Compound $A,{\text{C}}_8{\text{H}}_{10}\text{O}$, is found to react with $\text{NaOI}$ (produced by reacting $Y$ with $\text{NaOH}$) and yields a yellow precipitate with characteristic smell. $A$ and $Y$ are respectively:

1. Reacting with $\text{NaOI}$ to give a yellow precipitate (${\text{CHI}}_3$, Iodoform) indicates the presence of a methyl ketone ($-\text{CO}-{\text{CH}}_3$) or a methyl carbinol ($-\text{CH(OH)}-{\text{CH}}_3$).2. Compound $A$ has formula ${\text{C}}_8{\text{H}}_{10}\text{O}$.3. $1$ -Phenylethanol (${\text{C}}_6{\text{H}}_5-\text{CH(OH)}-{\text{CH}}_3$) has the required methyl carbinol group and matches the molecular formula.4. $Y$ must be ${\text{I}}_2$ because ${\text{I}}_2+\text{NaOH}\to \text{NaOI}+\text{NaI}+{\text{H}}_2\text{O}$.