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 labelled \(\mathrm{A}\) and \(\overline{\mathrm{B}}\) for testing. A and \(\mathrm{B}\) 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) \(\mathrm{A}\) was \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{I}\) (c) B was \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{I}\) (d) Addition of \(\mathrm{HNO}_{3}\) was unnecessary

Short answer

Statement (a) is true: A was \( \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{I} \)."

Step-by-step solution

Understand the Reaction with NaOH

When iodides such as \( \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{I} \) and \( \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{I} \) are boiled with \( \mathrm{NaOH} \), they undergo a nucleophilic substitution reaction, producing alcohols and releasing iodide ions \( \mathrm{I}^- \). This is important because the presence of iodide ions in solution is necessary for the subsequent steps.

Examine the Reaction with AgNO3

Adding \( \mathrm{AgNO}_3 \) to the acidic solution results in the formation of silver halides. Silver iodide (\( \mathrm{AgI} \)) precipitates as a yellow solid. In the test, the solution labeled \( \mathrm{B} \) gave a yellow precipitate, indicating the presence of iodide ions corresponding to \( \mathrm{AgI} \).

Analyze the Properties of C₆H₅I and C₆H₅CH₂I

\( \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{I} \) reacts faster due to its benzyl iodide's reactivity compared to phenyl iodide. The silver iodide precipitate is more likely to form quickly in the presence of \( \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{I} \), as it results in \( \mathrm{B} \) forming a yellow precipitate.

Determine Which is Label B

The yellow precipitate from solution \( \mathrm{B} \) indicates that it is \( \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{I} \), as this is likely to release iodide ions more rapidly under the conditions provided in the experiment.

Confirm the Correct Statement

Since \( \mathrm{B} \) is confirmed to be \( \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{I} \), statement (c) is false. Thus, statement (a) \(" A \) was \( \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{I} \)" is true."

Key concepts

Nucleophilic Substitution Reaction

Nucleophilic substitution reactions are a fundamental concept in organic chemistry, vital for understanding many reactions involving alkyl halides. In these reactions, a nucleophile, which is a chemical species with a pair of electrons to donate, attacks an electrophilic center of a molecule, replacing a leaving group. Here, the leaving group is typically a halide ion, such as iodide (
)In the context of \( \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{I} \) and \( \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{I} \), when these compounds are boiled with \( \mathrm{NaOH} \), the hydroxide ion \( \mathrm{OH}^- \) acts as the nucleophile.

• The iodide ion acts as the leaving group, resulting in the formation of alcohols.
• For \( \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{I} \), a benzyl halide, the hydroxyl substitution is more efficient due to the stronger reactivity of the benzylic position compared to phenyl.
• For \( \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{I} \), the reaction is generally slower, since the phenyl group is less reactive to nucleophilic attacks.

Understanding the different reactivities in this substitution reaction is essential for predicting outcomes and planning organic synthesis efficiently.

Silver Halide Precipitation

The precipitation of silver halides is a classic test to determine the presence of halide ions in a solution. Upon treating an acidic solution with \( \mathrm{AgNO}_{3} \), silver ions \( \mathrm{Ag}^+ \) react with halide ions to form silver halide precipitates. This principle is crucial for identifying (
)
different halides:

• Silver iodide (\( \mathrm{AgI} \)) forms a yellow precipitate, indicative of the presence of iodide ions.
• Silver bromide (\( \mathrm{AgBr} \)) gives a cream precipitate, while silver chloride (\( \mathrm{AgCl} \)) produces a white precipitate.

In the exercise, Solution B released iodide ions effectively, resulting in a yellow precipitate of silver iodide. This test is not only important for analytical chemistry but also for practical applications, like photographic films where the light sensitivity of silver halides is utilized.
By observing precipitate colors and reactions, chemists can verify the presence of particular ions, simplifying the identification process for unknown compounds.

Benzyl vs. Phenyl Reactivity

The difference in reactivity between benzyl and phenyl groups is an important aspect of organic chemistry. Benzyl groups, such as in \( \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{I} \), exhibit higher reactivity due to the benzylic position, which is adjacent to the stabilizing effect of a benzene ring. This position allows for easier substitutions and transformations. (
)

• The benzyl position benefits from electron delocalization and resonance with the phenyl ring. This makes it a more favorable site for nucleophilic attacks.
• In contrast, phenyl groups, like in \( \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{I} \), are less reactive due to the strong carbon-iodine bond within the aromatic system, offering resistance to nucleophilic substitution.
• The lone pair delocalization in aromatic systems prevents nucleophiles from easily attacking.

In experiments, this reactivity difference is crucial. It allows chemists to predict outcomes and behaviors of halides under given conditions. For instance, in the exercise, \( \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{I} \) more readily forms iodide ions upon reaction, aligning with its higher reactivity compared to \( \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{I} \). Understanding these differences helps in designing reactions and synthetic pathways effectively.