4.1

Question 1

4.1.1 a) How do you apply the IUPAC rules of nomenclature when systematically naming organic compounds?

Answer 1

1) Count the carbon atoms in the longest continuous chain - this gives you the stem, which is the name of the corresponding alkane (if there’s more than one longest chain, pick the one with the most side-chains). Remove the final letter -e from the alkane stem when the suffix starts with a vowel (e.g. hexan-2-ol).
2) You should number the longest chain so that the main functional group has the lowest possible number. The main functional group usually provides the suffix. For alcohols, alkenes, and ketones. add the number of the carbon atom the functional group is attached to, e.g. butan-2-ol, but-2-ene or pentan-2-one -. This isn’t necessary if there are no positional isomers, e.g. ethene, ethanol, propanone.
3 ) Any side-chains or less important functional groups are added as prefixes (put them in alphabetical order, after the number of the carbon atom each is attached to - if there’s more than one identical side-chain or functional group, use di, tri or tetra).

Question 2

4.1.1 a) What are the names and structural formulae of the first ten members of the alkanes homologous series and their corresponding alkyl group?

Answer 2

Methane: CH4, Methyl: CH3
Ethane: C2H6, Ethyl: C2H5 
Propane: C3H8, Propyl: C3H7
Butane: C4H10, Butyl: C4H9
Pentane: C5H12, Pentyl: C5H11
Hexane: C6H14, Hexyl: C6H13
Heptane: C7H16, Heptyl: C7H15
Octane: C8H18, Octyl: C8H17
Nonane: C9H20, Nonyl: C9H19
Decane: C10H22, Decyl: C10H21

Question 3

4.1.1 a) Name the appropriate prefix or suffix for common functional groups.

Answer 3

For an alkene, -C=C-, the suffix is -ene.
For an alcohol, -O-H, the prefix is hydroxy-, and the suffix is -ol.
For a haloalkane, -Cl/-Br/-I, the prefix is chloro-, bromo-, iodo-, respectively.
For an aldehyde, -(C=O)H, the suffix is -al.
For a ketone, -C(C=O)C-, the suffix is one.
For a carboxylic acid, -(C=O)OH, the suffix is -oic acid.
For an ester, -(C=O)O-C-, the suffix is -oate.
For an acyl chloride, -(C=O)-Cl, the suffix is -oyl chloride.
For an amine, -NH2, the prefix is amino-, and the suffix is -amine.
For a nitrile, -CN, the suffix is -nitrile.

Question 4

4.1.1 b i) What is a general formula?

Answer 4

i) General formula: the simplest algebraic formula for any member of a homologous series.
Alkanes: CnH(2n+2)
Alkenes: CnH(2n)
Alcohols: CnH(2n+1)OH
Carboxylic acid: CnH(2n)O2
Ketones: CnH(2n)O

Question 5

4.1.1 b ii) What is a structural formula?

Answer 5

ii) Structural formula: the minimal detail that shows the arrangement of atoms in a molecule.

Question 6

4.1.1 b iii) What is a displayed formula?

Answer 6

iii) Displayed formula: the relative positioning of atoms and the bonds between them.

Question 7

4.1.1 b iv) What is a skeletal formula?

Answer 7

iv) Skeletal formula: the simplified organic formula, shown by removing hydrogen atoms from alkyl chains, leaving just a carbon skeleton and associated functional groups.

Question 8

4.1.1 c) i) What is a homologous series?

Answer 8

i) Homologous series: a series of organic compounds having the same functional group but with each successive member differing by CH2.

Question 9

4.1.1 c) ii) What is a functional group?

Answer 9

ii) Functional group: a group of atoms responsible for the characteristic reactions of a compound.

Question 10

4.1.1 c) iii) What is an alkyl group?

Answer 10

iii) Alkyl group: of the formula C(n)H(2n+1)

Question 11

4.1.1 c) iv) What is an aliphatic compound?

Answer 11

iv) Aliphatic compound: a compound containing carbon and hydrogen joined together in straight chains, branched chains or non-aromatic rings.

Question 12

4.1.1 c) v) What is an alicyclic compound?

