the table is likely to be 1-butanol? Justify your answer. (2 marks)

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2. (2010, Q11) An organic liquid, when reacted with concentrated sulfuric acid, produces a compound that decolourises bromine water. What is the formula of the organic liquid? A) C6H12 B) C6H14 C) C6H11OH D) C5H11COOH

3. Write equations to show: (Davis, Disney, & Smith, 2018, p. 339) a) The dehydration of 1-propanol

b) The dehydration of 3-methyl-1-pentanol

c) The reaction of 2-butanol with hydrogen bromide

d) The reaction of 2-methyl-2-pentanol with hydrogen chloride

e) The oxidation of 2-propanol

f) The oxidation of 2-methyl-3-hexanol

g) The complete combustion of octan-1-ol

4. 31 Reactions of Alcohols (Schell & Hogan, 2018, pp. 54-55)


Production of Alcohols

Watch OC#20 Production of Alcohols

There are a number of different ways in which alkanols can be synthesised. A couple of examples are shown, with fermentation examined in more detail later in the module.

Synthesis of methanol Methanol is prepared industrially by the catalysed (copper/zinc oxide) reaction between hydrogen and carbon monoxide at 260oC and 10MPa.

CO(g) + 2H2(g) → CH3OH(g)

It is also possible for the carbon monoxide to be converted to carbon dioxide prior to being hydrogenated. This will also release water as a byproduct.

Synthesis of ethanol Ethanol is most commonly prepared through fermentation of glucose. This reaction is facilitated by yeast as a form of anaerobic respiration.

C6H12O6(aq) → C2H5OH(aq) + 2CO2(g) Ethanol, and many other alkanols, can also be synthesised through the hydration of an alkene, such as ethene. Hydration is the process of adding water molecules, or the equivalent of water molecules, to a substance. Ethylene can be hydrated to produce ethanol when heated with a dilute sulfuric acid (H2SO4) catalyst:

I can investigate the production of alcohols, including substitution reactions of halogenated organic compounds.


C2H4(g) + H2O(g) C2H5OH(g)

The hydration of ethylene is an example of an addition reaction. The double bond of ethene is electron rich, so it is liable to attack by any chemical species that is poor in electrons. The hydrogen ion, H+, is perhaps the simplest example of such a species, and for this reason strong acids are used to hydrate ethene. The H+ adds across the double bond, generating a cation, which receives two electrons from the HSO4- ion to form ethyl sulfate.

When the ethyl sulfate is removed from the concentrated sulfuric acid and added to water, the -O-SO3 group is replaced by the more powerful base -OH to give ethanol in high yield.

This is why the catalyst needs to be a dilute solution of sulfuric acid.

Synthesis of alcohols through substitution reactions The addition of water to a haloalkane can also facilitate a substitution reaction. The hydroxy group replaces the halogen to form an alkanol. This is because the C-halogen bond is highly polar and comparatively unstable compared to a C-H bond. This bond becomes more stable as we rise through the halogens, to the extent that the C-F bond is more stable than the C-H bond. This means a fluoroalkane will not undergo substitution to form an alkanol. Write a chemical equation to represent the reaction between 2-bromobutane and water. One of the products will be an alcohol. As with alkanols, halogenated alkanes can also be primary, secondary or tertiary (and for the same reason). As far as substitution reactions are concerned, tertiary haloalkanes are more likely to react, then secondary and finally primary, although the primary haloalkanes have a low reaction rate. This reaction is often carried out at higher temperatures than SLC and may involve reflux apparatus (a vertical condenser).


Question Set 20

1. Catalysts are required for the production of both ethanol from ethylene and ethylene from ethanol. (06, Q2)

What are the identities of Catalyst A and Catalyst B? Catalyst A Catalyst B A) Dilute H+ Conc H+ B) Dilute H+ Dilute H+ C) Conc H+ Conc H+ D) Conc H+ Dilute H+

2. Propose a method by which methyl-2-bromopropane could be used to produce an alcohol and name the alcohol. Include any relevant reagents, catalysts and conditions. (3 marks)

3. Which of the following options is required for the reaction shown below to take place? (Shenfield & Silove, 2018, p. 87) C4H9OH → C4H9Br + H2O A) NaOH B) Pd C) HNO3 D) UV light


Production of Alcohols by Fermentation

Watch OC#21 Fermentation

What is fermentation? The fermentation of sugars to ethanol is promoted by the following conditions:

The chemical equation for fermentation is Explain why industrial fermentation chambers cannot produce solutions above 15% alcohol v/v, however some spirits can be 40% alcohol v/v.

I can investigate the production of alcohols, including fermentation.


Experiment 22: Fermentation Reaction

Introduction

Fermentation is an anaerobic metabolic process that occurs in micro-organisms (fungi and bacteria) and produces energy by breaking down carbohydrates into simpler molecules. Fermentation was one of the first chemical processes to have been utilised (at least 5500 years ago) and its secrets intrigued a few great chemists (Lavoisier, Liebig, Pasteur). It wasn’t until 1939 that the complex biochemistry of this deceptively simple reaction was finally worked out. Today, fermentation is a vital process used in diverse industries (brewing, baking, pharmaceutical).

Planning Guidelines 1. Choose appropriate glassware to contain the fermenting solution. Think about size and

weight limitations. If your vessel is too large it will be unweighable. If your vessel is too small, the mass changes during fermentation will be negligible.

2. Prepare a 5% glucose solution. The volume of solution you prepare is determined by the size of the fermenting vessel. Add 1 g of dried yeast to the glucose solution and mix thoroughly. Add a pinch of Na2HPO4 or NaH2PO4 as a yeast nutrient.

3. Should your fermentation vessel be tightly capped? Study the fermentation equations given on the following pages for technical guidance.

4. How often should you monitor and record the changing mass during fermentation? Should you plot a graph of mass versus time? Note: As a rough estimate, your fermentation will continue for at least 24 hours.

Question Set 21

1. (2011, Q2) Which of the following shows two products that result from the fermentation of glucose? A) Cellulose and water B) Ethanol and oxygen C) Carbon dioxide and water D) Ethanol and carbon dioxide

2. (2010, Q15) What mass of ethanol is obtained when 5.68 g of carbon dioxide is produced during fermentation, at 25°C and 100 kPa? A) 2.95 g B) 5.95 g C) 33.6 g D) 147.2 g


3. Ethanol can be used as a replacement for some of the products currently developed with petroleum fractions. Using plastic credit cards as an example, draw a flowchart of the process by which a credit card can be produced industrially from biomass (cellulose).

4. (2009, Q13) In a fermentation experiment 6.50 g of glucose was completely converted to ethanol and carbon dioxide. What is the mass of carbon dioxide produced? A) 1.59 g B) 3.18 g C) 9.53 g D) 13.0 g

5. (2006, Q18, 4 marks) A student studying the mass change that occurs during fermentation added glucose, water and yeast to a flask and stoppered the flask with some cotton wool. The student measured the mass of the flask daily for seven days. The table shows the data collected.

