Problem 107 - A Complete Resource Book in Chemistry for JEE Main

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) \(\mathrm{A}\) was \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{I}\) (b) A was \(\mathrm{C}_{6} \mathrm{H}_{6} \mathrm{I}\) (c) B was \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{I}\) (d) addition of \(\mathrm{HNO}_{3}\) was unnecessary

Step-by-Step Solution

Answer

(a) A was mathrm{C}_{6} mathrm{H}_{5} mathrm{CH}_{2} mathrm{I}.

Step 1: Understanding the reaction setup

The problem involves testing two compounds labeled A and B by reacting them sequentially with NaOH, HNO₃, and AgNO₃. The identity of the compounds depends on the reaction outcomes, especially the formation of a precipitate.

Step 2: Reviewing halides precipitation with AgNO₃

Silver nitrate (AgNO₃) reacts with iodide ions (I⁻) to form a yellow precipitate of silver iodide (AgI). This indicates which compound produces free iodide ions upon reaction.

Step 3: Reactivity of aryl and benzyl iodides with NaOH

When boiled with NaOH, mathrm{C}_{6} mathrm{H}_{5} mathrm{I} (aryliodide) is less likely to liberate iodide ions than mathrm{C}_{6} mathrm{H}_{5} mathrm{CH}_{2} mathrm{I} (benzyliodide), which can form an alcohol and iodide ions. Thus, iodide ions are more likely liberated from benzyliodide.

Step 4: Analyzing which sample formed a yellow precipitate

Since solution B gave a yellow precipitate, compound B must be mathrm{C}_{6} mathrm{H}_{5} mathrm{CH}_{2} mathrm{I}. This is due to the formation of I⁻ ions which react with AgNO₃ to form yellow AgI.

Step 5: Concluding which statement is true

Since B gives the expected reaction for mathrm{C}_{6} mathrm{H}_{5} mathrm{CH}_{2} mathrm{I}, sample A that did not give a yellow precipitate must be mathrm{C}_{6} mathrm{H}_{5} mathrm{I}. Thus, statement (a) is true.

Key Concepts

Halide Reactions

In organic chemistry, halide reactions often involve the conversion of halogen-containing groups within molecules. A common analysis involves reacting these molecules with aqueous sodium hydroxide (NaOH) and then with silver nitrate (AgNO_3). Halide ions can react with AgNO_3 to form precipitates of varying colors, which help in identifying the specific halide present.
For instance, iodide ions (I^{-}) react with AgNO_3 to form a yellow precipitate of silver iodide (AgI). This characteristic yellow precipitate becomes an essential marker.
Thus, the presence of a yellow precipitate upon the addition of AgNO_3 indicates iodide ions were liberated, pointing towards a specific behavior of the initial organic compound undergoing the reaction.

• Halide ions react with silver nitrate.
• Precipitate color depends on the type of halide.
• Yellow precipitate indicates iodide ions.

Precipitation Reactions

Precipitation reactions involve the formation of a solid from a solution during a chemical reaction. In organic chemistry, these reactions are particularly useful for separating and identifying different ions in solution.
When you introduce silver nitrate (AgNO_3) to a solution containing iodide ions (I^{-}), a precipitation reaction occurs, producing silver iodide (AgI) as a yellow solid. This reaction serves as a diagnostic tool for detecting iodide ions.
This ability to form a yellow precipitate differentiates iodide ions from other halides, such as chloride (Cl^{-}), which yields a white precipitate. Using this visual difference, one can identify the presence of specific ions in a sample.

• Precipitation helps in identifying ions.
• AgI precipitate is yellow and indicates iodides.
• Different halides form visually distinct precipitates.

Aryl and Benzyl Iodides

Aryl and benzyl iodides are both organic compounds containing an iodine atom, but they differ in their reactivity due to their distinct structures.
Aryl iodides, such as C_6H_5I, have the iodine atom bonded directly to an aromatic ring (benzene). This bond is stable and less reactive, meaning they do not easily release iodide ions on reaction with bases like sodium hydroxide (NaOH).
Benzyl iodides, exemplified by C_6H_5CH_2I, possess the iodine atom attached to a benzyl group. This setup is more reactive, facilitating the release of iodide ions under basic conditions due to the formation of a benzyl alcohol.
The difference in reactivity between these iodides manifests in reactions used to identify them. Benzyl iodides readily release iodide ions, which react with AgNO_3 to form a yellow precipitate. Conversely, aryl iodides are far less reactive, so they usually do not exhibit such reactions as markedly.

• Aryl iodides: Stable, less reactive.
• Benzyl iodides: Reactive, more likely to form iodide ions.
• Reactivity differences help in identification.