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.
0.1 mole of compound S will weigh ....... g,
(given the molar mass in g mol\(^{-1}\) \( {C} = 12, \, {H} = 1, \, {O} = 16 )\):
If \( {A}_2{B} \) is 30% ionised in an aqueous solution, then the value of van't Hoff factor \( i \) is:
Match List - I with List - II:
Choose the correct answer from the options given below: