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).

Column-$I$ (Reactions)	Column-$II$ (Reagents)
$(A).$ Benzophenone $\to$ Diphenylmethane	$(1).$ $LiAlH_4$
$(B).$ Benzaldehyde $\to$ $1-$Phenylethanol	$(2).$ $DiBAL-H$
$(C).$ Cyclohexanone $\to$ Cyclohexanol	$(3).$ $Zn(Hg) / \text{Conc. } HCl$
$(D).$ Phenyl benzoate $\to$ Benzaldehyde	$(4).$ $CH_3MgBr$

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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