Write the final product(s) in each of the following reactions:
(a) H₃C−CH₃|C|−CH₃O−CH₃ + HI →
(b) H₃C−H₂C−HC|OH−CH₃ + Cu/573K →
(c) C₆H₅OH (i) CHCl₃ + Aq. NaOH → (ii) H⁺

(a) The reaction of tertiary alkyl ether with HI undergoes nucleophilic substitution. The reaction proceeds via SN1 mechanism. The iodide ion attacks the carbon atom bearing the ether linkage. The tertiary carbocation formed is more stable than the primary carbocation. Hence, the reaction proceeds to produce tert-butyl iodide and methanol.

H₃C−CH₃|C|−CH₃O−CH₃ + HI → H₃C−CH₃|C|−CH₃I + CH₃OH

(b) This reaction is an example of dehydrogenation of alcohol. The secondary alcohol is converted into an alkene upon treatment with Cu/573K. Specifically, 3-methylbutan-2-ol loses a molecule of hydrogen to form 3-methylbut-2-ene.

H₃C−H₂C−HC|OH−CH₃ →[Cu/573K] H₃C−H₂C−C=CH₂| CH₃

(c) This reaction sequence involves the Reimer-Tiemann reaction. Phenol reacts with chloroform and aqueous sodium hydroxide to form salicylaldehyde. The intermediate formed initially is dichlorocarbene, which reacts with phenol to give the salicylaldehyde. The subsequent acidification yields salicylaldehyde.

C₆H₅OH (i) CHCl₃ + Aq. NaOH → (ii) H⁺ o-Hydroxybenzaldehyde (Salicylaldehyde)

The detailed mechanism of the Reimer-Tiemann reaction involves the following steps:

1. Formation of dichlorocarbene:
   CHCl₃ + OH⁻ → ⁻CCl₃ + H₂O
   ⁻CCl₃ → :CCl₂ + Cl⁻

2. Nucleophilic attack by phenoxide ion:

   C₆H₅O⁻ + :CCl₂ → C₆H₅O−CCl₂⁻

3. Protonation and elimination of HCl:

   C₆H₅O−CCl₂⁻ + H₂O → C₆H₅O−CHCl₂ + OH⁻
   C₆H₅O−CHCl₂ + OH⁻ → C₆H₅O−CHO + 2HCl