CHEMISTRY MODULE 7
Organic Chemistry
CHEMISTRY MODULE 7 – ORGANIC CHEMISTRY
NOMENCLATURE
Inquiry question: How do we systematically name organic chemical compounds?
● investigate the nomenclature of organic chemicals, up to C8, using IUPAC conventions, including simple methyl and ethyl branched chains, including: (ACSCH127) – alkanes – alkenes – alkynes – alcohols (primary, secondary and tertiary) – aldehydes and ketones – carboxylic acids – amines and amides – halogenated organic compounds ● explore and distinguish the different types of structural isomers, including saturated and unsaturated hydrocarbons, including: (ACSCH035) – chain isomers – position isomers – functional group isomers
HYDROCARBONS
Inquiry question: How can hydrocarbons be classified based on their structure and reactivity?
● construct models, identify the functional group, and write structural and molecular formulae for homologous series of organic chemical compounds, up to C8 (ACSCH035) : – alkanes – alkenes – alkynes ● conduct an investigation to compare the properties of organic chemical compounds within a homologous series, and explain these differences in terms of bonding (ACSCH035) ● analyse the shape of molecules formed between carbon atoms when a single, double or triple bond is formed between them ● explain the properties within and between the homologous series of alkanes with reference to the intermolecular and intramolecular bonding present ● describe the procedures required to safely handle and dispose of organic substances (ACSCH075) ● examine the environmental, economic and sociocultural implications of obtaining and using hydrocarbons from the Earth
PRODUCTS OF REACTIONS INVOLVING HYDROCARBONS
Inquiry question: What are the products of reactions of hydrocarbons and how do they react?
● 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:
CHEMISTRY MODULE 7: ORGANIC CHEMISTRY PAGE 1 OF 65
– hydrogen (H2)
– halogens (X2) – hydrogen halides (HX)
– water (H2O) (ACSCH136) ● investigate, write equations and construct models to represent the reactions of saturated hydrocarbons when substituted with halogens
ALCOHOLS
Inquiry question: How can alcohols be produced and what are their properties?
● investigate the structural formulae, properties and functional group including: – primary – secondary – tertiary alcohols ● explain the properties within and between the homologous series of alcohols with reference to the intermolecular and intramolecular bonding present ● conduct a practical investigation to measure and reliably compare the enthalpy of combustion for a range of alcohols ● write equations, state conditions and predict products to represent the reactions of alcohols, including but not limited to (ACSCH128, ACSCH136): – combustion – dehydration – substitution with HX – oxidation ● investigate the production of alcohols, including: – substitution reactions of halogenated organic compounds – fermentation ● investigate the products of the oxidation of primary and secondary alcohols ● compare and contrast fuels from organic sources to biofuels, including ethanol
REACTIONS OF ORGANIC ACIDS AND BASES
Inquiry question: What are the properties of organic acids and bases?
● investigate the structural formulae, properties and functional group including: – primary, secondary and tertiary alcohols – aldehydes and ketones (ACSCH127) – amines and amides – carboxylic acids ● explain the properties within and between the homologous series of carboxylic acids amines and amides with reference to the intermolecular and intramolecular bonding present ● investigate the production, in a school laboratory, of simple esters ● investigate the differences between an organic acid and organic base ● investigate the structure and action of soaps and detergents ● draft and construct flow charts to show reaction pathways for chemical synthesis, including those that involve more than one step
CHEMISTRY MODULE 7: ORGANIC CHEMISTRY PAGE 2 OF 65
POLYMERS
Inquiry question: What are the properties and uses of polymers?
● model and compare the structure, properties and uses of addition polymers of ethylene and related monomers, for example: – polyethylene (PE) – polyvinyl chloride (PVC) – polystyrene (PS) – polytetrafluoroethylene (PTFE) (ACSCH136) ● model and compare the structure, properties and uses of condensation polymers, for example: – nylon – polyesters
CHEMISTRY MODULE 7: ORGANIC CHEMISTRY PAGE 3 OF 65
IUPAC NOMENCLATURE
OUTCOMES ● investigate the nomenclature of organic chemicals, up to C8, using IUPAC conventions, including simple methyl and ethyl branched chains, including: (ACSCH127) – alkanes – alkenes – alkynes – alcohols (primary, secondary and tertiary) – aldehydes and ketones – carboxylic acids – amines and amides – halogenated organic compounds ● analyse the shape of molecules formed between carbon atoms when a single, double or triple bond is formed between them
HYDROCARBONS
Hydrocarbons are compounds that contain only carbon and hydrogen. They are the main constituents of natural gas and crude oil.
Hydrocarbons can be saturated (in hydrogen) or unsaturated, depending on whether they contain just single bonds or double/triple bonds, respectively.
ALKANES
- Contain only C-C single bonds in backbone - Saturated hydrocarbon
- General formula for straight chained and branched alkanes = 퐶푛퐻2푛+2
Alkanes are called saturated hydrocarbons because no more hydrogen atoms can be added to the carbon atoms.
Alkanes are a homologous series: family of compounds that can be represented by one general molecular formula.
Name Molecular Formula MP BP
methane -182 -164 ethane -183 -89 propane -190 -42 butane -138 0 pentane -130 36 hexane -95 69 heptane -91 98 octane -57 126
CHEMISTRY MODULE 7: ORGANIC CHEMISTRY PAGE 4 OF 65
MOLECULAR STRUCTURE OF ALKANES
BRANCHED ALKANES
Branched alkanes are alkane compounds containing alkyl side groups.
Alkyl Groups Molecular Formula
Methyl
Ethyl
Propyl
Butyl
Pentyl
CHEMISTRY MODULE 7: ORGANIC CHEMISTRY PAGE 5 OF 65
STEM NOMENCLATURE
Longest Carbon Chain Prefix Number of identical groups Prefix
1 Meth- 5 Pent-
2 Eth- 6 Hex-
3 Prop- 7 Hept-
4 But- 8 Oct-
NAMING BRANCHED CHAIN ALKANES
1) Find the longest continuous chain of carbon atoms. Assign a parent name based on this number. 2) Find whatever groups that are not part of the longest continuous chain. Name these as prefixes and put them in alphabetical order. 3) Assign numbers to groups by counting from one end of the chain. A chain has two ends, and the end we start from is the one that gives the lowest set of numbers to the groups.
eg: 3-methylhexane
NAMING ALKANES WITH MULTIPLE ALKYL GROUPS
Number of identical groups Prefix Number of identical groups Prefix
2 Di 5 Penta
3 Tri 6 Hexa
4 Tetra 7 Hepta
LOWEST SET OF NUMBERS
To determine the lowest set of numbers:
1) Work out the numbers for the two possible names (going along the longest carbon chain in both directions) 2) List the numbers in each name in increasing order 3) Compare the first number of each name a. If there is a difference, the name with the lower number is correct b. If the numbers are the same, move onto the second number and so on until there’s a difference
CHEMISTRY MODULE 7: ORGANIC CHEMISTRY PAGE 6 OF 65
ALKENES
- Contains at least one C=C double bond in backbone - Unsaturated hydrocarbons
- General formula = 퐶푛퐻2푛
Carbon atoms with double and triple bonds can accept more hydrogen therefore alkenes and alkynes are unsaturated hydrocarbons.
The double bond carbon atoms are called functional groups because they are sites of chemical activity.
- A functional group is an atom, or group of atoms, which give a compound some characteristic physical and chemical properties.
It is necessary to assign a number to describe the location of the double bond. The bonds are numbered much like carbons. The lowest number is assigned to the bond.
ALKYNES
- Contain at least one C≡C triple bond in backbone - Unsaturated hydrocarbons
- General formula = 퐶푛퐻2푛−2
SHAPE
Bonds around Carbon Angle between Bonds Number of VSEPR groups Shape around carbon
Four Single Bonds 109.5 4 Tetrahedral
Two single bonds and one 120 3 Trigonal planar double bond
Two Double bonds or one 180 2 Linear triple bond and one single
CHEMISTRY MODULE 7: ORGANIC CHEMISTRY PAGE 7 OF 65
EXTENSION: CYCLIC HYDROCARBONS
Hydrocarbon compounds in which the carbon atoms have joined to form closed ring structures are called cyclic (or alicyclic) hydrocarbons. They may also be saturated or unsaturated:
- Those that contain only C-C bonds are called cycloalkanes. - Those that contain C=C bonds or C≡C bonds are called cycloalkenes and cycloalkynes respectively.
Due to the extra C-C bond that closes the ring structure, there are always two less hydrogen atoms than the respective straight chain.
Cyclohexane 3-methylcycloheptene EXTENSION: AROMATIC HYDROCARBONS
A unique type of cyclic structure is the aromatic hydrocarbons. These structures contain a ring of carbon atoms linked together such that they have delocalised electrons. Delocalised electrons are not held within bonds but are free to move around a structure.
Benzene
- The electrons of the double bond are free to move around the entire ring, forming a “cloud” of delocalised electrons above and below the plane of the ring. - The delocalisation is shown by drawing a circle around the inside of the ring. This depiction represents the equivalent nature of the six carbon-carbon bonds.
The delocalisation of the electrons in aromatic compounds gives them significantly different properties to alkenes. They have much greater stability and are often unreactive.
CHEMISTRY MODULE 7: ORGANIC CHEMISTRY PAGE 8 OF 65
PHYSICAL PROPERTIES
OUTCOMES ● conduct an investigation to compare the properties of organic chemical compounds within a homologous series, and explain these differences in terms of bonding (ACSCH035) ● explain the properties within and between the homologous series of alkanes with reference to the intermolecular and intramolecular bonding present PROPERTIES OF HYDROCARBONS
Members of the same homologous series have:
- A similar structure and the same general formula and functional group (each member of a homologous
series differs by a −퐶퐻2 − unit from the previous member) - A pattern to their physical properties - Similar chemical properties
MELTING AND BOILING POINTS
The melting and boiling points are measures of the thermal energy required to overcome intermolecular forces.
- As ↑ intermolecular forces → ↑ 퐸ℎ푒푎푡 → ↑ MP and ↑ BP
The packing of molecules also affects the boiling and melting point.
- Packing depends on molecular shape. Molecules that are small, symmetrical or unbranched tend to be able to pack more closely together. This results in stronger intermolecular forces. - The effect of packing on intermolecular forces strength is more significant for molecules in solid states (melting point).
MELTING AND BOILING POINTS OF ALKANES
Alkane molecules are nonpolar, the only intermolecular force is dispersion forces. As the length of the carbon chain increases, the overall force of attraction between molecules also increase. (Dispersion forces is proportional to molar mass). Because boiling and point is determined by the strength of intermolecular forces, boiling point increases as alkane chain length increases.
Molecular shapes also influence the strength of dispersion forces. Straight-chained alkanes are able to fit together more closely and tend to have higher boiling points than their corresponding chain isomers.
Melting point is affected by the strength of the dispersion forces, size and shape of the molecule.
Melting points of hydrocarbons follow the same general patterns as boiling points, with a few exceptions. The melting points of straight-chain hydrocarbons increase as the number of carbon increases. However, there are deviations in this trend, relating to whether the molecules have an even or odd number of carbon atoms.
Chains with even numbers of carbon atoms pack slightly more efficiently in the solid state than chains with odd numbers. The more efficient packing requires more energy to melt the compound resulting in a higher melting point.
CHEMISTRY MODULE 7: ORGANIC CHEMISTRY PAGE 9 OF 65
Name Molecular Formula Molecular Mass (품/풎풐풍) MP (℃) BP (℃)
methane 16.04 -182 -164
ethane 30.07 -183 -89
propane 44.09 -190 -42
butane 58.12 -138 0
pentane 72.15 -130 36
hexane 86.17 -95 69
heptane 100.2 -91 98
octane 114.2 -57 126
- BP (퐶1 − 퐶4) < 25℃ ∴ Gaseous at RTP - BP (퐶4 − 퐶8) > 25℃ ∴ Liquids at RTP - Generally, as ↑ molecular mass → ↑MP, however methane and ethane’s BP > propane. This is because ethane and methane’s shapes allow them to pack tighter relative to propane, increasing IMF and hence MP.
BOILING POINTS OF ALKENES AND ALKYNES
Alkenes and alkynes, like alkanes, are nonpolar hydrocarbons. Their molecules are also nonpolar so the forces of attraction between them are only weak dispersion forces. Members of these homologous series have relatively low boiling points similar to those observed for alkanes with the same number of carbon atoms.
CHEMISTRY MODULE 7: ORGANIC CHEMISTRY PAGE 10 OF 65
As with alkanes, the boiling points of alkenes and alkynes increase with molecular size as the strength of dispersion forces between molecules increase.
- The boiling points of alkenes are slightly lower than alkanes with the same number of carbon atoms. - Alkynes have higher boiling points than both alkenes and alkanes that have the same number of carbon atoms. This is due to the increase packing because of its linear shape.
Since alkenes have lower molecular mass compared to alkanes with the same number of carbon atoms, the strength of the dispersion forces is weaker. Hence the boiling points are lower.
MELTING POINTS OF ALKENES AND ALKYNES
Alkenes, alkynes and haloalkanes follow the similar pattern to alkanes. As the length of the carbon chain increases, the melting point increases. The alkenes ethene, propene and butene are all gases at room temperature, alkenes with 5-14 carbons are liquid, and longer-chained molecules are solid.
Alkynes follow the same general trend for melting points seen in alkanes and alkenes. The position of the triple bond can greatly affect the melting points as the shape of the molecule changes.
SOLUBILITY (LIKE DISSOLVE LIKE)
The solubility of a substance depends on the strength of the IMF within the solute and within the solvent (cohesive forces), in comparison to the IMF between the solute and solvent (adhesive forces).
The generalisation is that polar compounds tend to only dissolve well in polar compounds and non-polar compounds only dissolve in non-polar compounds.
Hydrocarbons are soluble in each other as well as in non-polar organic compounds such as benzene, diethyl ether and carbon tetrachloride (tetrachloromethane). This is because the cohesive forces within the solvent are also weak dispersion forces that are similar in strength to the dispersion forces within hydrocarbons.
DENSITY
The density of a substance is a measure of its mas per unit volume. (푔/푚퐿 or 푔/푐푚3)
Alkane Formula Density (품/풄풎ퟑ 풂풕 ퟐퟎ℃) Molecular Mass (품/풎풐풍)
Pentane 퐶5퐻12 0.626 72.15
Decane 퐶10퐻22 0.730 142.3
Pentadecane 퐶15퐻32 0.768 212.4
Isosane 퐶20퐻42 0.789 282.6
Alkanes are immiscible with water (cannot mix). When mixed, two layers form and the alkane will float on water. This occurs as the density of all alkanes < water.
CHEMISTRY MODULE 7: ORGANIC CHEMISTRY PAGE 11 OF 65
VOLATILITY
Volatility is the ability of a liquid (or solid) to escape and form a vapour. ↑ BP → ↑ IMF Strength → ↑ Volatility
Volatility is measured by vapour pressure, which is a measure of concentration in the gas phase above the liquid. It is constant at a constant temperature.
Hydrocarbons are non-polar and hence have dispersion forces as their only intermolecular force. Since the intermolecular forces are relatively weak, their bonds are easily overcome and hence the BP is relatively low. Therefore, hydrocarbons will be volatile.
↑ Molecular Mass → ↑ Strength of dispersion forces → ↑ 퐸ℎ푒푎푡 → ↓ Volatility at a constant temperature.
VISCOSITY
Viscosity refers to a substance’s resistance to fluid flow. Liquids with a relatively high resistance to flow have high viscosity. For a substance to flow, particles must flow over each other. Viscosity depends on:
- Strength of intermolecular forces. ↑ IMF → ↑ Viscosity - Size of molecules. ↑ Size → ↑ Viscosity - Temperature. ↑ Temperature → ↓ Viscosity
Viscosity decreases with increasing temperature. At higher temperatures, molecules have greater kinetic energy. Thus, the molecules move around more which increases the space between molecules. This causes the intermolecular forces to be weaker and hence viscosity decreases.
CHEMISTRY MODULE 7: ORGANIC CHEMISTRY PAGE 12 OF 65
FUNCTIONAL GROUPS
OUTCOMES ● investigate the nomenclature of organic chemicals, up to C8, using IUPAC conventions, including simple methyl and ethyl branched chains, including: (ACSCH127) – alcohols (primary, secondary and tertiary) – aldehydes and ketones – carboxylic acids – amines and amides – halogenated organic compounds FUNCTIONAL GROUPS
A functional group is an atom or group of atoms which give a compound some characteristic physical and chemical property.
A homologous series is a family of organic compounds with the same general formula or functional groups with similar chemical properties.
ALCOHOLS (ALKANOLS)
Alcohols are formed when a H atom on a hydrocarbon has been replaced with an -OH group.
For naming:
1) Find longest continuous chain of carbon atoms that contain the functional group. Assign a stem name and use the suffix “-ol”. 2) Name the substituents as prefixes in alphabetical order. 3) Number the chain so that the functional group at the end of the name gets the lowest possible number.
eg: 2-ethylbutan-1-ol
CHEMISTRY MODULE 7: ORGANIC CHEMISTRY PAGE 13 OF 65
Alcohols can also be classified according to the number of carbon atoms attached to the carbon bearing the -OH group.
- A primary (1°) alcohol is one in which the carbon bearing the -OH group is bonded to one other carbon atom. - A secondary (2°) alcohol is one in which the carbon bearing the -OH group is bonded to two other carbon atoms. - A tertiary (3°) alcohol is one in which the carbon bearing the -OH group is bonded to three other carbon atoms.
These three types of alcohols have different physical and chemical properties.
Butan-1-ol Hexan-3-ol 2-methylpentan-2ol Primary Alcohol Secondary Alcohol Tertiary Alcohol
ALDEHYDES AND KETONES (CARBONYLS)
A carbonyl functional group consists of carbon attached to an oxygen atom by a double bond.
The difference between aldehydes (alkanals) and ketones (alkanone) is the position of the carbonyl group along the carbon chain:
- If the carbonyl group is on a terminal carbon (at the end of a chain), it is called an aldehyde, which is given the suffix “-al”. - If the carbonyl group is in the middle of the carbon chain, it is called a ketone, which is given the suffix “- one”.
Aldehyde: Ketone:
CHEMISTRY MODULE 7: ORGANIC CHEMISTRY PAGE 14 OF 65
- When the aldehyde is the suffix, the aldehyde carbon is assigned the number 1. Since it is always carbon 1, “1” (terminal carbon) is omitted from the name. (“-al”) Eg. 2-methylbutanal
- When the ketone is the suffix, the carbon chain is numbered so that the ketone is assigned the lowest number. (“-one”) eg. Pentan-3-one.
