A hydrocarbon is a compound of carbon and hydrogen only. Saturated compounds contain only single carbon-carbon bonds (no double bonds).

CH₃CH₂CHCH₂ is but-1-ene and is unsaturated (but is a hydrocarbon). CH₃COCH₃ and CH₃CH₂OH both contain oxygen and therefore are not hydrocarbons.

CH₃CH₂CH(CH₃)CH₃ is methylbutane, a saturated hydrocarbon and is the correct answer.

Alkanes are relatively unreactive because they have strong carbon-carbon and carbon-hydrogen bonds that require a lot of energy to break. In addition, the bonds are all non-polar, which means they are not vulnerable to attack by most chemical reagents.

Alkanes do not have delocalised electrons.

Therefore 1 and 2 only is the correct answer.

C₅H₁₂ + 8O₂ → 5CO₂ + 6H₂O is the balanced equation, so the total sum is 20 - don't forget that C₅H₁₂ does have a '1' in front of it, although by convention we don't actually write that into the equation.

(When balancing combustion reactions of hydrocarbons, first balance out carbon and hydrogen atoms by adding coefficients (numbers) in front of CO₂ and H₂O, after that balance the oxygen atoms and the hydrocarbon as required.)

Propene contains a double carbon-carbon bond (C=C) that will readily undergo an addition reaction with bromine. The bromine molecule adds across the double bond, with one atom attaching to each carbon atom of the double bond.

Thus CH₃CHBrCH₂Br is the correct answer.

Ethanol is a primary alcohol. When oxidised primary alcohols produce an aldehyde initially (must be distilled off immediately as formed) that then readily reacts further to form a carboxylic acid (under reflux).

Ethanol will therefore form ethanoic acid (CH₃CO₂H).

CH₃CHO is an aldehyde (ethanal), CH₃CH₂CO₂H is propanoic acid (too many carbons), CH₃COOCH₃ is methyl ethanoate, an ester.

CH₃CO₂H is the correct answer.

Alkanes will completely combust in excess oxygen, producing carbon dioxide and water. If there is insufficient oxygen then carbon monoxide and carbon (soot) may be formed. Hydrogen gas will never be formed (it is too reactive especially at high temperature).

Therefore 1 and 3 only is the correct answer.

Typically in free radical substitution an alkane reacts with a halogen and hydrogen atoms are readily substituted with halogen atoms.

Initiation involves UV light breaking the halogen bond (homolytic fission) and producing two halogen free radicals (species with an unpaired electron - an odd number of electrons).

Propogation involves radicals (odd) reacting with non-radical species (even) to produce a radical (odd) and a non-radical (even).

Termination involves two free radicals (odd) coming together to produce a non-radical (even).

Therefore 1 and 2 only is the correct answer.

Ethene (CH₂CH₂₎ is an alkene with a double C=C bond that readily reacts through addition reactions.

1-chloroethane has formula CH₃CH₂Cl. HCl has been added to ethene, therefore HCl is the correct answer.

Addition of chlorine would produce 1,2-dichloroethane (CH₂ClCH₂Cl). Sodium chloride will not react with ethene (under any normal conditions) and exposure to UV light without a source of chlorine will not lead to a suitable product - even with chlorine this reaction would not be best used as the products would be very variable.

To produce poly(chloroethene) the monomer needs to be chloroethene. The formula of chloroethene is CHClCH₂, which is therefore the correct answer. This molecule has a double C=C bond and one chlorine atom.

The other options either have no double C=C bond (CH₂ClCH₃) or two chlorine atoms (CHClCHCl) or both (CH₂ClCH₂Cl).

Tertiary alcohols cannot be oxidised with acidified potassium dichromate(VI) so no reaction will take place, and there will be no colour change (the reaction mixture will remain orange).

Primary and secondary alcohols can be oxidised (to aldehydes or carboxylic acids and to ketones respectively) and teh colour change would be orange to green.

Therefore No change is the correct answer.

This is an esterification reaction. Esters are named by the 'alcohol bit' first then the 'acid bit', that is firstly the carbon chain section that is directly attached to the oxygen atom by a single bond, then the carbon chain section that includes the C=O.

Propanoic acid and ethanol will form the ester ethyl propanoate, the formula of which is CH₃CH₂CO₂CH₂CH₃. The other compounds are all esters, but propyl ethanoate (CH₃CO₂CH₂CH₂CH₃), methyl propanoate (CH₃CH₂CO₂CH₃) and ethyl ethanoate (CH₃CO₂CH₂CH₃).

Therefore the answer is CH₃CH₂CO₂CH₂CH₃.

Nucleophiles are electron pair donors. In order for a species to be a nucleophile it must have a lone (non-bonding) pair of electrons. The hydroxide ion (OH⁻) has three lone pairs on the oxygen atom, ammonia (NH₃) has one lone pair on the nitrogen atom. Both OH⁻ and NH₃ can therefore behave as nucleophiles.

Methane (CH₄) has no lone (non-bonding) pairs and therefore cannot behave as a nucelophile.

Therefore 1 and 2 only is the correct answer.

Halogenoalkanes will undergo nucleophilic substitution reactions will sodium hydroxide in water, to produce the corresponding alcohol. Thus chloroethane reacts with NaOH_(aq) in a nucleophilic substitution reaction to produce ethanol.

Halogenoalkanes have no double bonds and cannot undergo addition reactions. The reagent conc. sulfuric acid (H₂SO_4(aq)) is a common reagent (and catalyst) but will not react to produce the alcohol in this reaction.

Therefore the correct answer is NaOH_(aq) / nucleophilic substitution.

An electrophile is an electron pair acceptor. Electrophiles are electron deficient (often with a positive charge). Benzene will react, via an electrophilic substitution reaction with an electrophile which will replace a hydrogen atom on the benzene ring.

The nitration of benzene reaction: C₆H₆ + NO₂⁺ → C₆H₅NO₂ + H⁺

Therefore 1, 2 and 3 is the correct answer.