Construct reaction pathways linking the functional groups studied in Module 7 and apply retrosynthesis logic to plan multi-step syntheses, including reagents and conditions for each step

What this dot point is asking

NESA wants you to map all the functional-group conversions in Module 7 onto a single tree (alkane to alkene to alcohol to aldehyde to acid to ester, plus side branches to haloalkanes and amines), then plan a synthesis from a given starting material to a given target. The skill being tested is retrosynthesis: working backwards from the target, asking "what could have made this in one step?", and repeating until you reach the starting material.

The answer

The master synthesis tree (Module 7)

Going forward (oxidation direction or chain growth):

Each arrow has a specific reagent and condition.

Forward conversions (reagent reference)

Retrosynthesis: thinking backwards

To plan a synthesis, work backwards from the target. At each step, ask:

1. What functional group is on the target?
2. What is the most common one-step reaction that produces this functional group?
3. What is the precursor (the synthon) for that step?
4. Is that precursor accessible from the given starting material? If not, recurse.

Worked example: butan-2-ol from but-1-ene.

• Target: butan-2-ol ($CH_3CH_2CH(OH)CH_3$). Secondary alcohol.
• Backwards: a secondary alcohol comes from Markovnikov hydration of an alkene (water on the more substituted carbon). The alkene precursor with the C-OH at C2 is but-1-ene.
• Forward synthesis (one step): $CH_2=CHCH_2CH_3 + H_2O \xrightarrow{H_2SO_4} CH_3CH(OH)CH_2CH_3$.

Worked example: ethyl ethanoate from ethene.

• Target: ethyl ethanoate (ester). An ester comes from acid + alcohol with $H_2SO_4$.
• Precursors: ethanoic acid and ethanol.
• Ethanol comes from ethene by hydration.
• Ethanoic acid comes from ethanol by oxidation (reflux with $K_2Cr_2O_7$).
• Full forward synthesis: ethene to ethanol (split into two portions); oxidise half to ethanoic acid; esterify with the other half. Three forward steps.

Worked example: methyl propanoate from propan-1-ol.

• Target: methyl propanoate. From propanoic acid + methanol.
• Propanoic acid: oxidise propan-1-ol with excess acidified dichromate under reflux. We have propan-1-ol.
• Methanol: not given, must come from elsewhere (methane + Cl₂ to chloromethane, then NaOH; or accept the question wording that allows methanol as a reagent).
• Forward: oxidise propan-1-ol to propanoic acid, esterify with methanol and conc $H_2SO_4$.

Strategies for planning

Count the carbons

The target and starting material must have compatible carbon skeletons. HSC does not include carbon-skeleton-changing reactions (no $C-C$ bond formation), so the carbon count is preserved through every step.

Check the oxidation level

Alkane and alkene are at the same level for the carbon involved. Alcohol is one step up. Aldehyde and ketone are another step. Carboxylic acid is one more. Esters and amides are at the same level as carboxylic acids.

Identify the limiting step

Markovnikov hydration always gives the more substituted alcohol. To get the less substituted alcohol, use a haloalkane route (HBr addition followed by hydrolysis) or accept the Markovnikov product and work from there.

Use reflux when stated

Esterification, oxidation to acid, hydrolysis, and base hydrolysis all require reflux. Distillation only is for collecting an aldehyde or separating an ester after the reaction.

Common synthesis sequences

Examples in context

Example 1. Multi-step synthesis of ethyl ethanoate from ethene at Qenos Botany. A working flowsheet at the Botany Industrial Park converts ethene to ethyl ethanoate in three steps: (1) hydration of ethene with dilute sulfuric acid gives ethanol; (2) oxidation of half the ethanol stream with acidified dichromate gives ethanoic acid; (3) Fischer esterification combines the remaining ethanol with the ethanoic acid under concentrated $H_2SO_4$ catalyst. The plant uses the same retrosynthesis logic HSC students apply: work backwards from ester, identify alcohol plus acid, identify common alkene precursor. The carbon skeleton is conserved throughout because no $C-C$ bond is broken or formed.

Example 2. Synthesising 2-bromopropane from propan-2-ol in NSW HSC depth study. A common Stage 6 depth-study challenge is to convert propan-2-ol to 2-bromopropane. The two-step pathway is: (1) dehydrate the alcohol with concentrated $H_2SO_4$ at 170 degrees C to give propene; (2) add HBr across the double bond following Markovnikov to give 2-bromopropane (the H attaches to the carbon already bearing more H atoms). Students draw the synthesis as a flowchart with reagents and conditions on each arrow. NESA markers reward the correct intermediate and the correct conditions; an attempted direct substitution of OH with Br earns zero because HSC scope does not include that reaction.

Try this

Q1. Identify the reagent and condition required for each one-step conversion: (a) ethene to ethanol; (b) ethanol to ethanal; (c) ethanal to ethanoic acid. [3 marks]

• Cue. (a) Dilute $H_2SO_4$, water. (b) Acidified $K_2Cr_2O_7$, gentle heat with distillation. (c) Acidified $K_2Cr_2O_7$, reflux.

Q2. Plan a synthesis of propanoic acid starting from propan-1-ol. Calculate the mass of propanoic acid that could theoretically be made from 30.0 g of propan-1-ol. [3 marks]

• Cue. One-step oxidation with reflux acidified $K_2Cr_2O_7$; $n(\text{alcohol}) = 30.0 / 60.10 = 0.499$ mol; mass acid $= 0.499 \times 74.08 = 37.0$ g.

Q3. Plan a multi-step synthesis of methyl propanoate from propan-1-ol and methanol. (a) Draw the flowchart with intermediate. (b) State the reagent and condition for each step. (c) State the equilibrium constant constraint that limits Fischer esterification yield. [2+2+1 marks]

• Cue. (a) Propan-1-ol $\rightarrow$ propanoic acid $\rightarrow$ methyl propanoate. (b) Step 1: acidified $K_2Cr_2O_7$, reflux. Step 2: methanol with conc $H_2SO_4$ catalyst, reflux. (c) Equilibrium $K \approx 4$ means typical yield $\approx 67$ percent without removal of water.