Problem 107
Question
Bottles containing \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{I}\) and \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{I}\) lost their original labels. They were labeled \(\mathrm{A}\) and \(\mathrm{B}\) for testing. A and B were separately taken in test tubes and boiled with \(\mathrm{NaOH}\) solution. The end solution in each tube was made acidic with dilute \(\mathrm{HNO}_{3}\) and some \(\mathrm{AgNO}_{3}\) solution added. Solution B gave a yellow precipitate. Which one of the following statements is true for the experiment? (a) A was \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{I}\) (b) A was \(\mathrm{C}_{6} \mathrm{H}_{3} \mathrm{I}\) (c) \(\mathrm{B}\) was \(\mathrm{C}_{6} \mathrm{H}_{3} \mathrm{I}\) (d) addition of \(\mathrm{HNO}_{3}\) was unnecessary
Step-by-Step Solution
VerifiedKey Concepts
Silver Nitrate Test
- This test is useful in distinguishing substances that produce free halide ions from those that do not.
- The reaction is sensitive, making it a reliable indicator of iodide presence.
Substitution Reaction
When organic iodides such as benzyl iodide (\(\mathrm{C}_6\mathrm{H}_5\mathrm{CH}_2\mathrm{I}\)) are heated with sodium hydroxide (\(\mathrm{NaOH}\)), a substitution reaction takes place. The iodine atom is replaced by a hydroxyl group, converting benzyl iodide into benzyl alcohol (\(\mathrm{C}_6\mathrm{H}_5\mathrm{CH}_2\mathrm{OH}\)).
- This reaction releases free iodide ions into the solution, priming it for further testing.
Benzyl Iodide
Benzyl iodide readily undergoes a substitution with hydroxide ions from \(\mathrm{NaOH}\), releasing iodide ions into the solution.
- This property makes it detectable by the Silver Nitrate Test.
Iodobenzene
In the context of the Silver Nitrate Test, this means iodobenzene won't form a yellow precipitate, helping distinguish it from more reactive compounds like benzyl iodide.
- Its inertness under test conditions prevents iodide ion release.
Precipitation Reaction
In the experiment, when silver nitrate is added to a solution with free iodide ions, \(\mathrm{Ag}^+ \) and \(\mathrm{I}^- \) ions combine to form silver iodide (\(\mathrm{AgI}\)), a yellow precipitate.
- This reaction is immediate and visible, serving as a strong indicator of iodide presence.
- The distinct yellow color of \(\mathrm{AgI}\) confirms the completion of the precipitation reaction.