An organic compound having the molecular form | vedclass

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(D) The reaction with $I_2$ and dilute $NaOH$ is the iodoform test,which is given by compounds containing the $CH_3CO-$ group or the $CH_3CH(OH)-$ gro

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$1-$phenylethanol

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(D) The reaction with $I_2$ and dilute $NaOH$ is the iodoform test,which is given by compounds containing the $CH_3CO-$ group or the $CH_3CH(OH)-$ group.Among the given options,$1$-phenylethanol $(C_6H_5CH(OH)CH_3)$ contains the $CH_3CH(OH)-$ group.It undergoes oxidation to acetophenone $(C_6H_5COCH_3)$ in the presence of $NaOI$,which then reacts further to produce a yellow precipitate of iodoform $(CHI_3)$.The reaction sequence is:$C_6H_5CH(OH)CH_3$ $\xrightarrow[Oxidation]{NaOI} C_6H_5COCH_3$ $\xrightarrow[Haloform]{I_2+NaOH} C_6H_5COO^- + CHI_3 \downarrow$ (yellow precipitate).

An organic compound having the molecular formula $C_8H_{10}O$ on being heated with $I_2$ and dilute $NaOH$ gives a yellow precipitate. The expected compound is

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$A$ compound of molecular formula $C_8H_8O_2$ reacts with acetophenone to form a single cross-aldol product in the presence of base. The same compound on reaction with conc. $NaOH$ forms benzyl alcohol as one of the products. The structure of the compound is

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In the following reactions,
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$CH_3COCH_3 \xrightarrow[(ii) H_3O^+]{(i) NaBD_4} Y$
$X$ and $Y$ are

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