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Bottles containing C(6)H(5)I and C(6)H(5...

Bottles containing `C_(6)H_(5)I` and `C_(6)H_(5)-CH_(2)I` lost their original lables. They were labelled A and B for festing. A and B were separately taken in a test tube and boiled with `NaOH` solution. The end solution in each tube was made acidic with dilute `HNO_(3)` and then some `AgNO_(3)` solution was added. Substance B gave a yellow precipitate. Which one of the following statements is true for this experiment.

A

Addition of `HNO_3` was unnecessary

B

A was `C_6H_5CH_2I`

C

A was `C_6H_5CH_2I`

D

B was `C_6H_5I`

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem, we need to analyze the reactions of the compounds `C6H5I` (iodobenzene) and `C6H5-CH2I` (benzyl iodide) when they are treated with sodium hydroxide (NaOH) and silver nitrate (AgNO3) after being made acidic with dilute nitric acid (HNO3). ### Step-by-Step Solution: 1. **Identify the Compounds:** - A could be either `C6H5I` (iodobenzene) or `C6H5-CH2I` (benzyl iodide). - B could be the other compound. 2. **Reactions with NaOH:** - When `C6H5I` (iodobenzene) is treated with NaOH, it does not undergo nucleophilic substitution because it is an aryl halide. Therefore, it remains unchanged. - When `C6H5-CH2I` (benzyl iodide) is treated with NaOH, it undergoes nucleophilic substitution to form `C6H5-CH2O^-` (benzyl alcohol in its ionized form). 3. **Acidification with HNO3:** - The solution containing `C6H5I` remains unchanged after the reaction with NaOH. - The solution containing `C6H5-CH2O^-` (from benzyl iodide) will be converted back to benzyl alcohol when made acidic with dilute HNO3. 4. **Addition of AgNO3:** - When AgNO3 is added to the solution with benzyl alcohol, it forms a yellow precipitate of silver iodide (AgI) because benzyl alcohol can react with AgNO3. - The solution that initially contained `C6H5I` will not produce any precipitate because there is no reaction. 5. **Conclusion:** - Since substance B gave a yellow precipitate, it must be `C6H5-CH2I` (benzyl iodide), and substance A must be `C6H5I` (iodobenzene). ### Final Statements: - A was `C6H5I` (iodobenzene). - B was `C6H5-CH2I` (benzyl iodide). - The correct statement is: **B was `C6H5-CH2I`.**
To solve the problem, we need to analyze the reactions of the compounds `C6H5I` (iodobenzene) and `C6H5-CH2I` (benzyl iodide) when they are treated with sodium hydroxide (NaOH) and silver nitrate (AgNO3) after being made acidic with dilute nitric acid (HNO3). ### Step-by-Step Solution: 1. **Identify the Compounds:** - A could be either `C6H5I` (iodobenzene) or `C6H5-CH2I` (benzyl iodide). - B could be the other compound. ...
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