NCEA Level 2 Chemistry 91165 Organic free practice questions with answers

Question 1 (17 marks) [Naming, drawing and isomers]

(a) The condensed structural formula of a compound is CH₃CH₂CH₂CH₂NH₂. (i) State the molecular formula and name the compound using IUPAC rules. (ii) State the homologous series (family) it belongs to and the functional group present, and say how you can tell.

(a) Answer: C₄H₁₁N; name butan-1-amine; primary amine

• Evidence level: Merit (find the longest chain containing the functional group, number from the end giving the lowest locants, name branches and groups).

• (i) The molecular formula is C₄H₁₁N.

• The name is butan-1-amine.

• (ii) It is a primary amine because it contains an amino group, –NH₂.

• Achievement: Gives the correct family or functional group.

• Merit: Names the compound correctly (chain length, functional group and locants).

• Excellence: Names the compound fully correctly and explains how the numbering and the suffix/prefix are chosen: the chain is numbered so the principal group (or double bond) gets the lowest locant, then branches.

(b) Draw (give the condensed structural formula of) pentan-2-ol, and state its molecular formula. State, giving a reason, whether it is a saturated or an unsaturated compound.

(b) Answer: CH₃CH₂CH₂CH(OH)CH₃; C₅H₁₂O; saturated

• Evidence level: Achieved (translate a name into a structure).

• The stem and ending give the chain length and functional group; locants give where the group and any branches are placed.

• Condensed structural formula: CH₃CH₂CH₂CH(OH)CH₃. Molecular formula: C₅H₁₂O.

• Achievement: Gives a correct structure or molecular formula.

• Merit: Gives the correct structural formula and molecular formula.

• Excellence: (unsaturation) The compound is saturated because all carbon–carbon bonds are single bonds, so it cannot add more atoms.

(c) Name all the constitutional isomers of C₄H₉Cl that are haloalkanes, and classify each as a primary, secondary or tertiary compound.

(c) Answer: 4 isomers: 1-chloro-2-methylpropane; 1-chlorobutane; 2-chloro-2-methylpropane; 2-chlorobutane

• Evidence level: Excellence (systematically vary the carbon chain and the position of the functional group).

• Constitutional isomers have the same molecular formula but atoms joined in a different order. Work systematically: start with the straight chain, move the group along the chain, then shorten the chain and add branches.

• There are 4 constitutional isomers: 1-chloro-2-methylpropane (CH₃CH(CH₃)CH₂Cl); 1-chlorobutane (CH₃CH₂CH₂CH₂Cl); 2-chloro-2-methylpropane (CH₃C(Cl)(CH₃)CH₃); 2-chlorobutane (CH₃CH₂CH(Cl)CH₃).

• Check: each has formula C₄H₉Cl and no two are the same molecule.

• Achievement: Gives at least one correct isomer.

• Merit: Gives about half of the isomers with correct names.

• Excellence: Gives all 4 isomers with correct structures and names (a systematic method makes sure none are missed or repeated).

(d) The compounds 4-methylpent-2-ene (CH₃CH=CHCH(CH₃)CH₃), 3-methylpent-2-ene (CH₃CH₂C(CH₃)=CHCH₃), 2-methylprop-1-ene (CH₂=C(CH₃)CH₃) are alkenes. (i) State which of them can exist as cis and trans (geometric) isomers. (ii) For one that can, draw both isomers (cis and trans) and name them, and explain why the isomers exist.