Given below are two statements: Statement I: Aniline reacts with con. \( H_2SO_4 \) followed by heating at 453–473 K gives p-aminobenzene sulphonic acid, which gives blood red colour in the 'Lassaigne's test'. Statement II: In Friedel-Craft's alkylation and acylation reactions, aniline forms salt with the \( AlCl_3 \) catalyst. Due to this, nitrogen of aniline acquires a positive charge and acts as a deactivating group. In the light of the above statements, choose the correct answer from the options given below:
To resolve this inquiry, each assertion will be individually examined, followed by the substantiation of their veracity.
Statement I: The reaction of aniline with concentrated \(H_2SO_4\), subsequently heated to 453–473 K, yields p-aminobenzene sulphonic acid. This compound exhibits a blood-red coloration in the 'Lassaigne's test'.
Explanation:
Under conditions of concentrated \(H_2SO_4\) and heating within the 453–473 K range, aniline undergoes sulfonation to produce p-aminobenzene sulphonic acid, also identified as sulphanilic acid.
In the Lassaigne's test for sulfur, this compound generates sodium thiocyanate, which subsequently reacts with ferric chloride to produce a blood-red color.
This verifies the accuracy of Statement I.
Statement II: During Friedel-Craft's alkylation and acylation reactions, aniline forms a salt with the \(AlCl_3\) catalyst. Consequently, the nitrogen atom in aniline acquires a positive charge, rendering it a deactivating group.
Explanation:
Aniline, acting as a base, reacts with Lewis acids like \(AlCl_3\) to form a salt. The nitrogen atom in aniline becomes protonated, thereby acquiring a positive charge.
This positive charge on the nitrogen significantly diminishes the electron density of the aromatic ring, decreasing its reactivity in Friedel-Craft's reactions, thus functioning as a deactivating group.
This confirms the validity of Statement II.
Given that both statements are accurate, the appropriate selection is: Both Statement I and Statement II are true.
