Benzene can be obtained by heating either ben | vedclass

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Question

(B) Benzoic acid $(C_6H_5COOH)$ reacts with soda lime $(NaOH + CaO)$ to produce benzene via decarboxylation: $C_6H_5COOH + NaOH \xrightarrow{CaO} C_6H

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Soda lime and zinc dust

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(B) Benzoic acid $(C_6H_5COOH)$ reacts with soda lime $(NaOH + CaO)$ to produce benzene via decarboxylation: $C_6H_5COOH + NaOH \xrightarrow{CaO} C_6H_6 + Na_2CO_3$. Thus,$X$ is soda lime.Phenol $(C_6H_5OH)$ reacts with zinc dust to produce benzene via reduction: $C_6H_5OH + Zn \xrightarrow{\Delta} C_6H_6 + ZnO$. Thus,$Y$ is zinc dust.Therefore,$X$ and $Y$ are respectively soda lime and zinc dust.

Benzene can be obtained by heating either benzoic acid with $X$ or phenol with $Y$. $X$ and $Y$ are respectively

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