Answer 12

Alicyclic compound: an aliphatic compound arranged in non-aromatic rings with or without side chains.

Question 13

4.1.1 c) vi) What is an aromatic compound?

Answer 13

Aromatic compound: a compound containing a benzene ring.

Question 14

4.1.1 c) vii) Define the terms saturated and unsaturated.

Answer 14

Saturated: single carbon-carbon bonds only.
Unsaturated: the presence of multiple carbon-carbon bonds, including c=c, c≡c and aromatic rings.

Question 15

4.1.1 e) What are structural isomers?

Answer 15

Structural isomers are compounds with the same molecular formula (and so with the same number and type of atoms), but different structural formula (and therefore different arrangement of atoms).

Question 16

4.1.1 e) Give three different types of structural isomers, explaining their effect of chemical and physical properties.

Answer 16

Chain isomers: the carbon skeleton can be arranged differently, with alkyl groups in different places (e.g. as a straight chain, or branched in different ways). These isomers have similar chemical properties, but different physical properties.
Positional isomers: the carbon skeleton could be the same, with the functional group attached to different carbon atoms. These isomers also have different physical properties, but the chemical properties may be different too.
Functional group isomers: the same atoms can be arranged to give different functional groups. These isomers tend to have very different physical and chemical properties.

Question 17

4.1.1 f) i) What is homolytic fission?

Answer 17

i) Homolytic fission: when a covalent bond breaks, each bonding atom receives one electron from the bonded pair, forming two radicals.

Question 18

4.1.1 f) i) What is heterolytic fission?

Answer 18

ii) Heterolytic fission: when a covalent bond breaks, one bonding atom receives both electrons from the bonded pair (resulting in an cation and an anion).

Question 19

4.1.1 g) What are radicals?

Answer 19

Radical: a species with an unpaired electron (represented by ‘dots’ in mechanisms).

Question 20

4.1.1 h) What does a ‘curly arrow’ represent?

Answer 20

Curly arrow: show the movement of an electron pair, either in heterolytic fission or in the formation of a covalent bond (they should start from a bond, a lone pair of electrons, or a negative charge).

Question 21

4.1.2 a) What are alkanes?

Answer 21

Alkanes are saturated hydrocarbons, with the general formula C(n)H(2n+2), containing single C-C and C-H bonds as σ-bonds.

Question 22

4.1.2 a) What is sigma bond?

Answer 22

A sigma bond (σ-bond) is the overlap of orbitals directly between the bonding atoms (i.e. Two orbitals overlap, one from each bonding atom, and each orbital contains one electron. So the σ-bond has two electrons shared between the bonding atoms).
A σ-bond is a single covalent bond (e.g. C-C and C-H are sigma bonds).
A σ-bond has free rotation.

Question 23

4.1.2 b) Explain the shape and bond angle in alkanes.

Answer 23

Each carbon atom has four electrons in its outer shell. In an alkane molecule, each carbon atom therefore forms four single covalent bonds (four σ-bonds), and these four pairs of bonding electrons repel each other equally. This results in a tetrahedral shape around each carbon atom, with each bond angle at 109.5°. Because the σ-bond has free rotation, the shape of the whole molecule is not rigid, allowing them to rotate into different shapes.

Question 24

4.1.2 c) Explain the variations in boiling points of alkanes with different carbon-chain length and branching.

Answer 24

As an alkane chain gets longer, its relative molecular mass, and the number of electrons, increases - so as the chain length increases, the surface area also increases. With increased surface area contact between molecules, the induced dipole-dipole interactions are stronger. So more energy is needed to overcome this stronger intermolecular attraction, increasing the boiling point.
A branched-chain alkane has a lower boiling point than its straight-chained isomer (even though isomers of alkanes have the same molecular mass). Branched-chain alkanes can’t fit together as closely (or neatly) as straight-chained alkanes, and have fewer surface area points of contact between molecules. This results in fewer (and weaker) induced dipole-dipole interactions, which require less energy to overcome, and therefore lower the boiling point.

Question 25

4.1.2 d) Explain why alkanes have a low reactivity.

