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Organic reaction pathways: WACE Year 12 Chemistry

Syllabus dot point

“Describe the characteristic reactions of organic families, including substitution, addition, oxidation and esterification, and combine them into multi-step pathways”

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Quick answer

Each organic family has characteristic reactions: alkanes undergo substitution with halogens, alkenes undergo addition, alcohols can be oxidised to aldehydes/ketones then carboxylic acids and undergo esterification with acids, and haloalkanes undergo substitution. A reaction pathway chains these steps to convert a starting material into a target molecule.

What this dot point is asking

Each organic functional group reacts in characteristic ways. Knowing these reactions lets you predict products and design reaction pathways, sequences of steps that convert a starting material into a target compound.

Substitution (alkanes and haloalkanes)

Alkanes are unreactive but undergo substitution with halogens in ultraviolet light: an H atom is replaced by a halogen atom, e.g. CH4+Cl2→UVCH3Cl+HCl\text{CH}_4 + \text{Cl}_2 \xrightarrow{\text{UV}} \text{CH}_3\text{Cl} + \text{HCl}. This is a free-radical reaction that gives a mixture of products.

Haloalkanes undergo substitution with aqueous hydroxide to form alcohols: CH3CH2Br+OH−→CH3CH2OH+Br−\text{CH}_3\text{CH}_2\text{Br} + \text{OH}^- \rightarrow \text{CH}_3\text{CH}_2\text{OH} + \text{Br}^-. This is a useful way to introduce an OH group.

Addition (alkenes)

Alkenes are reactive because of the C=C double bond and undergo addition, where the double bond opens and atoms add across it:

  • Hydrogenation: CH2=CH2+H2→CH3CH3\text{CH}_2=\text{CH}_2 + \text{H}_2 \rightarrow \text{CH}_3\text{CH}_3 (nickel catalyst).
  • Halogenation: addition of Br2\text{Br}_2 (the rapid decolourising of bromine water is the test for a C=C).
  • Hydration: addition of water (steam, acid catalyst) gives an alcohol: CH2=CH2+H2O→CH3CH2OH\text{CH}_2=\text{CH}_2 + \text{H}_2\text{O} \rightarrow \text{CH}_3\text{CH}_2\text{OH}.
  • Hydrohalogenation: addition of HX\text{HX} gives a haloalkane.
Addition versus substitution

Alkenes react by addition (the double bond opens and nothing is expelled). Alkanes and haloalkanes react by substitution (one atom or group replaces another, releasing a small molecule such as HCl\text{HCl}).

Oxidation (alcohols)

Primary alcohols are oxidised (by acidified dichromate or permanganate) first to an aldehyde and then to a carboxylic acid. Secondary alcohols oxidise to a ketone. Tertiary alcohols are not readily oxidised, because the carbon bearing the OH has no H to remove. For example:

CH3CH2OH→[O]CH3CHO→[O]CH3COOH\text{CH}_3\text{CH}_2\text{OH} \xrightarrow{[O]} \text{CH}_3\text{CHO} \xrightarrow{[O]} \text{CH}_3\text{COOH}

The orange dichromate turning green is the visual sign of oxidation.

Esterification (alcohol plus carboxylic acid)

A carboxylic acid reacts with an alcohol, with a concentrated sulfuric acid catalyst, to form an ester and water. This is a reversible condensation reaction:

CH3COOH+CH3CH2OH⇌CH3COOCH2CH3+H2O\text{CH}_3\text{COOH} + \text{CH}_3\text{CH}_2\text{OH} \rightleftharpoons \text{CH}_3\text{COOCH}_2\text{CH}_3 + \text{H}_2\text{O}

Esters have characteristic sweet, fruity smells.

Esterification

The acid-catalysed, reversible condensation of a carboxylic acid with an alcohol to form an ester and water.

Building a pathway

A reaction pathway links these single steps. To plan one, identify the functional group of the start and target, then choose reactions that step from one family to the next. For example, to convert ethene to ethyl ethanoate, you would hydrate ethene to ethanol, oxidise some ethanol to ethanoic acid, then esterify the two.

A multi-step synthesis from ethene

Question

Outline a pathway from ethene (CH2=CH2\text{CH}_2=\text{CH}_2) to ethyl ethanoate, giving conditions for each step.

Solution

Step 1: ethene to ethanol (addition / hydration)
CH2=CH2+H2O→CH3CH2OH\text{CH}_2=\text{CH}_2 + \text{H}_2\text{O} \rightarrow \text{CH}_3\text{CH}_2\text{OH}, using steam and a phosphoric acid catalyst.
Step 2: ethanol to ethanoic acid (oxidation)
CH3CH2OH→[O]CH3COOH\text{CH}_3\text{CH}_2\text{OH} \xrightarrow{[O]} \text{CH}_3\text{COOH}, using acidified dichromate, heated under reflux to ensure full oxidation to the acid.
Step 3: esterification
CH3COOH+CH3CH2OH⇌CH3COOCH2CH3+H2O\text{CH}_3\text{COOH} + \text{CH}_3\text{CH}_2\text{OH} \rightleftharpoons \text{CH}_3\text{COOCH}_2\text{CH}_3 + \text{H}_2\text{O}, warmed with a concentrated sulfuric acid catalyst.

