Observe the following set of reactions. What | vedclass

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(C) $1$. The conversion of benzoyl chloride $(C_6H_5COCl)$ to benzaldehyde $(C_6H_5CHO)$ is the Rosenmund reduction,which uses $H_2 | Pd-BaSO_4$ as th

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$H_2 | Pd-BaSO_4 ; (CH_3)_2Cd ; C_6H_5CH=CH-C(=O)C_6H_5$

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(C) $1$. The conversion of benzoyl chloride $(C_6H_5COCl)$ to benzaldehyde $(C_6H_5CHO)$ is the Rosenmund reduction,which uses $H_2 | Pd-BaSO_4$ as the reagent. Thus,$X = H_2 | Pd-BaSO_4$.$2$. The conversion of benzoyl chloride $(C_6H_5COCl)$ to acetophenone $(C_6H_5COCH_3)$ is achieved using a dialkyl cadmium reagent,$(CH_3)_2Cd$. Thus,$Y = (CH_3)_2Cd$.$3$. The reaction between benzaldehyde $(C_6H_5CHO)$ and acetophenone $(C_6H_5COCH_3)$ in the presence of $OH^-$ at $293 \ K$ is a Claisen-Schmidt condensation reaction. The major product $Z$ is benzalacetophenone (chalcone),which has the structure $C_6H_5CH=CH-C(=O)C_6H_5$.

Observe the following set of reactions. What are $X, Y$ and $Z$ respectively?

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