State a reason why most halogenoalkanes are more reactive than alkanes.

Classify 1-bromopropane as a primary, secondary or tertiary halogenoalkane, giving a reason.

Explain the mechanism of the reaction between 1-bromopropane with aqueous sodium hydroxide using curly arrows to represent the movement of electron pairs.

State, giving your reason, whether the hydroxide ion acts as a Lewis acid, a Lewis base, or neither in the nucleophilic substitution.

Suggest two advantages of understanding organic reaction mechanisms.

polarity/polar «molecule/bond»
OR
carbon–halogen bond is weaker than C–H bond ✔

primary AND Br/bromine is attached to a carbon bonded to two hydrogens
OR
primary AND Br/bromine is attached to a carbon bonded to one C/R/alkyl «group» ✔

Accept “primary AND Br/bromine is attached to the first carbon in the chain”.

curly arrow going from lone pair/negative charge on O in HO^– to C ✔

curly arrow showing Br leaving ✔

representation of transition state showing negative charge, square brackets and partial bonds ✔

formation of organic product CH₃CH₂CH₂OH AND Br^– ✔

Do not allow curly arrow originating on H in HO^–.

Accept curly arrow either going from bond between C and Br to Br in 1-bromopropane or in the transition state.

Do not penalize if HO and Br are not at 180° to each other.

Do not award M3 if OH–C bond is represented.

«Lewis» base AND donates a pair of electrons ✔

Any two of:
choose «most» appropriate reaction «for preparing the target compound» ✔
design/discover new reactions/reagents ✔
apply this knowledge to other areas of chemistry/science ✔
«retro-»synthesis «more effective» ✔
control/predict «desired» products ✔
control rate of reaction «more effectively» ✔
satisfy intellectual curiosity ✔
predicting how changing reagents/conditions might affect reaction ✔
suggesting intermediates/transition states ✔

Accept other reasonable answers.