Day Mass (g) 1 381.05 2 376.96 3 373.42 4 370.44 5 370.42 6 370.40 7 370.39


a) Calculate the moles of CO2 released between days 1 and 7. (1 mark)

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b) Calculate the mass of glucose that underwent fermentation between days 1 and 7. Include a balanced chemical equation in your answer. (3 marks)

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6. 32 Production of Alcohols (Schell & Hogan, 2018, p. 56)


Oxidation of Alcohols

Watch OC#22 Oxidation of Alcohols

Oxidation is an important chemical reaction for alcohols as it can be used to help us identify whether an alcohol is a primary, secondary or tertiary alcohol. We need to select a strong oxidising agent, such as potassium permanganate or potassium dichromate. If we acidify these solutions, we can write both the oxidation and reduction half reactions which are taking place. In the two examples give, potassium is a spectator ion and does not take part in the reaction. However we can write the two reduction half equations (remember a strong oxidant CAUSES oxidation, so is hence reduced itself).

8H+(aq) + MnO4-(aq) + 5e- → 4H2O(l) + Mn2+(aq)

14H+(aq) + Cr2O72-(aq) + 6e- → 7H2O(l) + 2Cr3+(aq)

Oxidation of Primary Alcohols Depending on the quantity of the oxidising agent, primary alcohols may go through more than one oxidation step.

The first step is the formation of an alkanal. Ethanol has been chosen as an example with the permanganate ion as the oxidising agent.

5C2H5OH(aq) + 6H+(aq) + 2MnO4-(aq) → 5C2H4O(aq) + 2Mn2+(aq) + 8H2O(l)

I can investigate the products of the oxidation of primary and secondary alcohols.


The second step is the formation of an alkanoic acid from the alkanal. Ethanal has been chosen as an intermediate with the permanganate ion again acting as the oxidising agent.

5C2H4O(aq) + 6H+(aq) + 2MnO4-(aq) → 5CH3COOH(aq) + 2Mn2+(aq) + 3H2O(l)

NB. If the primary alcohol is methanol, this could further oxidise to carbon dioxide and water.

Oxidation of Secondary Alcohols Secondary alkanols can be oxidised by the same oxidising agents to form a ketone. Propan-2-ol has been chosen as an example with the permanganate ion as the oxidising agent.

5C3H7OH(aq) + 6H+(aq) + 2MnO4-(aq) → 5C3H6O(aq) + 2Mn2+(aq) + 8H2O(l)

Oxidation of Tertiary alcohols Tertiary alcohols do not undergo oxidation. This is a very useful, negative, test often used in their identification.

Question Set 22

1. Show how propan-1-ol can be oxidised to propanoic acid using acidified potassium dichromate solution. Show and name the intermediate. (4 marks)

2. Explain how you could form pentan-3-one from pentan-3-ol. Include any chemical equations relevant to your response. (2 marks)


Comparing Fuels

Watch OC#23 Comparing Fuels

Our reliance on fuels has enormous implications both economically and environmentally. The current reliance on fossil fuels (those fractions extracted from coal, crude oil and natural gas) is having a major impact economically (cost and control) and environmentally (high carbon dioxide emissions and global warming).

One idea to counter our heavy reliance on fossil fuels is to use biofuels. Bioenergy Australia defines biofuels as “liquid fuels that have been derived from other materials such as waste plant and animal matter”. (Bioenergy Australia (Forum) Ltd, 2016)

The natural plant polymer cellulose as a source of ethanol (via fermentation). Sources of cellulose include:

Specifically grown crops Agricultural residue Forestry residue

I can compare and contrast fuels from organic sources to biofuels, including ethanol.

http://www.giancruz.com/portfolio/ethanol/finalproject.swf


Solid council waste

Alternative sources that are renewable could be favourable as their use would prevent the potential problem of a finite and exhaustible fuel that petroleum will soon be facing.

This is no new idea. Ethanol already appears in our petrol as E10 a 10% mix of ethanol and petrochemicals, and has done so for 30 years. However the proportion of ethanol has not increased in Australia since that time as car engines need significant modifications if they are to use higher ratios of ethanol. Despite the fact that this is the case in some parts of the world, it is not so here in Australia.

There are both great advantages and great disadvantages associated with the use of ethanol as an alternative fuel to petroleum.

Advantages

Unlike petroleum, ethanol is a renewable resource. Ethanol burns more cleanly in air than petroleum, producing less carbon (soot) and

carbon monoxide. The use of ethanol as opposed to petroleum could reduce carbon dioxide emissions,

provided that a renewable energy resource was used to produce crops required to obtain ethanol and to distil fermented ethanol.

Disadvantages

Ethanol has a lower heat of combustion (per mole, per unit of volume, and per unit of mass) that petroleum.

Large amounts of arable land are required to produce the crops required to obtain ethanol, leading to problems such as soil erosion, deforestation, fertiliser run-off and salinity.

Major environmental problems would arise out of the disposal of waste fermentation liquors.

Typical current engines would require modification to use high concentrations of ethanol.

What has prompted the need for the addition of ethanol?

How does ethanol compare with octane in terms of oxygen requirements for complete combustion and energy production per mole and per gram?


Question Set 23

1. (2014, Q19)

An experimental car using ethanol as a fuel source requires

2270kJ of energy for every kilometre travelled.

Given that the heat of combustion of ethanol is 1360 kJ.mol−1, what is the maximum

distance that the car can travel on 1.0kilogram of ethanol?

A) 1.7 km B) 13 km C) 28 km D) 36 km

2. (2008, Q24, 5 marks) (modified)

The table shows four fuels and their various properties.

Property Petrol Kerosene Hydrogen Ethanol

Heat of combustion (kJ mol–1) 5460 10 000 285 1370

Boiling point (°C) 126 300 -253 78

Density (g mL–1) 0.69 0.78 n/a 0.78

Average molar mass (g mol–1) 114 210 2 46

(a) Which fuel provides the greatest amount of energy per gram? (1 mark)

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(b) A car has an 80 L petrol tank. Calculate the energy released by the complete combustion of one full tank if it ran on 100% ethanol. (2 marks)

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(c) How many litres of hydrogen gas at 25°C and 100 kPa would be needed to supply the same amount of energy as 80 L of ethanol? (2 marks)

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3. (2008, Q22, 5 marks) The following extract was taken from the website of a leading car manufacturer. Critically evaluate the extract with reference to ethanol being a ‘carbon-neutral’ fuel. Support your answer with relevant chemical equations.

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4. (2004, Q25, 5 marks)

The table shows properties of some fuels.

Fuel Main sources Heat of combustion

(kJ.g-1)

Boiling Point

(oC)

Methane • Petrochemical industry 55.6 -161.5

Propane • Petrochemical industry • Natural gas

50.3 -42.1

Octane • Refined from crude oil 47.9 125.7

Ethanol • Hydration of ethene • fermentation

29.7 78.3

The CO2 released from combustion of bio-ethanol is balanced by CO2 captured through photosynthesis. Therefore combustion of bio-ethanol does not increase the total amount of CO2 in the atmosphere. Thus bio-

ethanol has attracted attention as a carbon-neutral fuel – an energy source effective as a countermeasure to global warming.


