🧪 Ether Formation via Williamson Synthesis

Reactants:

• Phenylmethanol (PhCH₂OH)
• Sodium metal (Na)
• n-Propyl bromide (CH₃CH₂CH₂Br)

This reaction proceeds through the Williamson ether synthesis mechanism.

🔹 Step 1: Formation of Alkoxide

Sodium reacts with the hydroxyl group of phenylmethanol, forming:

• Sodium benzylate (PhCH2O− Na+)
• Hydrogen gas (H₂)

🔹 Step 2: Nucleophilic Substitution (SN2)

The benzylate ion acts as a nucleophile, attacking the electrophilic carbon of n-propyl bromide to form:

Final product: benzyl n-propyl ether (PhCH₂OCH₂CH₂CH₃)

Correct Option: C

❌ Why Other Options Are Incorrect

• Option A: Branched or secondary alkyl group — Not correct n-propyl
• Option B: Intermediate alkoxide, not final product — Incorrect
• Option D: Tert-butyl group present — Wrong alkylating agent

🧾 Structure Identification: i-BuBr (Isobutyl Bromide)

Acronym: i-BuBr = Isobutyl bromide

Structure: CH3–CH(CH3)–CH2–Br

Correct Option: A

❌ Other Incorrect Structures:

• Option B: sec-butyl bromide — Br on secondary carbon
• Option C: n-butyl bromide — linear, no branching
• Option D: tert-butyl bromide — Br on central carbon attached to three methyl groups

🟩 Final Answers Summary

• Ether Synthesis Product: Option C
• Isobutyl Bromide Structure: Option A