Answer
Step 1: Reaction with SOCl₂ (Thionyl Chloride)
- The lone pair on the oxygen atom of the carboxylic acid attacks the sulfur atom of SOCl₂.
- Chloride ion (Cl⁻) leaves, followed by a nucleophilic attack on the carbonyl carbon.
- This results in the formation of an acid chloride via a tetrahedral intermediate.
Reaction Type: Nucleophilic Acyl Substitution
Key Intermediate: Acid Chloride
Step 2: Reduction with LiAlH₄ (Lithium Aluminium Hydride)
- LiAlH₄ donates a hydride ion (H⁻) to the carbonyl carbon of the acid chloride.
- Chloride leaves, forming an aldehyde intermediate.
- Another hydride reduces the aldehyde to an alkoxide ion.
- Protonation yields a primary alcohol as the final product.
Reaction Type: Reduction
Final Product: Primary Alcohol
Step 3: Formation of Anhydride
- The oxygen of a carboxylic acid attacks the carbonyl carbon of the acid chloride.
- Chloride ion (Cl⁻) leaves, forming a tetrahedral intermediate.
- A base removes the extra proton, stabilizing the oxygen atom.
- The final product is an anhydride.
Reaction Type: Nucleophilic Acyl Substitution
Final Product: Anhydride
Key Concepts Recap
- Nucleophile: Electron-rich, seeks positive centers (e.g., lone-pair oxygen).
- Electrophile: Electron-poor, accepts electrons (e.g., carbonyl carbon).
- Reduction: Addition of hydrogen or removal of oxygen; here, converts acid chloride to alcohol.