Among the aromatic isomers with the molecular | vedclass

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(A) $1$. The molecular formula $C_8H_8O$ corresponds to several aromatic isomers.$2$. The iodoform test is given by compounds containing the $CH_3CO-$

Vedclass · Accepted Answer

$X = 	ext{Acetophenone}, Y = 	ext{Phenylacetaldehyde}$

Vedclass · Answer

(A) $1$. The molecular formula $C_8H_8O$ corresponds to several aromatic isomers.$2$. The iodoform test is given by compounds containing the $CH_3CO-$ group or the $CH_3CH(OH)-$ group. Among the isomers of $C_8H_8O$,$Acetophenone$ $(C_6H_5COCH_3)$ contains the $CH_3CO-$ group and thus gives a positive iodoform test. Therefore,$X = 	ext{Acetophenone}$.$3$. The Benedict's test is given by aldehydes. Among the isomers of $C_8H_8O$,$Phenylacetaldehyde$ $(C_6H_5CH_2CHO)$ is an aldehyde and gives a positive Benedict's test. Therefore,$Y = 	ext{Phenylacetaldehyde}$.$4$. Thus,$X = 	ext{Acetophenone}$ and $Y = 	ext{Phenylacetaldehyde}$.

Among the aromatic isomers with the molecular formula $C_8H_8O$,one isomer $X$ gives the iodoform test,while another isomer $Y$ gives the Benedict's test. What are the structures of $X$ and $Y$ respectively?

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