NCEA Level 2 Chemistry 91165 Properties of selected organic compounds: past papers, practice and marking
- NCEA Level 2
- Chemistry
- AS91165
- 4 credits
- External exam
About 91165
AS91165, Demonstrate understanding of the properties of selected organic compounds. Naming and drawing alkanes, alkenes, alkynes, haloalkanes, amines, alcohols and carboxylic acids, isomers, primary, secondary and tertiary, addition, substitution, elimination and oxidation, polymers, reaction schemes and the tests that tell compounds apart: the 4-credit external, in the seven question types it keeps reusing.
What 91165 covers
- NamingIUPAC names and structural formulae up to eight carbons in the longest chain, lowest locants, substituents in alphabetical order, and primary, secondary and tertiary alcohols and haloalkanes by the number of carbons bonded to the carbon carrying the group
- IsomersSame molecular formula with a different arrangement of atoms, drawing every isomer once, and cis and trans isomers: a C=C that prevents rotation and two different atoms or groups on each carbon of the double bond
- Addition reactionsAlkenes with H₂/Pt, Cl₂, Br₂, H₂O/H⁺ and hydrogen halides, the fast decolourising of bromine, and major and minor products from an asymmetric alkene with an asymmetric reagent
- PolymersPolymer from monomer and monomer from polymer, the C=C breaking to form new single bonds, the unsaturated monomer that reacts and the saturated polymer that does not, linked to the use
- Substitution and eliminationAlkanes with halogens in UV light, alcohols with HCl, PCl₃, PCl₅ and SOCl₂, haloalkanes with aqueous KOH and with ammonia, elimination with concentrated H₂SO₄ or alcoholic KOH, and Zaitsev for the major product
- Reaction schemesMissing compounds, reagents, conditions and reaction types in a scheme, oxidation of primary alcohols and alkenes with the start and end colours, and carboxylic acids and amines as acids and bases
- Identifying compoundsPlanning tests with bromine water, acidified permanganate or dichromate, litmus and carbonate, and physical properties (layers in water, boiling point) linked to the functional group and the chain length
91165 past papers
Open a 91165 past paper beside your own Ako page, then get it marked, part by part.
Work through a past paper All NCEA Level 2 Chemistry past papers
Practice questions
Question 1: Naming
A solvent cupboard holds four compounds. A: CH₃CH₂CH(CH₃)CH₂OH B: CH₃CH₂C(CH₃)(OH)CH₂CH₃ C: CH₃CH(Cl)CH₂CH₂CH₃ D: CH₃CH₂CH₂CH₂CH₂CH₂NH₂
- Give the IUPAC name of A, B, C and D.(up to Achieved)
- Classify A, B and C as primary, secondary or tertiary.(up to Achieved)
Check your answer
Part a: what gets the marks
- A is 2-methylbutan-1-ol and B is 3-methylpentan-3-ol.
- C is 2-chloropentane and D is hexan-1-amine.
A full answer: A: 2-methylbutan-1-ol. B: 3-methylpentan-3-ol. C: 2-chloropentane. D: hexan-1-amine.
Part b: what gets the marks
- A is a primary alcohol, B is a tertiary alcohol and C is a secondary haloalkane.
A full answer: A is primary, B is tertiary and C is secondary.
What gets each grade
Question 2: Naming
A student names the compound CH₃CH(CH₂CH₃)CH₂CH(OH)CH₃ as 2-ethylpentan-4-ol.
- Explain the two errors in this name, and give the correct IUPAC name.(up to Merit)
- Classify this alcohol as primary, secondary or tertiary, and explain your choice.(up to Merit)
- This alcohol is heated with acidified potassium dichromate. Predict what is seen, and explain your prediction.(up to Merit)
Check your answer
Part a: what gets the marks
- The correct name is 4-methylhexan-2-ol.
- The longest chain runs through the ethyl branch and has six carbons, so the parent is hexane and the branch is a methyl group; an ethyl on C2 shows the longest chain was missed.
- The chain must be numbered from the end nearer the OH group so the OH gets the lowest number, C2, not C4.
A full answer: The name must come from the longest chain. Following the chain through the ethyl branch gives six carbons, CH₃CH(OH)CH₂CH(CH₃)CH₂CH₃, so the parent is hexan- and the branch is a methyl. The chain is numbered from the end nearer the OH, so the OH is on C2 and the methyl on C4. The correct name is 4-methylhexan-2-ol.
Part b: what gets the marks
- The alcohol is secondary.
- The carbon bonded to the OH is bonded to two other carbon atoms, the CH₃ and the CH₂, so the alcohol is secondary.
A full answer: It is a secondary alcohol. The carbon bonded to the OH group is bonded to two other carbon atoms: the CH₃ at the end and the CH₂ next to the branch.
Part c: what gets the marks
- The dichromate turns from orange to green.
- A secondary alcohol is oxidised by hot acidified dichromate, and the orange Cr₂O₇²⁻ is reduced to green Cr³⁺.
A full answer: The solution turns from orange to green. The alcohol is secondary, so it is oxidised by the hot acidified dichromate, and the orange Cr₂O₇²⁻ is reduced to green Cr³⁺.
What gets each grade
Question 3: Naming
Three isomers of C₅H₁₁Br are each heated with aqueous KOH, and each product is then heated with acidified potassium dichromate. X: CH₃CH₂CH₂CH₂CH₂Br Y: CH₃CH₂CHBrCH₂CH₃ Z: (CH₃)₂CBrCH₂CH₃
- Name and classify X, Y and Z.(up to Merit)
- Justify which of the products from aqueous KOH will turn the acidified dichromate green and which will leave it orange.(up to Excellence)
Check your answer
Part a: what gets the marks
- X is 1-bromopentane, Y is 3-bromopentane and Z is 2-bromo-2-methylbutane.
- X is primary (the C bonded to Br is bonded to one other carbon), Y is secondary (two other carbons) and Z is tertiary (three other carbons).
A full answer: X: 1-bromopentane, primary (the carbon bonded to the Br is bonded to one other carbon). Y: 3-bromopentane, secondary (bonded to two other carbons). Z: 2-bromo-2-methylbutane, tertiary (bonded to three other carbons).
Part b: what gets the marks
- Aqueous KOH substitutes OH for Br on the same carbon, giving pentan-1-ol from X, pentan-3-ol from Y and 2-methylbutan-2-ol from Z, each with the same classification as its bromoalkane.
- So the alcohols are primary pentan-1-ol, secondary pentan-3-ol and tertiary 2-methylbutan-2-ol.
- When heated, primary pentan-1-ol is oxidised to pentanoic acid, CH₃CH₂CH₂CH₂COOH, and secondary pentan-3-ol is also oxidised, so both turn the dichromate from orange to green; tertiary 2-methylbutan-2-ol is not oxidised, so it stays orange.
A full answer: Aqueous KOH causes substitution: OH replaces Br on the same carbon. X gives pentan-1-ol (primary), Y gives pentan-3-ol (secondary) and Z gives 2-methylbutan-2-ol (tertiary).
When heated with acidified dichromate, pentan-1-ol is oxidised to pentanoic acid, CH₃CH₂CH₂CH₂COOH, and the solution turns from orange to green. Pentan-3-ol is also oxidised, so it turns green too. 2-methylbutan-2-ol is tertiary and is not oxidised, so the dichromate stays orange.