Given below are two statements:

Statement I : Aniline does not undergo Friedel-Crafts alkylation reaction.

Statement II : Aniline cannot be prepared through Gabriel synthesis.

In the light of the above statements, choose the correct answer from the options given below:

The correct option is: Both statement I and Statement II are true

Let us analyze both statements step-by-step to understand the underlying chemistry:

Analysis of Statement I:
Aniline ($C_6{H}_{5}N{H}_2$) contains an amino group ($-N{H}_2$) attached to the benzene ring. The nitrogen atom in the amino group has a lone pair of electrons, making aniline a strong Lewis base.
Friedel-Crafts alkylation reaction requires an anhydrous Lewis acid catalyst, such as aluminum chloride ($AlC{l}_3$).
When aniline is treated with an alkyl halide in the presence of anhydrous $AlC{l}_3$, the Lewis basic nitrogen atom of aniline readily donates its lone pair of electrons to the Lewis acidic $AlC{l}_3$ catalyst. This acid-base interaction forms a highly stable salt complex:
$C_6{H}_{5}N{H}_2+AlC{l}_3\to C_6{H}_5\stackrel{+}{N}{H}_2-\stackrel{-}{A}lC{l}_3$
In this salt complex, the nitrogen atom acquires a positive charge. The positively charged nitrogen becomes strongly electron-withdrawing, creating a powerful deactivating effect on the benzene ring (-I effect). Because of this extreme deactivation, the benzene ring becomes unreactive towards electrophilic attack. Hence, aniline does not undergo Friedel-Crafts alkylation. Thus, Statement I is true.

Analysis of Statement II:
Gabriel phthalimide synthesis is a standard method used for preparing primary aliphatic amines. The reaction involves the nucleophilic substitution ($S_{N}2$ mechanism) of an alkyl halide by the phthalimide anion.
To prepare aniline (a primary aromatic amine) via this route, one would require a halobenzene (like chlorobenzene or bromobenzene) as the substrate to react with potassium phthalimide.
However, aryl halides do not undergo nucleophilic substitution reactions easily under ordinary conditions. This is because the partial double-bond character of the carbon-halogen bond (due to resonance) and the steric hindrance of the aromatic ring prevent the nucleophile from attacking the sp² hybridized carbon. As a result, the phthalimide anion cannot displace the halogen atom from a benzene ring to form N-arylphthalimide. Therefore, aniline cannot be prepared by Gabriel phthalimide synthesis. Thus, Statement II is true.

Consequently, both Statement I and Statement II are correct.