Write the overall balanced chemical equation for the conversion of benzyl bromide (${\mathrm{C}}_6{\mathrm{H}}_5\mathrm{C}{\mathrm{H}}_2\mathrm{Br}$) to benzylamine (${\mathrm{C}}_6{\mathrm{H}}_5\mathrm{C}{\mathrm{H}}_2\mathrm{N}{\mathrm{H}}_2$) via the Gabriel synthesis. Also state the name of the intermediate formed after the first step.

Step-by-step solution

1. Step 1 – Nucleophilic substitution: Potassium phthalimide ($\mathrm{KPhth}$) attacks benzyl bromide, displacing bromide to give N‑benzylphthalimide and $\mathrm{KBr}$.
2. Step 2 – Hydrolysis with hydrazine: N‑benzylphthalimide reacts with excess hydrazine hydrate ($\mathrm{N}{\mathrm{H}}_2\mathrm{N}{\mathrm{H}}_2\cdot {\mathrm{H}}_2\mathrm{O}$). Hydrazine opens the imide ring, forming the free amine (benzylamine) and phthalhydrazide as a by‑product.
3. Step 3 – Combine the two steps: Adding the reactants of both steps yields the overall balanced equation:
4. ${\mathrm{C}}_6{\mathrm{H}}_5\mathrm{C}{\mathrm{H}}_2\mathrm{Br}+\mathrm{KPhth}+\mathrm{N}{\mathrm{H}}_2\mathrm{N}{\mathrm{H}}_2\cdot {\mathrm{H}}_2\mathrm{O}\to {\mathrm{C}}_6{\mathrm{H}}_5\mathrm{C}{\mathrm{H}}_2\mathrm{N}{\mathrm{H}}_2+\mathrm{KBr}+\mathrm{Phthalhydrazide}$.
5. The key intermediate after the first step is N‑benzylphthalimide.

Concept tested: Gabriel synthesis overall transformation

Published 11 September 2026 · How this solution was prepared and checked