Assess the potential of ethanol as an alternative fuel, making use of data from the table.

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5. 33 Biofuels (Schell & Hogan, 2018, pp. 57-58)

6. 35 Bioethanol and Biodiesel (Schell & Hogan, 2018, pp. 60-61)

7. 36 Comparing Fossil fuels and Biofuels (Schell & Hogan, 2018, pp. 62-63)


Reactions of Organic Acids and Bases


Concept Map

Inquiry question 5•What are the

properties of organic acids and bases?


Comparing Properties of different Organic Groups

Watch OC#24 Comparing Properties of Organic Groups

The high melting points and boiling points in alkanols is due to ………………………………

…………………………………………………………………………………………………………., as shown below.

The double bonded oxygen in aldehydes and ketones (carbonyl bond) is another source of polarity but as there is no O-H bond, there can only be dipole-dipole interactions between adjacent molecules. The location of the double bond within the molecule of a ketone, increases the polarity of the bond but only slightly, hence comparable ketones have a slightly higher boiling point than their corresponding aldehyde (propanal 48oC vs propanone 56oC, butanal 75oC vs butanone 80oC).

I can investigate the structural formulae, properties and functional groups including primary, secondary and tertiary alcohols, aldehydes and ketones, amines and amides and carboxylic acids.


The ability of the COOH group in carboxylic acids to be involved in two hydrogen bonds gives an alkanoic acid an even higher boiling point than that of a similar sized alkanol. Two hydrogen bonds can occur between a pair of alkanoic acid molecules as shown below.

Size is also important as larger molecules require additional energy for motion. So we would expect an increase in melting point and boiling point as we _ _ _ _ _ _ _ _ molecular weight (ie, increase the length of the chain or number of carbons) and increase the number of electrons per molecule.

The presence of nitrogen in amines and amides also affects both the chemical reactivity of the molecules and their physical properties. Compete the table below (Aylward & Findlay, 2008, pp. 102-112):

Table 2 Comparing the properties of a range of organic compounds

Compound Homologous Series

Molar Mass (g.mol-1) Reactivity MP (oC) BP (oC) Bonding

(Inter/Intra)

1-butanol Alcohol (P) 74.1 -89.8 117.7

2-butanol Alcohol (S) 74.1 -114.7 99.5

2-methyl-2-propanol Alcohol (T) 74.1 25.7 82.6

1-butylamine Amine 73.1 -49.1 77.9

Propanoic acid

Carboxylic acid

74.1 -20.7 140.8

Propanamide Amide 73.1 77.5 213

Butanal Aldehyde 72.1 -99 74.8

Butanone Ketone 72.1 -86.6 79.6


Question Set 24

1. Draw the structural and molecular formulae for each of the compounds listed in Table 2 (12 marks).

2. (2010, Q14) The table shows information about three carbon compounds.

What is the best estimate for the boiling point of compound Z?

(A) 31°C (B) 101°C (C) 114°C (D) 156°C


3. (2008, Q11) In which of the following alternatives are the three compounds listed in order of increasing boiling point? (A) Pentane, butan-1-ol, propanoic acid (B) Propanoic acid, butan-1-ol, pentane (C) Propanoic acid, pentane, butan-1-ol (D) Butan-1-ol, propanoic acid, pentane

3. (03, Q25, 4 marks) Explain the trends in boiling points shown in the graph.

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4. Worksheet 7.3 Families examined – properties and structure (Commons, 2018, pp. 123-124)

5. 39 Aldehydes and Ketones (Schell & Hogan, 2018, p. 68)


6. 40 Amines and Amides (Schell & Hogan, 2018, p. 69) 7. 41 Carboxylic Acids (Schell & Hogan, 2018, pp. 70-71) 8. 7 Organic Chemistry II: Boiling Points of some Organic Compounds (Dobinson, 2010,

p. 11)


Properties and Bonding

Video OC#25 Properties and Bonding

The carboxylic acid functional group, -COOH, in alkanoic acids can lose a hydrogen ion and behave as a weak acid.

The diagram right shows a common way to illustrate the characteristic structure of the members of a homologous series, in this case, carboxylic (or alkanoic) acids. The R group represents a carbon chain of unspecified length (for our purposes, between 0 and 7 carbons). Figure 37 shows a skeletal formula as the carbon atom within the functional group (COOH) is not labelled with a C. In this type of formula the junction of bonds with no letter indicating the atom present is assumed to be a carbon and hydrogens bonded directly to carbon atoms are not shown.

However, whilst you may see this short-hand version, or a condensed structural formula, eg CH3CH2COOH, in different text books or web sites, it is best in HSC Chemistry Exams to always draw full (or graphic) structural formulae where possible.

The functional group is the key to the chemical properties of the compound, but it is also relevant when discussing the physical properties. In figure 37, there is a non-polar region,

I can explain the properties within and between the homologous series of carboxylic acids, amines and amides with reference to the intermolecular and intramolecular bonding present.

Figure 37 Carboxylic acid functional group


indicated by the R, and two other features of interest; C=O and C-O-H. Both of these groups contain polar covalent bonds, with the C=O bond especially so, and hence both affect the bonding which occurs within the members of an homologous series but also between carboxylic acids and other organic compounds.

Important properties (within or between series shown): Density (within): This is a function of how tightly packed the molecules are in the solid or liquid phase. Whilst stronger intermolecular forces, ie hydrogen bonds, may hold molecules closer together than the weaker dispersion forces, side branches can affect the density as they mean the molecules may not be as tightly packed. We can compare a secondary and tertiary alcohol of comparable molar mass, we find the tertiary alcohol, with the side branch has a lower density (ρ):

2-butanol ρ = 0.802 g.cm-3

2-methyl-2-propanol ρ = 0.781 g.cm-3

Melting Point (within): Melting point indicates the transition from solid to liquid. This involves sufficient energy to break the bonds between molecules which are keeping them rigid and allowing the molecules to move past one another in the liquid state. Intermolecular forces must be broken here, so the stronger the bonds, the more energy will be required and hence the higher the MP (see Figure 38 (Davis, Disney, & Smith, 2018, p. 277)). Hydrogen bonds between molecules require the most energy for separation and hence the higher mp for the carboxylic acids and amides (See Table 2). Aldehydes and ketones are slightly lower as there are no hydrogen bonds, only dipole-dipole interactions. Amines and amides contain nitrogen atoms which are also highly electronegative and form polar bonds with carbon atoms, however as the C-N bond is not as strongly polar as the C-O bond, the hydrogen bonding in amines and amides is weaker than that in a corresponding alcohol.

Boiling Point (within): Boiling point indicates the transition from liquid to gas. This involves sufficient energy to break the bonds between molecules which are holding the molecules close enough for them to move past one another in the liquid state to allow each molecule to move independently. Intermolecular forces must be broken here, so again, the stronger the bonds, the more energy will be required and hence the higher the BP. Hydrogen bonds between molecules require the most energy for separation and hence the higher bp for the carboxylic acids and amides (See Table 2). Aldehydes and ketones are slightly lower as there are no hydrogen bonds, only dipole-dipole interactions. In amines, the presence of the N atom provides opportunities for H bonding and even more so for amides, where dimers, similar to those between carboxylic acid molecules, can form, as shown in Figure 37 (Davis, Disney, & Smith, 2018, p. 284).