Answer 25

The σ-bond in alkanes has a high electron density, which is concentrated directly between the bonding atoms, resulting in a strong electrostatic attraction between the nuclei and the shared pair of electrons. This requires a large amount of energy to overcome, and the low reactivity of alkanes is a result of the high bond enthalpy of the σ-bonds.
The low polarity of σ-bonds (due to the similar electronegativity between carbon and hydrogen in a C-H bond, and the same electronegativity in a non-polar C-C bond) makes it difficult to attract nucleophiles and electrophiles, further resulting in low reactivity of alkanes.

Question 26

4.1.2 e) Outline the complete combustion of alkanes, and why they are used as fuels.

Answer 26

All alkanes react with a plentiful supply of oxygen, undergoing complete combustion to produce carbon dioxide and water. This combustion reaction gives out heat, allowing alkanes to be used as fuels. Alkanes tend to be a good source of fuel as they are readily available, easy to transport, and burn in a plentiful supply of oxygen without releasing toxic products (although carbon dioxide is a greenhouse gas).

e. g. C3H8 (g) + 5O2 (g) → 3CO2 (g) + 4H2O (l)
e. g. C6H14 (g) + 9,1/2O2 (g) → 6CO2 (g) + 7H2O (l)

Question 27

4.1.2 e) Explain how incomplete combustion of alkane fuels can be potentially hazardous.

Answer 27

Alkane fuels will undergo incomplete combustion in a limited supply of oxygen, not only producing carbon dioxide and water, but also the colourless, odorless, and toxic gas, carbon monoxide (CO). Carbon monoxide binds to haemoglobin (which transports oxygen in the bloodstream), potentially leading to oxygen deprivation.
e.g. C7H16 (g) + 7,1/2O2 (g) → 7CO (g) + 8H2O (l)
Sometimes, carbon itself is formed as well, as soot.
e.g. C7H16 (g) + 4O2 (g) → 7C (g) + 8H2O (l)

Question 28

4.1.2 f) Consider the production of chloromethane to explain how alkanes form haloalkanes.

Answer 28

Alkanes react with chlorine or bromine in a radial substitution reaction to form haloalkanes (where a hydrogen atom is substituted by a halogen). Consider the production of chloromethane:
CH4 + Cl2 → CH3Cl + HCl.
Initiation - UV light (in a photochemical reaction) provides energy to break the Cl-Cl covalent bond, forming two free radicals in homolytic fission:
Cl2 → 2Cl•
Propagation - the desired product is formed in this stage, along with other radicals (including another Cl•):
CH4 + Cl• → CH3• + HCl
CH3• + Cl2 → CH3Cl + Cl•
Termination - two radicals react to form a stable molecule (there are a number of possible termination steps):
2Cl2• → Cl2
CH3• + Cl• → CH3Cl
CH3• + CH3• → C2H6

Question 29

4.1.2 g) What are the limitations of radical substitution?

Answer 29

A mixture of organic products is produced by:
- the random collision of radicals, causing further substitution to happen
- or when a mixture of isomers are formed, due to radical substitution taking place along any point of the carbon chain
Only one product is usually desirable, so this reaction has a low atom economy (as well as requiring money and time to separate the mixture).

Question 30

4.1.3 a) What are alkenes?

Answer 30

Alkenes are unsaturated hydrocarbons containing a least one double carbon, C=C, bond (a C=C bond comprises of a π-bond and a σ-bond). Aliphatic alkenes that contain one double bond have the general formula C(n)H(2n). Cyclic alkenes and alkenes with more than one double bond do not obey the general formula.

Question 31

4.1.3 a) Explain the nature of a double carbon bond (C=C).

Answer 31

A double carbon bond (C=C) comprises of a σ-bond (an overlap of orbitals directly between the bonding atoms) and a π-bond (a sideways overlap of adjacent p-orbitals, above and below the bonding C atoms).
i.e. In a π-bond, two p-orbitals, one from each bonding atom, overlap above and below the plane of the carbon atoms, with each p-orbital containing one electron. So the π-bond has two electrons shared between the bonding atoms, with the electron density concentrated above and below the bonding atoms).
The π-bond restricts rotation around the double bond.

Question 32

4.1.3 b) Explain the shape and bond angles in alkenes.

