Aniline does not undergo Friedel-Crafts reaction. Why?
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Solution and Explanation
Due to salt formation with AlCl\(_3\), aniline becomes a Lewis base, deactivating the benzene ring.
Step 1: Role of Lewis Acid Catalyst in Friedel-Crafts Reaction - Friedel-Crafts alkylation and acylation require a Lewis acid catalyst like AlCl\(_3\) to generate the electrophile.
Step 2: Interaction Between Aniline and AlCl\(_3\) - Aniline (\( \text{C}_6\text{H}_5\text{NH}_2 \)) has a lone pair on nitrogen, which interacts with AlCl\(_3\), forming a salt. \[ \text{C}_6\text{H}_5\text{NH}_2 + AlCl_3 \rightarrow [\text{C}_6\text{H}_5\text{NH}_2]^+ [AlCl_3]^- \] - This deactivates the benzene ring, making it less reactive toward electrophilic substitution.
Step 3: Conclusion Due to this salt formation, aniline does not undergo Friedel-Crafts reaction.
Top Questions on Amines
- Identify correct statements: (A) Primary amines do not give diazonium salts when treated with \({NaNO}_2\) in acidic condition.
(B) Aliphatic and aromatic primary amines on heating with \({CHCl}_3\) and ethanolic \({KOH}\) form carbylamines.
(C) Secondary and tertiary amines also give carbylamine test.
(D) Benzenesulfonyl chloride is known as Hinsberg’s reagent.
(E) Tertiary amines react with benzenesulfonyl chloride very easily.
Choose the correct answer from the options given below: Amines have a lone pair of electrons on the nitrogen atom, due to which they behave as Lewis bases. Greater the value of \( K_b \) or smaller the value of \( pK_b \), stronger is the base. Amines are more basic than alcohols, ethers, esters, etc. The basic character of aliphatic amines should increase with the increase of alkyl substitution. However, it does not occur in a regular manner, as a secondary aliphatic amine is unexpectedly more basic than a tertiary amine in aqueous solutions. Aromatic amines are weaker bases than ammonia and aliphatic amines. Electron-releasing groups such as \( -CH_3 \), \( -NH_2 \), etc., increase the basicity, while electron-withdrawing substituents such as \( -NO_2 \), \( -CN \), halogens, etc., decrease the basicity of amines. The effect of these substituents is more pronounced at the para-position than at the meta-position.
(a) Arrange the following in increasing order of their basic character. Give reason:
\[ \includegraphics[width=0.5\linewidth]{ch29i.png} \]- Amines have a lone pair of electrons on nitrogen atom due to which they behave as Lewis base. Greater the value of \( K_b \) or smaller the value of \( pK_b \), stronger is the base. Amines are more basic than alcohols, ethers, esters, etc. The basic character of aliphatic amines should increase with the increase of alkyl substitution. But it does not occur in a regular manner as a secondary aliphatic amine is unexpectedly more basic than a tertiary amine in aqueous solutions. Aromatic amines are weaker bases than ammonia and aliphatic amines. Electron releasing groups such as \(–CH_3\), \(–NH_2\), etc., increase the basicity while electron-withdrawing substituents such as \(–NO_2\), –CN, halogens, etc., decrease the basicity of amines. The effect of these substituents is more at the p₊ than at m₋ position.
- Assertion (A): Aromatic primary amines cannot be prepared by Gabriel Phthalimide synthesis. Reason (R): Aryl halides do not undergo nucleophilic substitution reaction with the anion formed by phthalimide.
- Give plausible explanation for each of the following:
Amides are less basic than amines.
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