Question

An aromatic organic compound A yields B and C when it reacts with CHCl$_3$ and KOH. On distillation with Zn dust, D is formed. On oxidation, D yields compound E having molecular formula C$_7$H$_6$O$_2$. Identify A, B, C, D and E. Write the chemical equation of each reaction also.

Hint:

Phenol is a key starting compound in aromatic chemistry. Remember: Reimer–Tiemann → aldehyde, Zn dust → benzene, oxidation → benzoic acid.

Answer 1

Step 1: Reaction of A with CHCl$_3$ and KOH.
This is the
Reimer–Tiemann reaction. Phenol reacts with chloroform and alkali to form salicylaldehyde (ortho-hydroxy benzaldehyde, B) and para-hydroxy benzaldehyde (C). \[ A = C_6H_5OH \, (Phenol) \] \[ C_6H_5OH + CHCl_3 + 3KOH \;\longrightarrow\; o\text{-}OH{-}C_6H_4{-}CHO \,(B) + p\text{-}OH{-}C_6H_4{-}CHO \,(C) + 3KCl + 2H_2O \] Step 2: Distillation of A with Zn dust.
Phenol on distillation with Zn dust gives benzene. \[ C_6H_5OH \;\xrightarrow{Zn}\; C_6H_6 + ZnO \] \[ D = C_6H_6 \, (Benzene) \] Step 3: Oxidation of D.
Benzene on oxidation yields benzoic acid. \[ C_6H_6 \;\xrightarrow{[O]}\; C_6H_5COOH \] \[ E = C_7H_6O_2 \, (Benzoic \, acid) \] Final Identification:
\[ A = Phenol, \quad B = o\text{-}Hydroxybenzaldehyde, \quad C = p\text{-}Hydroxybenzaldehyde, \quad D = Benzene, \quad E = Benzoic \, acid \] Conclusion:
The aromatic compound A is phenol. Through Reimer–Tiemann reaction it forms aldehydes, through reduction with Zn it forms benzene, and further oxidation of benzene produces benzoic acid.