Answer 32

There are three regions of electron density around each of the carbon atoms in a double bond: two single covalent bonds, and one double bond (with four covalent bonds in total). Because there are no lone pairs, each of the three electron densities repel by the same amount, i.e. as far apart as possible, and the bond angle around each carbon atom is 120°. This results in a trigonal planar arrangement. Carbon atoms in a C=C double bond, and the atoms bonded directly to these carbons, all lie in the same plane.

Question 33

4.1.3 c) i) What are stereoisomers?

Answer 33

Stereoisomers: compounds with the same structural formula, but with a different arrangement in space. There are two types of stereoisomerism: E/Z isomerism and optical isomerism.

Question 34

4.1.3 c) i) What are E/Z isomers?

Answer 34

E/Z isomers are a type of stereoisomer. They are a result of the restricted rotation around the C=C bond (due to the π-bond’s electron density being above and below the plane of a σ-bond). This results in the groups attached to each carbon atom being fixed relative to each other. In an E/Z isomer, there must be two different groups attached to each carbon in a C=C double bond.

Question 35

4.1.3 c) i) What are cis-trans isomers?

Answer 35

cis-trans isomers, a type of E/Z isomers, exist when two of the substituent groups attached to each carbon atom of the C=C group are the same (i.e. the two carbon atoms have one group in common).
- ‘cis’ means that the same groups are on the same side of the double bond.
- ‘trans’ means that the same groups are on opposite sides of the double bond.
The use of E as equivalent to trans and Z as equivalent to cis is only consistently correct when there is an H on each carbon atom of the C=C bond.

Question 36

4.1.3 c) ii) How are Cahn-Ingold-Prelong (CIP) priority rules used to identify E and Z stereoisomers?

Answer 36

The Cahn-Ingold-Prelong (CIP) priority rules require assigning each substituent a priority based on their relative mass: the molecule with the highest mass (on one carbon) is given the highest priority (on that carbon) - do this for both carbons in the C=C double bond (if the atoms directly bonded to one carbon atom are the same, then look further along the chain). If the two highest priority groups (for each carbon) are on the same side of the C=C double bond, then the isomer is Z. If the highest priority groups are on different sides of the C=C bond, then the isomer is E.
e.g. 2-bromo-1-chloropropene: The left-hand carbon in the double bond is bonded to a H and a Cl (higher priority of the two). The right-hand carbon is bonded to a CH3 and Br (higher priority of the two). If the two higher-priority groups (Cl and Br) are diagonally opposite each other, then the molecule is called E-2-bromo-1-chloropropene. If they are on the same side, then the molecule is called Z-2-bromo-1-chloropropene.
Remember that for cis-trans isomers, the use of E as equivalent to trans and Z as equivalent to cis is only consistently correct when there is an H on each carbon atom of the C=C bond.

Question 37

4.1.3 e) Explain why alkenes have a high reactivity.

Answer 37

Alkenes are much more reactive than alkanes - this is due to the presence of a π-bond. The π-bond in a C=C double bond has a relatively low bond enthalpy: the electron density is spread out above and below the nuclei, and the π-electrons (on the outside of the double bond) are more exposed than the electrons in a σ-bond. This results in a weaker electrostatic attraction between the nuclei and the shared pair of electrons in a π-bond, which in turn requires less energy to break (especially when compared to the high bond enthalpy of a σ-bond). This is why the π-bond breaks and the σ-bond remains intact when alkenes react.
The C=C double bond also has a high electron density, which attracts electrophiles and makes it more likely to react.

Question 38

4.1.3 g) What are electrophiles?

Answer 38

Electrophiles are electron-pair acceptors. Examples include cations (positive ions), and molecules with a δ+ region.

Question 39

4.1.3 h) What are electrophilic addition reactions?

Answer 39

Electrophilic addition reactions bond two or more molecules into one product. In alkenes, the π-bond breaks, and a small molecule is added across the carbon atoms to make a saturated organic product. Electrophilic addition of alkenes usually requires heterolytic fission of a small molecule.

Question 40

4.1.3 f) i) Which reaction forms alkanes from alkenes?

