What are X and Y respectively in the followin | vedclass

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(C) The first reaction is the reduction of benzoic acid $(C_6H_5COOH)$ to benzaldehyde $(C_6H_5CHO)$. This cannot be done directly with $SOCl_2$ and $

Vedclass · Accepted Answer

$(i) \ C_2H_5OH|H^+, (ii) \ DiBAL-H, (iii) \ H_2O; \ C_6H_5-CH=CH-CO-C_6H_5$

Vedclass · Answer

(C) The first reaction is the reduction of benzoic acid $(C_6H_5COOH)$ to benzaldehyde $(C_6H_5CHO)$. This cannot be done directly with $SOCl_2$ and $H_2|Ni$ (which reduces acid chlorides to aldehydes,but the conditions are specific). The standard laboratory method to reduce a carboxylic acid to an aldehyde involves converting it to an ester first,followed by reduction with $DiBAL-H$.Step $1$: $C_6H_5COOH + C_2H_5OH \xrightarrow{H^+} C_6H_5COOC_2H_5 + H_2O$ (Esterification).Step $2$: $C_6H_5COOC_2H_5 \xrightarrow{(i) \ DiBAL-H, (ii) \ H_2O} C_6H_5CHO$ (Reduction of ester to aldehyde).Thus,$X$ is $(i) \ C_2H_5OH|H^+, (ii) \ DiBAL-H, (iii) \ H_2O$.Step $3$: The second reaction is an Aldol condensation between benzaldehyde $(C_6H_5CHO)$ and acetophenone (implied,though the prompt shows $C_6H_5COOH$ as a reactant,the product $Y$ is clearly the Claisen-Schmidt condensation product,benzalacetophenone or chalcone,$C_6H_5-CH=CH-CO-C_6H_5$).Therefore,the correct option is $C$.

What are $X$ and $Y$ respectively in the following reactions?
$C_6H_5COOH \xrightarrow{X} C_6H_5CHO$
$C_6H_5CHO \xrightarrow[OH^-, 293 \ K]{C_6H_5COOH} Y \text{ (Major product)}$

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What are $X$ and $Y$ respectively in the following reactions?$C_6H_5COOH \xrightarrow{X} C_6H_5CHO$$C_6H_5CHO \xrightarrow[OH^-, 293 \ K]{C_6H_5COOH} Y \text{ (Major product)}$