SCSA WACE 2020-style practice: An organic compound has a molecular ion peak at m/z = 60 in its mass spectrum, a broad infrared absorption near 3000\ cm^-1 together with a strong band near 1710\ cm^-1, and two peaks in its proton NMR. Deduce the structure and justify each piece of evidence.

A 6 mark question rewards using each technique to build and confirm the structure. The molecular ion at m/z = 60 gives a molar mass of 60\ g mol^-1, consistent with C_2H_4O_2 (ethanoic acid, 2(12) + 4(1) + 2(16) = 60). The broad infrared band near 3000\ cm^-1 indicates an O-H of a carboxylic acid (broadened by hydrogen bonding), and the strong band near 1710\ cm^-1 indicates a C=O carbonyl, together pointing to a -COOH group. The proton NMR shows two signals, consistent with two hydrogen environments: the -CH_3 protons and the acidic -COOH proton. All three lines of evidence agree on ethanoic acid, CH_3COOH. Markers reward M = 60 giving the formula, the O-H and C=O assignments, the two NMR environments, and the consistent conclusion of ethanoic acid.

§-Syllabus dot point

WAChemistrySyllabus dot point

How do chemists design efficient syntheses and use instrumental techniques to identify and measure substances?

Evaluate chemical synthesis using percentage yield and atom economy and green chemistry principles, and interpret instrumental analysis data to identify substances

A focused answer to the WACE Year 12 Chemistry dot point on percentage yield, atom economy, green chemistry and instrumental analysis, with a worked yield calculation and common mistakes.

Generated by Claude Opus 4.89 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking

Designing a chemical synthesis means choosing reactions that convert available starting materials into a target product efficiently, safely and with minimal waste. Two quantitative measures judge that efficiency, and modern practice is guided by green chemistry.

Percentage yield

The percentage yield compares the amount of product actually obtained with the maximum predicted by stoichiometry (the theoretical yield):

\%\,\text{yield} = \frac{\text{actual yield}}{\text{theoretical yield}} \times 100

Real syntheses never reach 100 percent because of side reactions, incomplete reactions (equilibria), losses during purification, and impurities in the reactants.

Atom economy

Atom economy measures how much of the total mass of all reactants ends up in the desired product, rather than in by-products:

\%\,\text{atom economy} = \frac{\text{molar mass of desired product}}{\text{total molar mass of all products}} \times 100

A reaction can have a high yield but poor atom economy if it produces large amounts of unwanted by-products. Addition reactions tend to have high atom economy (everything ends up in one product); substitution and elimination reactions have lower atom economy because they release small molecules.

Green chemistry

Green chemistry designs syntheses to reduce environmental impact: maximising atom economy and yield, using renewable feedstocks, avoiding toxic or hazardous reagents and solvents, using catalysts rather than stoichiometric reagents, minimising energy use, and designing products and by-products that are safe and degradable. The goal is to prevent waste rather than to clean it up afterwards.

Instrumental analysis

Chemists identify and measure substances using instrumental techniques. The WACE course expects you to interpret data from:

Mass spectrometry (MS): measures the mass of the molecule (molecular ion peak gives the molar mass) and the masses of fragments, which help deduce structure.
Infrared (IR) spectroscopy: absorption bands at characteristic wavenumbers reveal functional groups (for example a broad band near $3300\ \text{cm}^{-1}$ for O-H, a strong band near $1700\ \text{cm}^{-1}$ for C=O).
Nuclear magnetic resonance (NMR) spectroscopy: the number of signals, their chemical shifts and the relative peak areas indicate the different proton (or carbon) environments and how many of each.

Combining these techniques lets a chemist confirm the identity and purity of a synthesised product.

When answering synthesis questions in the WACE examination, identify the limiting reagent before computing the theoretical yield, keep yield and atom economy separate, and quote IR and NMR evidence specifically (state the wavenumber or shift and the group it indicates).

Exam-style practice questions

Practice questions written in the style of SCSA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

WACE 20227 marksEthene can be converted to ethanol by two routes: (i) hydration,

\text{C}_2\text{H}_4 + \text{H}_2\text{O} \rightarrow \text{C}_2\text{H}_5\text{OH}

; (ii) reaction with concentrated sulfuric acid then water, which overall also gives ethanol but produces some by-products. (a) Calculate the atom economy of route (i). (b) A hydration plant uses

2.80\ \text{kg}

of ethene and obtains

3.45\ \text{kg}

of ethanol. Calculate the percentage yield. (

M(\text{C}_2\text{H}_4)=28.0

M(\text{C}_2\text{H}_5\text{OH})=46.0\ \text{g mol}^{-1}

.) (c) State why route (i) is preferred on green-chemistry grounds.

Show worked answer →

A 7 mark question rewards the atom economy, the yield, and the green-chemistry comparison.

(a) In route (i) the only product is ethanol, so all reactant mass ends up in the product:

\text{atom economy} = \frac{46.0}{28.0 + 18.0} \times 100 = \frac{46.0}{46.0} \times 100 = 100\%.

(b) Moles of ethene $= \dfrac{2800}{28.0} = 100\ \text{mol}$ , so theoretical ethanol $= 100\ \text{mol}$ , theoretical mass $= 100 \times 46.0 = 4600\ \text{g} = 4.60\ \text{kg}$ .

\%\,\text{yield} = \frac{3.45}{4.60} \times 100 = 75\%.

(c) Route (i) is a direct addition with $100\%$ atom economy (no by-products) and uses a reusable acid catalyst rather than stoichiometric concentrated sulfuric acid, so it generates less waste and uses fewer hazardous reagents, both key green-chemistry aims.

Markers reward $100\%$ atom economy, $75\%$ yield (from the limiting ethene), and a green-chemistry reason citing atom economy or waste.

WACE 20206 marksAn organic compound has a molecular ion peak at

m/z = 60

in its mass spectrum, a broad infrared absorption near

3000\ \text{cm}^{-1}

together with a strong band near

1710\ \text{cm}^{-1}

, and two peaks in its proton NMR. Deduce the structure and justify each piece of evidence.

Show worked answer →

A 6 mark question rewards using each technique to build and confirm the structure.

The molecular ion at $m/z = 60$ gives a molar mass of $60\ \text{g mol}^{-1}$ , consistent with $\text{C}_2\text{H}_4\text{O}_2$ (ethanoic acid, $2(12) + 4(1) + 2(16) = 60$ ).

The broad infrared band near $3000\ \text{cm}^{-1}$ indicates an $\text{O-H}$ of a carboxylic acid (broadened by hydrogen bonding), and the strong band near $1710\ \text{cm}^{-1}$ indicates a C=O carbonyl, together pointing to a $\text{-COOH}$ group.

The proton NMR shows two signals, consistent with two hydrogen environments: the $\text{-CH}_3$ protons and the acidic $\text{-COOH}$ proton.

All three lines of evidence agree on ethanoic acid, $\text{CH}_3\text{COOH}$ .

Markers reward $M = 60$ giving the formula, the $\text{O-H}$ and C=O assignments, the two NMR environments, and the consistent conclusion of ethanoic acid.