Phenols

Reimer–Tiemann reaction: phenol is heated with chloroform and aqueous NaOH (or KOH) at about 340 K. The product is acidified to give salicylaldehyde (2-hydroxybenzaldehyde). A $-CHO$ group enters at the ortho position.

$$C_6H_5OH + CHCl_3 + 3NaOH \xrightarrow{340\ K} o\text{-}HOC_6H_4CHO + 3NaCl + 2H_2O$$

    OH                          OH
    |                           |  
   / \     CHCl3, aq. NaOH     / \__CHO
  |   |   ---------------->   |   |
   \ /       340 K; H+         \ /
 phenol                   salicylaldehyde

(b) Benzenediazonium chloride

Benzenediazonium chloride is unstable. When boiled with water (dilute acid), it is hydrolysed to phenol and $N_2$ is evolved:

$$C_6H_5N_2Cl + H_2O \xrightarrow{\Delta} C_6H_5OH + N_2\uparrow + HCl$$

Chlorobenzene needs NaOH at 623 K and 300 atm (Dow's process). Benzenesulphonic acid needs fusion with NaOH.

Kolbe's (Kolbe–Schmitt) carboxylation reaction: sodium phenoxide is heated with $CO_2$ at about 400 K and 4–7 atm. This gives sodium salicylate, which is acidified to give salicylic acid (2-hydroxybenzoic acid).

$$C_6H_5ONa + CO_2 \xrightarrow[4\text{-}7\ atm]{400\ K} o\text{-}HOC_6H_4COONa \xrightarrow{H^+ (dil.\ HCl)} o\text{-}HOC_6H_4COOH$$

Dow's process is the manufacture of phenol from chlorobenzene by hydrolysis with aqueous NaOH at high temperature and pressure. The sodium phenoxide formed is then acidified:

$$C_6H_5Cl + 2NaOH \xrightarrow{623\ K,\ 300\ atm} C_6H_5ONa + NaCl + H_2O$$ $$C_6H_5ONa + HCl \rightarrow C_6H_5OH + NaCl$$

(c) Reimer-Tiemann reaction

When phenol is heated with $CHCl_3$ and alkali, dichlorocarbene ($:CCl_2$) attacks the ortho position. Hydrolysis then gives o-hydroxybenzaldehyde (salicylaldehyde):

$$C_6H_5OH + CHCl_3 + 3KOH \rightarrow o\text{-}HOC_6H_4CHO + 3KCl + 2H_2O$$

(d) Sodalime and Zn dust

Benzoic acid loses $CO_2$ when heated with sodalime (decarboxylation). Phenol is reduced to benzene when heated with zinc dust:

$$C_6H_5COOH + 2NaOH \xrightarrow{CaO,\ \Delta} C_6H_6 + Na_2CO_3 + H_2O \qquad C_6H_5OH + Zn \xrightarrow{\Delta} C_6H_6 + ZnO$$

(d) Liebermann's test

Phenol reacts with $NaNO_2$ in conc. $H_2SO_4$ to give a deep green or blue colour. On dilution with water the colour turns red, and adding NaOH brings back the blue/green colour. This is Liebermann's nitroso reaction, a test for phenols.

D can be made by boiling benzenediazonium chloride with dilute $H_2SO_4$, so D is phenol. Working back and forth from D:

$$\underset{(A)}{C_6H_5NO_2} \xrightarrow{Sn/HCl} \underset{(B)}{C_6H_5NH_2} \xrightarrow[0\text{-}5^\circ C]{NaNO_2 + HCl} \underset{(C)}{C_6H_5N_2^+Cl^-} \xrightarrow{H_2O,\ \Delta} \underset{(D)}{C_6H_5OH}$$

Reactions

1. Reduction: $C_6H_5NO_2 + 6[H] \xrightarrow{Sn/HCl} C_6H_5NH_2 + 2H_2O$
2. Diazotisation: $C_6H_5NH_2 + NaNO_2 + 2HCl \xrightarrow{0\text{-}5^\circ C} C_6H_5N_2Cl + NaCl + 2H_2O$
3. Hydrolysis: $C_6H_5N_2Cl + H_2O \xrightarrow{\Delta} C_6H_5OH + N_2 + HCl$
4. Reimer–Tiemann reaction, then reduction:

$$C_6H_5OH \xrightarrow{CHCl_3,\ aq.\ KOH} o\text{-}HOC_6H_4CHO \xrightarrow{LiAlH_4} o\text{-}HOC_6H_4CH_2OH$$

$LiAlH_4$ reduces the $-CHO$ group of salicylaldehyde to $-CH_2OH$.

Benzene sulphonic acid is fused with solid NaOH at about 573 K, giving sodium phenoxide, which is then acidified to liberate phenol:

$$C_6H_5SO_3H + 3NaOH \xrightarrow{\text{fused, }573\,K} C_6H_5ONa + Na_2SO_3 + 2H_2O$$ $$C_6H_5ONa + HCl \rightarrow C_6H_5OH + NaCl$$

(d) $NaHCO_3$

Carboxylic acids are stronger acids than carbonic acid, so they decompose $NaHCO_3$ with brisk effervescence of $CO_2$:

$$RCOOH + NaHCO_3 \rightarrow RCOONa + H_2O + CO_2\uparrow$$

Phenol is weaker than carbonic acid and gives no effervescence. KOH, Na and NaOH react with both, so they cannot tell them apart.

a) Phenol from chlorobenzene (Dow's process)

Chlorobenzene is heated with aqueous NaOH at about 623 K and 300 atm to give sodium phenoxide, which is acidified:

$$C_6H_5Cl + 2NaOH \xrightarrow{623\ K,\ 300\ atm} C_6H_5ONa + NaCl + H_2O$$ $$C_6H_5ONa + HCl \rightarrow C_6H_5OH + NaCl$$

b) Why phenol is nitrated more easily than benzene

The –OH group has a lone pair on oxygen which is delocalised into the ring (+M effect). This raises the electron density, especially at the ortho and para positions, so the ring is activated towards electrophilic attack. Phenol is therefore nitrated with dilute $HNO_3$ at room temperature to give o- and p-nitrophenol, while benzene needs a hot conc. $HNO_3$/conc. $H_2SO_4$ mixture.

c) Phenol to phenolphthalein

Two molecules of phenol are condensed with phthalic anhydride in the presence of conc. $H_2SO_4$ (or anhydrous $ZnCl_2$) on heating:

$$2C_6H_5OH + \text{phthalic anhydride} \xrightarrow{conc.\ H_2SO_4,\ \Delta} \text{phenolphthalein} + H_2O$$