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  4. In the reaction, C6H5OH->[NaOH](A)->[CO2][140°C,(4-7)atm](B)->[HCl](C), the compound (C) is

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Q. In the reaction, $\ce{C6H5OH->[NaOH](A)->[CO_2][140^°C,(4-7)atm](B)->[HCl](C)}$, the compound $(C)$ is

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Solution:

Treatment of sodium salt of phenol with $CO_2$ under pressure brings about substitution of the $-COOH$ group for the hydrogen of the ring. This is called as Kolbe's reaction

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