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Q.(a) Why does aniline not participate in the Friedel-Crafts reaction?

(b) Carry out the conversion: Ethanamine -> Ethanoic acid.
(c) CH3COOH --NH3/heat--> A --Br2 + NaOH--> B. Identify compounds 'A' and 'B'. (1+1+1=3)
Tripura TbseHigher Secondary (+2 Stage) Examination 2025Subjective· 3mImportance★★★★★
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(a) AlCl3 poisons aniline by binding to its lone pair. (b) Diazotisation/hydrolysis then oxidation converts the amine to the acid. (c) Ammonolysis then Hofmann bromamide degradation shortens the chain by one carbon.

(a) Why aniline does not undergo Friedel-Crafts reaction.

Friedel-Crafts reactions need a Lewis acid catalyst such as anhydrous AlCl3. Aniline (C6H5NH2) has a lone pair of electrons on nitrogen that is a good Lewis base, so it readily coordinates with (donates its lone pair to) the AlCl3 catalyst, forming a salt-like complex, C6H5-NH2->AlCl3. This makes the nitrogen positively charged, converting -NH2 (normally a strong activating, ortho/para-directing group) effectively into a strongly deactivating, meta-directing group (like -NH3+). With the catalyst tied up and the ring deactivated, the Friedel-Crafts alkylation/acylation simply does not proceed on aniline.

(b) Ethanamine -> Ethanoic acid.

Step 1 (diazotisation, even for a primary aliphatic amine): CH3CH2NH2 + NaNO2 + HCl (273-278 K) forms an unstable aliphatic diazonium salt that decomposes immediately, releasing N2 gas and giving ethanol: CH3CH2NH2 --NaNO2/HCl, 273-278K--> CH3CH2OH + N2 + HCl (this step is often just summarised as amine -> alcohol via nitrous acid).

Step 2 (oxidation): CH3CH2OH --[O] (alkaline KMnO4 or acidified K2Cr2O7)--> CH3COOH (ethanoic acid).

(c) CH3COOH --NH3/heat--> A --Br2+NaOH--> B. …

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