Figure 38 Intermolecular bonding between two acid molecules


Figure 39 Intermolecular bonding between amides.

Solubility in water (between): Our solubility rule is like dissolves like. So we can assume molecules with some polarity will interact with polar water molecules and will show some degree of miscibility, see Figure 40 (Davis, Disney, & Smith, 2018, p. 279). This is particularly high for acids, alcohols and amines, less so for aldehydes and ketones. However, unless the functional group occurs several times throughout the molecule, the presence of a hydrocarbon chain will not attract water molecules, hence the solubility of polar molecules will decrease with chain length as shown in Table 9.3 below (Davis, Disney, & Smith, 2018, p. 270).

Solubility in non-polar media (between): Again, based on our solubility rule of like dissolves like, we can assume molecules with some polarity will interact with other polar

Figure 40 Hydrogen bonding between acid and water molecules


molecules and will show some degree of miscibility. Likewise, those with long non-polar regions can interact with other non-polar molecules. As for solubility in water, polar solvents will be miscible with acids and alcohols, but are less miscible with aldehydes and ketones. Unless the functional group occurs several times throughout the molecule, the presence of a hydrocarbon chain will not attract polar molecules, hence the solubility of polar molecules will increase in non-polar solvents with chain length, ie even with a polar head, the longer the non-polar tail, the more miscible the substance in a non-polar solvent. Long molecules with a polar head and non-polar tail can act as surfactants. This is a term we will explore later in this module.

Brønsted-Lowry classification (between): Carboxylic acids have the ability to donate a proton. The highly polar O-H bond can be made to ionise, fulfilling the definition of a BL acid, although organic acids are weak acids. Amines can likewise accept a proton by donating their electron pair (Lewis definition) to a hydrogen atom, making them BL bases.

Question Set 25

1. Compare and explain the solubility of aldehydes and ketones in water (Hegarty, 2018, p. 56). (3 marks)

2. Draw labelled diagrams to show the intermolecular forces between neighbouring ketone molecules and between a ketone and water (Hegarty, 2018, p. 58). (4 marks)


3. Look at the diagram below (Hegarty, 2018, p. 60).

What is the difference between the amide on the left and the one on the right in terms of their solubility in water? (4 marks)

4. Worksheet 7.4 Observing organics (Commons, 2018, p. 125) 5. Comparing Homologous Groups (Schell & Hogan, 2018, pp. 72-74)


Esterification

Video OC#26 Esterification

What is an ester?

An ester is

The reaction of alcohols with carboxylic acids requires the presence of H2SO4(aq) catalyst to produce an ester.

eg.

Hydrolysis of esters

Under the influence of bases eg. NaOH, esters are hydrolysed to alcohols and alkanoate ions. The alkanoic acids may be obtained from the reaction by acidification with mineral acids, e.g. HCl.

I can investigate the production, in a school laboratory, of simple esters.


Acid catalysis

Esterification is catalysed by the addition of a small amount of _ _ _ _ . Esterification is called a condensation reaction because a _ _ _ _ _ molecule condenses out (recall condensation polymerisation).

Only a few drops of concentrated acid needs to be added to a mixture of alkanol and alkanoic acid to catalyse the reaction.

If concentrated sulfuric acid is added in large amounts, say 5% to 10% of the reaction volume, it can have a significant effect on the position of equilibrium. Concentrated sulfuric acid is a dehydrating agent, that is, it has a strong affinity for water. If a significant amount of sulfuric acid is present, it will shift the equilibrium position to the right by absorbing water.

alcohol + acid ⇌ ester + water

This will _ _ _ _ _ _ _ _ the yield of ester. However, using large amounts of sulfuric acid is wasteful, uneconomic and complicates the separation of ester from the reaction mixture.

Refluxing

Esterification requires heat for the reaction to reach equilibrium quickly, ie within an hour rather than after many days.

When the reaction mixture is heated, volatile components, such as the reactant alcohol and the product ester, could escape. This problem is overcome by refluxing the reaction mixture.

A condenser is placed on top of the reaction vessel so that any volatile components pass into the condenser, as shown in Figure 39. The condenser can be water or air-cooled and causes the volatile components to condense back to liquid and fall back into the reaction mixture.

Refluxing also improves the safety of the operation, as the volatile components are flammable.

Figure 41 Refluxing apparatus


Experiment 23: Esterification

Introduction

Esters are a class of organic compounds of great industrial importance. Their production involves the simple and direct reaction of an alkanol and an alkanoic acid. Esterification is performed using a technique called refluxing. Refluxing allows for the safe, prolonged heating of volatile and/or flammable reactants.

Aim

The aim of this experiment is to prepare butyl ethanoate. Butyl ethanoate (known commercially as butyl acetate) is used in the manufacture of plastics and safety glass, and as a solvent in lacquers.

Question Set 26

1. The condensed structural formula of an organic compound is: CH3CH2CH2CH2COOCH2CH2CH3 The name of the organic compound is: A) Pentylpropanoate C) Propylpentanoate B) Octanoic acid D) Octan-3-one

(Science Teachers' Association of NSW, 2019, p. 54)

2. (2004, Q1)

Ethanol can be reacted with ethanoic acid to produce ethyl ethanoate. What type of reaction is this? (A) Esterification (B) Hydration (C) Polymerisation (D) Reduction

3. (2009, Q10) Which of the following is the main organic product resulting from the reaction of butanoic acid and pentanol?


4. (2002, Q9)

The following list of steps refers to an experimental plan for making an ester in a flask. Some of the steps in the list may NOT be required for this experiment. The steps are NOT in the correct sequence. 1. Heat the mixture under reflux. 2. Add three drops of concentrated sulfuric acid. 3. Add 1 mL of ethanol. 4. Add 1 mL of ethene. 5. Add 1 mL of ethanoic acid. 6. Distil the mixture. 7. Add three drops of phenolphthalein indicator. Which alternative is the best sequence for making an ester? a) 3, 5, 7, 1 b) 4, 3, 7, 6 c) 5, 4, 2, 6 d) 5, 3, 2, 1

5. (2002, Q10) Which equation represents esterification?


6. (2010, Q22, 6 marks) A student prepared the compound methyl propanoate in a school laboratory. (a) Give a common use for the class of compounds to which methyl propanoate belongs. (1 mark)

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(b) In the preparation of this compound a few drops of concentrated sulfuric acid were added to the starting materials. The mixture was then refluxed for a period of time. Why was it necessary to reflux the mixture? (2 marks)

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(c) Name the TWO reactants used in preparing the methyl propanoate and draw their structural formulae. (3 marks)

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7. (2009, Q16, 3 marks) Describe how to prepare an ester in the school laboratory. Include a specific safety precaution in your answer.

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8. (2007, Q23, 3 marks) When hexanoic acid and ethanol are mixed together under certain conditions, esterification occurs. Describe the conditions necessary for this reaction and give the structural formulae and names of the products.

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9. (2005, Q20, 7 marks) The flow chart shows a series of steps involved in the production of ethyl butanoate. Describe the chemistry and procedure involved in each of these steps, using diagrams where appropriate.

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glucose Ethyl butanoate Pure ethanol Mixture containing ethanol


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10. (2003, Q21, 5 marks) You performed a first-hand investigation to prepare an ester by reflux. (a) Identify the products formed when propanoic acid and butanol are refluxed with an acid catalyst. (1 mark)

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(b) Draw a fully labelled diagram of the equipment assembled for use. (2 marks)


(c) Outline the advantages of using reflux to prepare the ester. (2 marks)

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11. Acids, alcohols and esters (de Vreeze & McMicking, 1998, pp. 41-42) 12. Refluxing and distillation (de Vreeze & McMicking, 1998, p. 43) 13. 43 Esters (Schell & Hogan, 2018, pp. 75-76) 14. 44 Esterification (Schell & Hogan, 2018, p. 77)


Organic Acids and Bases

Video OC#27 Organic acids and bases

Whilst we have had some experience with organic acids and know the acidic nature of carboxylic acids, it is also important to be aware of the fact that the amines are good organic bases.

The best way to analyse organic substances to distinguish between their acidic, neutral or basic nature is to use our Brønsted-Lowry or Lewis definitions of acids and bases. If we can find a proton which could be donated (or a lone pair of electrons which could be accepted), we have an acidic substance (eg Figure 42). Carboxylic acids can donate a proton, according to the Brønsted-Lowry definition of acids as shown below. Ethanoic acid can donate a proton (H+) to form the ethanoate ion.

C C

O

OH

H

H

H

C C

O

O–

H

H

H

Figure 42 Ethanoic (acetic) acid (CH3COOH)

I can investigate the differences between an organic acid and organic base.


Likewise if we find a structure which could accept a proton, we have a basic substances. The presence of the lone pair on the nitrogen atom of an amine is an indicator that these organic compounds could attract and accept a proton and hence act as a BL base (eg Figure 43). This is a result of the lone pair of electrons on the N atom. Hence it can accept a proton (BL definition). Methylamine can accept a proton (H+) to form the protonated methylamine ion (methyl ammonium).

CH

H

H

N

H

H

C N+H

H

H H

H

H

Figure 43 Methylamine (CH3NH2)

Whilst we can generalise and suggest that most organic acids and bases are weak, to determine the actual strength, we need to look at the Ka and Kb values.

Question Set 27

1. Use an equation to show why butyric acid (butanoic acid) fulfils the Brønsted-Lowry criteria to act as an acid. (2 marks)

2. Primary, secondary and tertiary amines can act as Brønsted-Lowry bases. Use a diagram, or series of diagrams, to explain why this is the case. (3 marks)


3. The four nitrogenous bases which make up rungs in the DNA spiral staircase are shown below (Davis, Disney, & Smith, 2018, p. 350). Circle and identify any functional groups you recognise and show why any one of these bases can be classified as a base. (4 marks)

4. Proteins are made up of amino acids. A peptide bond is formed between adjacent amino acids. Draw a diagram to show the formation of a peptide bond using two (named) amino acids.

5. 45 Organic Acids and Bases (Schell & Hogan, 2018, p. 78)


Saponification

Video OC#28 Saponification

Saponification is the conversion in basic solution of fats and oils to glycerol and salts of fatty acids. Esters can be converted to alkanols and salts of alkanoic acids by alkaline hydrolysis in the reaction known as saponification. This is the process responsible for the production of soaps.

eg. methyl octadecanoate + NaOH ⇌ sodium octadecanoate + methanol

(a soap)

More commonly, a fat or oil when combined with a base (sodium hydroxide) form glycerol (an alcohol) and a salt (soap).

The salts of long chain alkanoic acids are called soaps. They can lather in water and dissolve oils and grease which are normally immiscible in water.

Structure of a soap

Soaps consist of a charged region (anionic head) and a non-polar region (tail) as shown in the diagram below (Davis, Disney, & Smith, 2018, p. 353).

I can investigate the structure and action of soaps and detergents.


Cleaning Action of Soaps

In order to understand the action of soap as a cleaning product we need to look at its chemical structure. We also need to understand the term surfactant. Surface active agents (surfactants) are chemical substances which decrease the surface tension of water. This enables them to “dissolve” dirt and grease. Cleaning agents such as soap are surfactants.

Soap has a long tail made up of many carbon-carbon bonds (with hydrogens attached). This part of the molecule is hydrophobic (water-fearing). However, as it is non-polar it will dissolve other non-polar substances like oil or wax.

The other end of the soap, the head, is anionic. This end readily dissolves in water, it is hydrophilic.

As one part of the soap is dissolved in water while the other end is attached to oil, the soap molecule forms a bridge between the oil and the water. Hence it is able to move oily substances away from fabrics, fibres, ceramics or even skin. Several soap molecules attack the oily substance, lifting it away from the fabric. More soap molecules come in surrounding the oil molecule with their tails attached to the oil and their heads sticking out forming a structure known as a micelle. This keeps the oil molecules separate from one another as each is surrounded by a sphere of negative “heads” each repelling other similar complexes. The decrease in surface tension allows the water to combine with the oil molecules.

Comparing Soaps and Detergents

Synthetic surfactants include detergents. These differ from soaps both in terms of their structure and chemical composition and their action in water.

Molecular Structure Both soaps and detergents have a similar structure consisting of a hydrophobic tail and hydrophilic head, however while soaps have and anionic head, synthetic detergents have

Figure 44 Formation of a micelle (Davis, Disney, & Smith, 2018, p. 357)


either anionic, cationic or non-ionic (polar) heads.

Chemical Composition Soaps are the metallic salts of long chain fatty acids (sodium or potassium salts), while detergents are hydrocarbons with either a sulfate (anionic), ammonium (cationic) or ethoxy (non-ionic) heads.

As mentioned above, detergents may be one of the following types:

Figure 45 Anionic, Cationic and Non-ionic detergents (Davis, Disney, & Smith, 2018, p. 354)

Anionic detergents: The original and still most widely used group of detergents. The structure of the molecules is very similar to natural soaps. They have a long, non-polar tail and an anionic head (a sulfonate C-O-SO2-O-, similar to SO42-). They work just like soaps but are slightly more effective. Used in laundry detergents and dishwashing liquids. These also produce a lot of foam which has no particular impact on the cleaning ability of the surfactant.

Cationic detergents: These have a region which is a derivative of ammonium (H replaced with alkyl groups, at least one, if not two of which are long chains.) The long chains are non-polar while the nitrogen region is water soluble. These are the preferred cleaner for plastics as well as hair conditioners and fabric softeners as they are absorbed into the fibres reducing friction and static. They are also a component of many disinfectants and antiseptics.

Non-ionic detergents: These have a similar tail to the other types of surfactants but the head is different (again!)


This time it contains a sequence of oxy groups with a terminal alcohol. This creates polarity within the head end and hence is water soluble. They are molecules not ions. These produce less foam and are used in cosmetics, paints, adhesives, pesticides, etc.

Experiment 24: Saponification

Introduction

Saponification is the simple process of making soap. The ingredients to make soap (fat or oil and NaOH) are readily obtainable and cheap. The history of soap-making goes back many centuries and we can speculate that soap was discovered by accident, perhaps when hot animal fat dripped into the alkaline ashes of a wood fire during cooking.

Method

1 Warning: Wear safety glasses!

2 Warning: Sodium hydroxide is caustic. Avoid skin contact. Clean up spills immediately.

3 Weigh 4 g of olive oil into a 150 mL beaker.

4 Add 10 mL of water, 1 g of NaOH and 10 mL of ethanol to the beaker.

5 Set up a water bath using a tripod, wire gauze, 250 mL beaker and Bunsen burner.

6 Heat the reaction mixture in the water bath and stir the mixture every five minutes. Note: After about 30 minutes of heating, the mixture will suddenly become creamy and then start to curdle. This signals that soap has been made.

7 While you are waiting, prepare a saturated salt solution for Step 9. Dissolve about 18 g of NaCl in 50 mL of water. Also set up a filtration apparatus for Step 9.

8 Stop heating after the soap forms. Carefully remove the reaction beaker and allow it to cool.

9 Add the salt solution and stir. Spoon out the curdles of soap into the folded filter paper.

10 Wash the raw soap with water, using a wash bottle. Spoon out the rinsed soap onto a pad of paper towels. Gently pat to absorb excess water.

Testing your Home-made Soap

11 Weigh out 0.2 g of home-made soap and 0.2 g of soap shaved from a bar and transfer

into separate test tubes. Add 5 mL of water to each. Shake both vigorously (at the same

time) for 15 seconds.


12 Observe and record:

a the amount of suds b the longevity of the suds.

Question Set 28

1. (09, Q27 d) 4 marks)

i) Explain the cleaning action of soap in terms of its molecular

structure. (2 marks)

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ii) Soap is one product of saponification. Name the other product and draw its

structural formula. (2 marks)

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2. (02, Q28, a) i) 1 mark)

Define saponification.

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3. (01, Q28 d) i) 1 mark)

Name the chemical process used to make soap.

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The equation represents a reaction that can be performed in a school laboratory.


i) Identify both this type of reaction and the reactant A. (2 marks)

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5. (13, Q32 a) 3 marks)

The diagram shows a sequence of steps in the removal of grease from a surface.

Explain the process shown in these steps.

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6. (06, Q29 b), 6 marks)

The diagram represents how one class of molecules assembles in water to form a

structure called a micelle.


i) Identify the class of molecules shown. (1 mark)

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ii) Account for the formation of a micelle. (2 marks)

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iii) Explain what happens when oil is added to water containing these molecules

(3 marks)

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7. (02, Q28 a) ii) 3 marks)

Account for the cleaning action of soap.

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…….……………….……………………………………………………………………… 8. (12, Q33 e) 7 marks)

Initially soap was the only product of the surfactant industry. Due to societal pressures

and chemical developments, production in this industry has evolved to include a wide

range of products.

Account for these changes over time with reference to the structure and uses of

surfactants.

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9. (04, Q28 b) 6 marks)

The table shows the structures of three different classes of detergents.


(i) Account for the cleaning action of anionic detergents. (2 marks)

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(ii) Assess the environmental impacts of the different classes of detergents. (4 marks)

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10. (03, Q30 c) 5 marks)

Analyse how an understanding of the structure and cleaning action of soaps led to the

development of synthetic detergents.

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11. 46 Hydrolysis of Esters (Schell & Hogan, 2018, p. 79) 12. 47 Soaps (Schell & Hogan, 2018, pp. 80-82) 13. 48 Detergents (Schell & Hogan, 2018, pp. 83-84) 14. 23 Saponification (Gribben & St Germain, 2005, p. 33) 15. 24 Properties of Soaps (Gribben & St Germain, 2005, p. 34)


Organic Reaction Pathways

Video OC#29 Organic Reaction Pathways

We now have a very large number of organic substances which can be converted into other organic compounds. A flow chart is a convenient way of summarising this information and allowing us, at a glance, to identify the key functional groups and organic series and how they are related to one another. A simplified example is shown below.

C CH

H

H

H

H

H C CH

H

H

H

H

ClCl2

C CH

H

H

H

O

H

H

NaOH

C CH

H

H

OH

O

H+ / MnO4-

CC

O

O

CH

HH H

HH

CH3OH / conc. H2SO4

I can draft and construct flow charts to show reaction pathways for chemical synthesis, including those that involve more than one step.


Here is a more complex example (Davis, Disney, & Smith, 2018, p. 360).

Question Set 29

1. Devise reaction pathways for the following reactions (Chan, et al., 2019, p. 370) (12 marks) a) Ethanamine from ethene

b) Butan-1-ol from butane c) Pentanoic acid from 1-chloropentane


d) Methanoic acid from methane e) Ethanol from ethene

f) Propyl methanoate from propane and methane

2. Complete Chapter 13 Review Questions 18 and 19 (Chan, et al., 2019, p. 372) 3. Worksheet 7.5 Converting chemicals – organic reaction pathways (Commons, 2018,

pp. 126-127) 4. Worksheet 7.6 Focus on Functional Groups (Commons, 2018, pp. 128-130) 5. 49 Organic Reaction Pathways (Schell & Hogan, 2018, pp. 85-86)


Polymers


Concept Map

Inquiry question 7•What are the

properties and uses of polymers?


Some Common Polymers – Polyethylene (PE)

Watch OC#30 Polyethylene

There are two types of polymerisation processes: addition polymerisation and condensation

polymerisation.

Addition polymerisation:

Addition polymerisation occurs between

monomers of the type CH2=CHX where

these add together by breaking one of the

double bonds and forming long chains.

There are no atoms lost in the process.

eg. n(CH2=CHX) → —(CH2-CHX)n—

Note that while the monomer is unsaturated, the polymer is saturated. Some commercially significant examples of addition polymers include: polythene (polyethene), polystyrene (polyethylbenzene), PVC (polychloroethene), polyacrylonitrile (polycyanoethene), teflon (PTFE or polytetrafluoroethene) and polypropylene (polypropene).

I can model and compare the structure, properties and uses of addition polymers of ethylene and related monomers, for example polyethylene (PE).

https://www.e-education.psu.edu/files/matse081/animations/lesson07/u07_polymerSynthF.swf


Polymerisation - Production of Polythene (polyethylene)

n(CH2=CH2) (-CH2-CH2-)n (reaction conditions affect polymer product)

Model the polymerisation process for ethene to polyethene using the molymod kits. There are two industrially important polymers made from polyethylene. Contrast the structure and properties for each polymer 1. Low Density Polyethylene, LDPE:

Uses include: Wrapping materials, Carry bags, Milk bottles, Squeeze bottles

2. High Density Polyethylene, HDPE: Uses include: Kitchen utensils, Toys, Grocery carry bags, Building materials

There are a number of properties which need to be taken into account when we are selecting a polymer for a particular use. These include:

Melting or softening point Stability to heat and/or light Chemical stability Mechanical strength

http://alevelchem.com/swf/addn_polymerisation.swf


Flexibility or rigidity These properties are affected by the structure of the polymer which is characterised by

Average molecular weight (the longer the chain, the higher the melting point and the harder the polymer)

Chain branching (unbranched chains tend to intertwine and line up very closely producing a highly crystalline arrangement. This increases density, melting point and hardness. Higher degrees of branching lead to a more amorphous structure with lower m.p., low density, softness and flexibility.)

