NCEA Level 3 Chemistry 91391 Organic Chemistry: past papers, practice and marking
- NCEA Level 3
- Chemistry
- AS91391
- 5 credits
- External exam
About 91391
AS91391, Demonstrate understanding of the properties of organic compounds. Functional groups, isomerism, chemical tests and reaction schemes. The standard where knowing which test distinguishes what is worth the most marks.
What 91391 covers
- NamingIUPAC names and structural formulae (amines: primary only)
- IsomerismConstitutional isomers from properties, cis–trans, enantiomers drawn in 3D
- Chemical testsPositive tests with start and end colours, the order of tests, deducing structures
- Reaction schemesDevise a route or complete a flowchart, with reagents and conditions
- Condensation and hydrolysisEsters, amides, dipeptides, polyesters and polyamides; hydrolysis in acid and base
- Reaction conditionsReflux versus distillation, aqueous versus alcoholic, purifying an ester
91391 past papers
Open a 91391 past paper beside your own Ako page, then get it marked, part by part.
Work through a past paper All NCEA Level 3 Chemistry past papers
Practice questions
Question 1: Naming
Name the following compounds: (i) CH₃CH₂CH₂COOH (ii) CH₃CH₂COOCH₂CH₃ (iii) CH₃CH₂CONHCH₃ (iv) CH₃CH(NH₂)COOH
- Give the IUPAC name of each and identify the functional group.(up to Merit)
- Explain how you can tell from the formula CH₃CH₂COOCH₂CH₃ which part came from the acid and which from the alcohol.(up to Merit)
Check your answer
Part a: what gets the marks
- (i) is butanoic acid, a carboxylic acid: the –COOH carbon is C1 of a four-carbon chain.
- (ii) is ethyl propanoate, an ester: the alcohol part (ethyl, two carbons) is named first and the acid part (propanoate, three carbons including the C=O) second.
- (iii) is N-methylpropanamide, an amide: three carbons from the acid part and a methyl group on the nitrogen.
- (iv) is 2-aminopropanoic acid, a carboxylic acid with an amine group on C2.
A full answer: (i) butanoic acid, carboxylic acid. (ii) ethyl propanoate, ester. (iii) N-methylpropanamide, amide. (iv) 2-aminopropanoic acid, which has an amine group and a carboxylic acid group (an amino acid).
Part b: what gets the marks
- The C=O carbon always comes from the acid, because the acid supplies the whole –COOH group except its –OH.
- In CH₃CH₂–C(=O)–O–CH₂CH₃ the single-bonded O joins the C=O carbon to the CH₂CH₃, so CH₃CH₂CO– is the acid part and –CH₂CH₃ the alcohol part.
- The name reverses the written order: the alcohol part is named first as the alkyl group (ethyl) and the acid part second with the -oate ending (propanoate).
A full answer: Locate the carbonyl first. The C=O carbon always comes from the acid. This is because esterification takes the whole –COOH group and removes only its –OH, leaving the carbonyl carbon and its double-bonded oxygen intact and attached to the rest of the acid's carbon chain.
Follow the connectivity.
CH₃CH₂–C(=O)–O–CH₂CH₃
The acid part is everything up to and including the C=O carbon: CH₃CH₂–CO–, three carbons in total, from propanoic acid.
The single-bonded oxygen is the bridge. It came from the alcohol: specifically, it is the oxygen of the alcohol's –OH, with its hydrogen removed.
The alcohol part is everything beyond that oxygen: –CH₂CH₃, from ethanol.
The rule to remember: the single-bonded oxygen connects to the alcohol carbon. Find the C=O, look at the oxygen next to it, and everything on the far side came from the alcohol.
Where the water came from.
CH₃CH₂COOH + HO–CH₂CH₃ → CH₃CH₂COOCH₂CH₃ + H₂O
The –OH from the acid and the –H from the alcohol combine to form the water, which is why the ester's bridging oxygen is the alcohol's, not the acid's.
The naming consequence. The name reverses the written order: the alcohol part is named first as the alkyl group, and the acid part second with the -oate ending.
CH₃CH₂COOCH₂CH₃ = ethyl propanoate
So the name reads right-to-left across the formula. This is the standard trap: writing "propyl ethanoate", which is a different compound entirely (CH₃COOCH₂CH₂CH₃). Identifying the C=O first, then working outward, makes the assignment unambiguous every time.
What gets each grade
Question 2: Reaction schemes
Butanone, CH₃COCH₂CH₃, is converted into the ester butan-2-yl ethanoate (1-methylpropyl ethanoate), CH₃COOCH(CH₃)CH₂CH₃, in two steps.
- Give the reagents and conditions for each step, and the structural formula of the intermediate.(up to Merit)
- Identify the type of reaction in each step.(up to Achieved)
Check your answer
Part a: what gets the marks
- Butanone is reduced with NaBH₄ to butan-2-ol, CH₃CH(OH)CH₂CH₃.
- Butan-2-ol is heated under reflux with ethanoic acid and concentrated H₂SO₄ to form the ester.
A full answer: Step 1: butanone + NaBH₄ → butan-2-ol, CH₃CH(OH)CH₂CH₃. Step 2: butan-2-ol + ethanoic acid, CH₃COOH, with concentrated H₂SO₄, heat under reflux → CH₃COOCH(CH₃)CH₂CH₃.
Part b: what gets the marks
- Step 1 is reduction, and step 2 is condensation (esterification, a substitution), with water released.
A full answer: Step 1: reduction. Step 2: condensation (esterification), releasing water.
What gets each grade
Question 3: Isomerism
Compound P is 3-aminobutanoic acid, CH₃CH(NH₂)CH₂COOH.
- Explain how the two enantiomers could be distinguished.(up to Merit)
- Compound Q is 2,3-dihydroxybutanal, CH₃CH(OH)CH(OH)CHO. Identify every asymmetric carbon atom in Q, justifying each.(up to Merit)
Check your answer
Part a: what gets the marks
- The two enantiomers have identical melting points, boiling points and solubility, so these cannot tell them apart.
- They rotate the plane of plane-polarised light by the same amount in opposite directions, so a polarimeter tells them apart.
A full answer: Pass plane-polarised light through a solution of each. One enantiomer rotates the plane of the light clockwise and the other anticlockwise, by the same amount. Their other physical properties are identical, so this is how they are distinguished.
Part b: what gets the marks
- C2 of Q is asymmetric: it carries four different groups, –H, –OH, –CHO and –CH(OH)CH₃.
- C3 of Q is asymmetric: it carries four different groups, –H, –OH, –CH₃ and –CH(OH)CHO.
- C1 (the CHO carbon) has only three groups and C4 (CH₃) carries three H atoms, so neither is asymmetric.
A full answer: Two asymmetric carbons: C2 and C3. C2 is bonded to –H, –OH, –CHO and –CH(OH)CH₃; C3 is bonded to –H, –OH, –CH₃ and –CH(OH)CHO. Each has four different groups. C1 (the aldehyde carbon) and C4 (a CH₃) are not asymmetric.