Worked Solutions
Module 7: Organic Chemistry — Worked Solutions (HSC Chemistry)
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Worked examples for HSC Chemistry Module 7: Organic Chemistry. Each shows where the marks are awarded, the key idea, and the full solution explained by your choice of tutor — Stella, Ella or Cassie.
How to use these
Try each question first, then check your working. Use the tutor tabs to read the full solution in the style that suits you: Stella is direct and challenging, Ella is warm and explains the why, and Cassie is concise and analytical.
Example 1 — Nomenclature and structural isomers
Question
A saturated hydrocarbon has the molecular formula $\text{C}_4\text{H}_{10}$. Draw or describe the two structural isomers and give the correct IUPAC name of each.
Solution
$\text{C}_4\text{H}_{10}$ is a saturated alkane (it fits $\text{C}_n\text{H}_{2n+2}$), so the isomers differ only in carbon-chain branching.
Isomer 1 — butane: an unbranched chain of four carbons, $\text{CH}_3\text{-CH}_2\text{-CH}_2\text{-CH}_3$.
Isomer 2 — 2-methylpropane: a three-carbon chain with a methyl branch on the middle carbon, $(\text{CH}_3)_3\text{CH}$.
Two isomers only. Both share $\text{C}_4\text{H}_{10}$ but have different connectivity — that's what makes them structural isomers. Name from the longest chain and number to give the branch the lowest locant.
First, what does "saturated" tell us? It means all carbon–carbon bonds are single, so this is an alkane and fits the general formula $\text{C}_n\text{H}_{2n+2}$ — and $\text{C}_4\text{H}_{10}$ checks out. Structural isomers then come from rearranging how those four carbons are joined.
Isomer 1 is the straight chain: $\text{CH}_3\text{-CH}_2\text{-CH}_2\text{-CH}_3$. The longest chain is four carbons, so it's butane.
Isomer 2 branches: take a three-carbon chain and attach a $\text{-CH}_3$ (methyl) group to the central carbon, $(\text{CH}_3)_3\text{CH}$. The longest chain is now three carbons (propane) with a methyl branch on carbon 2, giving 2-methylpropane.
These two are the only possibilities for $\text{C}_4\text{H}_{10}$. They're structural isomers because they share a molecular formula but differ in how the atoms are connected.
Saturated → alkane, $\text{C}_n\text{H}_{2n+2}$. Two isomers:
- Butane: $\text{CH}_3\text{-CH}_2\text{-CH}_2\text{-CH}_3$ (straight chain)
- 2-methylpropane: $(\text{CH}_3)_3\text{CH}$ (methyl branch on central C)
Same formula, different connectivity → structural isomers.
Where the marks go
- 1 mark: Correctly draws/describes the straight-chain isomer
- 1 mark: Correctly draws/describes the branched isomer
- 1 mark: Correct IUPAC names: butane and 2-methylpropane
Key idea
Structural isomers share a molecular formula but differ in connectivity; name from the longest carbon chain with branches as substituents at the lowest locant.
Example 2 — Reactions of functional groups
Question
Ethene, $\text{C}_2\text{H}_4{}_{(g)}$, can be converted to ethanol, which can then be oxidised. (i) Write the balanced equation for the hydration of ethene to ethanol and name the reaction type. (ii) Write the balanced equation for the complete combustion of ethanol, $\text{C}_2\text{H}_5\text{OH}_{(l)}$.
Solution
(i) Hydration adds water across the $\text{C=C}$ double bond — an addition reaction (acid-catalysed):
$$\text{C}_2\text{H}_4{}_{(g)} + \text{H}_2\text{O}_{(g)} \rightarrow \text{C}_2\text{H}_5\text{OH}_{(l)}$$
It's an addition because the double bond opens and the molecule gains atoms without losing any.
(ii) Complete combustion gives $\text{CO}_2$ and $\text{H}_2\text{O}$ only. Balance C, then H, then O:
$$\text{C}_2\text{H}_5\text{OH}_{(l)} + 3\text{O}_2{}_{(g)} \rightarrow 2\text{CO}_2{}_{(g)} + 3\text{H}_2\text{O}_{(l)}$$
Check the oxygen last — the $\text{OH}$ oxygen is easy to forget when you count.
(i) Ethene has a carbon–carbon double bond, and that bond is reactive: it can "open up" and let new atoms join on. When we add water across it we get ethanol, and because atoms are being added with nothing lost, we call it an addition reaction (it needs an acid catalyst):
$$\text{C}_2\text{H}_4{}_{(g)} + \text{H}_2\text{O}_{(g)} \rightarrow \text{C}_2\text{H}_5\text{OH}_{(l)}$$
(ii) "Complete" combustion means plenty of oxygen, so the only products are carbon dioxide and water. I balance carbon first (2 C → $2\text{CO}_2$), then hydrogen (6 H → $3\text{H}_2\text{O}$), and finally oxygen. The right side now needs $2(2) + 3 = 7$ oxygen atoms; one comes from the $\text{OH}$ in ethanol, leaving 6 from $\text{O}_2$, which is $3\text{O}_2$:
$$\text{C}_2\text{H}_5\text{OH}_{(l)} + 3\text{O}_2{}_{(g)} \rightarrow 2\text{CO}_2{}_{(g)} + 3\text{H}_2\text{O}_{(l)}$$
(i) Hydration (addition across $\text{C=C}$):
$$\text{C}_2\text{H}_4{}_{(g)} + \text{H}_2\text{O}_{(g)} \rightarrow \text{C}_2\text{H}_5\text{OH}_{(l)}$$
- Reaction type: addition
(ii) Complete combustion → $\text{CO}_2 + \text{H}_2\text{O}$:
$$\text{C}_2\text{H}_5\text{OH}_{(l)} + 3\text{O}_2{}_{(g)} \rightarrow 2\text{CO}_2{}_{(g)} + 3\text{H}_2\text{O}_{(l)}$$
Where the marks go
- 1 mark: Correct balanced hydration equation forming ethanol
- 1 mark: Identifies the reaction as an addition reaction
- 1 mark: Correct products ($\text{CO}_2$ and $\text{H}_2\text{O}$) for complete combustion
- 1 mark: Correctly balanced combustion equation with $3\text{O}_2$
Key idea
Addition across a $\text{C=C}$ double bond opens it to gain atoms (hydration → alcohol); complete combustion of an alcohol gives only $\text{CO}_2$ and $\text{H}_2\text{O}$.