Answer 40

Hydrogenation is an addition reaction, where hydrogen is added across a C=C double bond - it saturates the alkene to produce an alkane. Conditions include a temperature of 150° and a suitable catalyst (e.g. nickel). 
e.g. consider ethene to ethane:
H2C=CH2 + H2 → CH3CH3
e.g. consider buta-1,3-diene to butane
H2C=(CH2)2=CH2 + 2H2 → CH3(CH2)2CH3

Question 41

4. 1.3 f) ii) Which reaction forms dihaloalkanes from alkenes? How is this used as a test for saturation?
   h) Outline this using a reaction mechanism.

Answer 41

Halogenation is an electrophilic addition reaction - a halogen (X2) is added across a C=C double bond. Th alkene becomes saturated, and a dihaloalkane is produced. This can be used as a test for saturation: when orange bromine water is mixed with an unsaturated compound, it decolourises.
e.e. consider ethene reacting with bromine:
H2C=CH2 + Br2 → CH2BrCH2Br
1) A dipole is induced in the Br2 molecule when it nears the electron dense C=C double bond (this is because the double bond repels the electrons in Br2, polarising it). The electrons from the π-bond move to the Brδ+ atom (represented in a curly arrow going from the double bond to the Brδ+ atom), which results in a bond forming between the bromine atom and a carbon atom. Heterolytic fission of the Br2 molecule simultaneously takes place, where the electron pair is donated to Brδ- (represented in a curly arrow going from Br-Br bond to the Brδ- atom).
2) This leaves a negative :Br- ion, and a carbocation (C+) intermediate on the alkene. Two electrons from the negative bromine ion are shared with the carbocation (represented in a curly arrow going from electron on :Br- to C+), forming a second bond, and thereby making a stable product (1,2-dibromoethane).

Question 42

4. 1.3 f) iii) Which reaction forms haloalkanes from alkenes?
   h) Outline this using a reaction mechanism.

Answer 42

When hydrogen halides are added to alkenes, haloalkanes form in electrophilic addition reactions.
e.g. consider ethene reacting with hydrogen bromide:
H2C=CH2 + HBr → CH2BrCH3
The polar HBr molecule (a result of the large difference in electronegativity between the bromine and hydrogen atoms) contains the dipole H(δ+)-Br(δ-). The Hδ+ atom in hydrogen bromide is attracted to the electron dense C=C double bond in ethene, and vice versa (i.e. the electron pair in the π-bond is attracted to the Hδ+ atom). The electrons from the π-bond move to the Hδ+ atom, causing the C=C double bond to break (represented in a curly arrow going from the double bond to the Hδ+ atom). This results in a bond forming between the hydrogen on hydrogen bromide and the carbon on ethene. Heterolytic fission of the HBr molecule will then take place simultaneously (this is represented in a curly arrow going from H-Br bond to the Brδ- atom).
2) This leaves a negative :Br- ion, and a carbocation (C+) intermediate on the alkene. Two electrons from the negative bromine ion are shared with the carbocation (represented in a curly arrow going from electron on :Br- to C+), forming a second bond, and making a stable product (2-bromobutane).

Question 43

4.1.3 f) iv) Which reaction forms alcohols from alkenes?

Answer 43

Hydration is an addition reaction, where steam, H2O(g), is added to a gaseous alkene, in the presence of an acid catalyst (such as H3PO4), to make an alcohol.
e.g consider the formation of ethanol from ethene:
H2C=CH2(g) + H2O(g) → C2H5OH

Question 44

4.1.3 i) How can Markownikoff’s rule predict the formation of a major organic product in addition reaction of H-X to unsymmetrical alkenes?

Answer 44

When a hydrogen halide is added to an unsymmetrical alkene, there are two possible products - experimentally, one product is more likely to be produced than another product (i.e. this is the major product).
Markownikoff’s rule predicts the formation of a major organic product: it states that the hydrogen (from the hydrogen halide) becomes attached to the carbon with the most hydrogen atoms to start with. This means that the other carbon atom in the double bond (which has more alkyl groups attchaed to it) is the carbon that will go on to form the carbocation. The reasoning for this is that carbocations with more alkyl groups are more stable than those with hydrogens (each alkyl group donates electrons towards the positive carbocation, making it more stable). The hydrogen (from the hydrogen halide) is therefore more likely to attach itself to the carbon that forms the least stable carbocation - this allows the carbocation that actually forms to be as stable as possible.