CARBOXYLIC ACIDS (ALKANOIC ACIDS)
Carboxylic acids contain the carbonyl group, C=O, connected to a hydroxyl group. The carboxyl group is abbreviated as -COOH, and is given by the suffix “-oic acid”.
- Since the carboxylic acid uses up 3 bonds on the carbon atom, it must be situated at the end of the chain (terminal carbon).
Eg: 2,2,3-trimethylbutanoic acid
CHEMISTRY MODULE 7: ORGANIC CHEMISTRY PAGE 15 OF 65
IUPAC Name Common Name
Methanoic acid (푯푪푶푶푯) Formic acid
Ethanoic acid (푪푯ퟑ푪푶푶푯) Acetic acid
Propanoic acid (푪푯ퟑ푪푯ퟐ푪푶푶푯) Propionic acid
Butanoic acid (푪푯ퟑ푪푯ퟐ푪푯ퟐ푪푶푶푯) Butyric acid
Pentanoic acid (푪푯ퟑ푪푯ퟐ푪푯ퟐ푪푯ퟐ푪푶푶푯) Valeric acid
2-Hydroxypropanoic acid (푯푪ퟑ푯ퟓ푶ퟔ) Lactic acid
2-Hydroxypropane-1,2,3-tricarboxylic acid (푯ퟑ푪ퟔ푯ퟓ푶ퟕ) Citric acid AMINES AND AMIDES
Amines and amides both contain nitrogen. Amides contain the group. They use the suffix “-amine” or the prefix “amino-”.
Amides are carboxylic acids where the OH group has been replaced with an amine. They use the suffix “amide”. When they quoted as the prefix, the carbonyl and the amine are named separately (using the prefixes “oxo” and “amino” respectively).
Eg. 3-methylpentan-2-amine
Eg. 5-amino-3-methyl-oxopentanoic acid
CHEMISTRY MODULE 7: ORGANIC CHEMISTRY PAGE 16 OF 65
HALOGENATED ORGANIC COMPOUNDS
Halogenated organic compounds have hydrogen atoms replaced with a halogen (Group 17 element).
The halogen functional groups are named using prefixes placed in front of their name of the alkane:
• Br: bromo • Cl: chloro • F: fluoro • I: iodo
Eg. 1,2-dibromo-3,4-difluoropentan-3-one
DIFFERENT FUNCTIONAL GROUPS
The highest priority functional group takes the suffix, and the other functional groups are placed earlier in the name.
Aldehyde > Ketone > Alcohol > Alkyne = Alkene
Name Structure Suffix Prefix
Carboxylic Acid (Alkaonic acid) -oic acid
Amide -amide oxo- & amino- (The carbonyl & amine groups are named as if they were separate)
Aldehyde (Alkanal) -al oxo-
Ketone (Alkanone) -one oxo-
Alcohol (Alkanol) -ol hydroxy-
Amine -amine -amino
Alkene -ene -en-
CHEMISTRY MODULE 7: ORGANIC CHEMISTRY PAGE 17 OF 65
STRUCTURE ISOMERS
OUTCOMES ● explore and distinguish the different types of structural isomers, including saturated and unsaturated hydrocarbons, including: (ACSCH035) – chain isomers – position isomers – functional group isomers ISOMERS
Structural isomers are molecules that have the same molecular formula, but their atoms are arranged in different ways, giving rise to different structural formulae.
Although isomers have the same molecular formula, they are different compounds with different chemical and physical properties, as well as different names.
CHAIN ISOMERS
Chain isomers involve rearrangement of the carbons in the backbone, resulting in a different number of carbons in the longest chain or different branching in the carbon chain.
EXAMPLE – HEXANE
An alkane with the molecular formula 퐶6퐻14 has 5 chain isomers. Each isomer has the same molecular formula but has a different arrangement and a different name.
Hexane
2-methylpentane 3-methylpentane
2,2-dimethylbutane 2,3-dimethylbutane
CHEMISTRY MODULE 7: ORGANIC CHEMISTRY PAGE 18 OF 65
POSITION ISOMERS
Position isomers result when molecules have the same carbon chain but the functional group is at a different location.
Position isomers only exist for molecules that contain functional groups, where the chain is long enough for the functional group to occupy different positions. (Eg. Not possible for ethene or ethane).
EXAMPLE – BUTENE
But-1-ene But-2-ene
FUNCTIONAL GROUP ISOMERS
Functional group isomers result when the atoms in the molecules are arrange in different ways that lead to the isomers having different functional groups.
EXAMPLE
Propanoic acid and 1-hydroxypropan-2-one are functional group isomers as both molecules have the formula
퐶3퐻6푂2, but propanoic acid contains a carboxyl functional group while 1-hydroxypropan-2-one contains carbonyl and hydroxyl functional groups.
Propanoic acid 1-hydroxypropan-2-one
CHEMISTRY MODULE 7: ORGANIC CHEMISTRY PAGE 19 OF 65
GEOMETRIC ISOMERS
Two atoms joined by a single bond can rotate freely around the single bond. Geometric isomers can occur when there is restricted rotation somewhere in the molecule. Restricted rotation can occur about a C=C or a ring.
CHEMISTRY MODULE 7: ORGANIC CHEMISTRY PAGE 20 OF 65
HYDROCARBON REACTIONS
OUTCOMES ● 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)
– water (H2O) (ACSCH136) COMBUSTION REACTIONS
All combustion reactions are exothermic. It may be complete or incomplete.
For complete combustion to occur, excess oxygen must be readily available, and the products are always carbon dioxide and water.
25 퐶 퐻 + 푂 → 8퐶푂 + 9퐻 푂 8 18(푙) 2 2(𝑔) 2(𝑔) 2 (푙)
Incomplete combustion occurs when there is insufficient oxygen. The products are water and three different oxidations of carbon, soot, carbon monoxide and carbon dioxide.
Incomplete combustion results in less energy being produce per mole of fuel combustion, making it less efficient. This is due to the reduction of C=O bonds being formed. The formation of C=O releases a large amount of energy.
Alkenes and alkynes tend to burn with a more sooty flame compared to alkanes due to the higher percentage of carbon atoms. Some of the carbon may not combine with oxygen.
- Standard molar heat of combustion is the energy released. Therefore, it is positive. - Standard enthalpy of combustion is always negative as combustion is exothermic. It is the change in enthalpy.
REACTIONS OF UNSATURATED HYDROCARBONS
STABILITY OF CARBON BONDS
The bond energies dictate the overall reactivity of hydrocarbon compounds. Subsequent carbon-carbon bonds in a multiple bond are less stable and weaker than the original single covalent bond.
- This means that double and triple bonds are highly reactive and can break open more easily and allow atoms to join (saturate). - This makes the alkenes and alkynes highly reactive compared to alkanes. - Alkenes and alkynes are able to react with a number of chemical reactions called addition reactions.
CHEMISTRY MODULE 7: ORGANIC CHEMISTRY PAGE 21 OF 65
ADDITION REACTIONS OF ALKENES
HYDROGENATION
Alkenes react with hydrogen gas in the presence of a metal catalyst to form a saturated alkane.
푁𝑖 퐴푙푘푒푛푒 + 퐻2(𝑔) → 푆푎푡푢푟푎푡푒푑 퐴푙푘푎푛푒
HALOGENATION (BROMINATION AND CHLORINATION)
Bromine adds to almost all alkenes very rapidly at RTP to give a compound that has two bromines on adjacent carbons. The reaction occurs spontaneously at RTP.
This is used as in indicator for alkenes and alkynes.
ADDITION OF HX (HYDROHALOGENATION)
Any of the hydrogen halides (HF, HCl, HBr and HI) can add to the double bond of an alkane to give the corresponding alkyl halide. The double bond is converted into a single bond.
CHEMISTRY MODULE 7: ORGANIC CHEMISTRY PAGE 22 OF 65
PREDICTING A MAJOR PRODUCT
When an asymmetric reagent, such as HBr is added to an asymmetric alkene, more of one isomer is produced than the other.
- The predominant isomer is called the major product and the other isomer(s) is called the minor product(s). - In some reactions, only the major product will be formed.
The major product obtained from an addition reaction can be predicted using Markovnikov’s rule.
In addition reactions involving unsymmetrical alkenes, the hydrogen atom will predominantly bond to the carbon atom bearing the greater number of hydrogen atoms.
ADDITION OF WATER (HYDRATION)
Water alone does not react with alkenes, but if an aqueous acid catalyst (eg. Dilute H2SO4) is added and the mixture is heated, water adds to the C=C double bond to give an alkanol.
ADDITION REACTIONS OF ALKYNES
Alkynes undergo addition reactions in a similar fashion to alkenes.
- One of the three bonds in the triple bond is broken, and two new bonds form. The original triple bond is converted into a double bond. - The second addition reaction can be stopped by controlling equivalents of reagent used or by using a specialised reagent.
CHEMISTRY MODULE 7: ORGANIC CHEMISTRY PAGE 23 OF 65
HYDRATION OF ALKYNES
The addition of water to alkynes is different.
- Alkynes do not react with water with only an acid catalyst. A mercury(II) catalyst, such as mercury(II) sulfate, must also be present. - A carbonyl (ketone or aldehyde) is produced instead of an alcohol. The alcohol forms as an intermediate which quickly rearranges to form the carbonyl.
SUBSTITUTION REACTIONS OF ALKANES
A substitution reaction occurs when an atom or functional group in a molecule is replaced or substituted by another atom or group.
Alkanes are far less reactive than alkanes and alkynes as C-C single bonds are relatively strong. However, their hydrogen atoms can be substituted by halogens. These reactions do not occur spontaneously at RTP.
Substitution of alkanes can only be carried out with chlorine and bromine (fluorine reacts too explosively, and iodine does not react) and required energy in the form of ultraviolet (UV) radiation.
Chloromethane is called the substitution product of this reaction.
OUTCOME ● describe the procedures required to safely handle and dispose of organic substances (ACSCH075)
All organic waste should be discarded into appropriately labelled waste containers.
CHEMISTRY MODULE 7: ORGANIC CHEMISTRY PAGE 24 OF 65
PHYSICAL PROPERTIES OF ALCOHOLS
OUTCOMES ● investigate the structural formulae, properties and functional group including: – primary – secondary – tertiary alcohols ● explain the properties within and between the homologous series of alcohols with reference to the intermolecular and intramolecular bonding present INTERMOLECULAR AND INTRAMOLECULAR BONDING IN ALCOHOLS
Alcohols contain a highly electronegative oxygen atom. This creates a polar bond in the molecule.
- The hydrocarbon chain of the alcohol is non-polar (alkyl chain) and can form dispersion forces. It is hydrophobic (repelled from water). - The -OH functional group is polar and can form hydrogen bonds, dipole-dipole and ion-dipole bonds. It is hydrophilic (attracted to water).
As the chain length of an alcohol increases, the strength of the dispersion force increases. However, the extent of hydrogen bonding does not change.
- The strength of the hydrogen bonding depends primarily on the molecule’s shape and the number of hydrogen bond donors and acceptors available.
Thus the length of the hydrocarbon chain influences the properties within the homologous series of alcohols, and the hydroxyl group influences the differences in properties between alcohols and other homologous series.
CHEMISTRY MODULE 7: ORGANIC CHEMISTRY PAGE 25 OF 65
MELTING AND BOILING POINT OF ALCOHOLS
Alkanol Molecular Mass (g/mol) Melting Point (℃) Boiling Point (℃)
Methanol 32.04 -98 65
Ethanol 46.07 -114 78
Propan-1-ol 60.09 -126 97
Butan-1-ol 74.12 -90 118
Pentan-1-ol 88.15 -79 138
Hexan-1-ol 102.17 -52 157
Heptan-1-ol 116.20 -34 176
Octan-1-ol 130.22 -16 195
- ↑ Chain length → ↑Molecular mass → ↑ Strength of dispersion forces → ↑BP and ↑ MP - However, the shape of ethanol and methanol allow them to pack more closely together than propan-1-ol, thus resulting in stronger intermolecular forces and a higher MP.
CHEMISTRY MODULE 7: ORGANIC CHEMISTRY PAGE 26 OF 65
The position of the hydroxyl group within the molecule can also affect the melting and boiling points of alcohols.
For example, three alcohols with the same molecular formula: 퐶4퐻9푂퐻
Type of Alcohol Molecular Model BP (℃) Alcohol
Primary Butan-1-ol 118
Secondary Butan-2-ol 100
Tertiary 2-methylpropan-2-ol 82
Secondary and tertiary alcohols have lower boiling points than primary alcohols with the same number of carbon atoms.
- Since the three alcohols are isomers, their dispersion forces will be the same. - However, the hydrogen bonding is weaker in secondary and tertiary alcohols. The alkyl group adjacent to the OH group hinders the OH groups from getting closer together, restricting their ability to form strong hydrogen bonds. - Hence the lower boiling points arise from weaker hydrogen bonding in secondary and tertiary alcohols.
CHEMISTRY MODULE 7: ORGANIC CHEMISTRY PAGE 27 OF 65
ALKANES VS ALKANOLS
- Hydrogen bonding allows the alkanols to have a higher boiling point than similar molecular mass alkanes.
SOLUBILITY IN WATER
Alkanol Solubility in Water at RTP (g/100mL)
Methanol Miscible
Ethanol Miscible
Propan-1-ol Miscible
Butan-1-ol 7.9
Pentan-1-ol 2.3
Hexan-1-ol 0.59
Small alcohols dissolve well in water. However, solubility in water decreases with increasing carbon chain length.
The solubility of alcohols in water is dictated by size because the opposing effects of the polar and non-polar portions of the molecule.
- The polar hydroxyl group is hydrophilic (“water loving”) and the non-polar hydrocarbon chain is hydrophobic (“water-hating”)
CHEMISTRY MODULE 7: ORGANIC CHEMISTRY PAGE 28 OF 65
Water molecules cannot solvate the large non-polar carbon chains in long alcohols.
- The -OH group can form hydrogen bonds with water, allowing solvation of this portion of the molecule. - However, to solvate the large non-polar carbon chain, many strong hydrogen bonds between water molecules need to be broken. Since the alkyl chain has no strong attraction to water, these cohesive hydrogen bonds cannot be broken. - Thus, the alkyl end remains unsolvated, and solubility in water is drastically decreased.
A very general rule for solubility in water, if it has less than a 4:1 carbon : oxygen ratio, it is soluble.
Small alcohols are good solvents for dissolving both polar and non-polar substances due to the presence of both polar and non-polar areas in the molecule.
- The polar hydroxyl group can form polar interactions such as ion-dipole, hydrogen bonding and dipole- dipole forces with other polar and ionic substances - The non-polar hydrocarbon chain can form dispersion forces with other non-polar substances.
SOLUBILITY IN ORGANIC SOLVENTS
Alcohols become more soluble in non-polar organic solvents such as hexane, benzene and toluene, as the length of the carbon chain increases.
- The large alkyl chain can form strong dispersion forces with other non-polar substances. These are strong enough to disrupt hydrogen bonds holding the alcohol together. - Therefore, the alcohol molecules separate and disperse throughout the solvent.
CHEMISTRY MODULE 7: ORGANIC CHEMISTRY PAGE 29 OF 65
REACTIONS OF ALCOHOLS
OUTCOMES ● write equations, state conditions and predict products to represent the reactions of alcohols, including but not limited to (ACSCH128, ACSCH136): – combustion – dehydration – substitution with HX – oxidation ● investigate the products of oxidation of primary and secondary alcohols
DEHYDRATION
When alcohols are heated with concentrated sulfuric or phosphoric acid as a catalyst, an OH and a H atom on the adjacent carbon will be eliminated from the alcohol to give an alkene and water.
The reactivity and rate of reaction varies depending on the type of alcohol.
- Tertiary alcohols are the most reactive and always react the fastest. Dehydration occurs readily at room temperature. - Primary and secondary alcohols require higher temperatures. Primary alcohols are less reactive and react slower than secondary alcohols.
SUBSTITUTION REACTION OF HALOALKANES
Alcohols undergo substitution in the presence of a hydrogen halide (HCl, HBr, HI) to give the corresponding alkyl halide and water.
CHEMISTRY MODULE 7: ORGANIC CHEMISTRY PAGE 30 OF 65
The trend in reactivity is the same as for dehydration. For primary alcohols, the rate of reaction is also dependent on the hydrogen halide used. HCL is the slowest whilst HI is the fastest.
OXIDATION
- Alcohols can be oxidised with strong oxidising agents, such as acidified solutions of permanganate (MnO4 ) or 2- dichromate (Cr2O7 ) to give carbonyl compounds. A change in oxidation states can be seen for the carbon atoms.
OXIDATION OF PRIMARY ALCOHOLS
Primary alcohols are oxidised into carboxylic acids.
- The oxidation of alcohols is driven by a simultaneous reduction reaction, usually of inorganic reagents. - 2- Either acidified permanganate (MnO4 purple in colour) or dichromate (Cr2O7 orange in colour) can be used as the oxidant. - The oxidation occurs stepwise: the alcohol is first oxidised to the aldehyde, which is then oxidised into the carboxylic acid.
Aldehydes are very reactive and the second oxidation generally occurs too rapidly for it to be separated practically.
- MnO4 : Purple → colourless
2- Cr2O7 : Orange → green
OXIDATION OF SECONDARY ALCOHOLS
Secondary alcohols can be oxidised to produce ketones.
- 2- - Either acidified permanganate (MnO4 purple in colour) or dichromate (Cr2O7 orange in colour) can be used as the oxidant. - Since there are no hydrogen atoms attached to the ketone carbon, no further oxidation can occur. - Secondary alcohols are generally less reactive than primary alcohols, thus require higher temperatures and longer reaction times to be oxidised.
Slower decolourisation of the acidified oxidants.
CHEMISTRY MODULE 7: ORGANIC CHEMISTRY PAGE 31 OF 65
OXIDATION OF TERTIARY ALCOHOLS
Tertiary alcohols cannot be oxidised to give any carbonyl compound. The carbon bearing the hydroxyl group has no hydrogen atoms, thus it cannot be oxidised.
COMBUSTION
Alcohols can readily combust in the same manner as hydrocarbons. They can undergo complete and incomplete combustion.
