Question

Bottles containing C₆H₅I and C₆H₅CH₂I lost their original labels. They are labelled as A and B. A and B were separately taken in a test tubes and boiled with NaOHsolution.Then in each test tube Dilute HNO₃ and AgNO₃ was added. Substance B gave a yellow precipitate. Which one of thefollowingstatements is true for this experiment?

A. Addition of HNO₃ was not necessary
B. A was C₆H₅CH₂I
C. A was C₆H₅Cl
D. B was C₆H₅I

Answer 1

The yellow precipitate in Substance B after reaction with NaOH, HNO₃, and AgNO₃ indicates the presence of iodide ions. B was C₆H₅I because silver iodide is yellow. The necessary addition of dilute HNO₃ ensures that the observed precipitate is due to halide ions.

Step-by-step explanation:

The student's question involves a chemistry procedure to identify two compounds, C₆H₅I and C₆H₅CH₂I, which have lost their labels and are now labeled as A and B. They were tested by boiling with NaOH solution followed by reaction with dilute HNO₃ and AgNO₃. Substance B produced a yellow precipitate, a characteristic reaction of iodide ions with AgNO₃ in the presence of nitric acid.

From this information, we can determine that the yellow precipitate is the result of the reaction between silver nitrate and iodide ions, forming silver iodide, which is yellow. Hence, B must contain iodide ions (I⁻) and is therefore likely C₆H₅I. The purpose of adding HNO₃ is to ensure the precipitate formation is due to halide ions and not due to other anions such as carbonate or hydroxide, which would also precipitate with silver nitrate. Therefore, the addition of HNO₃ is necessary to prevent false positives.

Given this evidence, the true statement regarding this experiment is: B was C₆H₅I.