CH_3COOH xrightarrow{text{Sodalime}} (A) xrig | vedclass

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(B) Step $1$: Decarboxylation of $CH_3COOH$ with sodalime $(NaOH + CaO)$ gives methane $(CH_4)$.$CH_3COOH + NaOH \xrightarrow{\Delta} CH_4 + Na_2CO_3$

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$CH_3OH$

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(B) Step $1$: Decarboxylation of $CH_3COOH$ with sodalime $(NaOH + CaO)$ gives methane $(CH_4)$.$CH_3COOH + NaOH \xrightarrow{\Delta} CH_4 + Na_2CO_3$ $(A = CH_4)$Step $2$: Free radical bromination of methane gives methyl bromide $(CH_3Br)$.$CH_4 + Br_2 \xrightarrow{hv} CH_3Br + HBr$ $(B = CH_3Br)$Step $3$: Nucleophilic substitution of methyl bromide with aqueous $KOH$ gives methanol $(CH_3OH)$.$CH_3Br + KOH_{(aq)} \rightarrow CH_3OH + KBr$ $(C = CH_3OH)$Thus,the final product $(C)$ is $CH_3OH$.

$CH_3COOH$ $\xrightarrow{\text{Sodalime}} (A)$ $\xrightarrow{Br_2 / hv} (B)$ $\xrightarrow[KOH]{\text{Aq.}} (C)$ is $:-$

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