CCl_4 xrightarrow{Aq. KOH} A xrightarrow[ii) | vedclass

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(B) Step $1$: $CCl_4$ reacts with aqueous $KOH$ to form potassium carbonate $(K_2CO_3)$,which upon acidification gives carbonic acid $(H_2CO_3)$,which

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Gives Haloform reaction

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(B) Step $1$: $CCl_4$ reacts with aqueous $KOH$ to form potassium carbonate $(K_2CO_3)$,which upon acidification gives carbonic acid $(H_2CO_3)$,which decomposes to $CO_2$ and $H_2O$. However,in the context of this reaction sequence,$CCl_4$ with $KOH$ typically leads to the formation of $CO_2$ or carboxylic acid derivatives. Given the reaction with $CH_3MgBr$,$A$ is $CO_2$. Step $2$: $CO_2 + CH_3MgBr \rightarrow CH_3COOMgBr$. Step $3$: Upon hydrolysis $(H_3O^+)$,$CH_3COOMgBr$ gives acetic acid $(CH_3COOH)$. Step $4$: $B$ is $CH_3COOH$. Acetic acid contains a methyl group attached to a carbonyl carbon $(CH_3-C=O)$,which gives a positive iodoform test (a type of Haloform reaction).

$CCl_4$ $\xrightarrow{Aq. KOH} A$ $\xrightarrow[ii) H_3O^+]{i) CH_3MgBr} B$
Which of the following is correct about $B$?

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$CCl_4$ $\xrightarrow{Aq. KOH} A$ $\xrightarrow[ii) H_3O^+]{i) CH_3MgBr} B$ Which of the following is correct about $B$?