Question

Method by which aniline cannot be prepared is

• A. Reduction of nitrobenzene with $H_2/Pd$ in ethanol
• B. Potassium salt of phthalimide treated with chlorobenzene followed by hydrolysis with aqueous $NaOH$ solution
• C. Degradation of benzamide with bromine in alkaline solution
• D. Hydrolysis of phenylisocyanide with acidic solution

Answer 1

The correct option is B Potassium salt of phthalimide treated with chlorobenzene followed by hydrolysis with aqueous $NaOH$ solution

$\left(a\right)\text{Reduction of nitrobenzene in}{H}_2/Pd$ in ethanol gives aniline. The reaction is given as sollows,


$\left(b\right)$ Potassium salt of phthalimide treated with chlorobenzene followed by hydrolysis with aqueous $NaOH$ solution. This reaction is Gabriel phthalimide synthesis. According to the reaction, only primary amines are formed. Aniline is a primary amine. But one of the drawbacks of this reaction is that aryl halides, that is, chlorobenzene will not undergo nucleophilic substitution reaction with the anion of phthalimide. Therefore, this reaction will not proceed further.
General reaction of Gabriel synthesis is given as,


$\left(c\right)$ Hydrolysis of phenyl isocyanide solution with acidic solution gives N-phenyl formamide which on further hydrolysis in acidic conditions gives aniline and formic acid.


$\left(d\right)$ Degradation of benzamide with bromine in alkaline solution gives aniline. This reaction is Hoffmann bromamide degradation.


Therefore, the only reaction in which aniline is not formed is Gabriel phthalimide synthesis.

Hence, the answer is option $\left(b\right).$