Write the reaction involved in Kolbe's reaction.
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Solution and Explanation
Step 1: Formation of sodium phenoxide Phenol reacts with sodium hydroxide to form sodium phenoxide. \[ C_6H_5OH + NaOH \rightarrow C_6H_5ONa + H_2O \]
Step 2: Reaction with carbon dioxide Sodium phenoxide is heated with carbon dioxide under pressure. \[ C_6H_5ONa + CO_2 \xrightarrow[4-7\,atm]{373\,K} o\text{-}HOC_6H_4COONa \] The carboxyl group enters predominantly at the ortho position due to activation of the ring by the phenoxide ion.
Step 3: Acidification The sodium salt formed is treated with dilute hydrochloric acid. \[ o\text{-}HOC_6H_4COONa + HCl \rightarrow o\text{-}HOC_6H_4COOH + NaCl \]
Step 4: Product obtained The final product is salicylic acid (2-hydroxybenzoic acid), an important industrial and pharmaceutical compound. \[ \boxed{ o\text{-}HOC_6H_4COOH } \]
Overall Reaction \[ \boxed{ C_6H_5ONa + CO_2 \xrightarrow[4-7\,atm]{373\,K} o\text{-}HOC_6H_4COONa \xrightarrow{H^+} o\text{-}HOC_6H_4COOH } \]
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Kolbe's reaction - Assertion (A): The presence of -OH group in phenols directs the incoming group to ortho and para positions.
Reason (R): -OH group in phenols activates the aromatic ring towards electrophilic substitution reaction. - Assertion (A) : Phenol is more acidic than 4-methylphenol. Reason (R) : The presence of an electron releasing group in phenol makes it more acidic.
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