Alcohols are good fuels as the combustion of alcohol is highly exothermic and since they are already oxygenated, they are more prone to complete combustion.
CHEMISTRY MODULE 7: ORGANIC CHEMISTRY PAGE 32 OF 65
PRODUCTION OF ALCOHOLS
OUTCOMES ● investigate the production of alcohols, including: – substitution reactions of halogenated organic compounds – fermentation PRODUCTION BY ADDITION (HYDRATION)
Water alone does not react with alkenes, but if an aqueous acid catalyst (eg. Dilute H2SO4) is added and the mixture is heated, water adds to the C=C double bond to give an alkanol.
PRODUCTION BY SUBSTITUTION
Alcohols can also be prepared from substitution reactions of haloalkanes.
This reaction occurs by heating the haloalkane with a solution of sodium hydroxide or potassium hydroxide. The hydroxide ion from the aqueous base replaces the halogen atom to generate an alcohol and halide salt.
X: Cl, Br, I. Fluoroalkanes will not react as the C-F bond requires too much energy to break.
This reaction occurs due to the highly polarised carbon-halogen bond, which produces a partial positive charge on the carbon atom.
- The partially positive carbon atom can be easily “attacked” by a negatively charged hydroxide ion. - This results in the formation of a covalent carbon-oxygen bond. In the process, the negative charge is donated to the electronegative halogen atom, which leaves as a halide ion.
The rate of this reaction is dependent on the type of haloalkane and the halogen atom that leaves the molecule.
- Haloalkanes can be categorised as primary, secondary and tertiary. The reactivity is highest for primary, followed by secondary and then tertiary. - This is because the presence of alkyl groups greatly hinders the ability of the hydroxide ion to approach the partially positive carbon and thus slows the reaction.
CHEMISTRY MODULE 7: ORGANIC CHEMISTRY PAGE 33 OF 65
The type of halogen atom leaving the molecule also has an effect on the rate. The reaction occurs the fastest with iodide, followed by bromide and then chloride. This is because the carbon-halogen bonds are of different strengths.
Bond Average Energy (kJ/mol)
C-F 485
C-Cl 338
C-Br 276
C-I 238
- The lower the bond energy, the easier it is to break the bond.
It is also possible for haloalkanes to undergo substitution reactions with water to form alcohols. This reaction occurs much more slowly, and the reactivity is highest for tertiary.
PRODUCTION BY FERMENTATION
Fermentation is a process that involves the conversion of carbohydrates into simple alcohols by the action of enzymes. This is a natural process used by microorganisms to extract energy.
Carbohydrates have the molecular formula of: 퐶푥(퐻2푂)푦. Carbohydrates are abundant in plant material. They are also called saccharides.
- The simplest carbohydrates are monosaccharides. They are the building blocks of more complex carbohydrates, such as disaccharides like sucrose and polysaccharides like cellulose.
The fermentation of monosaccharides, such as glucose and fructose is the simplest form of fermentation which process ethanol and carbon dioxide.
This process relies on the presence of zymase, an enzyme found in yeast.
CONDITIONS FOR FERMENTATION
1) Zymase is present – found in yeast 2) Warm temperatures (30 − 40℃ but depends on the yeast strain) 3) Anaerobic environment (oxygen limited environment) 4) Aqueous solution of sugar
CHEMISTRY MODULE 7: ORGANIC CHEMISTRY PAGE 34 OF 65
- Fermentation of monosaccharides must be catalysed by zymase. Since zymase is a biological catalyst, it is sensitive to temperature. - Yeast can produce ethanol to concentrations up to only about 15% v/v. This is due to ethanol being toxic and around 15% v/v, yeast will start to die. - To produce higher alcohol contents, it is necessary to distil the liquid: - If the aqueous mixture from a fermentation process is subjected to fractional distillation, 95% ethanol can be obtained. - To obtain 100% ethanol, more elaborate procedures such as molecular sieving are required.
CHEMISTRY MODULE 7: ORGANIC CHEMISTRY PAGE 35 OF 65
ENVIRONMENTAL IMPACTS OF HYDROCARBONS
OUTCOMES ● examine the environmental, economic and sociocultural implications of obtaining and using hydrocarbons from the Earth SOURCES OF HYDROCARBONS
The primary source of hydrocarbons is from petroleum. Petroleum is a mixture of hundreds and thousands of different alkanes, ranging from methane up to alkanes with 40 or more carbons.
- The mixture of gases found in petroleum is called natural gas and the mixture of liquid components is called crude oil. - Petroleum is found within pores of rocks deep in the ground.
The complex mixture is separated into fractions according to their boiling points using fractional distillation.
- The petroleum is heated to about 400℃ to produce hot liquid/vapour mixture that enters the fractioning tower. - Inside the tower are horizontal trays, each which contains many bubble caps upon which alkanes condense - Fractions which have lower boiling points will rise higher in the column before condensing. Fraction which have higher boiling points will not rise as high and will condense towards the bottom of the column.
CHEMISTRY MODULE 7: ORGANIC CHEMISTRY PAGE 36 OF 65
Differences in physical and chemical properties of each petroleum fraction mean that they are suitable for different purposes. Generally, light fractions (LPG, petrol, naphtha) are more useful and are in higher demand than heavy fractions (heavy fuel oil, lubricating oil, wax and asphalt).
Some of the longer alkanes are further processed through cracking, which involves heating alkanes to high temperatures in the absence of oxygen. This causes them to split and form shorter, more useful alkanes as well as alkenes.
- A zeolite catalyst, which consists of Al, Si and O, may be employed to allow this reaction to be carried at lower temperatures.
USES OF HYDROCARBONS
The major use of petroleum is transport. Hydrocarbons are excellent fuels and the combustion of hydrocarbons is the primary source of energy production globally.
Unsaturated hydrocarbons are highly reactive and can undergo addition reactions. This makes them extremely important as raw materials for the production of other organic chemicals, such as haloalkanes and alcohols, and commercially valuable goods such as plastic.
THE PROBLEMS WITH USING PETROLEUM
Petroleum deposits in the ground are formed by the burial and decomposition of prehistoric living organisms over millions of years. Thus, petroleum is a finite and non-renewable resource.
As the world’s crude oil diminishes, there will be enormous negative economic and sociocultural consequences. Such as the instability of world markets and increase costs of goods.
Another huge problem that arises is that the combustion of petroleum releases huge amounts of carbon dioxide into the atmosphere.
- Carbon dioxide is a greenhouse gas, so it absorbs infrared radiation from the atmosphere and keeps our Earth warm. - The extra carbon dioxide produces through combustion is a major contributor to the enhanced greenhouse effect which causes global warming. - The consequences for global warming include rising sea levels which in the long term will result in land loss and flooding, more frequent and intense extreme weather events, warming of the oceans and disruptions to the feeding behaviour of wildlife.
The higher concentrations of carbon dioxide has also resulted in the acidifications of oceans which is threatening the survival of aquatic life.
CHEMISTRY MODULE 7: ORGANIC CHEMISTRY PAGE 37 OF 65
BIOFUELS
OUTCOMES ● compare and contrast fuels from organic sources to biofuels, including ethanol ALTERNATIVE FUEL SOURCES
A possible solution to the reliance on non-renewable crude oil is to use biofuels. Biofuels are fuels derived from biomass, which is biological material from living or recently living organisms such as wood, crops, wet waste and animal waste.
BIOETHANOL
Bioethanol is ethanol produced from the fermentation of monosaccharides such as glucose and fructose.
Monosaccharides are the building blocks of more complex carbohydrates which make up plant material. Thus monosaccharides can be sourced from:
- Sucrose based feedstock (eg. Sugarcane and fruits) - Starch based feedstock (eg. Grains like wheat and corn) - Cellulose based feedstock (eg. Wood residues)
Fermentation usually involves mashing up grains, sugarcanes or fruits with water to create an aqueous solution of sugar to which yeast is added.
- Sucrose (퐶11퐻22푂11) and starch are readily hydrolysed into monosaccharides during the fermentation process as yeast contains the necessary enzymes required to catalyse the breakdown.
- Hydrolysis: 퐶11퐻22푂11(푎푞) + 퐻2푂(푙) → 2퐶6퐻12푂6(푎푞) 푦푒푎푠푡 - Fermentation: 퐶6퐻12푂6(푎푞) → 2퐶2퐻5푂퐻(푎푞) + 2퐶푂2(𝑔) - Cellulose is difficult to break down to its component sugars as the enzyme cellulase is not readily available for industrial use.
Bioethanol has been developed as a substitute for petroleum-based ethanol and as an alternative to petrol. It has the potential to be used as standalone fuel to completely replace petrol but is usually used as an additive to petrol.
- In Australia, most cars post-1986 can use up to 10% ethanol (E10 fuel).
BIOGAS
Biogas consists of mixtures of gases, such as methane, carbon monoxide and hydrogen, released from the natural breakdown of organic matter by anaerobic bacteria.
- The organic matter is sourced from natural wastes from agriculture and households, such as manure, human sewage, food processing wastes and crop wastes. - To produce biogas, the waste is placed in a large enclosed tank, called the digester, containing anaerobic bacteria. The gas released from the decay is collected by gas outlets.
The biogas collected can be combusted as a fuel to generate electricity or to heat boilers from industrial processes and for cooking and heating water in homes.
CHEMISTRY MODULE 7: ORGANIC CHEMISTRY PAGE 38 OF 65
The biofuel production represents a carbon cycle, where plants absorb carbon dioxide during growth, recycling the carbon dioxide released during combustion.
The use of biogas also helps reduce the enhanced greenhouse effect as methane is a greenhouse gas that has a larger effect than carbon dioxide. By collecting it and using it to produce electricity, less is released into the atmosphere. Another environmental benefit is that bioethanol and biodiesel are cleaner fuels than petroleum fuels.
- They are oxygenated, so complete combustion is more likely to occur and they do not contain sulfur impurities. - They are non-toxic and biodegradable, thus do not pose as severe a threat to the environment in the event of a spill.
OBSTACLES IN THE PRODUCTION OF BIOFUELS
In Australia, biomass sources currently being used are waste residues.
- Bioethanol is made from sugar cane molasses (waste) and waste from starch and red sorghum production. - Biogas is generated from the treatment of waste water - Biodiesel is produced from waste vegetable oil from restaurants and industrial food producers
However, this only produces a small percentage of Australia’s fuel needs. It is not possible to manufacture enough biofuel from these sources to replace all petroleum fuels used today.
1) Large amounts of fertile land would be required to grow the crops. a. This requires clearing of forests and bushland which will contribute to an increase in carbon dioxide levels in the atmosphere and to habitat loss. Fertile land is also limited thus it competes with food crops, resulting in increase in food prices. 2) Intensive farming can lead to land degradation and erosion. 3) A large amount of fertiliser would be required to replace the nutrients taken from the ground 4) A large amount of water would be required 5) There is also a large amount of energy required for harvesting of crops and processing biofuels, energy that is currently derived from fossil fuels. 6) Large scale commercial agriculture leads to the reduction of biodiversity due to the loss of important organic matter from that crop.
The potential of biofuels lies in making it financially viable compared to conventional petroleum-based fuels.
CHEMISTRY MODULE 7: ORGANIC CHEMISTRY PAGE 39 OF 65
ORGANIC ACIDS AND BASES
OUTCOMES ● investigate the differences between an organic acid and organic base CARBOXYLIC ACID
Organic acids are acids that have a carbon-based structure. The most common type of carboxylic acids are alkanoic acids. They occur abundantly in nature.
- Propanoic acid is found in cheese - Methanoic acid is produce in ant venom - Lactic acid is in milk - Citric acid is in citric fruits
All carboxylic acids are weak acids thus they will partially ionise in water to produce hydrogen or hydronium ions.
- Each carboxylic acid group is monoprotic - The covalent bond between oxygen and hydrogen is highly polar. The bonding electrons are strongly attracted to oxygen so the hydrogen can be drawn easily by a base, leaving behind a negative charge on the oxygen, - If more than one carboxylic acid group is present in the molecule, these acids will be polyprotic. For
example, citric acid is triprotic (퐻3퐶6퐻5푂7).
Different carboxylic acids will have different strengths, ionising to different extents. Their strengths can be compared by comparing their acid dissociation constant Ka.
Acid Formula Ka1 Ka2 Ka3
Methanoic 퐻퐶푂푂퐻 1.8 × 10−4
−5 Ethanoic (acetate) 퐶퐻3퐶푂푂퐻 1.8 × 10
−5 Propanoic 퐶퐻3퐶퐻2퐶푂푂퐻 1.3 × 10
−3 Fluoroethanoic 퐶퐻2퐹퐶푂푂퐻 2.2 × 10
−4 −5 −7 Citric 퐻3퐶6퐻5푂7 7.4 × 10 1.7 × 10 4.1 × 10
CHEMISTRY MODULE 7: ORGANIC CHEMISTRY PAGE 40 OF 65
The strength of a carboxylic acid is influenced by the length and substitution of the hydrocarbon chain.
- As the chain length increases, the strength of the acid decreases. - This is because the alkyl groups are capable of donating electron density. Carbon is more electronegative, so it pulls electron density towards itself and away from hydrogen. The additional electron density allows carbon to “donate” additional negative charge to neighbouring carbon atoms. - This increases the negative charge on the -COO- group, making it more attracted to the H+.
- Substituting a highly electronegative atom, such as a halogen, onto the hydrocarbon chain increases the strength of the acid. The strong electron-withdrawing power of the substituent helps weaken the oxygen- hydrogen bond. This makes it easier for the hydrogen to be dissociated.
- As the number of electronegative atoms increases, so does the strength of the acid. - Fluorine has the biggest electron-withdrawing effect, then chlorine, bromine and iodine.
AMIDES (WEAK B/L BASES)
Organic bases are organic compounds that are characterised by the presence of an atom with a lone pair of electrons that can accept an H+.
- Nitrogen containing compounds such as amines are the most common organic bases.
Many amine bases exist in nature. The most important being the four nitrogenous DNA bases: adenine, cytosine, guanine and thymine, and the 20 natural amino acids used to make proteins in living organisms.
CHEMISTRY MODULE 7: ORGANIC CHEMISTRY PAGE 41 OF 65
Amino acids are molecules that contain both a carboxylic acid and a basic amine group.
Simpler amines are made by substitution reactions of haloalkanes with ammonia.
Amines act as a base in an analogous manner to ammonia. The lone pair of electrons on the nitrogen can accept a proton, forming an ammonium ion. Like ammonia, they are weak bases.
The base dissociation constant Kb can be used to compare the strength of different amine bases.
Base Formula Kb
−5 Ammonia 푁퐻3 1.8 × 10
−4 Methylamine 퐶퐻3푁퐻2 4.4 × 10
−4 Ethylamine 퐶퐻3퐶퐻2푁퐻2 5.4 × 10
−4 Propylamine 퐶퐻3퐶퐻2퐶퐻2푁퐻2 6.9 × 10
- ↑ Chain length → ↑ Partial negative → ↑ H+ acceptor → ↑ Base strength
Alkyl groups are capable of donating electron density. This results in a build-up of partial negative charge on the electronegative nitrogen atom, allowing it to pick up H+ more readily and also stabilise the positive charge on the ammonium ion. The longer the chain, the greater the electron density donate.
CHEMISTRY MODULE 7: ORGANIC CHEMISTRY PAGE 42 OF 65
Amides are also nitrogen containing compounds with a lone pair of electrons.
- Amides are neutral compounds. The presence of the highly electronegative oxygen atom in the C=O group pulls electron density away from the nitrogen atom, which makes it more difficult to accept a H+ and stabilise a positive charge if nitrogen was to accept it.
REACTIONS OF CARBOXYLIC ACIDS AND AMINES
- Carboxylic acid + reactive metal → salt + hydrogen gas - Carboxylic acid + metal hydroxide/oxide → salt + water - Carboxylic acid + metal carbonate/bicarbonate → salt + carbon dioxide + water - Amine + acid → ammonium salt
When carboxylic acids react with amines, the product formed will depend on the conditions. At low temperatures, a proton transfer (neutralisation reaction) proceeds between the carboxylic acid and amine, forming a carboxylate ion (acid) and an alkyl (base).
At higher temperatures or in the presence of a suitable catalyst, an amide is produced with the elimination of water (condemnation reaction), which forms the OH group on the acid and a hydrogen from the amine.
- Condensation reaction: Where two or more molecules combine to form a larger molecule with the simultaneous elimination of a small molecule such as water or methanol.
CHEMISTRY MODULE 7: ORGANIC CHEMISTRY PAGE 43 OF 65
PHYSICAL PROPERTIES OF CARBOXYLIC ACIDS, AMINES AND AMIDES
OUTCOMES ● explain the properties within and between the homologous series of carboxylic acids amines and amides with reference to the intermolecular and intramolecular bonding present ● investigate the structural formulae, properties and functional group including: – aldehydes and ketones (ACSCH127) – amines and amides – carboxylic acids BOILING POINT
Carboxylic acids, amines and amides are all polar molecules. They can form hydrogen bonds with other molecules.
- ↑ Alkyl chain → ↑ MM → ↑ Dispersion forces → ↑ BP
Amides do not exhibit a linear relationship between boiling point and molecular weight. This is because the hydrogen bonding exhibited by the amides is extensive and are more complex.
Amides exhibit the highest boiling points compared to carboxylic acids and amines. Carboxylic acids have higher boiling points than amines.
- Carboxylic acids and amines have less atoms that can form hydrogen bonds. Each carboxylic acid molecule can form two hydrogen bonds with another acid molecule through the double bonded oxygen and OH group, forming a dimer (existing in pairs).
- Each amine molecule can only form one hydrogen bond with another amine molecule.
CHEMISTRY MODULE 7: ORGANIC CHEMISTRY PAGE 44 OF 65
Amines and amides can be classified into primary, secondary and tertiary depending on the number of alkyl groups attached to the nitrogen.
- Primary amines have the highest boiling points, followed by secondary and then tertiary amines. The same trend occurs with amides.
SOLUBILITY
Small amines, amides and carboxylic acids dissolve completely in water. However, solubility decreases as the hydrocarbon chain increases.
- ↑ Carbon chain length → ↑ Non-polar nature of compound → ↑ Dispersion forces domination → Water cannot solvate the long hydrocarbon chain due to cohesive bonds → ↓ Solubility
Similar to alcohols, the trend is reversed for their solubility in organic solvents.
