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Chemistry Question 63 – NEET-UG 2024

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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:

Recall that Friedel-Crafts reactions require a Lewis acid catalyst, and consider how the basic amino group of aniline interacts with such catalysts.

Step 1: Analyze Statement I (Friedel-Crafts Alkylation)✦ Active

Aniline contains a basic amino group (−NH2). In Friedel-Crafts reactions (alkylation or acylation), a Lewis acid catalyst like AlCl3 is used. The amino group of aniline reacts with the Lewis acid to form a salt, C6H5NH2+−AlCl3−. This anilinium ion is a strong deactivating group and a meta-director, which makes Friedel-Crafts reactions unfavorable or impossible under normal conditions. Therefore, Statement I is correct.

💡 Teacher's Secret Hint

The formation of the anilinium ion deactivates the benzene ring towards electrophilic substitution.

Step 2: Analyze Statement II (Gabriel Synthesis)○ Expand

Gabriel phthalimide synthesis is a method used for the preparation of primary aliphatic amines. It involves the nucleophilic substitution of an alkyl halide by the phthalimide anion. Aromatic amines like aniline cannot be prepared by this method because aryl halides (e.g., bromobenzene) do not undergo nucleophilic substitution with the phthalimide anion under the reaction conditions typically employed for Gabriel synthesis. Therefore, Statement II is correct.

💡 Teacher's Secret Hint

Aryl halides are generally unreactive towards nucleophilic substitution due to the partial double bond character of the C-X bond and the instability of the aryl carbocation.

Step 3: Conclusion○ Expand

Since both Statement I and Statement II are correct, the option that states 'Both Statement I and Statement II are true' is the correct answer.

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