Question 45

4.1.3 j) Which process forms polymers from alkene monomers?

Answer 45

A polymer is a very long macromolecule chain, formed from multiple repeat units of smaller molecules (known as monomers). Unsaturated alkene molecules act as monomers, and undergo addition polymerisation to produce long saturated chains containing no double bonds. During addition polymerisation of alkenes, the monomers have their π-bond broken, and an electron from each π-bond is donated to form a σ-bond with a neighbouring carbon atom on a different monomer.

Question 46

4. 1.3 j) i) Deduce the repeat unit of an addition polymer, from the given monomer: H2C=CH2.
   ii) Deduce the monomer from the given section of an addition polymer: -CH2-CH2-CH2-CH2-

Answer 46

The repeat unit of an addition polymer can be deduced from a given monomer:
e.g. nH2C=CH2 → [-CH2-CH2-]n
The monomer can be deduced from a section of an addition polymer:
e.g. -CH2-CH2-CH2-CH2-
the repeat unit: [-CH2-CH2-]
the monomer: H2C=CH2

[Note: A repeat unit is the specific arrangement of atoms in a polymer that is repeated over and over again. A repeat unit is always written in square brackets, and the letter n is placed after the square brackets to show there is a large number of repeats]

Question 47

4.1.3 k) Why do plastics need to be processed in different ways (rather than in landfills)?

Answer 47

Most polymers are oil-derived plastics, and while cheap, they are non-biodegradable - this is a result of their lack of reactivity, which, while beneficial at times, is a problem when disposing plastics. Alongside this, plastics are usually produced from crude oil and are therefore non-renewable.
There are benefits for sustainability when processing waste polymers in different ways (as opposed to using landfills, where plastics become dangerous to the wildlife, can’t be reused and ruin the landscape).

Question 48

4.1.3 k) Outline some methods for processing waste polymers.

Answer 48

Plastic waste can be reused is various ways, e.g. they can be recycled (once sorted, the plastics are melted and then remoulded into other products) - this conserves finite fossil fuels and decreases the amount of waste going to landfill. Other processes that you need to know include:

i) The combustion of plastics for energy production - the heat produced can be used to manufacture steam, which in turn drives a turbine that generates electricity. However, carbon dioxide (a greenhouse gas) is often produced as a result.
iii) While burning plastics provides a good method of disposal, often toxic waste products are produced. For example, when halogenated plastics (e.g. PVC) are combusted, HCl (a toxic gas) is produced. This can be removed using gas scrubbers (such as CaO), which neutralise the acidic gas by allowing it to react with a base. So the development of products that remove toxic waste is vital when disposing plastic.
ii) Polymers can be used as organic feedstock (by being cracked into monomers) for the production of plastics and other organic chemicals.

Question 49

4.1.3 l) Explain what biodegradable polymers are, outlining their advantages and disadvantages.

Answer 49

Biodegradable polymers can be decomposed (e.g. by the action of microorganisms). While biodegradable polymers can be mixed with an addition polymer, a bioplastic is a material made solely from a renewable, source that is biodegradable. Despite being renewable, biodegradable polymers are expensive. They also need the right conditions before they’ll decompose (such as oxygen and moisture - which you won’t necessarily find in a landfill), so you need to collect and separate the biodegradable polymers from non-biodegradable plastics (which requires time and can be costly). However, they do reduce dependency on finite resources and alleviate problems from disposal of persistent plastic waste.

Question 50

4.1.3 l) Explain what photodegradable polymers are, outlining their advantages and disadvantages.

Answer 50

Photodegradable polymers are decomposed when exposed to light (or energy with wavelengths similar to light). However, photodegradable plastics in landfills may not be exposed to sufficient light, and therefore won’t degrade - this means they need to be collected and sorted (which requires time and money) from non-biodegradable plastics. Despite being expensive, they do reduce dependency on finite resources and alleviate problems from disposal of persistent plastic waste.