CHEMISTRY MODULE 7: ORGANIC CHEMISTRY PAGE 45 OF 65
ESTERS
OUTCOMES ● investigate the production, in a school laboratory, of simple esters FORMATION OF ESTERS
Esters are organic compounds with the functional group −퐶푂푂 −.
Esters are formed from the reaction of a carboxylic acid and an alcohol. The OH group on the acid is replaced with an OR group from the alcohol. This reaction is called esterification (a condensation reaction).
The reaction is extremely slow. Concentrated H2SO4 (dehydrating agent) is used to remove the water and catalyse the reaction.
[퐻 푂][푒푠푡푒푟] Because it is all liquids, it has a equilibrium constant: 퐾 = 2 푒푞 [푎푙푘푎푛표𝑖푐 푎푐𝑖푑][푎푙푘푎푛표푙]
Esters are everywhere in nature and are used in a number of industrial applications.
- Short chain esters are known for their distinctive, fruit like odours and many occur naturally in fruits and the essential oils for plants. - Due to their pleasant odours, they are commonly used as flavouring (banana lollies) agents in processed foods, as well as fragrances in perfumes and cosmetics. - Fats and oils are also naturally occurring triesters derived from glycerol and fatty acids.
IUPAC NOMENCLATURE OF ESTERS
1) The alkanol is changed to “alkyl” and is the first word of the esters name 2) The alkanoic acid becomes the alkanoate and is the second word of the esters name
CHEMISTRY MODULE 7: ORGANIC CHEMISTRY PAGE 46 OF 65
For example, an ester that smells like banana.
pentanol + ethanoic acid → pentyl ethanoate + water
ESTERIFICATION
To increase rate of reaction:
- ↑ Temperature - ↑ Acid catalyst
To increase yield:
- Remove water
- Concentrated H2SO4 (catalyst and dehydrating agent)
REFLUX
Esterification must be carried out under reflux.
Reflux is a technique that involves heating a reaction mixture in a vessel fitted with a cooling condenser so that the volatile reactants and products are returned to the reaction mixture without any loss.
CHEMISTRY MODULE 7: ORGANIC CHEMISTRY PAGE 47 OF 65
PURPOSE OF REFLUXING
Component Purpose
Reaction flask Contains volatile reactants and products
Condenser Prevents the volatile reactant or product from escaping before the reaction has reached equilibrium by cooling the reactant vapour into a liquid. Water enters at the base and leaves from the top
Boiling chips To provide a surface upon which bubbles form, promoting even boiling
Open top To avoid dangerous pressure build-up inside the apparatus
CHEMISTRY MODULE 7: ORGANIC CHEMISTRY PAGE 48 OF 65
SOAPS
OUTCOMES ● investigate the structure and action of soaps and detergents STRUCTURE OF SOAPS
Soaps are surfactants which, when dissolve in water, help to remove dirt, oil and foreign matter from surfaces.
Soaps are salts of fatty acids
- Fatty acids are carboxylic acids with long hydrocarbon chains (10+) - The salt of a fatty acid consists of a negatively charged carboxylate ion (called the head) with a long hydrocarbon chain (called the tail) and a positively charged ion.
In water the sodium or potassium ions float free and do not play a part in the cleaning actions of soaps. It is the negatively charged fatty acid ion which is responsible for the cleaning action.
- The charged head is hydrophilic (water loving) due to its polar nature - The tail is hydrophobic (water hating) due to its non-polar nature
SAPONIFICATION
Soaps are produced from the hydrolysis of fats and oils (lipids) in a basic solution such as NaOH.
Fats and oils are known as fatty esters, triglycerides or triesters.
CHEMISTRY MODULE 7: ORGANIC CHEMISTRY PAGE 49 OF 65
When esters are heated in the presence of a strong base such as NaOH or KOH, the ester is broken down to give the alcohol and a carboxylate ion.
- This reaction is known as saponification. It is a type of hydrolysis reaction, involving the breaking of a chemical bond by the addition of water
The general reaction for saponification of a triglyceride is:
HOW SOAPS WORK
Soap is a surfactant (surface active agents)
- This means it functions by reducing the surface tension of water and binding to grease and dirt to emulsifies them.
The hydrophobic part of the soap molecule is long, non-polar hydrocarbon chain. It is strongly repelled by water molecules.
When soap molecules are added into water, they form an oriented monolayer (with tails sticking out of the water) at the surface in order to satisfy the interaction of both the hydrophobic and hydrophilic portions of the soap molecule. This effectively breaks the hydrogen bonding between molecules of water and thus reduces the surface tension of water.
The components self-assemble into the most stable arrangement, which consists of spherical structures with the carboxylate groups forming a negatively-charged spherical surface, with the hydrocarbon chains inside the core of the sphere.
CHEMISTRY MODULE 7: ORGANIC CHEMISTRY PAGE 50 OF 65
The spheres are called micelles. The hydrophobic tails are shielded from the water by the polar heads, which minimises the repulsive forces in the system.
CHEMISTRY MODULE 7: ORGANIC CHEMISTRY PAGE 51 OF 65
1. DISSOLUTION
Soap molecules must be first dissolve in water. The hydrophilic head of a soap ion interacts with water molecules via ion-dipole interactions and hydrogen bonding.
2. ADSORPTION
Grease and oil consists of non-polar molecules.
- The hydrophobic tails of the soap ions dissolve in the grease due to dispersion forces and orientates themselves.
Surfactant molecules continue to absorb into the grease, decreasing the surface tension of water at the interface between the grease and water.
The hydrophilic heads interacting with water via ion-dipole forces effectively pull the grease off the surface.
3. EMULSIFICATION
With agitation, the grease layer breaks into smaller, spherical droplets (micelles), with the hydrophilic surfactant head groups interacting with the water via ion-dipole forces, and the hydrophobic surfactant tails adsorbed into the grease. This forms a dispersion of grease droplets in water (an emulsion).
The negative charged heads on the soap repel each other, preventing the grease and dirt from joining together and keeping them dispersed throughout the solution. Therefore, the grease and oil can be simply rinsed away, leaving a clean surface.
CHEMISTRY MODULE 7: ORGANIC CHEMISTRY PAGE 52 OF 65
SYNTHETIC DETERGENTS
OUTCOMES ● investigate the structure and action of soaps and detergents PROBLEMS WITH SOAP
- The soap ion must be dissolved in water to have a cleaning effect - As salts of weak acids, in low pH, soaps are converted into uncharged fatty acids
− + 퐶퐻3(퐶퐻2)16퐶푂푂 푁푎 (푎푞) + 퐻퐶푙(푎푞) → 퐶퐻3(퐶퐻2)16퐶푂푂퐻 (푠) + 푁푎퐶푙(푎푞)
Soluble fatty acid ion + acid → Insoluble fatty acid molecule + ionic salt
- Soaps can also form insoluble salts (scum) in hard water (water with high concentration of divalent metal ions such as Ca2+ and Mg2+)
− + 2+ − 2+ + 2퐶퐻3(퐶퐻2)16퐶푂푂 푁푎 (푎푞) + 푀푔 (푎푞) → [퐶퐻3(퐶퐻2)16퐶푂푂 ]2푀푔 (푠) + 2푁푎 (푎푞)
Soluble fatty acid ion + divalent metal ion → insoluble salt + sodium or potassium ion
SYNTHETIC DETERGENTS
- All such synthetic surfactants are known as detergents - They have a similar structure to soaps, with a long non-polar hydrocarbon tail and a polar head. However, they vary in the structure of the polar head group.
ANIONIC DETERGENTS
All anionic detergents have a negatively charged polar head.
CHEMISTRY MODULE 7: ORGANIC CHEMISTRY PAGE 53 OF 65
− The main structural difference between anionic detergents and soap is the presence of a sulfate (푅 − 푆푂2 − 푂 ) or − − sulfonate group (푅 − 푂 − 푆푂2 − 푂 ) instead of a carboxylate (푅 − 퐶푂푂 ).
EFFECTIVENESS IN HARD OR ACIDIC WATER
Anionic detergents don’t form insoluble precipitates. They form salts that are all soluble. However, the effectiveness is still reduced in hard and acidic water. The positive ions in hard water/acidic water will attract the negative head of the detergent.
Phosphate is an example of a builder that can be added to soften the water.
CATIONIC DETERGENTS
All cationic detergents have a positively charged polar head. The positive head is usually a quaternary ammonium ion. They are also known as fatty amine salts.
Generally cationic detergents are not very good cleaning agents due to the strong attraction of cationic detergents to negatively charged surfaces. (Most surfaces are negatively charged).
This attraction however can be beneficial in certain situations.
- Many fabrics acquired acquire a negative charge when they become wet. - Cationic detergents are used in fabric softeners and hair conditioners.
The strong attraction of cationic detergents to negatively charged surfaces can be detrimental in other situations.
- Cationic detergents are particularly toxic to microorganisms. They are attracted to the negative surface of bacteria and damage or kill bacteria that are involved in their decomposition. They therefore have very low biodegradability.
CHEMISTRY MODULE 7: ORGANIC CHEMISTRY PAGE 54 OF 65
NON-IONIC DETERGENTS
All non-ionic detergents have uncharged polar heads. The polar heads consists of polar groups such as ethoxylates. Although these surfactants are uncharged, the polar head groups are still attracted to the highly polar water molecules forming numerous hydrogen bonds.
Non-ionic detergents don’t foam as much as other detergents. Hence, they can be used in dish washers.
ENVIRONMENTAL IMPACTS OF SURFACTANTS
Detergents are synthetic, whereas soaps are made from naturally occurring biological materials (fats and oils). Thus detergents are less biodegradable.
- The enhanced stability of detergents means they persist in the environment, causing damage to the mucus membrane in wildlife and resulting in excessive frothing in the water ways. This leads to less sunlight penetration. - Toxic to aquatic life.
Anionic detergents often contain builders such as sodium triphosphate (푁푎5푃3푂10).
- The builders react with minerals in hard water and form soluble molecules. - High levels of phosphates entering the rivers and waterways can lead to eutrophication (turning a lake into swamp, algae etc)
CHEMISTRY MODULE 7: ORGANIC CHEMISTRY PAGE 55 OF 65
REACTION PATHWAYS
OUTCOMES ● draft and construct flow charts to show reaction pathways for chemical synthesis, including those that involve more than one step FLOWCHART OF REACTIONS
[Deleted sorry. Mod 8 might have some flowcharts for rxns]
CHEMISTRY MODULE 7: ORGANIC CHEMISTRY PAGE 56 OF 65
POLYMERS
OUTCOMES ● model and compare the structure, properties and uses of addition polymers of ethylene and related monomers, for example: – polyethylene (PE) – polyvinyl chloride (PVC) – polystyrene (PS) – polytetrafluoroethylene (PTFE) (ACSCH136) ● model and compare the structure, properties and uses of condensation polymers, for example: – nylon – polyesters POLYMER
A polymer is a long chain molecule made up of repeating units, called monomers joined by covalent bonds. The process of linking monomer units is called polymerisation.
Polymers may be natural or synthetic:
- Natural polymers are made by living organisms. Eg. Hair, starch, cellulose, DNA and silk - Synthetic polymers are manufactured. Eg. Plastics like polyethylene, polyvinyl chloride and nylon.
ADDITIONAL POLYMERS
Addition polymers are polymers made by adding unsaturated molecules to each other, without the elimination of any atoms.
- Additional polymerisation is a type of addition reaction, in which one of the bonds in the C=C double bond is broken to form two new single bonds.
A simple way of representing polymers is by writing the repeat units in square brackets followed by the subscript n where n is the number of monomer units in the polymer.
Addition polymers are all synthetic, they do not exist in nature.
CONDENSATION POLYMERS
Condensation polymers are polymers formed through the condensation reaction of difunctional monomers with the elimination of a small molecule such as water or methanol in the process.
CHEMISTRY MODULE 7: ORGANIC CHEMISTRY PAGE 57 OF 65
- Monomers used to synthesise condensation polymers usually contain groups such as alcohol, carboxylic and amine.
Condensation polymers are found everywhere in nature.
- Polysaccharides such as cellulose and starch are condensation polymers made from glucose monomers.
POLYESTERS
Polyesters are condensation polymers in which the repeating units are joined by ester links.
Polymers made from two different monomers are called copolymers.
CHEMISTRY MODULE 7: ORGANIC CHEMISTRY PAGE 58 OF 65
POLYAMIDES
Polyamides are condensation polymers in which the repeating units are joined by amide links.
The nylon class refers to polyamides that have linear carbon chains in the repeating units.
PROPERTIES OF POLYMERS
Physical properties of polymers are important in determining their uses. These include:
- Melting point (softening point) - Mechanical strength - Flexibility.
This is due to the chemical structure of the polymers which leads to different strengths of intermolecular forces that is dependent on:
- Molecular weight or chain length - Extent of chain branching - Presence of side groups (eg. -OH)
CHAIN LENGTH
Polymers are extremely large covalent molecules. The dominant intermolecular force is dispersion.
The length of a polymer (and its molecular weight) depends on the number of monomers the polymer contains. The melting point, rigidity and hardness of a polymer increases with an increase in chain length.
CHAIN BRANCHING
Polymers are able to form branched and unbranched chains. Unbranched chains are able to pack more closely in an orderly fashion, forming a rigid crystalline solid.
CHEMISTRY MODULE 7: ORGANIC CHEMISTRY PAGE 59 OF 65
Branched polymer chains are unable to align with each other, forming an amorphous solid that has weak intermolecular forces between the chain.
CRYSTALLINITY AND AMORPHOUS
If the polymer chains have few branches, as in the case with HDPE, the molecules can sometimes line up in a regular arrangement, creating crystalline regions. The regular arrangements brings the polymer chains closer together. The IMF between closely packed chains are stronger, and the presence of crystalline regions strengthens the material overall.
Crystalline regions in a polymer prevents the transmission of light through the material, making it appear cloudy or opaque.
Amorphous region will form where the polymer chains are randomly tangled and unable to pack very closely. In some polymer materials, the entire solid is amorphous. Amorphous polymers are usually more flexible and weaker and are often transparent. (LDPE) Increasing the percentage crystallinity of a material makes it stronger and less flexible. This also makes the material less transparent because crystalline regions scatter light. There are more crystalline regions in unbranched polymers
There are crystalline and amorphous regions in all polymers.
SIDE GROUPS
Side groups can be introduced to make a material more rigid and brittle. This results in a harder polymer.
CROSS-LINKING
Polymer chains are held together by intermolecular forces or they can be linked by covalent bonds called cross-links to form a large extended network.
CHEMISTRY MODULE 7: ORGANIC CHEMISTRY PAGE 60 OF 65
Polymers with only intermolecular forces between their chains are called thermoplastic polymers. They soften when heated as the intermolecular forces are relatively weak and easily broken to allow the chains to move between one another. This property allows polymers to be remoulded
Polymers with cross-links are called thermosetting polymers. Since covalent bonds are very string, cross links limit movement between polymer chains, making the polymer more rigid, hard and heat resistant. These polymers cannot be remoulded (like XLDPE).
CHEMISTRY MODULE 7: ORGANIC CHEMISTRY PAGE 61 OF 65
TYPES OF POLYMERS
POLYETHYLENE (PE)
Polyethylene (polyethene) is the most popular plastic in the world. It has a very simple structure.
Ethene is an unsaturated molecule because a double carbon-carbon bond. When ethene polymerases, the double bond breaks and new covalent bond are formed between carbon atoms on nearby monomers. The polyethene formed does not contain any double bonds.
Ethene is one of the most simple and versatile monomers. It is easily able to undergo addition polymerisation
LOW DENSITY POLYETHYLENE (LDPE) (4)
LDPE is produced under high temperatures and pressures. Under these harsh conditions, the polymer is formed too rapidly for the molecules to be neat and symmetrical. The products usually contain too many small chains (branches) that divide off the main polymer.
The molecules in the polymer cannot pack closely together thus reducing the dispersion forces.
The arrangement of the polymer molecules can be described as disordered or non-crystalline.
- Flexible, chemically inert, good elongation - Low melting point/thermoplastic - Lightweight, good puncture resistance - Waterproof
Used in: Milk carton lining, bowls, flexible water pipes, bottles
HIGH DENSITY POLYETHYLENE (HDPE) (2)
Highly specialised transition metal catalyst, known as Ziegler-Natta catalysts are used to avoid the need for high pressure. Due to the polymer being produced under a lower pressure, the conditions are milder and there are fewer branches.
The lack of branches allows the molecules to pack together tightly increasing the density and the hardness of the polymer formed. The arrangement of the polymer molecules is more ordered, resulting in crystalline sections.
Used in: Food packaging, dustbins, crates, drums, water pipes
CHEMISTRY MODULE 7: ORGANIC CHEMISTRY PAGE 62 OF 65
POLYVINYL CHLORIDE (PVC)
PVC (polychloroethene) is made out of vinyl chloride monomers. The chlorine atoms introduce dipoles into the long molecules. This increases the IMF between molecules, which leads to a higher melting point. A PVC item burning in a flame will not continue to burn when it is removed from the flame. It is used in products such as conveyor belts, cordial bottles, water pipes and the covering of electrical wires.
Pure PVC is very hard and brittle. Additives are incorporated into PVC to improve its flexibility, thermal stability and UV stability. In a fire, PVC decomposes to form toxic and corrosive hydrogen chloride.
POLYSTYRENE (PS)
First commercial production by IG Farben; used in disposable household products, plastic model kits, laboratory containers, insulation and packaging.
Benzene rings are covalently bonded to every second carbon atom in the polymer chain. This causes polystyrene to be a hard but quite brittle plastic with a low density. It is used to make food containers, picnic sets, refrigerator parts, and CD and DVD cases. Polystyrene is made from styrene (ethylbenzene monomers).
Polystyrene (polyethylbenzene) is commonly manufactured as a foam. Foamed polymers are formed by blowing a gas through melted polymer materials. Foaming can drastically change the physical properties of a polymer material. 95-98% air
Polystyrene foam is produced by introducing pentane into melted polystyrene beads. The beads swell up to produce the lightweight, insulating, shock-absorbing foam that is commonly used for takeaway hot drink containers, bean bag beans, packaging materials and safety helmet linings. Once polystyrene has been converted to a foam, it is difficult to recycle.
CHEMISTRY MODULE 7: ORGANIC CHEMISTRY PAGE 63 OF 65
POLYTETRAFLUOROETHYLENE (PTFE)
Polytetrafluoroethene is used in cookware fabrics, wiper blades, nail polish, industrial coatings. (aka Teflon, or Fluon). Made out of tetrafluoroethene monomers.
