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Compound A, $C_8H_{10}O$, is found to react with NaOI (produced by reacting Y with NaOH) and yields a yellow precipitate with characteristic smell. A and Y are respectively
Explanation
Only the $CH_3CH(OH)-$ alcohol gives the iodoform test.
Detailed Solution
1-Phenylethanol is a secondary alcohol which on oxidation gives phenyl methyl ketone (acetophenone). This on reaction with $I_2$ and NaOH forms iodoform and sodium benzoate.
$2NaOH + I_2 \rightarrow NaOI + NaI + H_2O$
$C_6H_5CH(OH)CH_3\ (A) \xrightarrow{NaOI} C_6H_5COCH_3$ (acetophenone)
$C_6H_5COCH_3 \xrightarrow{I_2/NaOH} C_6H_5COONa + CHI_3$ (iodoform, yellow ppt)
So A is $C_6H_5CH(OH)CH_3$ and Y is $I_2$.
$2NaOH + I_2 \rightarrow NaOI + NaI + H_2O$
$C_6H_5CH(OH)CH_3\ (A) \xrightarrow{NaOI} C_6H_5COCH_3$ (acetophenone)
$C_6H_5COCH_3 \xrightarrow{I_2/NaOH} C_6H_5COONa + CHI_3$ (iodoform, yellow ppt)
So A is $C_6H_5CH(OH)CH_3$ and Y is $I_2$.