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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

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

Vedclass · Accepted Answer

$K_2CO_3$

Vedclass · Answer

(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 :

Similar Questions

The major product of the reaction is:

Which halide will be most reactive towards $SN^2$ reaction?

Which of the following represents the Wurtz-Fittig reaction?

Identify the major product $X$ in the following reaction: $CH_3-CH_2-CHF-CH_3 \xrightarrow[\Delta]{alc. KOH} X$ (Major)

What is the major product obtained in the reaction of $1-$phenyl$-2-$bromobutane with $NaOMe$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module