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(C) $1$. Decarboxylation of $C_6H_5COONa$ with soda lime $(NaOH/CaO)$ gives $X = C_6H_6$ (Benzene).$2$. Gattermann-Koch reaction of benzene with $CO/H

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Sodium salt of benzoic acid

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(C) $1$. Decarboxylation of $C_6H_5COONa$ with soda lime $(NaOH/CaO)$ gives $X = C_6H_6$ (Benzene).$2$. Gattermann-Koch reaction of benzene with $CO/HCl$ in the presence of anhydrous $AlCl_3$ gives $Y = C_6H_5CHO$ (Benzaldehyde).$3$. The reaction of benzaldehyde with concentrated $NaOH$ is a Cannizzaro reaction.$4$. In the Cannizzaro reaction,$C_6H_5CHO$ undergoes disproportionation to give $A = C_6H_5CH_2OH$ (Benzyl alcohol,the reduction product) and $B = C_6H_5COONa$ (Sodium benzoate,the oxidation product).

Consider the following sequence of reactions.
$C_6H_5COONa$ $\xrightarrow[\Delta]{NaOH/CaO} X$ $\xrightarrow[\text{Anhy. } AlCl_3]{CO/HCl} Y$ $\xrightarrow[NaOH]{\text{conc. }} A + B$.
If $A$ is the reduction product of $Y$,what is $B$?

Similar Questions

Product $(B)$ is :

The structure of the main product $(B)$ of the following reaction will be:

The product $(4)$ in the given reaction sequence is:

The reaction of ethyl methyl ketone with $Cl_2$ / excess $OH^{-}$ gives the following major product:

$CH_3-CH(OH)-CH_2-CH_2-CO-CH_3 \rightarrow CH_3-CH(OH)-(CH_2)_3-CH_3$
The above conversion can be achieved by:

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Consider the following sequence of reactions.$C_6H_5COONa$ $\xrightarrow[\Delta]{NaOH/CaO} X$ $\xrightarrow[\text{Anhy. } AlCl_3]{CO/HCl} Y$ $\xrightarrow[NaOH]{\text{conc. }} A + B$.If $A$ is the reduction product of $Y$,what is $B$?