Some ethanol is kept aside for the final esterification rather than all being oxidised.

When answering pathway questions in the WACE examination, give the reagents and conditions for every step, write balanced equations, and identify the functional group formed at each stage.

Over-oxidising or skipping conditions

A common error is omitting reaction conditions (catalyst, UV, reflux, acidified oxidant); the conditions carry marks. Another frequent slip is oxidising a primary alcohol only to the aldehyde when the question wants the carboxylic acid (reflux is needed), or claiming a tertiary alcohol oxidises (it does not). A third is using addition language for an alkane reaction, which is substitution.

Exam-style questions

Questions in the style of SCSA exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

WACE 2022-style8 marks
Outline a synthesis of 1-chloropropane starting from propene CH3CH=CH2\text{CH}_3\text{CH=CH}_2, where the first step must produce propan-1-ol. (a) Give the reagents, conditions and equation for each step. (b) Identify the reaction type at each step. (c) Suggest one reason the overall yield of 1-chloropropane is less than 100%100\%.
Show worked answer →

An 8 mark question rewards a valid two-step route with reagents, conditions, equations, types, and a yield comment.

Step 1 (propene to propan-1-ol): to obtain the primary alcohol (anti-Markovnikov product), this is best framed as hydration to give propan-1-ol; conditions are steam with an acid catalyst (or, in practice, a route via hydroboration, but at WACE level state addition of water across the double bond):

CH3CH=CH2+H2O→CH3CH2CH2OH.\text{CH}_3\text{CH=CH}_2 + \text{H}_2\text{O} \rightarrow \text{CH}_3\text{CH}_2\text{CH}_2\text{OH}.

Reaction type: addition (hydration).

Step 2 (propan-1-ol to 1-chloropropane): substitution of the -OH\text{-OH} group. React the alcohol with a chlorinating agent (for example HCl\text{HCl} with a catalyst, or PCl3\text{PCl}_3/SOCl2\text{SOCl}_2):

CH3CH2CH2OH+HCl→CH3CH2CH2Cl+H2O.\text{CH}_3\text{CH}_2\text{CH}_2\text{OH} + \text{HCl} \rightarrow \text{CH}_3\text{CH}_2\text{CH}_2\text{Cl} + \text{H}_2\text{O}.

Reaction type: substitution.

(c) The yield is below 100%100\% because of competing reactions (for example formation of the secondary alcohol/chloride from Markovnikov addition, or elimination to propene), incomplete reaction, and losses during separation and purification.

Markers reward the addition step with conditions and equation, the substitution step with conditions and equation, the two reaction types, and a sensible yield-loss reason.

WACE 2020-style6 marks
Compound P\text{P} is a haloalkane CH3CH2Br\text{CH}_3\text{CH}_2\text{Br}. Describe how P\text{P} can be converted in successive steps into (i) ethanol, (ii) ethanoic acid, and (iii) ethyl ethanoate. Give the reagents and conditions for each step and write the equation for step (iii).
Show worked answer →

A 6 mark question rewards each conversion with correct reagents and conditions plus the final equation.

(i) P\text{P} to ethanol (substitution)
warm CH3CH2Br\text{CH}_3\text{CH}_2\text{Br} with aqueous sodium hydroxide; the hydroxide replaces the bromide: CH3CH2Br+OH−→CH3CH2OH+Br−\text{CH}_3\text{CH}_2\text{Br} + \text{OH}^- \rightarrow \text{CH}_3\text{CH}_2\text{OH} + \text{Br}^-.
(ii) ethanol to ethanoic acid (oxidation)
heat under reflux with acidified potassium dichromate; the primary alcohol is oxidised fully to the carboxylic acid, CH3CH2OH→[O]CH3COOH\text{CH}_3\text{CH}_2\text{OH} \xrightarrow{[O]} \text{CH}_3\text{COOH} (orange to green).
(iii) esterification
warm the ethanoic acid with ethanol and a concentrated sulfuric acid catalyst:
CH3COOH+CH3CH2OH⇌CH3COOC2H5+H2O.\text{CH}_3\text{COOH} + \text{CH}_3\text{CH}_2\text{OH} \rightleftharpoons \text{CH}_3\text{COOC}_2\text{H}_5 + \text{H}_2\text{O}.

Markers reward the hydroxide substitution, the reflux-with-dichromate oxidation to the acid, and the acid-catalysed reversible esterification equation.

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