Tetrafluoroethene is formed when all the hydrogen atoms in ethene are replaced by highly electronegative fluorine atoms.
It has quite exceptional properties that are very different from those of polyethene. It can be used to make non-stick frying pans, medical implants, gears and clothing. The electronegative fluorine atoms reduce the strength of intermolecular bonds with other substances.
- Non-stick - Heat resistant - Chemical resistant - Good mechanical properties - Low friction coefficient - Flame resistant - High melting point POLYETHYLENE TEREPHTHALATE (PET)
Polyesters are a class of polymers that are formed through the process of condensation polymerisation. Polyesters are formed by combining monomers that contain carboxylic acid and hydroxyl functional groups. They are typically formed by reacting a dicarboxylic acid monomer with a diol monomer.
This is the most often polymer used to make polyester fabric. PET is synthesised by reacting benzene-1,4-dioic acid monomers with ethane-1,2-diol monomers.
PET has a range of uses including recyclable drink bottles and food packaging. PET is a strong material because the ester groups are polar, so that there are dipole-dipole attractions between the polymer chains. Benzene rings make it stiff and strong, resistant to deformation.
CHEMISTRY MODULE 7: ORGANIC CHEMISTRY PAGE 64 OF 65
NYLON 6,6
Nylon stockings created shopping frenzy in the USA in the 1940s. It is also used in clothing, parachutes, kitchen utensils, toothbrushes, fishing lines, guitar strings, seatbelts.
Nylon is formed when a monomer containing an anime group on each end reacts with a monomer with a carboxyl group on each end, a polyamide can form.
The term ‘nylon’ refers to the group of polyamides, in which the monomers are linear carbon chains. A common example is nylon-6,6 which is named so because the dicarboxylic acid monomer has a chain of 6 carbons and the diamine monomer also has a chain of 6 carbon atoms.
Nylon can be easily drawn into fibres that have high tensile strength. These fibres are used to produce strong lightweight material for clothes.
CHEMISTRY MODULE 7: ORGANIC CHEMISTRY PAGE 65 OF 65
CHEMISTRY MODULE 8
Applying Chemical Ideas
CHEMISTRY MODULE 8 – APPLYING CHEMICAL IDEAS
ANALYSIS OF INORGANIC SUBSTANCES
Inquiry question: How are the ions present in the environment identified and measured?
● analyse the need for monitoring the environment ● conduct qualitative investigations – using flame tests, precipitation and complexation reactions as appropriate – to test for the presence in aqueous solution of the following ions: – cations: barium (Ba2+), calcium (Ca2+), magnesium (Mg2+), lead(II) (Pb2+), silver ion (Ag+), copper(II) (Cu2+), iron(II) (Fe2+), iron(III) (Fe3+) – – – – – 2– – anions: chloride (Cl ), bromide (Br ), iodide (I ), hydroxide (OH ), acetate (CH3COO ), carbonate (CO3 ), 2– 3– sulfate (SO4 ), phosphate (PO4 ) ● conduct investigations and/or process data involving: – gravimetric analysis – precipitation titrations ● conduct investigations and/or process data to determine the concentration of coloured species and/or metal ions in aqueous solution, including but not limited to, the use of: – colourimetry – ultraviolet-visible spectrophotometry – atomic absorption spectroscopy
ANALYSIS OF ORGANIC SUBSTANCES
Inquiry question: How is information about the reactivity and structure of organic compounds obtained?
● conduct qualitative investigations to test for the presence in organic molecules of the following functional groups: – carbon–carbon double bonds – hydroxyl groups – carboxylic acids (ACSCH130) ● investigate the processes used to analyse the structure of simple organic compounds addressed in the course, including but not limited to: – proton and carbon-13 NMR – mass spectroscopy (ACSCH19) – infrared spectroscopy (ACSCH130)
CHEMICAL SYNTHESIS AND DESIGN
Inquiry question: What are the implications for society of chemical synthesis and design? ● evaluate the factors that need to be considered when designing a chemical synthesis process, including but not limited to: – availability of reagents – reaction conditions (ACSCH133) – yield and purity (ACSCH134) – industrial uses (eg pharmaceutical, cosmetics, cleaning products, fuels) (ACSCH131) environmental, social and economic issues
CHEMISTRY MODULE 8: APPLYING CHEMICAL IDEAS PAGE 1 OF 48
ION IDENTIFICATION
OUTCOMES ● analyse the need for monitoring the environment
CHEMICAL MONITORING AND MANAGEMENT
Chemical monitoring and management are needed to ensure the health and wellbeing of both plants and animals. Monitoring also allows for more effective management.
For example, the destruction of UV-protective ozone, caused by widespread use of chlorofluorocarbons (CFCs)
- Hazardous ions in the environment such as mercury have also been identified and monitored using chemical techniques. - Nutrients required by humans and other living organisms have also been discovered through environmental monitoring.
OUTCOMES ● conduct qualitative investigations – using flame tests, precipitation and complexation reactions as appropriate – to test for the presence in aqueous solution of the following ions: – cations: barium (Ba2+), calcium (Ca2+), magnesium (Mg2+), lead(II) (Pb2+), silver ion (Ag+), copper(II) (Cu2+), iron(II) (Fe2+), iron(III) (Fe3+) – – – – – 2– – anions: chloride (Cl ), bromide (Br ), iodide (I ), hydroxide (OH ), acetate (CH3COO ), carbonate (CO3 ), 2– 3– sulfate (SO4 ), phosphate (PO4 ) QUALITATIVE ION ANALYSIS
- Precipitate: Insoluble ionic salt formed when 2 ionic solutions are mixed together. - Qualitative analysis: Identification of the constituents of components in a sample. - Quantitative analysis: Measurement/determination of the amount (expressed in a concentration) of the constituent. - Destructive testing: Test that irreversibly alter the composition of a sample.
CHEMISTRY MODULE 8: APPLYING CHEMICAL IDEAS PAGE 2 OF 48
PRECIPITATION REACTIONS
SOLUBILITY RULES
Precipitation reactions are useful for the rapid identification of ions.
- Solutions of specific ions are added to the unknown aqueous sample. - The formation or absence of a precipitate can be used to determine the ion present.
Rule Exceptions
+ + + + + + All Group 1 (Li , Na , K , Rb , Cs ) and ammonium (NH4 ) are NONE soluble
- - All nitrate (NO3 ) and acetate (CH3COO ) salts are soluble NONE
All chloride (Cl-) and bromide (Br-) salts are soluble Ag+, Pb+ (Hg+)
All iodide (I-) salts are soluble Ag+, Pb+, Hg+
2- + 2+ 2+ 2+ 2+ + All sulfate (SO4 ) salts are soluble Ag , Pb , Ba , Ca , Sr , Hg
- + 2+ 2+ All hydroxide (OH ) salts are insoluble Group 1 and NH4 salts are soluble. Ba , Ca are slightly
2- 3- + All carbonate (CO3 ) and phosphate (PO4 ) are insoluble Group 1 and NH4 salts are soluble
2− 2− −2 − − − − − Solution 푃푂4 푆푂4 퐶푂3 푂퐻 퐶푙 퐵푟 퐼 퐶퐻3퐶푂푂 퐵푎2+ White insoluble White White White White White soluble insoluble soluble soluble soluble 퐶푎2+ Colourless White insoluble White White White White White soluble insoluble soluble soluble soluble 푀푔2+ White White White White White White White White soluble insoluble soluble insoluble insoluble soluble soluble soluble 푃푏2+ White insoluble White White White Bright White soluble insoluble insoluble insoluble Yellow 퐶푢2+ Blue Blue Blue soluble Green Blue Blue Blue Not stable Blue soluble insoluble insoluble insoluble soluble soluble 퐹푒2+ Pale green Green Green Grey Green Green Green Green Green soluble insoluble soluble insoluble insoluble soluble soluble soluble 퐹푒3+ Yellow Pink Yellow Not Brown Brown Brown Not stable Brown soluble insoluble soluble stable insoluble soluble soluble 퐴푔+ Colourless Yellow White Yellow Not White Pale yellow insoluble White soluble insoluble insoluble insoluble stable insoluble
+ - 퐴푔 halides precipitates decompose with exposure to light to form 퐴푔(푠). This is observed as darkening of the solid.
CHEMISTRY MODULE 8: APPLYING CHEMICAL IDEAS PAGE 3 OF 48
COMPLEXATION REACTIONS
Complexation reactions are commonly used to identify transition metal cations (due to their d orbitals).
- This includes 퐴푔+, 퐶푢2+, 퐹푒2+, 퐹푒3+ and 푃푏2+
A complexation reaction occurs when a coordination compound or complex forms.
- These consist of a central metal ion (typically a transition metal) coordinated to molecules or ions (polar & lone pair of electron molecules). - The coordinating species are called ligands. They are typically neutrally or negatively charged. (E.g. − − 퐻2푂, 푁퐻3, 퐶푙 , 퐶푁 )
Coordination complexes are usually coloured and are indicated by the formula being enclosed in square 2+ 2− brackets. E.g. [퐶푢(푂퐻2)6] and [퐶푢퐶푙4] .
COMPLEXATION REACTIONS AND ION IDENTIFICATION
HYDRATED TRANSITIONAL METAL ION
When transition metal ions dissolve in water, a complexation reaction occurs:
2+ − 퐹푒퐶푙2(푎푞) + 6퐻2푂(푙) → [퐹푒(푂퐻2)6] (푎푞) + 2퐶푙 (푎푞)
- Water molecules (ligands) coordinate to the transition metal centre - Charge stays the same as the original cation - Hydrated transition metal centres often have characteristic colours that can be used to identify them.
IDENTIFICATION WITH COMPLEXATION
Complexation reactions can be also used in conjunction with precipitation reactions to identify ions.
Sliver halides can be distinguished due to their different complexation reactions with ammonia:
+ − - 퐴푔퐶푙 is soluble in dilute ammonia: 퐴푔퐶푙(푠) + 2푁퐻3(푎푞) ⇌ [퐴푔(푁퐻3)2] (푎푞) + 퐶푙 (푎푞) + − - 퐴푔퐵푟 is soluble in concentrated ammonia: 퐴푔퐵푟(푠) + 2푁퐻3(푎푞) ⇌ [퐴푔(푁퐻3)2] (푎푞) + 퐵푟 (푎푞) - 퐴푔퐼 is insoluble in any concentration of ammonia
+ 2+ This test can be used to distinguish between Ag and Pb . (푃푏퐶푙2 does not dissolve in ammonia)
CHEMISTRY MODULE 8: APPLYING CHEMICAL IDEAS PAGE 4 OF 48
Complexation reactions can also be used as confirmatory tests:
- Iron (III) ions react with thiocyanate (SCN-) to produce blood red [Fe(SCN)]2+
- Copper (II) ions react with ammonia to form pale blue Cu(OH)2(s) which then dissolves to form deep blue 2+ [Cu(NH3)4] - + − o Ammonia initially reacts with water to produce the OH : 푁퐻3(푎푞) + 퐻2푂(푙) → 푁퐻4 (푎푞) + 푂퐻 (푎푞)
FLAME TESTING
A flame test is used to qualitatively identify some metal ions in a sample. It can be used to distinguish between cations that undergo similar precipitation reactions. For example, Ba2+ and Ca2+ or Group 1 metals. However, not all metals produce a diagnostic flame colour.
ATOMIC EMISSION
The flame colour arises from atomic emissions of the metal.
- The heat of the flame excites electrons into higher energy levels - Atomic emission occurs when an electron in a higher energy level returns to a lower energy level (relaxation) - During relaxation, a wavelength of light matching the energy difference between the levels is emitted.
Wavelength is inversely proportional to energy:
ℎ푐 퐸 = 휆
- The greater difference → shorter 휆
The colour of the flame produced in the flame test results from a combination of the most intense wavelengths emitted in the visible regions.
Metals that do not produce a characteristic flame colour typically have atomic emissions that fall outside the visible region.
CHEMISTRY MODULE 8: APPLYING CHEMICAL IDEAS PAGE 5 OF 48
CHARACTERISTIC FLAME COLOUR
Metal Ion Colour
퐵푎2+ Yellow-green (apple green)
퐶푎2+ Orange-red (brick red)
푀푔2+ No characteristic flame colour
푃푏2+ Grey-blue
퐶푢2+ Green (halide = blue green)
퐹푒2+/퐹푒3+ Orange-brown
퐴푔+ No characteristic flame colour
LIMITATIONS OF THE FLAME TEST
- Flame tests are only useful for some metal cations. - Destructive - Can be ambiguous which metal ion is present o The flame colours may be similar (e.g. Rb and Cs produce the same colour as K) o Different oxidation states of the same metal cannot be distinguished (e.g. Fe3+ and Fe2+) - Inconclusive results can be caused by interference created by other metal ions present in the sample. (e.g. Ba flame masks Cu, Li and Sr) o Interference can sometimes be removed by using coloured glass filters. For example, blue cobalt glass can be used to filter out the bright yellow light produced by Na, a common impurity.
ADDITIONAL TESTS FOR ION IDENTIFICATION
SOLUTION PH
Conjugates of strong acids/bases will dissolve to produce a solution with neutral pH. Conjugate bases of weak acids/bases will produce a non-neutral pH.
Basic Ion (pH > 7) Neutral Ion (pH = 7)
Hydroxide (OH-) Chloride (Cl-)
- - Acetate (CH3COO ) Bromide (Br )
2- - Carbonate (CO3 ) Iodide (I )
3- 2- Phosphate (PO4 ) Sulfate (SO4 )
ACID CARBONATE REACTION
2- Carbonate anions (CO3 ) can be identified through the addition of an acid. Bubbles will be observed. A lime water test can be used to confirm the presence of CO2(g).
+ 2− 2퐻 (푎푞) + 퐶푂3 (푎푞) → 퐻2푂(푙) + 퐶푂2(푔)
CHEMISTRY MODULE 8: APPLYING CHEMICAL IDEAS PAGE 6 OF 48
FLOWCHARTS
- To create a flowchart effectively, go from least soluble to most soluble. - It is important to include ‘in excess’ for flowcharts with more than 1 ion in a sample. o Include filtering out precipitates as well.
CATIONS
CHEMISTRY MODULE 8: APPLYING CHEMICAL IDEAS PAGE 7 OF 48
ANION
CHEMISTRY MODULE 8: APPLYING CHEMICAL IDEAS PAGE 8 OF 48
GRAVIMETRIC ANALYSIS
OUTCOMES ● conduct investigations and/or process data involving: – gravimetric analysis WHAT IS GRAVIMETRIC ANALYSIS
Gravimetric analysis uses the mass of product (precipitate) formed through a precipitate reaction to quantify a target ion. It is used in applications such as monitoring water quality, determining the mineral composition of soil and product testing.
Gravimetric analysis is typically performed with the following steps:
1) A known quantity of sample is dissolved in water 2) The target ion is precipitated from the solution using a suitable reagent 3) The precipitate is collected by filtration, then washed, dried and weighed
The reagent used in the precipitation reaction must react selectively with the target ion and form a highly insoluble precipitate in order to give accurate results. The selective reagent can be chosen using the solubility rules, while the solubility of the resulting precipitate can be found by examining the solubility product constant (Ksp).
Undesired precipitation reactions can be prevented through experimental design.
The amount of the target ion present is usually expressed as a mass percentage of the original sample.
GRAVIMETRIC ANALYSIS ASSUMPTIONS (VALIDITY)
- Only the target ion precipitates out o Selective reagent o Remove or prevent the precipitation of other salts - All the target ion precipitates out
o Selective reagent (Highly insoluble precipitate, really small Ksp) o Added in excess
CHEMISTRY MODULE 8: APPLYING CHEMICAL IDEAS PAGE 9 OF 48
- All of the precipitate is collected and in a pure form o Filtration (quantitative filter paper) o Wash precipitate with warm distilled water multiple times o Ensure it is dry when measuring its mass
SULFATE CONTENT OF FERTERLISER
Fertilisers (a solid mixture) commonly contain ammonium sulfate (NH4)2SO4, as a sulfur source for plants. The sulfate 2+ content of fertiliser can be determined gravimetrically through the addition of Ba to form a highly insoluble BaSO4.
2- - 3- - Barium can precipitate with CO3 , Cl , PO4 and OH anions as well. To precipitate out:
2- - CO3 : Add acid in excess 3 - - PO4 and OH : Acidify solution (Don’t use H2SO4) - - Cl : Add AgNO3 in excess
If other anions precipitate with Ba, the mass of the residue collected will increase. Therefore, a higher w/w % will be calculated than actually present.
EXAMPLE QUESTION 1
The following results were obtained for the gravimetric analysis of sulfate ions in several fertiliser samples
Sample A B
Mass of fertiliser sample used (g) 13.5 1.5
Mass of BaSO4 collected (g) 1.642 0.0870
There are two methods:
A B 푚(퐵푎푆푂4) = 1.642 푔 푚(퐵푎푆푂4) = 0.0870 푔 푛(퐵푎푆푂4) = 0.00703663719 푚표푙 푚푚(푆푂4) 96.07 % 표푓 푆푂4 𝑖푛 퐵푎푆푂4 = = = 41.17% 푛(푆푂4) = 푛(퐵푎푆푂4) = 0.00703663719 푚표푙 푚푚(퐵푎푆푂4) 233.37 푚(푆푂4) = 0.679520932 푔 푚(푆푂4) = 0.0870 × 41.17% = 0.035814768 푔 푚(푆푂4) 푚(푆푂4) % 표푓 푆푂4 𝑖푛 푓푒푟푡𝑖푙𝑖푠푒푟 = = 5.01% % 표푓 푆푂 𝑖푛 푓푒푟푡𝑖푙𝑖푠푒푟 = = 2.4% 13.5 4 1.5
PROS AND CONS OF GRAVIMETRIC ANALYSIS
ADVANTAGES
- It is very accurate, if the gravimetric analysis is carefully designed (assumptions) and conducted. - It does not require speciality equipment to perform. - It does not require the construction of a calibration curve, which can be time consuming
DISADVANTAGES
- The target ion of interest must be able to form a sufficiently insoluble precipitate for accurate results (A limitation of the technique)
CHEMISTRY MODULE 8: APPLYING CHEMICAL IDEAS PAGE 10 OF 48
- The precipitate must be highly pure o Impurities can be adsorbed to the surface of the precipitate or trapped within the precipitate, resulting in increased mass and inaccurate results - It is extremely time consuming compared to modern analytical techniques - Concentration of the target ion must be high enough to produce a precipitate that can be accurately weighed.
