(a) REIMER-TIEMANN REACTION

Step 1 - Statement.
When phenol is treated with CHLOROFORM in the presence of AQUEOUS SODIUM HYDROXIDE at
340 K, a $-CHO$ group is introduced at the ORTHO position of the ring. Acidification of
the product gives SALICYLALDEHYDE.

Step 2 - The reaction.
$$C_6H_5OH \xrightarrow[\text{340 K}]{CHCl_3 + \text{aq. NaOH}} \text{(sodium salt of salicylaldehyde)} \xrightarrow{H_3O^+} \text{2-Hydroxybenzaldehyde}$$

Step 3 - Mechanism note.
The electrophile in this reaction is DICHLOROCARBENE $(:CCl_2)$, generated from
chloroform by the action of NaOH.
$$CHCl_3 + OH^- \rightarrow :CCl_2 + Cl^- + H_2O$$

(b) KOLBE'S REACTION

Step 1 - Statement.
Sodium phenoxide (obtained by treating phenol with NaOH) is heated with CARBON DIOXIDE
under pressure (400 K, 4-7 atm). Acidification of the product gives SALICYLIC ACID
(2-hydroxybenzoic acid).

Step 2 - The reaction.
$$C_6H_5OH \xrightarrow{NaOH} C_6H_5O^-Na^+ \xrightarrow[\text{400 K, 4-7 atm}]{CO_2} \text{sodium salicylate} \xrightarrow{H_3O^+} \text{2-Hydroxybenzoic acid}$$

Step 3 - Why the phenoxide is used.
$CO_2$ is a WEAK electrophile. The PHENOXIDE ion is more electron-rich (more strongly
activated) than phenol itself, so it is reactive enough to attack $CO_2$.

(c) FRIEDEL-CRAFTS ACYLATION OF ANISOLE

Step 1 - Statement.
Anisole $(C_6H_5OCH_3)$ reacts with an acyl chloride (e.g. $CH_3COCl$) in the presence of
ANHYDROUS $AlCl_3$ (a Lewis acid catalyst) to give a methoxy-substituted aromatic ketone.

Step 2 - The reaction.
$$C_6H_5OCH_3 + CH_3COCl \xrightarrow[CS_2]{\text{anhyd. } AlCl_3} \text{4-Methoxyacetophenone (major, para)} + \text{2-Methoxyacetophenone (minor, ortho)} + HCl$$

Step 3 - Why para is the major product.
The $-OCH_3$ group is activating and ORTHO/PARA directing (it releases electrons by
resonance). The PARA product predominates because the bulky acyl group experiences less
STERIC HINDRANCE at the para position than at the ortho position, which is crowded by
the neighbouring $-OCH_3$ group.