Question

Bottles containing $PhI$ and $Ph{CH}_{2}I$ lost their original labels. They were labelled as $\left(A\right)$ and $\left(B\right)$ for testing. $\left(A\right)$ and $\left(B\right)$ were separately taken in test tubes and boiled with $NaOH$ solutions. The end solution in each tube was made acidic with dilutie $H{NO}_3$ and some $Ag{NO}_3$ solution was added. Substance $\left(B\right)$ gave a yellow precipitate. Which of the following statements is true for this experiment?

• A. Addition of $H{NO}_3$ was unnecessary
• B. $\left(A\right)$ was $PhI$
• C. $\left(A\right)$ was $Ph{CH}_{2}I$
• D. $\left(B\right)$ was $PhI$

Answer 1

The correct option is B $\left(A\right)$ was $PhI$

$PhI+aq.NaOH\to Noreaction$

$PhC{H}_{2}I+Aq.NaOH\to PhC{H}_{2}OH+I^{\ominus }\stackrel{AgN{O}_3}{\to }AgI(Yellowppt.)$

(A) was iodobenzene $PhI$ and (B) was benzyl iodide $PhC{H}_{2}I$.

Since (A) is aromatic iodide, it will not give free iodide ions. Hence, no precipitate will be obtained with silver nitrate.

(B) is alkyl iodide. It will give free iodide ions. Hence, yellow precipitate will be obtained with silver nitrate.