PRECIPITATION TITRATIONS
OUTCOMES ● conduct investigations and/or process data involving: – gravimetric analysis – precipitation titrations TITRATION
Precipitation titration are commonly used in the quantification of halides (Cl-, Br-, I-). In these procedures, silver ions are usually the titrant and consequently they are known as argentometric titrations.
The main three titrations used are:
- Mohr’s method - Volhard’s method - Fajan’s method
MOHR’S METHOD (DIRECTION TITRATION)
Mohr’s method quantifies chloride, bromide and cyanide ions using direct titration with a standard silver nitration solution.
+ − 퐴푔 (푎푞) + 푋 (푎푞) → 퐴푔푋(푠)
Yellow potassium chromate (퐾2퐶푟푂4) is used as the indictor:
- Silver preferentially precipitates with halide ions - When excess silver is added, it will react with chromate ions to produce a reddish-brown precipitate
퐴푔2퐶푟푂4(푠)
+ 2− 2퐴푔 (푎푞) + 퐶푟푂4 (푎푞) ⇌ 퐴푔2퐶푟푂4(푠)
LIMITATIONS OF MOHR’S METHOD
The presence of multiple anions in a sample can lead to invalid results. Carbonate and phosphate ions will precipitate with silver.
+ 2− 2퐴푔 (푎푞) + 퐶푂3(푎푞) → 퐴푔2퐶푂3(푠)
+ 3− 3퐴푔 (푎푞) + 푃푂4(푎푞) → 퐴푔3푃푂4(푠)
CHEMISTRY MODULE 8: APPLYING CHEMICAL IDEAS PAGE 11 OF 48
The conical flask must be thoroughly stirred during the titration to ensure the Ag+ ions are evenly distributed, so
AgCrO4 will not precipitate early.
The titration must be conducted at a pH 6.5-9.0
+ - - At pH < 6.5, the chromate ion (a weak base) accepts H to become HCrO4 , so there will not be enough - chromate in solution to produce the end point colour change. HCrO4 is also bright orange, which obscures the endpoint. - At pH > 9.0, the concentration of OH- is too high. It will form a precipitate with Ag to produce AgOH which is a brown precipitate. It also obscures the endpoint and increases the silver needed to reach it.
Mohr’s method is suitable for the analysis of chloride and bromide ions, but not iodide or thiocyanate ions. This is because the chromate ions are strongly adsorbed to the surface of AgI and AgSCN precipitates, making the endpoint indistinct.
In this precipitation titration, the concentration of the indicator used is quite low as adding too much chromate indicator will lead to an intense yellow colour which masks the end point. Hence the excess titrant required to cause a colour change is relatively large.
+ 2− - 2퐴푔 (푎푞) + 퐶푟푂4(푎푞) ⇌ 퐴푔2퐶푟푂4(푠) 2- - Small concentrations of CrO4 means that equilibrium lies towards the LHS. To shift equilibrium towards the RHS, a relatively large amount of Ag+ needs to be added.
This excess silver required to reach the end point leads to a systematic error, which can be corrected using a blank titration.
- The same quantity of chromate indictor is to be used in the titration is added to a suspension of calcium carbonate (which imitates the white AgX precipitate).
- AgNO3 is added to the solution until a colour change occurs.
- The volume of AgNO3 required to cause the colour change in the blank is the excess volume required. This volume is subtracted from the titration volumes.
VOLHARD’S METHOD (BACK TITRATION)
Volhard’s method quantifies anions used in back titration procedure.
- A known excess quantity of AgNO3 is added to the sample to precipitate all of the analyte anions: + − o 퐴푔 (푎푞) + 푋 (푎푞) → 퐴푔푋(푠)
- The excess AgNO3 is determined by back titration against KSCN(aq) standard solution + − o 퐴푔 (푎푞) + 푆퐶푁 (푎푞) → 퐴푔푆퐶푁(푠) - A small amount of Fe3+ is added as the indictor. When excess SCN- is present, there will be a permanent colour change. 3+ − 2+ o 퐹푒 (푎푞) + 푆퐶푁 (푎푞) → [퐹푒(푆퐶푁)] (푎푞)
LIMITATIONS OF VOLHARD’S METHOD
3+ The Volhard titration must be performed in low pH to prevent the precipitation of Fe indictor as Fe(OH)3. At low - pH, there is a relatively low [OH ], therefore the precipitate Fe(OH)3 will not form.
The Volhard method is less valid if the first precipitate formed is more soluble than AgSCN.
CHEMISTRY MODULE 8: APPLYING CHEMICAL IDEAS PAGE 12 OF 48
−10 −12 - For example, AgCl (퐾푠푝 = 1.77 × 10 ) is more soluble than the AgSCN (퐾푠푝 = 1.03 × 10 ). This means that AgCl can dissolve during the titration leading to a greater titration volume required.
To avoid this problem, the procedure can be modified.
- The first precipitate can be removed via filtration before titration. However, this can lead to inaccuracies and is less convent . - A dense organic liquid like nitrobenzene or chloroform can be added to the conical flask to act as a barrier between the precipitate and aqueous layer.
FAJAN’S METHOD (DIRECT TITRATION AND ADSORPTION INDICATOR)
In the analysis of chloride, an anionic dye is added to the sample.
- A standard solution of AgNO3 is added and the precipitation occurs: + − o 퐴푔 (푎푞) + 퐶푙 (푎푞) → 퐴푔퐶푙(푠) - Before the equivalence point, a small excess of chloride ions is incorporated into the precipitate, so it carries a negative charge. The dye is repelled and remains in solution. - After the equivalence point, there is a small excess of silver ions. Therefore, the precipitate becomes slightly positively charged, and the dye adsorbs onto it, resulting in a precipitate colour change.
The most suitable indicator depends on the analyte.
PROS AND CONS OF FAJAN’S METHOD
The Fanjan’s method relies on a large precipitate surface area, to allow sufficient dye to adsorb so the colour change is easily visible.
- Dextrin (starch) can be added to prevent silver halide precipitates from aggregating - A high concentration of spectator ions can also caused aggregation, so this method cannot be used with all samples. - The analyte must also be concentrated enough, as the colour will not be seen if there is too little precipitate.
The pH of the titration reaction must be carefully controlled as anionic dye indicators are conjugate bases of weak acids. This means it will react with free H+ and will not adsorb onto the surface of the precipitate.
CHEMISTRY MODULE 8: APPLYING CHEMICAL IDEAS PAGE 13 OF 48
ATOMIC ABSORPTION SPECTROSCOPY
OUTCOMES ● conduct investigations and/or process data to determine the concentration of coloured species and/or metal ions in aqueous solution, including but not limited to, the use of: – atomic absorption spectroscopy ATOMIC ABSORPTION
The Bohr model of the atom where electrons are contained in shells around the nucleus of the atom:
- The further away from the nucleus, the higher the energy level of the shell - These shells have discrete or quantised energy levels - The lowest energy electronic configuration of an atom is called the ground state
When electrons in the ground state (gas) absorb light energy (photons) corresponding to a difference between levels, they are excited to a higher energy level.
The light that passes through the sample without being absorbed can be collected as an absorption spectrum.
- The absorbed wavelengths of light can be seen as black lines on a bright background. - The energies of the absorbed wavelengths match energy gaps between two shells.
As each element has a unique set of energy levels, the wavelengths of light absorbed by each element is characteristic.
ATOMIC ABSORPTION SPECTROSCOPY (AAS)
Atomic absorption spectroscopy is a sensitive and highly selective technique that can be used to measure small concentrations of metal ions. It has an important role in the detection of toxic heavy metals.
In AAS, the concentration of the analyte is calculated from the amount of wavelength of light absorbed by the sample.
- The wavelength used in the analysis is based on atomic absorptions of the analyte. - The amount of light absorbed is called the absorbance. It is determined from the relative intensities of light before and after the sample
퐼0 퐴 = log 10 퐼
Where:
- A: Absorbance which has no units as its dimensionless
- I0: The intensity of light before the sample - I: The intensity of light after the sample
The measured absorbance is directly proportional to the concentration of analyte present in the sample.
CHEMISTRY MODULE 8: APPLYING CHEMICAL IDEAS PAGE 14 OF 48
HOW DOES IT WORK
1) An aqueous solution of sample is drawn up through a fine capillary tube into a nebuliser which turns it into a fine mist. 2) Carrier gases (air and acetylene) sweeps the droplets into the furnace, which burns the solvent off and atomises the sample (to their ground states). 3) The hollow cathode lamp emits wavelengths of light matching the energy gaps of the element being analysed through the atomised sample 4) Analyte atoms in the sample with absorb a fraction of light 5) The remaining light passes to a monochromator which selects a wavelength of light for analysis. 6) The photomultiplier detector measures the intensity of light and replays the data to the analysis software.
The hollow cathode lamp contains the element which being tested. It emits a certain spectral pattern – unique to each metal.
CALIBRATION USING STANDARD SOLUTIONS
The calibration curve is used to establish the relationship between measured absorbance and the analyte concentration for a particular instrument. It is produced by measuring the absorbance of standard solutions.
Absorbance is proportional to concentration for dilute samples in AAS.
- The best results are obtained when absorbance is 0.1- 0.8 - At higher concentrations, interaction between analyte atoms affect the absorbance values, leading to inaccurate data - At lower concentrations, the instrument response may not be sensitive enough
CHEMISTRY MODULE 8: APPLYING CHEMICAL IDEAS PAGE 15 OF 48
APPLICATIONS OF AAS
- It is quick, easy, accurate and highly sensitive and is commonly used to determine the concentrations of over 65 elements. - AAS provides a means of investigating phenomena which could not be studied before, such as trace nutrients and heavy metal pollution. o Trace elements work in organisms by helping enzymes to function o Concentration normally range between 1 – 100 ppm o The discovery of cobalt deficiency in seemingly good pastureland in coastal south-western Australia where animal health could not be maintained. o The discovery of a molybdenum deficiency in the soils of arid parts of Victoria where legumes crops could not be supported.
ADVANTAGES AND DISADVANTAGES OF AAS
Advantages
- Simple procedure - Very sensitive technique (can measure ppb) - Extremely accurate - Specific (can analyse a single ion in a mixture) - Although destructive, uses a very small sample size - Very fast analysis (1 minute per element) - Little sample preparation required
Disadvantages
- High initial costs - Can only test for one atom or ion at a time with most instruments - Only metals and a few non-metals can be tested - Cannot distinguish between different oxidation states of the metal (e.g. Fe2+ and Fe3+) - Destructive analysis - Only test aqueous solutions
CLASS NOTES FOR AAS
Some elements produce very intense spectral lines that serve as a unique characteristic marker for their presence. This can be used as a qualitative indicator. When a flame is looked at through a spectrometer it can be further analysed to determine the concentration of the substance.
AAS is a technique used to identify the presence and concentration of substances by analysing the spectrum produced when a sample is vaporised and absorbs certain frequencies of light. It is primarily used to determine the concentrations of cations.
AAS measures the amount of light absorbed by the sample.
CHEMISTRY MODULE 8: APPLYING CHEMICAL IDEAS PAGE 16 OF 48
COLOURIMETRY
OUTCOMES ● conduct investigations and/or process data to determine the concentration of coloured species and/or metal ions in aqueous solution, including but not limited to, the use of: – colourimetry COLOURIMETRY
Colourimetry is a technique for determining the analyte concentration based on the absorbance of a coloured solution.
- The amount of light that is absorbed by a sample is compared to the amount absorbed by a blank or reference - The blank contains all components of the measured sample except the analyte
A coloured solution mostly absorbs its complementary colour, which is directly opposite in the colour wheel.
The wavelength of light used in colourimetry corresponds to the strongest absorption of the analyte of interest, known as the absorption maximum (휆푚푎푥)
- This wavelength is used as sensitivity increases with the intensity of absorption, allowing for more accurate data to be collected - Requires an intensely coloured analyte
The main coloured species analysed by colourimetry are transitional metal complexes. Transition metal ions are often weakly coloured in solution, so they are commonly converted to intensely coloured transition metal complexes to allow for quantitative analysis. The colour of complex ions depends on:
- Central metal ion - Ligands
CHEMISTRY MODULE 8: APPLYING CHEMICAL IDEAS PAGE 17 OF 48
HOW COLOURIMETRY WORKS
1) A light source provides a continuous source of white light which is narrowed and aligned into a beam using a slit. 2) A coloured filter allows a small range of wavelengths to pass through the sample, and blocks the other wavelengths 3) The light beam passes through the sample, which absorbs a fraction of the light 4) The remaining light is transmitted through the sample and reaches the detector, which converts the amount of light into an electrical signal.
The calculation of absorbance is similar to the method used in atomic absorption spectroscopy (AAS).
- However, the amount of light absorbed by the sample is calculated relative to the amount of light absorbed by a blank sample. - This is to remove the influence of the solvent and other substances present in the sample on the measured absorbance
퐼0 퐴 = log 10 퐼
- A: Absorbance which has no units as its dimensionless. Typically, between 0.3 and 2.5 (AU)
- I0: The intensity of light passing through the blank sample - I: The intensity of light passing through the analyte sample
퐴 = 휀푙푐
- 휀: Molar absorptivity (also known as extinction coefficient) (: 퐿 푐푚−1 푚표푙−1) - 푙: Path length of the sample in cm - 푐: Concentration of the substance in the sample in mol/L
Many factors can influence the absorbance readings therefore, it is always more accurate to construct a calibration curve by measuring the absorbance of standard solutions under the sample experimental conditions. This helps to reduce systematic error.
Similar to AAS, absorbance is proportional to concentration for samples with moderate absorbances (0.3 – 2.5). Samples should be concentrated or diluted so that measured absorbances are in the range where the relationship is linear.
CHEMISTRY MODULE 8: APPLYING CHEMICAL IDEAS PAGE 18 OF 48
DETERMINATION OF IRON CONTENT USING COLOURIMETRY
3+ Due to the pale green colour of Fe (aq), it cannot be directly measured using colourimetry, and requires conversion to the intensely coloured [Fe(SCN)]2+ complex.
- Fe2+ is first oxidised to Fe3+ using acidified hydrogen peroxide:
2+ + 3+ 2퐹푒 (푎푞) + 2퐻 (푎푞) + 퐻2푂2(푎푞) → 2퐹푒 (푎푞) + 4퐻2푂(푙)
- Fe3+ is then reacted with thiocyanate (SCN-) to form the deep red iron (III) thiocyanate complex:
3+ − 2+ 퐹푒 (푎푞) + 푆퐶푁 (푎푞) → [퐹푒(푆퐶푁)] (푎푞)
The complex ion strongly absorbs blue light at a wavelength of 447 nm.
DETERMINING STOICHIOMETRY USING COLOURMETRY
Colourimetry can be used to determine the stoichiometric ration of reactants in the formation of an ionic species such as a coordination complex. The absorbances of a series of solutions containing different ratios of reactants are measured at a wavelength absorbed by the product. The correct stoichiometry of reactants is given at the maximum absorbance.
The measured absorbances of the solutions are plotted against the volume of a reactant used, and the maximum absorbance is found using lines of best fit.
The ratio of the reactants in the product corresponds to the moles combined to achieve the maximum absorbance.
OUTCOMES ● conduct investigations and/or process data to determine the concentration of coloured species and/or metal ions in aqueous solution, including but not limited to, the use of: – ultraviolet-visible spectrophotometry SPECTROSCOPY OF ORGANIC COMPOUNDS
The wavelengths absorbed by an organic molecule can be matched to its structural fragments. A structural fragment that absorbs a characteristic wavelength is called a chromophore.
The overall spectrum can also be used as a signature or fingerprint of the whole molecule. By comparing the collected spectrum with a database of standards, the compound can be identified.
UV-VISIBLE SPECTROPHOTOMETRY
UV-visible spectrophotometers can also be used to measure the absorbance of a sample over a range of wavelengths.
- The prism is rotated so that different wavelengths (190 – 700 nm) pass through the monochromator to the sample then to the detector. - The absorbance at each wavelength is recorded. - The UV-visible spectrum plots absorbance against wavelength
A UV-visible spectrum is often used to find the absorption maximum of a sample.
CHEMISTRY MODULE 8: APPLYING CHEMICAL IDEAS PAGE 19 OF 48
ULTRAVIOLET-VISIBLE SPECTROSCOPY OF ORGANIC COMPOUNDS
It is primarily used in detecting the presence of conjugation within organic molecules.
Conjugation is where molecules have alternating double or triple and single bonds.
The UV-Vis spectra of organic compounds arise as a result of electronic transitions, just as they do for metals and transition metal complexes.
Organic compounds can absorb UV and visible radiation as the wavelength in these regions correspond to the energy gaps between molecular orbitals.
- Similar to the absorption of light by electronic transitions between atomic orbitals in flame test and atomic absorption spectroscopy (AAS). - When a covalent bond forms, the atomic orbitals (the orbitals in the individual atoms) combine to produce new molecular orbitals which contain the electron pair shared in the bond.
- When an appropriate wavelength of light matching the energy gap is absorbed by an organic compound, an electron can be excited from a lower energy molecular orbital to a higher energy molecular orbital.
The energy gap, and consequently the wavelength required for an electronic transition, depends on the type of bonding.
CHEMISTRY MODULE 8: APPLYING CHEMICAL IDEAS PAGE 20 OF 48
- For C-C and C-H single bonds, the energy gaps are very large and require shorter wavelengths in the far UV region (<150 nm) - For double bonds or triple bonds (C=C, C=O, C=N) or bonds with heteroatoms that have lone electron pairs (C-Cl, C-O), the energy gaps are smaller therefore light of longer wavelengths is absorbed (150-200 nm) - When conjugation exists in a molecule, the energy gaps are even smaller, resulting in longer wavelengths being absorbed (>200 nm) o In summary: ↑Conjugation → ↓Energy Gaps → ↑ Wavelength absorbed
- If the conjugation is extended further, the energy gap will be small enough for the molecule to absorb wavelengths in the visible region (400-800 nm), resulting in a highly coloured compound.
In practice, the range from 190 to 700 nm is detectable by a typical commercial UV-Vis spectrometer.
- UV-Vis spectroscopy of organic compounds is limited in the most part to conjugated systems - For measurements below 190nm, a vacuum spectrometer and a suitable UV light source are required.
INTERPRETING UV-VIS SPECTRA
A UV-Vis spectrum is a plot of absorbance versus absorbed wavelengths. The diagram below is the UV-Vis spectrum for buta-1,3-diene.
