CCl_4 + KOH (excess) rightarrow end product o | vedclass

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Question

(A) The reaction of carbon tetrachloride $(CCl_4)$ with excess aqueous potassium hydroxide $(KOH)$ proceeds as follows:$1$. Initially,$CCl_4$ undergoe

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$K_2CO_3$

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(A) The reaction of carbon tetrachloride $(CCl_4)$ with excess aqueous potassium hydroxide $(KOH)$ proceeds as follows:$1$. Initially,$CCl_4$ undergoes nucleophilic substitution where chlorine atoms are replaced by hydroxyl groups to form an unstable intermediate,$C(OH)_4$.$2$. $C(OH)_4$ is highly unstable due to the presence of four hydroxyl groups on a single carbon atom,leading to the loss of two water molecules $(2H_2O)$ to form carbonic acid $(H_2CO_3)$.$3$. Since the reaction occurs in the presence of excess $KOH$ (a strong base),the carbonic acid is neutralized to form potassium carbonate $(K_2CO_3)$ and water.$4$. The overall reaction is: $CCl_4 + 6KOH \rightarrow K_2CO_3 + 4KCl + 3H_2O$.

$CCl_4 + KOH$ (excess) $\rightarrow$ end product of the reaction is :

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