Identify the reaction involved

$N$-ethylphthalimide is a derivative of phthalimide. To obtain a primary amine from an $N$-alkylphthalimide, the standard method is the Gabriel phthalimide synthesis.

Key principle of Gabriel synthesis

In Gabriel synthesis:

• Potassium phthalimide reacts with an alkyl halide to form $N$-alkylphthalimide.
• Then, cleavage of the phthalimide group gives the primary amine.

This cleavage is commonly done using hydrazine $(NH_2NH_2)$.

Apply to the given compound

Here, the compound is $N$-ethylphthalimide. On treatment with hydrazine:

$N\text{-ethylphthalimide} \xrightarrow{NH_2NH_2} C_2H_5NH_2 + \text{phthalhydrazide}$

Thus, ethylamine is liberated.

Check the options

• A: $CaH_2$ — drying/reducing-type reagent, not used here.
• B: $H_2O$ — simple water will not cleave $N$-ethylphthalimide to ethylamine.
• C: $NaBH_4$ — mild reducing agent; not used for Gabriel deprotection.
• D: $NH_2NH_2$ — correct reagent for hydrazinolysis of $N$-alkylphthalimide.

Conclusion

The correct answer is:

$\boxed{D\; (NH_2NH_2)}$