The interpretation of the UV-Vis spectrum of organic compounds is straightforward. By comparing 휆푚푎푥 with a database of standards, the type of bonding in a compound can be identified.
CHEMISTRY MODULE 8: APPLYING CHEMICAL IDEAS PAGE 21 OF 48
Chromophore 흀풎풂풙 (풏풎)
퐶 − 퐻 122
퐶 − 퐶 135
퐶 = 퐶 162
퐶 ≡ 퐶 137, 178, 196, 222
퐶 − 퐶푙 173
퐶 − 퐵푟 208
The exact wavelengths absorbed depends on the rest of the molecule, hence compounds have distinctive UV-Vis spectra.
The presence of strong absorption bands in the spectrum above 200nm can generally be used to indicate the presence of conjugation in a molecule. The greater the extent of conjugation, the longer the wavelength absorbed.
CHEMISTRY MODULE 8: APPLYING CHEMICAL IDEAS PAGE 22 OF 48
CHEMICAL TESTS FOR ORGANIC ANALYSIS
OUTCOMES ● conduct qualitative investigations to test for the presence in organic molecules of the following functional groups: – carbon–carbon double bonds – hydroxyl groups – carboxylic acids (ACSCH130) CARBON-CARBON DOUBLE BONDS
All halogens are coloured (bromide, chloride, fluoride and iodide). The addition reaction, halogenation, can be used as an indicator for alkenes and alkynes through the bromine test.
Once bromide is added to a substance, it is an alkene or alkyne when there is a rapid spontaneous decolourisation of
Br(l). This should be done in the absence of UV light.
HYDROXYL GROUPS (ALCOHOLS)
Alcohols have hydroxyl (-OH) functional groups. The presence of the hydroxyl functional group can be detected by reacting a dry sample of the alcohol with a small piece of active metal such as sodium.
- Alcohols react readily with metallic sodium, forming a sodium alkoxide salt and hydrogen gas.
- This occurs for all degrees of alcohol, with reactivity decreasing from primary to tertiary.
The evolution of a gas indicates a reaction has occurred. The gas evolved can be collected and tested with a lit splint.
If a ‘pop’ sound occurs, H2 has evolved and a hydroxyl group is present.
Alternatively, the pH of the alkoxide salt solution formed can be tested with an indicator. Alcohols are neutral. The alkoxide ion produces a strong base.
− − 푅푂(푎푞) + 퐻2푂(푙) → 푅푂퐻(푎푞) + 푂퐻 (푎푞)
CHEMISTRY MODULE 8: APPLYING CHEMICAL IDEAS PAGE 23 OF 48
Once the presence of a hydroxyl group is confirmed, the type of alcohol can be determined by reacting the alcohol with an oxidising agent.
The reduction of permanganate or dichromate in acidic solutions results in distinctive colour changes, which can be used to distinguish between the types of alcohol
− 2+ 푀푛푂4 → 푀푛 (Purple → Colourless) 2− 3+ 퐶푟2푂7 → 퐶푟 (Orange → Green)
CARBOXYL GROUP
Carboxylic (alkanoic) acids contain the carboxyl (-COOH) functional group. They are weak acids as the hydrogen atom bonded to the oxygen atom will partially dissociate in water.
The presence of a carboxyl group can be determined by an indictor test.
- An aqueous solution of the carboxylic acid can be tested with blue litmus paper. The presence of the carboxyl group will turn the litmus paper red.
Alternatively, a carbonate test can be performed. Carboxylic acids react with metal carbonates and metal hydrogen carbonates to produce a salt, carbon dioxide and water. For example:
2퐶퐻3퐶퐻2퐶푂푂퐻(푎푞) + 푁푎2퐶푂3(푎푞) → 2푁푎퐶퐻3퐶퐻2퐶푂푂(푎푞) + 퐶푂2(푔) + 퐻2푂(푙)
+ 2− 2퐻 (푎푞) + 퐶푂3 (푎푞) → 퐶푂2(푔) + 퐻2푂(푙)
Effervescence (bubbles) indicates that a reaction has occurred. The gas evolved can be collected and confirmed as a carbon dioxide by bubbling it through limewater (calcium hydroxide).
퐶푎(푂퐻)2(푎푞) + 퐶푂2(푔) → 퐶푎퐶푂3(푠) + 퐻2푂(푙)
CHEMISTRY MODULE 8: APPLYING CHEMICAL IDEAS PAGE 24 OF 48
ANALYSING A SAMPLE
INFRARED SPECTROSCOPY
Infrared spectroscopy analyses the interaction of molecules with IR light. It primarily gives information about the bonds or functional groups present in a molecule and can be used for fingerprinting purposes.
The absorption of IR radiation by an organic compound is associated with a change in the vibrational energy levels of the molecule. IR induces molecular vibrations.
A molecule can be thought of as a group of spherical masses (atoms) connected by strings (covalent bonds).
- At room temperature, the atoms within the molecules are constantly in motion, vibrating back and forth. - The fundamental vibrational motions include stretching and bending.
However, unlike masses on a spring, the energy of molecular vibration is quantised. This means every vibration can only occur at specific frequency that is characteristic to the type of motion, the atoms attached and the bond strength.
When a molecule is irradiated with energy that matches the energy gap for one of its vibrational modes, it will absorb the energy and be promoted to a higher vibrational energy level. This results in increased amplitude for the vibration.
- The energy difference between the vibrational levels corresponds to the frequencies in the infrared region
Since different types of bonds give rise to different vibrational frequencies, each functional group will absorb in a different IR range. This absorption frequency can be used as a diagnostic marker for the presence of functional groups.
CHEMISTRY MODULE 8: APPLYING CHEMICAL IDEAS PAGE 25 OF 48
An important aspect of IR spectroscopy is that not all vibrations can be detected.
- IR active vibrations are those that cause a change in the dipole moments of a molecule. - IR inactive vibrations do not cause changes in the dipole moment of the molecule. For example, vibrations of
symmetrical non-polar molecules like H2 do not result in the absorption of IR energy. Thus, no peak will be observed for these vibrations. - The reason for this requirement involves the mechanism by which the radiation transfer its energy to the molecule.
INFRARED SPECTROMETER
- The IR radiation source is an inert solid that is heated electronically to cause thermal emission of radiation in the infrared region. - This incident light is split into two equivalent beams that are passed through a reference and the sample, where some energy is absorbed, and some is transmitted. - The beam chopper alternatively passes the beams to the monochromator which filters out a single wavelength of light. - The detector calculates the ratio between the intensities of the reference and sample beams to determine the absorbance.
CHEMISTRY MODULE 8: APPLYING CHEMICAL IDEAS PAGE 26 OF 48
INTERPRETATION OF INFRARED SPECTRUM
An infrared spectrum is constructed by plotting transmittance vs wavenumber.
- The transmittance is plotted, hence absorbed wavelengths are shown by troughs pointing downwards. - Infrared absorption wavelength is reported in wavenumbers which is a type of frequency measurement. It is used because the numbers are more convenient. - The x-axis is not a linear scale. It changes at 2000 cm-1
CHEMISTRY MODULE 8: APPLYING CHEMICAL IDEAS PAGE 27 OF 48
- 3600 − 2300 푐푚−1: stretching of single bonds to hydrogen - 2300 − 1900 푐푚−1: stretching of triple and consecutive double bonds - 1900 − 1500 푐푚−1: stretching of carbonyls and double bonds - 1500 − 666 푐푚−1: fingerprint region. Many overlapping signals are present in this region, leading to a unique spectrum for different molecules. Peaks in the region are not normally interpreted in detail but comparison of this region with a database of standards allows for the identification of specific compounds.
The exact frequencies absorbed depends on the rest of the molecule, hence compounds have unique infrared absorption spectra. Infrared absorption beaks are described qualitatively as strong (s), medium (m), or weak (w). Additionally, broad (br) troughs may be observed.
ADVANTAGES AND DISADVANTAGES OF IR SPECTROSCOPY
ADVANTAGES
- Infrared spectroscopy requires milligram-sized samples, but these can usually be recovered, depending on the technique used. - Infrared spectra are very easy and quick to run, and are relatively inexpensive - Infrared spectra can be used for fingerprinting purposes
DISADVANTAGES
- Functional groups can be identified from the spectrum, but other structural features cannot be determined. - It can be difficult to resolve spectra from complex mixtures. The presence of multiple organic substances can result in masking and distortion of characteristic absorptions. - The sample must be very dry. Water present in samples will make it appear as if the compound contains O-H bonds.
CHEMISTRY MODULE 8: APPLYING CHEMICAL IDEAS PAGE 28 OF 48
ELEMENTAL ANALYSIS
OUTCOMES ● investigate the processes used to analyse the structure of simple organic compounds addressed in the course, including but not limited to: – mass spectroscopy (ACSCH19) COMBUSTIVE ANALYSIS
Elemental analysis of organic compounds is the determination of mass fractions or percentages of carbon, hydrogen and heteroatoms in a sample. This information can be used to determine the empirical formulae of unknown compounds.
- Empirical formula: simplest whole number mole ratio - Molecular formula: moles of atoms in the molecule
The mass of each element in a compound can be obtained experimentally by means of combustion analysis.
- This process involves burning a weighed sample of the organic compound in excess oxygen, then collecting and weighing the combustion products - From these mass measurements, the percentage composition of the sample can be determined
DETERMINING EMPIRCAL FORMULA
1) Assume that you have 100g of the compound. 2) Calculate the moles of each element 3) Find the simplest whole number ratio
DETERMINING MOLECULAR FORMULA
A molecular formula can always be determined by multiplying the subscripts in the empirical formula by an integer.
To determine the molecular formula, the number of empirical units would need to be calculated. Therefore, you need to know the approximate molar mass.
푀표푙푎푟 푀푎푠푠 푁표. 표푓 푒푚푝𝑖푟𝑖푐푎푙 푢푛𝑖푡푠 (푥) = 퐸푚푝𝑖푟푐푎푙 푀푎푠푠
MASS SPECTROMETRY
Mass spectrometry is an analytical technique for studying the chemicals present in samples, and for probing the molecular structure of compounds. A mass spectrometer measures the masses of molecules and atoms by volatising then ionising them.
The most common type of mass spectrometer is the electron ionisation magnetic sector instrument
- A very small (picogram to nanogram) sample is injected into the mass spectrometer and vaporised into its gaseous state. - The sample is then ionised by bombardment by a high energy stream of electrons, which knock valence electrons from the molecules in the sample, creating cations.
CHEMISTRY MODULE 8: APPLYING CHEMICAL IDEAS PAGE 29 OF 48
− + − 푀(푔) + 푒 → 푀 (푔) + 2푒
− Many of these cations are unstable and will decompose into fragment (daughter) ions and radicals.
+ + 푀 (푔) → 푋 (푔) + 푌(푔)
- The cations are then accelerated into a magnetic field which causes the ions to travel on a curved path. Uncharged radicals cannot be accelerated - The curvature depends on the mass-to-charge (m/z) ratio of the particles. The larger the ion, and the lower the charge, the lower the deflection. - By varying the strength of the magnetic field, ions of differing mass can be brought to focus on the detector - The detector is composed of a conductive metal. Like an anode, it is a source of electrons. When the positive ions collide with it, they gain an electron and become neutralised. A small current is produced, amplified and recorded as a spectrum by a computer.
1) Vaporisation 2) Ionisation 3) Acceleration 4) Deflection 5) Detection
CHEMISTRY MODULE 8: APPLYING CHEMICAL IDEAS PAGE 30 OF 48
INTERPRETING MASS SPECTRA
The mass spectrum records the abundances of the fragments of different m/z ratio relative to the most intense peak (the base peak).
- Each peak represents an ion with a specific m/z ratio. Since the charge of these fragments are usually +1, the value of m/z is usually the same as mass.
The mass spectrum can be used to determine the molecular formula of a substance.
- The ions with the greatest mass will correspond to a molecule that has only lost a single electron (parent peak), giving relative molecular mass of the intact molecule - This can be combined with data from elemental analysis to calculate molecular formula
The masses of the detected fragments can be used to deduce structural information about the molecule. Working backwards, the identified fragments can be reassembled to generate the original molecule.
CHEMISTRY MODULE 8: APPLYING CHEMICAL IDEAS PAGE 31 OF 48
The fragment ions that are observed and the abundance of each fragment (height of each peak) is dependent on the stability of the ion
- The more stable the fragment ion, the higher its abundance - The base peak represents the fragment ion with the highest stability
Different types of fragments have different stabilities, which result in unique fragmentation patterns. The overall fragmentation pattern can be used for fingerprint purposes to identify the molecule by comparison with the spectra of known compounds from a library.
Mass Fragment Mass Fragment
15 −퐶퐻3 31 −푂퐶퐻3
17 −푂퐻 43 −퐶퐻2푂퐻
29 −퐶퐻2퐶퐻3 45 −퐶푂푂퐻 −퐶푂퐻 ISOTOPES IN MASS SPECTRA
Some elements have more than one isotope in high abundance. In particular, chlorine and bromine have isotopes that differ by a mass of 2.
Isotope Natural Abundance
ퟑퟓ ퟏퟕ푪풍 76%
ퟑퟕ ퟏퟕ푪풍 24%
ퟕퟗ ퟑퟓ푩풓 51%
ퟖퟏ ퟑퟓ푩풓 49%
The presence of these isotopes can be observed in the mass spectra of chlorinated or brominated compounds.
USES OF MASS SPECTROMETRY
Due to the sensitivity of mass spectrometers, the mass spectra of compounds can be used as fingerprints for identifying compounds.
- Mass spectrometry can be used to unambiguously identify a substance as no two substance produce the same fragmentation pattern - Experimentally obtained spectra can be compared with databases of known substances
Mass spectrometry is often used in conjugation with chromatographic separation and is routinely used to analyse a wide range of industrial, environmental and forensic samples. It is also used for:
- Detect and identify the use of steroids in athletes - Identify compounds in illicit drug samples - Monitor the breath of patients by anaesthetists during surgery - Determined whether honey is adulterated with corn syrup
CHEMISTRY MODULE 8: APPLYING CHEMICAL IDEAS PAGE 32 OF 48
- Locate oil deposits by measuring petroleum precursors in rock - Monitor fermentation processes for the biotechnology industry - Detect dioxins in contaminated fish - Establish the elemental composition of semiconductor materials
ADVANTAGES OF MASS SPECTROMETRY
- Can be used to determine a substances molecular formula - The masses of fragment ions can be used to identify structural fragments in the molecule - Isotopes can be determined, and used to match a sample to a particular source - It is quick, highly sensitive technique that can be performed on very small samples
DISADVANTAGES OF MASS SPECTROMETRY
Mass spectrometry is less useful in the analysis of very complex mixtures
- When hundreds of compounds are present in a sample, there may be so many peaks present that it becomes impossible to determine what fragments are present - In these cases, mass spectrometry can be used to determine molecular masses and molecular formula, but cannot be used to determine the structure of components - Overshowed by the large quantity of structural information available in NMR spectroscopy.
CHEMISTRY MODULE 8: APPLYING CHEMICAL IDEAS PAGE 33 OF 48
NMR SPECTROSCOPY
OUTCOMES ● investigate the processes used to analyse the structure of simple organic compounds addressed in the course, including but not limited to: – proton and carbon-13 NMR NUCLEAR SPIN
Nuclear magnetic resonance (NMR) spectroscopy is one of the most powerful techniques for the analysis of organic compounds.
If a nucleus contains an odd number of protons and/or odd number of neutrons, it can exhibit spin (similar toe electron spin). This means that the nucleus will behave like a magnet in a magnetic field, and can be detected by NMR (NMR-active or spin-active).
The spins of the nuclei in a sample are not oriented, and are initially degenerate (same energy). However, when placed in an external magnetic field, the spin-active nuclei will adopt one of two states:
1) A higher energy spin state aligned against the magnetic field (antiparallel) 2) A lower energy spins state aligned with the magnetic field (parallel)
The difference in energy between the two spin states is Δ퐸, and this amount of energy lies within the radiofrequency region of the electromagnetic spectrum.
CHEMISTRY MODULE 8: APPLYING CHEMICAL IDEAS PAGE 34 OF 48
One way of producing an NMR spectrum:
- The sample is placed in a magnetic field - Radio waves of different frequencies are sent through the sample - When the energy of the wave matches Δ퐸 for a particular nucleus, it can absorb the energy and transition from the lower energy spin state to the higher energy spin state. The nucleus can release the energy again to return to the lower spin state. Switching between these two states is called resonance. - The frequency at which resonance occurs is recorded as a signal in the NMR spectrum
CARBON-13 NMR SPECTRA
Most carbon atoms in a sample will usually be carbon-12, which cannot be analysed using NMR. However, approximately 1% of a typical sample of carbon will consist of the carbon-13 isotope, which is spin-active and can be detected in NMR spectroscopy.
A carbon-13 NMR spectrum gives information about the number and type of carbon environments in a molecule.
The frequencies that are absorbed in an NMR experiment are proportional to the size of Δ퐸. Electrons are moving charged particles, so they will create a small magnetic field around the nucleus which will change the size of Δ퐸, and hence the absorbed frequencies. This is known as electron shielding.
Electron shielding creates unique chemical environments in a molecule.
- Electron shielding changes depending on the atoms and bonds around the nucleus (indicated by the position of the signal) - The number of unique chemical environments for a particular type of nucleus will determine the number of unique Δ퐸, which will match the number of signals in an NMR spectrum.
CARBON ENVIRONMENTS
If the carbon nuclei can be interchanged via bond rotation or rotation in space, they will be in the same chemical environment and produce the same signal.
CHEMISTRY MODULE 8: APPLYING CHEMICAL IDEAS PAGE 35 OF 48
CHEMICAL SHIFT
The resonance frequencies in an NMR spectrum are converted to values known as chemical shift, measured in parts per million (ppm).
- Converting to chemical shift removes the impact of magnetic field strength, which increases Δ퐸 - This allows the comparison of spectra collected in spectrometers of different strengths
The chemical shift (position of the signal) that a nucleus produces depends on electron shielding (what is bonded to the carbon).
- A nucleus surrounded by low electron density is less shielded, so it experiences more of the magnetic field and resonates at higher frequencies (higher ppm). - A nucleus surrounded by high electron density is more shielded and will resonate at a lower frequency (lower ppm).
Tetramethylsilane (TMS) is usually added to samples as a standard.
