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:
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:
- Statement I is false but Statement II is true
- Both Statement I and Statement II are false
- Statement I is true but Statement II is false
- Both Statement I and Statement II are true
The Correct Option is D
Approach Solution - 1
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.
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.
Thus, both statements are true.
Top Questions on Amines
- How will you convert nitrobenzene to 2, 4, 6-Tribromo aniline?
- (i) The reaction in which an amide is converted into a primary amine by the action of Br2 and alcoholic NaOH is known as ________.
(ii) How is a primary amine distinguished from a secondary amine using the Hinsberg test? - Write the structures of (A), (B), and (C) in the following reaction:
\[ C_6H_5NO_2 \xrightarrow{Fe/HCl} (A) \xrightarrow[\text{273 K}]{NaNO_2 + HCl} (B) \xrightarrow[\Delta]{H_2O/H^+} (C) \] - What happens when aniline reacts with the following:
(A) Bromine water
(B) $(CH_3CO)_2O$ / Pyridine
(C) $HNO_2 + HCl (0^\circ - 5^\circ C)$ - Why is \(-NH_2\) group of aniline acetylated before carrying out nitration?
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