Bottles containing $${C_6}{H_5}I$$  and $${C_6}{H_5}C{H_2}I$$   lost their original labels. They were labelled $$A$$ and $$B$$ for testing. $$A$$ and $$B$$ were separately taken in test tubes and boiled with $$NaOH$$  solution. The end solution in each tube was made acidic with dilute $$HN{O_3}$$  and then some $$AgN{O_3}$$  solution was added. Substance B gave a yellow precipitate. Which one of the following statements is true for this experiment ?

Solution :
\[\begin{align} & {{C}_{6}}{{H}_{5}}I\xrightarrow{NaOH}{{C}_{6}}{{H}_{5}}ONa\xrightarrow{HN{{O}_{3}}/{{H}^{+}}} \\ & {{C}_{6}}{{H}_{5}}OH\xrightarrow{AgN{{O}_{3}}}\text{No}\,\,\text{yellow}\,\,ppt. \\ & {{C}_{6}}{{H}_{5}}C{{H}_{2}}I\xrightarrow{NaOH}{{C}_{6}}{{H}_{5}}C{{H}_{2}}ONa\xrightarrow{HN{{O}_{3}}/{{H}^{+}}} \\ & {{C}_{6}}{{H}_{5}}C{{H}_{2}}OH\xrightarrow{AgN{{O}_{3}}}.\,\text{yellow}\,\,ppt. \\ \end{align}\]
Since benzyl iodide gives yellow ppt. hence this is compound $$B$$ and $$A$$ was phenyl iodide \[\left( {{C}_{6}}{{H}_{5}}I \right).\]