Different types of carbon environments have characteristic chemical shifts, generally between 0 – 200 ppm.
CHEMISTRY MODULE 8: APPLYING CHEMICAL IDEAS PAGE 36 OF 48
PROTON NMR SPECTRA
The number of signals in the spectrum matches with the number of unique hydrogen environments.
Different types of hydrogen environments have characteristic chemical shifts
- Decreased electron shielding leads to downfield shifts (higher ppm) - TMS is also used as a standard and has a chemical shift of 0.0ppm
The chemical shifts can be divided into general regions:
- 0-1.5 ppm: protons on saturated carbon centres, or on carbons attached to saturated centres - 1.5-2.5 ppm: protons on carbons attached to unsaturated centres (alkenes, alkynes) - 2.5-4.5 ppm: protons on carbons attached to electronegative atoms - 4.5-6.5 ppm: protons on alkene carbons - 6.5-8.0 ppm: protons on aromatic rings (benzene)
Additional information can be obtained from higher resolution H-NMR spectra: from signal splitting and the area under each integral.
HYDROGEN ENVIRONMENTS
Substitute ‘X’ a dummy atom and see how many different isomers there are.
푛 =number of hydrogens on the adjacent carbon
CHEMISTRY MODULE 8: APPLYING CHEMICAL IDEAS PAGE 37 OF 48
INTEGRALS AND SPIN-SPIN SPLITTING IN H-NMR
INTEGRALS (AREA UNDER CURVE)
The area under the signal corresponds to the number of hydrogen nuclei in that environment. Use a ruler to measure the ratio.
SPLITTING (MULTIPLICITY)
In H-NMR each signal can be split into multiple peaks, called a multiplet. Splitting arises from the interaction of the spins of nearby nuclei, called coupling.
Splitting follows the n + 1 rule: protons that have n protons in a different chemical environment on immediately adjacent carbons show n + 1 peaks in their signal.
The heights of the peaks in multiplet matches Pascal’s triangle, with taller peaks in the middle and a symmetrical structure.
Neighbouring H Peaks in signal Multiplicity Ratio of Heights
0 1 Singlet 1
1 2 Doublet 1: 1
2 3 Triplet 1: 2: 1
3 4 Quartet 1: 3: 3: 1
4 5 Quintet 1: 4: 6: 4: 1
5 6 Sextet 1: 5: 10: 10: 5: 1
7 7 Septet 1: 6: 15: 20: 15: 6: 1
CHEMISTRY MODULE 8: APPLYING CHEMICAL IDEAS PAGE 38 OF 48
ADVANTAGES AND DISADVANATGES OF NMR SPECTROSCOPY
NMR spectroscopy by far is the most powerful technique for determining and confirming the structures of organic compounds, due to the large amount of structural information given in a spectrum.
NMR is less sensitive than other techniques, so the sample sizes required are larger than for other techniques. However, NMR spectroscopy is non-destructive for samples can be recovered.
Highly complex, large molecules will produce a spectrum that are difficult to interpret.
Compounds must be dissolved in a suitable solvent to produce a clear NMR spectrum
NMR spectrometers maintenance requires technical support.
CHEMISTRY MODULE 8: APPLYING CHEMICAL IDEAS PAGE 39 OF 48
CHEMICAL SYNTHESIS AND DESIGN
OUTCOMES ● evaluate the factors that need to be considered when designing a chemical synthesis process, including but not limited to: – availability of reagents – reaction conditions (ACSCH133) – yield and purity (ACSCH134) – industrial uses (eg pharmaceutical, cosmetics, cleaning products, fuels) (ACSCH131) environmental, social and economic issues CHEMICAL SYNTHESIS
Chemical synthesis refers to the purposeful use of chemical reactions to obtain a desired product. A synthesis reaction is when smaller compounds undergo a chemical reaction to produce a larger compound (product).
The development of chemical syntheses for desirable product compounds has been of fundamental importance to our standard of living and has addressed issues such as:
- Replenishment of depleting resources - Large-scale production of compounds that are scarce in nature - Development of unique chemicals to address specific needs
DESIGNING A CHEMICAL SYNTHESIS PROCESS
CONSIDERATIONS FOR REACTION PATHWAY DESIGN
Desired products typically cannot be produced in a single step from the available starting materials. Instead, a multi- step chemical synthesis is required.
A reaction pathway is commonly designed by working backwards from the desired product and identifying the intermediate products and reaction conditions that would result in the formation.
The analysis of a compound to devise a suitable reaction pathway is called a retrosynthetic analysis.
- A retrosynthetic analysis can often identify a number of alternative reactions pathway to produce a desired product. This can be seen for the synthesis of butanone.
A number of factors are considered when selecting the most desirable reaction pathway:
- Availability of reactants and reagents - Yield and purity of the product - Reaction conditions - Economic factors such as industrial uses of the products and by-products, the cost of reagents and the efficiency of the process - Environmental impacts of reagents, products and waste generated by the process itself - Social impacts of the synthesis such as the health hazards associated with reagents, products and waste,
CHEMISTRY MODULE 8: APPLYING CHEMICAL IDEAS PAGE 40 OF 48
AVAILABILITY OF REAGENTS
The availability of reagents is a significant factor in the selection of a reaction pathway.
- Hard to obtain - ↓Supply → ↑Price → ↑Cost of production
YIELD AND PURITY
The yield of a reaction is the amount of product obtained compared to the theoretical maximum expressed as a percentage. The theoretical maximum amount of product that can be formed from a reaction is based on the amount of the limiting reagent.
The amount of product that is obtained from a reaction (actual yield) is commonly expressed as a percentage of the theoretical maximum (theoretical yield).
퐴푐푡푢푎푙 푌𝑖푒푙푑 (푔) 푃푒푟푐푒푛푡푎푔푒 푌𝑖푒푙푑 (%) = 푇ℎ푒표푟푒푡𝑖푐푎푙 푌𝑖푒푙푑 (푔)
An additional consideration in the selection of a reaction pathway is the number of steps required to convert a given starting material to the desired product compound.
- The overall yield of a reaction pathway typically decreases with the number of steps in the reaction sequence - A reaction pathway with a large number of single steps reactions will result in a low yield.
The overall yield for a multi-step reaction pathway can be calculated from the yield for each of the individual steps
푌𝑖푒푙푑 표푓 푓𝑖푟푠푡 푠푡푒푝 (%) 푌𝑖푒푙푑 표푓 푠푒푐표푛푑 푠푡푒푝 (%) 푂푣푒푟푎푙푙 푌𝑖푒푙푑 (%) = × 100 100
LINEAR AND CONVERGENT SYNTHESIS
A linear synthesis is where each intermediate product was the reactant for the next subsequent step.
One strategy to improve the overall product yield of a multi-step reaction pathways is convergent synthesis.
- A convergent synthesis reduces the number of linear steps in a sequence by using individual reaction pathways to synthesis the intermediate products - The intermediate products are then combined in a single step to generate the final desired product.
A convergent reaction pathway typically has a higher overall yield compared to the same reaction pathway performed as a linear sequence.
CHEMISTRY MODULE 8: APPLYING CHEMICAL IDEAS PAGE 41 OF 48
SELECTIVITY
An additional consideration in the selection of a reaction pathway is the selectivity of each of the individual steps to form the desired product. TLDR: basically, consider the major and minor products.
PURITY
Purity refers to the amount of desired product in a sample, expressed as a percentage.
- Product purity is commonly used as a measure of quality of a product. - Typically, higher purity chemicals are more labour intensive to produce than lower purity chemicals due to the need for additional purification processes.
Product purity is continually monitored throughout chemical manufacture to ensure that the product meets specifications. This process is commonly referred to as quality control.
푀푎푠푠 표푓 푑푒푠𝑖푟푒 푝푟표푑푢푐푡 (푔) 푃푒푟푐푒푛푡푎푔푒 푃푢푟𝑖푡푦 (%) = 푇표푡푎푙 푚푎푠푠 표푓 푠푎푚푝푙푒 (푔)
REACTION CONDITIONS
Reaction conditions must be considered when designing syntheses.
- Conditions are adjusted to ensure the desired product can be obtained in high yield and at a reasonable reaction rate. - This is particularly important in the chemical industry, where small gains in yield and rate can dramatically improve profitability.
To increase the viability of an industrial process:
- Maximise yield - Maximise rate - Minimise waste o Use the by-product of the reaction for something else o Recycle energy - Safety
Reaction conditions that are commonly optimised to improve the yield and/or rate of a chemical process include:
- Concentration of reactants and products - Temperature - Pressure - Catalyst
However, there are other considerations in addition to product yield and reaction rate when selecting optimal temperature and pressure conditions:
- High temperatures can be unsafe (increase in costs) - Specialised equipment and greater energy is required to create and maintain high pressures - Side reactions are more likely at higher temperatures - Reactants, products and catalysts often decompose at high temperatures
CHEMISTRY MODULE 8: APPLYING CHEMICAL IDEAS PAGE 42 OF 48
- Catalysts are often very expensive relative to the other reagents in the process.
ENVIRONMENTAL ISSUES
Most industrial processes consume large amounts of energy. For example, the production of ammonia via the Haber process accounts for around 1.2% of the global energy consumption each year. Issues of high energy consumption may include:
- 퐶푂2(푔) + 퐻2푂(푙) (Greenhouse gases)
- 퐶푂(푔) (Toxic gas)
- 푁푥푂푦(푔) (Associated with photochemical smogs)
- 푆푂2(푔) (Associated with the formation of acid rain)
Whenever possible, energy should be recycled to minimise environmental impact and reduce costs.
THERMAL POLLUTION
Many industrial processes use high temperatures processes, which means cooling the products is often necessary. Water is commonly used as a coolant as there are huge water bodies and due to its high heat capacity.
Discharging warm water into the environment may cause thermal pollution. The temperature of water body may increase, causing the concentration of oxygen to decrease. This potentially kills off aquatic life.
Holding ponds can be used to allow water to cool before it is discharged. Releasing warm water into the ocean is also a feasible solution, as the ocean is large enough to dissipate heat quickly.
TOXIC EMISSIONS
Chemical processes should be designed to use substances that minimise hazards to users and the environment. For example, the production of sulfuric acid is associated with fugitive emission of SO2(g). It is a toxic gas and forms an acidic solution in water.
Toxic emissions should be carefully monitored to reduce environmental impact.
WASTE DISPOSAL
Depending on the chemical process, different types of waste are generated which must be appropriately handled.
- Acidic wastes are generated by many industrial processes. - Solid wastes, if inert, are frequently discharged into the ocean. - Toxic waste disposal is often expensive, and there are usually government regulations that dictate how they must be handled. - Non-toxic ionic wastes are often discharged into the ocean.
CHEMISTRY MODULE 8: APPLYING CHEMICAL IDEAS PAGE 43 OF 48
CASE STUDY: PRODUCTION OF SULFURIC ACID
Sulfuric acid is manufactured using the contact process (as the reaction occurs on the surface of the catalyst).
푆(푠) → 푆푂2(푔) ⇌ 푆푂3(푔) → 퐻2푆2푂7 → 퐻2푆푂4
The key step in the contact process is the conversion of SO2 to SO3. It is the yield determining step as it is a reversible process.
2푆푂2(푔) + 푂2(푔) ⇌ 2푆푂3(푔) Δ퐻 = −197 푘퐽/푚표푙
The industrial yield is typically 90-95%.
- 30-50% excess oxygen is present in the mixture. (Oxygen enriched air) o To drive the equilibrium towards the RHS to increase yield.
- The reaction mixture is passed over beds of vanadium oxide catalyst pellets (V2O5) o In pellets to increase surface area and rate of reaction. - 400 − 550℃ moderate temperature. The reaction mixture is progressively cooled as the reaction is exothermic. o The forward reaction is exothermic. Therefore, if temperature decreases, the RHS is favoured
- SO3 is removed from the reaction mixture before it is passed over a final bed of vanadium oxide. o To decrease pressure to drive the equilibrium to the RHS - A low pressure of 1-2 atm is used o High pressures favour the side with the least amount of gas moles. However, all the other conditions already make the yield 90-95%. It is not economically viable to use high pressure in this case.
Contact Process - Reactions
푆(푠) + 푂2(푔) → 푆푂2(푔) Δ퐻 < 0
2푆푂2(푔) + 푂2(푔) ⇋ 2푆푂3(푔) Δ퐻 < 0
푆푂3(푔) + 퐻2푆푂4(푙) → 퐻2푆2푂7(푙)
퐻2푆2푂7(푙) + 퐻2푂(푙) → 2퐻2푆푂4(푙)
퐻2푆푂4 is used in many other chemical processes such as the production of fertilisers, synthetic detergents and HCl.
CHEMISTRY MODULE 8: APPLYING CHEMICAL IDEAS PAGE 44 OF 48
CONTACT PROCESS CONSIDERATIONS
푆푂2(푔) 푎푛푑 푆푂3(푔) forms acidic rain in the atmosphere. They are both toxic gases as well.
CASE STUDY: HABER PROCESS
The Haber process is the main source of ammonia
- Ammonia is an important raw material for nitrogen-based fertilisers - Modern agriculture is dependent on replenishing nitrogen in the soil with fertilisers, as nitrogen is required for protein production and leaf growth.
−1 N2(g) + 3H2(g) ⇋ 2NH3(g) ∆퐻 = −92푘퐽 푚표푙
TEMPERATURE
A moderate temperature of 400 − 500℃ is used in industry.
- ↑ Temperature → Favours the endothermic reaction (LHS) → ↓ Yield % - ↓ Temperature → Favours the exothermic reaction (RHS) → ↑ Yield%
A moderate temperature is considered a compromise between rate a yield as it achieves an economically viable rate and yield.
PRESSURE
A pressure of around 200 atm is used in industry
- ↑ Pressure → Favours the side with the least gas mols (RHS) → ↑ Yield% - A high pressure also benefits rate of reaction as there are significantly more collisions and potential effective collisions resulting in reactions.
However, high pressures are expensive as energy is required to create high pressure, and equipment must be reinforced to contain increased pressure. High pressure can also pose a safety risk
CATALYST
A magnetite catalyst containing iron (II/III) oxide (퐹푒3푂4) is used. The catalyst provides an alternative energy path that is lower than the original. It increases the rate of reaction and does not affect yield.
REMOVAL OF PRODUCT
The ammonia can be separated from the unreacted reactants through liquifying it by cooling the system to −33℃. The ammonia can be removed, and the unreacted gases are recycled back into the reaction chamber to reduce waste.
YIELD
Yield is typically 10-20% in the Haber Process.
CHEMISTRY MODULE 8: APPLYING CHEMICAL IDEAS PAGE 45 OF 48
OTHER CONDITIONS/CONSIDERATIONS
- Oxygen must be excluded to avoid explosions - Heat produced by the exothermic reaction is recycled to pre-warm the input gases - Carbon monoxide and sulfur contaminants in the reactants must be monitored as they will poison the catalyst and decrease efficiency.
ENVIRONMENTAL IMPACTS
The production of ammonia via the Haber process accounts for around 1.2% of the global energy consumption per year. The source of energy is from the combustion of fossil fuels/
- 퐶푂2(푔) + 퐻2푂(푙) are both greenhouse gases
- 퐶푂(푔) is a toxic gas
- 푁푥푂푦 is associated with photochemical smog
- 푆푂2(푔) is associated with the formation of acid rain
Since the Haber process is an exothermic reaction, the heat produced from the reaction can be recycled.
When warm water is discharged into the environment it can be problematic as oxygen has a lower solubility at high temperatures.
PRODUCTION OF ETHENE
Crude oil → Cracked to smaller fragments
CHEMISTRY MODULE 8: APPLYING CHEMICAL IDEAS PAGE 46 OF 48
ATOM ECONOMY
The principle of atom economy seeks to design chemical reactions such that the greatest number of atoms from reagents are incorporated into the desired product.
The atom economy of a process is calculated as a percentage based on the molecular mass of the desired product and the sum of the molecular masses of the reactants
푚표푙푎푟 푚푎푠푠 표푓 푝푟표푑푢푐푡 퐴푡표푚 퐸푐표푛표푚푦 (%) = 푠푢푚 표푓 푚표푙푎푟 푚푎푠푠 표푓 푟푒푎푐푡푎푛푡푠
CHEMISTRY MODULE 8: APPLYING CHEMICAL IDEAS PAGE 47 OF 48
THINGS TO REMEMBER ‘LOL’
Rule Exceptions
+ + + + + + All Group 1 (Li , Na , K , Rb , Cs ) and ammonium (NH4 ) are NONE soluble
- - All nitrate (NO3 ) and acetate (CH3COO ) salts are soluble NONE
All chloride (Cl-) and bromide (Br-) salts are soluble Ag+, Pb+ (Hg+)
All iodide (I-) salts are soluble Ag+, Pb+, Hg+
2- + 2+ 2+ 2+ 2+ + All sulfate (SO4 ) salts are soluble Ag , Pb , Ba , Ca , Sr , Hg
- + 2+ 2+ All hydroxide (OH ) salts are insoluble Group 1 and NH4 salts are soluble. Ba , Ca are slightly
2- 3- + All carbonate (CO3 ) and phosphate (PO4 ) are insoluble Group 1 and NH4 salts are soluble
2− 2− −2 − − − − − Solution 푃푂4 푆푂4 퐶푂3 푂퐻 퐶푙 퐵푟 퐼 퐶퐻3퐶푂푂 퐵푎2+ White insoluble White White White White White soluble insoluble soluble soluble soluble 퐶푎2+ Colourless White insoluble White White White White White soluble insoluble soluble soluble soluble 푀푔2+ White White White White White White White White soluble insoluble soluble insoluble insoluble soluble soluble soluble 푃푏2+ White insoluble White White White Bright White soluble insoluble insoluble insoluble Yellow 퐶푢2+ Blue Blue Blue soluble Green Blue Blue Blue Not stable Blue soluble insoluble insoluble insoluble soluble soluble 퐹푒2+ Pale green Green Green Grey Green Green Green Green Green soluble insoluble soluble insoluble insoluble soluble soluble soluble 퐹푒3+ Yellow Pink Yellow Not Brown Brown Brown Not stable Brown soluble insoluble soluble stable insoluble soluble soluble 퐴푔+ Colourless Yellow White Yellow Not White Pale yellow insoluble White soluble insoluble insoluble insoluble stable insoluble
CHEMISTRY MODULE 8: APPLYING CHEMICAL IDEAS PAGE 48 OF 48