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(B) $1$. The first step is the Gattermann-Koch reaction where benzene reacts with $CO$ and $HCl$ in the presence of anhydrous $AlCl_3/CuCl$ to form $A

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Acetophenone

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(B) $1$. The first step is the Gattermann-Koch reaction where benzene reacts with $CO$ and $HCl$ in the presence of anhydrous $AlCl_3/CuCl$ to form $A$,which is benzaldehyde $(C_6H_5CHO)$.$2$. The second step is the Claisen-Schmidt condensation reaction between benzaldehyde $(A)$ and a ketone $(B)$ in the presence of a base $(OH^-)$ to form the product,which is chalcone $(C_6H_5-CH=CH-CO-C_6H_5)$.$3$. For the formation of chalcone,the ketone $B$ must be acetophenone $(C_6H_5COCH_3)$.$4$. Therefore,the substance $B$ is acetophenone.

Consider the following sequence of reactions:
$Benzene$ $\xrightarrow{CO, HCl, \text{anhydrous } AlCl_3/CuCl} A$ $\xrightarrow{B, OH^-, 293 K} \text{Chalcone (1,3-diphenylprop-2-en-1-one)}$
The substance $B$ is:

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In the reaction sequence,
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Consider the following sequence of reactions:$Benzene$ $\xrightarrow{CO, HCl, \text{anhydrous } AlCl_3/CuCl} A$ $\xrightarrow{B, OH^-, 293 K} \text{Chalcone (1,3-diphenylprop-2-en-1-one)}$The substance $B$ is: