Principle of Gabriel phthalimide synthesis

Gabriel phthalimide reaction is used to prepare primary alkyl amines by nucleophilic substitution of potassium phthalimide with an alkyl halide.

General reaction: $\text{C}_6\text{H}_4(\text{CO})_2\text{NK} + \text{R-X} \rightarrow \text{N-alkyl phthalimide} \xrightarrow{\text{hydrolysis}} \text{R-NH}_2$

Important limitation

This reaction works well only with primary alkyl halides that undergo $S_N2$ reaction.

It does not work well for:

• tertiary alkyl halides (do not undergo $S_N2$)
• aryl halides
• cases where severe steric hindrance blocks substitution
• it gives only primary amines, not secondary/tertiary amines

Check each option

A: neo-pentylamine

neo-pentylamine is: $(CH_3)_3C-CH_2-NH_2$ To prepare it by Gabriel synthesis, we would need neopentyl halide: $(CH_3)_3C-CH_2-X$ Although this is formally a primary halide, it is highly sterically hindered at the $\beta$-carbon and reacts very poorly in $S_N2$. Therefore, it is not prepared effectively by Gabriel phthalimide reaction.

B: n-butylamine

n-butylamine is: $CH_3CH_2CH_2CH_2NH_2$ It can be prepared from a normal primary alkyl halide such as 1-bromobutane: $CH_3CH_2CH_2CH_2Br$ This undergoes $S_N2$ readily with potassium phthalimide. Hence, this can be prepared by Gabriel synthesis.

C: t-butylamine

t-butylamine is: $(CH_3)_3C-NH_2$ This would require tert-butyl halide: $(CH_3)_3C-X$ Tertiary halides do not undergo $S_N2$ reaction, so Gabriel synthesis fails. Therefore, cannot be prepared.

D: triethylamine

triethylamine is: $(C_2H_5)_3N$ Gabriel synthesis gives only primary amines, not tertiary amines. Therefore, cannot be prepared.

Conclusion

Only n-butylamine can be prepared by Gabriel phthalimide reaction.

Therefore, the correct option is: $\boxed{\text{B}}$