IB DP Chemistry -Reactivity 3.4 Electron-pair sharing reactions - IB Style Questions For HL Paper 1A -FA 2025
▶️ Answer/Explanation
Methylpropanoic acid is a carboxylic acid. To obtain a high yield of a carboxylic acid from an alcohol, a primary alcohol must be fully oxidized under reflux with an oxidizing agent.
• \(2\)-methylpropan-\(1\)-ol is a primary alcohol, which can be oxidized first to an aldehyde and then to the carboxylic acid under reflux.
• \(2\)-methylpropan-\(2\)-ol is a tertiary alcohol, which is not readily oxidized under normal conditions, so it will not give a good yield of the acid.
• Distillation is used to stop oxidation early (at the aldehyde stage), so it does not give the highest yield of the acid.
Therefore, the correct combination is the primary alcohol \(2\)-methylpropan-\(1\)-ol under reflux.
✅ Answer: (B)
▶️ Answer/Explanation
Benzene contains a stable delocalized π-electron ring. It resists reactions that would break this aromatic stability.
- Electrophilic substitution allows benzene to react while keeping the aromatic ring intact.
- Electrophilic addition would destroy aromaticity, so it is not favoured.
- Nucleophilic substitution is highly unlikely due to the electron-rich ring.
- Free radical substitution does not occur under normal conditions.
Therefore, benzene most readily undergoes:
\(\boxed{\text{Electrophilic substitution}}\)
✅ Answer: (B)
▶️ Answer/Explanation
Propene is an alkene, and alkenes typically react with halogens through electrophilic addition. This involves the π bond attacking a halogen molecule.
In the dark, iodine does not undergo photochemical homolysis, so no free radicals form. Therefore, free radical substitution does not occur.
The reaction proceeds by:
- iodine acting as an electrophile (polarized by the alkene π electrons)
- the π bond opening to form a di-iodo product
Thus, the correct mechanism is:
\(\boxed{\text{electrophilic addition}}\)
✅ Answer: (A)