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Write Brief Answer · Q17

Q.Complete the following reactions:

(i) CH3-CH2-OH --PBr3--> A --aq.NaOH--> B --Na--> C
(ii) C6H5-OH --Zn dust--> A --CH3Cl/Anhydrous AlCl3--> B --acidified KMnO4--> C
(iii) Anisole --t-butylchloride/AlCl3--> A --Cl2/FeCl3--> B --HBr--> C
(iv) 1-methyl-1-(1-hydroxyethyl)cyclohexane [a cyclohexane ring carbon bearing both -CH3 and -CH(OH)CH3] --H+--> A --(i) O3
(ii) H2O--> B
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Step 1. (i) CH3CH2OH + PBr3 gives A = CH3CH2Br (bromoethane); + aq.NaOH gives B = CH3CH2OH (ethanol, regenerated by substitution); + Na gives C = CH3CH2ONa (sodium ethoxide) + 1/2 H2.

Step 2. (ii) C6H5OH + Zn dust gives A = C6H6 (benzene, section 11.15.1); + CH3Cl/anhydrous AlCl3 (Friedel-Crafts alkylation) gives B = C6H5-CH3 (toluene); + acidified KMnO4 (benzylic oxidation) gives C = C6H5-COOH (benzoic acid).

Step 3. (iii) Anisole + t-butylchloride/AlCl3 (Friedel-Crafts alkylation, directed para by the strongly activating -OCH3) gives A, predominantly 1-tert-butyl-4-methoxybenzene; + Cl2/FeCl3 (further ring chlorination, directed by both the methoxy and tert-butyl groups) gives B, a ring-chlorinated tert-butylanisole; + HBr (which cleaves the ARYL METHYL ether specifically, SN2 attack of Br- on the methyl carbon, since the aryl-O bond itself does not break) gives C, the corresponding substituted phenol + CH3Br. …

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