Chain stiffening (addition of a large side group into a chain reduces its flexibility. Addition of chlorine (PVC) or benzene (polystyrene) produces more rigidity in the polymer.

Crosslinks (joining of two or more linear chains to increase the hardness of the polymer. Vulcanised rubber has sulfur bridges between the chains for strength.

Question Set 30

1. If you wanted to produce a strong, rigid polymer, what would be the major considerations in terms of structure and bonding?

2. How do chain branching and cross-linking affect the properties of different polymers?

3. Describe the process of addition polymerisation to produce Low Density Polyethylene (LDPE)

https://www.e-education.psu.edu/files/matse081/animations/lesson07/u07_polymerSynthF.swf


4. (03, Q11) Which polymer is made by the polymerisation of methyl methacrylate?

5. (2003, Q17, 5 marks) The flowchart shows the production of polyethylene.

A sample of polyethylene was produced by Process Y. The

following graph shows the distribution of molecular weights of

polymer molecules in the sample.

(c) Why is a range of molecular weights observed? (1 mark)

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6. 51 Monomers and Polymers (Schell & Hogan, 2018, p. 90) 7. 52 Polyethylene and its production (Schell & Hogan, 2018, pp. 91-92)


Polyvinyl chloride (PVC)

Watch OC#31 Polyvinylchloride

In addition to the different forms of polyethylene, there are a number of other commercially significant polymers. We will look at two of them in a little more detail:

Vinyl Chloride:

Vinyl choride (or chloroethene) is the monomer used in the production of polyvinyl chloride (PVC). PVC is water resistent and flame resistent and used in the manufacture of electrical insulation, appliance leads, sewerage and drain pipes and garden hoses.

I can model and compare the structure, properties and uses of addition polymers of ethylene and related monomers, for example polyvinyl chloride (PVC)

Cl H

C == C

H H


Polymers with few polar bonds are insoluble in water (hydrophobic). The nature of the C bonding provides stability in the polymers. An exception is PVC where the C-Cl bond can be attacked by uv light and needs to be protected. Heating to high temperatures can be a problem, especially in fires. PVC releases HCl, while acrylics produce hydrogen cyanide.

Question Set 31

1. Vinyl chloride monomers undergo addition polymerisation to form PVC. Draw the structural formula of the monomer and a tetramer section of the polymer. (Science Teachers' Association of NSW, 2019, p. 256)

2. Justify 2 named uses for an addition polymer which has been formed from the following monomer: CH2=CHCl (3 marks)


Polystyrene (PS)

Watch OC#32 Polystyrene

Styrene:

C CC C

C C

C C

H

H

H

H

H H

H H

Styrene (ethenyl benzene) is the monomer used in the production of polystyrene. Polystyrene is transparent and when expanded by the addition of gas, has a low density and is a good insulator. Polystyrene is used in the manufacture of tool handles, CD and DVD cases, foam drinking cups, foam packing and containers and insulation.

I can model and compare the structure, properties and uses of addition polymers of ethylene and related monomers, for example polystyrene (PS).


Question Set 32

1. (2010, Q6) The diagram shows a section of a polymer.

What is the systematic name of the monomer?

A. Polybenzene B. Benzylethene C. Ethylbenzene D. Ethenylbenzene

6. (2007, Q26, 4 marks)

Explain how the structure and properties of polyethylene and polystyrene relate to the way each is used.

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Polytetrafluoroethylene (PTFE)

Watch OC#33 Polytetrafluoroethylene

Tetrafluoroethene

C C

F

F

F

F

Tetrafluoroethene is the monomer used in the production of the addition polymer polytetrafluoroethene (PTFE). The best known PTFE is Teflon. Teflon is renowned for its low friction, non-stick properties and high chemical resistance, resulting in its application to cookware. PTFE is now used as an insulator for electrical wiring, in bearings and gears as well as for cookware. It is also used in Gore-Tex textiles and as a membrane separating the half-cells for the electrolytic production of sodium hydroxide.

I can model and compare the structure, properties and uses of addition polymers of ethylene and related monomers, for example polytetrafluoroethylene (PTFE).


Question Set 33

1. (2004, Q17, 5 marks) The structures of two commercially significant monomers are shown.

(a) Identify the common name of ONE of the monomers. (1 mark) ..................................................................................................................................................... (b) The uses of polymers are dependent on their properties. Discuss this statement with reference to a polymer made from one of the above monomers. (3 marks)

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(c) Draw the structure of a polymer made from one of the above monomers. (1 mark).

2. Polymer banknotes have been around for a while now, but their history is an interesting one. What is the significance of the following banknote?

Figure 46 A $7 bank note https://csiropedia.csiro.au/polymer-banknotes/


3. Complete the following table.

4. 53 Other Addition Polymers (Schell & Hogan, 2018, pp. 93-94)

Monomer Polymer Properties Uses Ethene Low density

polyethylene

Ethene High density polyethylene

Vinyl chloride (chloroethene)

Polyvinyl chloride

Styrene (ethenyl benzene)

Polystyrene

Tetrafluoroethene Polytetrafluoroethene (PTFE)


Condensation Polymers - Nylon

Watch OC#34 Condensation Polymers

Condensation polymerisation reactions occur between monomers each having two or more complimentary functional groups. During the reaction a small molecule such as water is eliminated (released).

The formation of cellulose is an example of a condensation polymerisation reaction:

n(HO–C6H10O4–OH) → H–(O–C6H10O4)n–OH + (n–1)H2O

In the formation of cellulose shown above, n glucose molecules are combined to form the cellulose chain and (n–1) molecules of water.

When two glucose monomers react, a hydroxyl group from each monomer combine to release a water molecule. This leaves an oxygen atom linking the two monomers. This process repeats to form the cellulose chain.

Cellulose is therefore a polymer made up of many glucose monomers. When glucose units

I can model and compare the structure, properties and uses of condensation polymers, for example nylon.


combine one must be inverted relative to the other. The geometry of the glucose rings produce a polymer molecule which is very linear in structure. Cellulose is a major component of biomass. Cellulose contains the basic carbon-chain structures needed to build petrochemicals.

Nylon

Another important condensation polymer is nylon. Nylon was first produced in 1938 and marketed in toothbrushes. The following year “nylon stockings were exhibited to the public at the New York World’s Fair by creator Du Pont, where they were touted as being ‘as strong as steel, as fine as a spider’s web’. On 15 May 1940, nylon stockings went on sale throughout the USA, and in New York City alone four million pairs were sold in a matter of hours. Sixty-four million pairs sold in the first year” (Selinger & Barrow, 2017, p. 250).

One form of nylon is nylon-6,6. It is formed through the condensation polymerisation of 1,6-hexanediamine and adipoyl dichloride.

C C C C C COO

Cl ClH

H H

H H

H

H

H

N C C C C C C N

H

H H

H H

H H

H H

H H

H H

H H

H

Figure 47 adipoyl chloride (COCl(CH2)4COCl) Figure 48 1,6-hexanediamine (NH2(CH2)6NH2)

The critical bond formed during the polymerisation process is an amide bond, with HCl the small molecule which is released in the process.

Draw a dimer in the space below to show how these monomers join together.

Question Set 34

1. Nylon is a complex condensation polymer with identifiable amide bonds. Complete the table below for nylon. (3 marks)


Structure of nylon

Properties of nylon

Uses of nylon

2. Glucose (C6H12O6) is a monomer that can form naturally occurring polymers.

The approximate atomic weights for the elements which make up glucose are shown in the table.

Using data from the table, what would be the approximate molecular weight of a polymer made from 5 glucose monomers? A) 810 B) 828 C) 882 D) 900

3. (2014, Q18) This is a representation of a segment of the polymer nylon 6,6.

Which of the following represents the two monomers that are used to produce nylon 6,6?

Element Approximate atomic weight Carbon 12

Hydrogen 1

Oxygen 16


4. (2002, Q18a, 1 mark)

Name the type of polymerisation shown in the following reaction.

..................................................................................................................................................

5. Worksheet 7.8 Consumer Products – polymers and soaps (Commons, 2018, pp. 134-136)

6. 54 Condensation Reactions (Schell & Hogan, 2018, p. 95)


Condensation Polymers - Polyesters

Watch OC#35 Polyester

Polyesters have great strength, elasticity and electrical resistance and are widely used. The most common type of polyester is polyethylene terephthalate. “When it is used to manufacture bottles, it is marketed as PET; when it is sold as a fibre, it goes by the name of Terylene or Dacron” (Selinger & Barrow, 2017, p. 222). Polyesters may be used in textile manufacture due to their “low water absorption and minimal shrinkage after washing” (Thickett, 2018, p. 178).

In PET, there are two monomers, just as there are for nylon, however this time, they are terephthalic acid (benzene-1,4-dicarboxylic acid) and ethylene glycol (ethan-1,2-diol). The ester bond is formed when an acid and alcohol are combined to release a water molecule. The same thing is happening here, it is just a long chain which forms.

I can model and compare the structure, properties and uses of condensation polymers, for example polyesters.


C

C

C

C

C

C

C

O

O

C

O

OH

H

H

H

H

H

CCO

O

H

HH

H

H

H

Figure 49 benzene-1,4-dicarboxylic acid Figure 50 ethan-1,2-diol

Question Set 35

1. Polyester is a condensation polymer with identifiable ester bonds. Complete the table below for polyester. (3 marks)

Structure of polyester

Properties of polyester

Uses of polyester

2. Practical Activity 7.5 Modelling polymers (Commons, 2018, pp. 152-153) 3. 55 Condensation Polymers (Schell & Hogan, 2018, pp. 96-97)


References Aylward, G. H., & Findlay, T. (2008). SI Chemical Data (6th ed.). Milton, QLD: John Wiley &

Sons Australia.

Bioenergy Australia (Forum) Ltd. (2016). What are Biofuels? Retrieved from Bioenergy Australia: http://biofuelsassociation.com.au/biofuels/

Chan, D., Sturgiss, J., Hillier, K., Lennard, L., Waldron, P., Hogendoorn, B., . . . Bolt, R. (2019). Pearson Chemistry 12 Student Book. Melbourne: Pearson Australia.

Commons, P. (2018). Pearson Chemistry NSW 12 Skills and Assessment Book. Melbourne: Pearson Australia.

Crisp, P. (2019). Hexane. RiskAssess. Sydney, NSW, Australia: Ecosolve Australia.

Davis, A., Disney, A., & Smith, D. (2018). Chemistry in Focus Year 12 Student Book. Melbourne: Nelson Cengage Learning.

de Vreeze, D., & McMicking, K. (1998). Instant lessons in Chemistry Book 3. Sydney: Emerald City Books.

Dobinson, G. (2010). Interpreting data in Senior Chemistry. Melbourne: Blake Education Pty Ltd.

Favre, H. A., & Powell, W. H. (2013). Nomenclature of Organic Chemistry: IUPAC Recommendations and Preferred Names 2013. Cambridge: International Union of Pure and Applied Chemistry.

Gribben, P. (1999). Interpreting Data in Chemistry Book 2. Sydney: Emerald City Books.

Gribben, P. (2010). Maximising Exam Marks Senior Chemistry Book 2. Sydney: Blake Education.

Gribben, P., & St Germain, W. (2005). Assignments in Chemistry Book 1. Melbourne: Blake Education.

Gribben, P., & St Germain, W. (2005). Assignments in Chemistry Book 2. Melbourne: Blake Education.

Hardinger, S. A. (2019, April 24). Sigma bond. Retrieved from Illustrated Glossary of Organic Chemistry: www.chem.ucla.edu/~harding/IGOC/S/sigma_bond.html?fbclid=IwAR3UdyssgfPpIJyj9mPl8-LqMSOpx7zCNOpNFMd4OJHdXo0lWNdb9JPCTgQ

Hegarty, B. (2018). Year 12 Chemistry Student Workbook Book 3. Brisbane: Bronwen Hegarty.

International Union of Pure and Applied Chemistry. (2017). Nomenclature. Retrieved from IUPAC: https://iupac.org/what-we-do/nomenclature/


Odlum, C. (1999). 2 unit Chemistry HSC Course Past HSC Questions and Answers 1982-1998. Sydney: Odlum & Garner.

Schell, M., & Hogan, M. (2018). Surfing NSW Chemistry 7 & 8. Sydney: Science Press.

Schiller, M. (2013). Examples of Esters. Retrieved September 23, 2015, from Easy Chem: http://www.easychem.com.au/the-acidic-environment/esterification/examples-of-esters

Science Teachers' Association of NSW. (2019). Excel Success One HSC Chemistry. Sydney: Pascal Press.

Selinger, B., & Barrow, R. (2017). Chemistry in the Marketplace (6th ed.). Melbourne: CSIRO Publishing.

Shenfield, A., & Silove, J. (2018). ATAR Notes Chemistry Topic Tests. Caulfield North, VIC: InStudent Publishing.

Silove, J. (2018). ATAR Notes Chemistry Complete Course Notes NSW Year 12 2019-2020. Melbourne: InStudent Publishing Pty Ltd.

Thickett, G. (2018). Excel HSC Chemistry. Sydney: Pascal Press.

University of Illinois. (2019, April 24). Alkyl Halides and Alkenes from Alcohols. Retrieved from Organic Chemistry: https://archives.library.illinois.edu/erec/University%20Archives/1505050/Organic/Alcohols/Chapter%206/sec6-6/6-6.htm

Walker, L. (2019, April 15). Sigma Bonds. (C. R. Harrison, Interviewer)

student name:

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