What is the product obtained when CCl_4 is bo | vedclass

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(D) When $CCl_4$ is boiled with an aqueous solution of $KOH$,the chlorine atoms are replaced by hydroxyl groups to form an unstable intermediate,$C(OH

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Carbon dioxide

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(D) When $CCl_4$ is boiled with an aqueous solution of $KOH$,the chlorine atoms are replaced by hydroxyl groups to form an unstable intermediate,$C(OH)_4$.This intermediate quickly loses a water molecule to form carbonic acid $(H_2CO_3)$,which further decomposes into $CO_2$ and $H_2O$.The overall reaction is: $CCl_4 + 4KOH$ $\rightarrow C(OH)_4 + 4KCl$ $\rightarrow H_2CO_3 + 4KCl$ $\rightarrow CO_2 + 2H_2O + 4KCl$.

What is the product obtained when $CCl_4$ is boiled with $KOH$?

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