11.29 Give equations of the following reactions: (i) Oxidation of propan-1-ol with alkaline KMnO4 solution. (ii) Bromine in CS2 with phenol. (iii) Dilute HNO3 with phenol. (iv) Treating phenol with chloroform in presence of aqueous NaOH.

11.29 Oxidation of propane-1-ol with alkaline KMnO4 solution gives propanoic acid as the product. As the oxidation of primary alcohol gives carboxylic acid as the major product in the presence of a strong oxidizing reagent. And here KMnO4 is a very strong oxidizing agent. A mixture of o-bromo phenol and p-bromo phenol is formed. The formation of 2 products depends totally on the reaction conditions. Dilute HNO3 with phenol. Only dilute acid will be required for the nitration of phenol, nitric acid contains a small amount of nitrous acid which because of the activation of the ring will be more than enough to nitrate the phenol. Two products are formed 2-nitrophenol (ortho) and 4-nitrophenol (para). Treating Phenol with chloroform in presence of aqueous NaOH results in the formation of salicylaldehyde (Reimer-Tiemann reaction)

11.29 Give equations of the following reactions:

(i) Oxidation of propan-1-ol with alkaline KMnO₄ solution.

(ii) Bromine in CS₂ with phenol.

(iii) Dilute HNO₃ with phenol.

(iv) Treating phenol with chloroform in presence of aqueous NaOH.

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Alok Kumar Singh | Contributor-Level 10

10 months ago

11.29

Oxidation of propane-1-ol with alkaline KMnO₄ solution gives propanoic acid as the product. As the oxidation of primary alcohol gives carboxylic acid as the major product in the presence of a strong oxidizing reagent. And here KMnO₄ is a very strong oxidizing agent.

A mixture of o-bromo phenol