Q.(i) How would you convert (Give only arrow head reaction): Cumene -> Phenol?

The cumene process converts cumene to phenol (+ acetone) via cumene hydroperoxide; the Lucas test distinguishes alcohols by how fast turbidity (alkyl chloride) appears, tracking carbocation stability.

(i) Cumene → Phenol:

$\text{Cumene} \xrightarrow[\text{(ii) dil. } H_2SO_4/H_3O^+]{\text{(i) } O_2/\text{air}} \text{Phenol} + \text{Acetone (by-product)}$

(via the unstable cumene hydroperoxide intermediate, which is acid-cleaved to give phenol and acetone.)

(ii) Lucas test (conc. HCl + anhydrous ZnCl$_2$):

• 3° alcohol: turbidity (insoluble chloride) appears immediately at room temperature — readily forms a stable 3° carbocation (S$_N$1).
• 2° alcohol: turbidity appears after about 5 minutes (sometimes needs gentle warming).
• 1° alcohol: no turbidity at room temperature; reacts only very slowly even on prolonged heating (1° carbocation is too unstable to form readily).

OR:

(i) Reimer–Tiemann reaction example:

$\text{Phenol} + CHCl_3 + NaOH(aq) \xrightarrow{60\text{–}70^\circ C, \text{ then } H_3O^+} \text{Salicylaldehyde (2-hydroxybenzaldehyde)}$

(via a dichlorocarbene intermediate, giving ortho-formylation as the major product.)

(ii) The reaction $C_2H_5ONa + (CH_3)_3C\text{-}Cl \rightarrow (CH_3)_3C\text{-}OC_2H_5$ is unsuitable because tert-butyl chloride is a 3° alkyl halide; with a nucleophile/base like ethoxide, substitution (S$_N$2) is sterically blocked at the crowded tertiary carbon, so elimination (E2) dominates instead.

…