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 solve this problem, let's analyze each statement individually and provide the correct reasoning behind why both statements are true.
Statement I: Aniline reacts with concentrated \(H_2SO_4\) followed by heating at 453–473 K to form p-aminobenzene sulphonic acid, which gives a blood red color in the 'Lassaigne's test'.
Explanation:
In concentrated \(H_2SO_4\), aniline undergoes sulfonation to form p-aminobenzene sulphonic acid (also known as sulphanilic acid) when heated in the range of 453–473 K.
This compound, upon Lassaigne's test for sulphur, forms sodium thiocyanate, which reacts with ferric chloride to give a blood-red color.
This establishes that Statement I is true.
Statement II: In Friedel-Craft's alkylation and acylation reactions, aniline forms a salt with the \(AlCl_3\) catalyst. Due to this, the nitrogen of aniline acquires a positive charge and acts as a deactivating group.
Explanation:
Aniline is a base and reacts with Lewis acids such as \(AlCl_3\) to form a salt. The nitrogen atom of aniline gets protonated, acquiring a positive charge.
This positively charged nitrogen significantly reduces the electron density of the aromatic ring, making it less reactive in the Friedel-Craft's reactions, hence acting as a deactivating group.
This confirms that Statement II is also true.
Since both statements are true, the correct option is: Both Statement I and Statement II are true.
Was this answer helpful?
0
0
Hide Solution
Verified By Collegedunia
Approach Solution -2
Statement I is correct because aniline does react with concentrated sulfuric acid and, upon heating, forms p-aminobenzene sulfonic acid, which gives a blood red color in Tassaigne’s test.
Statement II is also correct as in Friedel-Crafts alkylation and acylation reactions, aniline forms a salt with the \(\text{AlCl}_3\) catalyst. This interaction results in a positive charge on nitrogen, causing it to act as a deactivating group, making further substitution reactions difficult on the benzene ring.
