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 orig- inal labels. They were labelled A and 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? \([2003]\) (a) \(\mathrm{A}\) is \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{I}\) (b) \(\mathrm{A}\) is \(\mathrm{C}_{6} \mathrm{H}_{3} \mathrm{I}\) (c) \(\mathrm{B}\) is \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{I}\) (d) addition of \(\mathrm{HNO}_{3}\) was unnecessary

Short answer

B is C6H5CH2I; A is C6H5I.

Step-by-step solution

Analyze the Reaction with NaOH

When phenyl iodide ( C_6H_5I ) or benzyl iodide ( C_6H_5CH_2I ) is boiled with NaOH , it can undergo a nucleophilic substitution reaction to form alcohols. Specifically, C_6H_5I typically forms C_6H_5OH (phenol), while C_6H_5CH_2I forms C_6H_5CH_2OH (benzyl alcohol). However, the iodide ion will be liberated, which can be confirmed in the next steps.

Add HNO3 and AgNO3

To each tube, dilute HNO_3 is added followed by AgNO_3 . This is a common test for halides: iodide ions, if present, will react with AgNO_3 to form a precipitate of silver iodide ( AgI ). The characteristic of the precipitate can help identify the compound.

Interpret the Precipitate Observation

Solution B gives a yellow precipitate upon the addition of AgNO_3 , indicating the presence of AgI , which forms from the iodide ions. AgI is known for its yellow color. This suggests that B contained an iodide that released free iodide ions readily, such as from benzyl iodide ( C_6H_5CH_2I ), which readily hydrolyzes.

Conclusion Based on Precipitate

Since solution B formed a yellow precipitate, solution B must have contained C_6H_5CH_2I , which readily reacted to release I^- ions. Therefore, B is C_6H_5CH_2I and A is C_6H_5I . Among the given options, (d) indicating the addition of HNO_3 was unnecessary, is misleading as it's crucial for the proper test conditions, but relevant to the experiment analysis is that A is actually C_6H_5I .

Key concepts

Nucleophilic Substitution Reaction

Nucleophilic substitution reactions are a category of reactions where a nucleophile selectively bonds with or attacks the positive or partially positive charge of an atom or a group of atoms in a molecule. This leads to the substitution of the reacting group. In our experiment, both phenyl iodide (C_6H_5I) and benzyl iodide (C_6H_5CH_2I) undergo nucleophilic substitution when treated with NaOH.

During this reaction, the hydroxide ions (OH-) from NaOH act as the nucleophile, attacking the carbon atom bonded to iodine. This results in the displacement of the iodide ion (I-) and the formation of phenol (C_6H_5OH) or benzyl alcohol (C_6H_5CH_2OH), depending on the starting material. The key takeaway here is that nucleophilic substitution reactions can help in altering and rearranging the molecular structure, facilitating the identification of substances based on their behavior.

Halide Identification Test

The halide identification test is essential in determining the presence of halide ions, such as chloride, bromide, and iodide. In the described experiment, the test was employed to identify iodide ions released in the solution after the nucleophilic substitution reaction with NaOH.

After acidifying the solution using dilute HNO_3, AgNO_3 is added. This specific sequence is crucial as it limits the formation of other precipitates that may cause false positives. When iodide ions are present, they interact with AgNO_3 to form a precipitate of silver iodide (AgI). The test serves as a reliable mechanism to confirm the presence or absence of iodide ions.

Precipitate Formation

A precipitate is an insoluble solid that emerges from a liquid solution. It occurs when the reactants in a solution form a compound that is not able to dissolve in the solvent, thus forming a distinct solid substance.

In the experiment, when AgNO_3 interacts with the iodide ions, silver iodide (AgI) forms as a yellow precipitate. This yellow coloring is distinctive and helps in identifying the specific halide involved. Precipitate formation is significant in analytical chemistry as it provides direct visual cues for qualitative analysis, helping to easily distinguish between different substances based on their reactivity and visual outcomes.

Phenyl Iodide

Phenyl iodide, known chemically as C_6H_5I, is an aromatic compound where an iodine atom is attached to a phenyl group. This compound is less prone to nucleophilic substitution than its benzyl counterpart due to the stability and resonance of the aromatic ring.

In the experiment, phenyl iodide reacts differently from benzyl iodide due to these stability factors, making it crucial for students to understand its chemical behavior. Such information would help decipher the results of the halide identification test, as phenyl iodide does not readily release iodide ions compared to benzyl iodide, influencing the formation of precipitates.

Benzyl Iodide

Benzyl iodide, or C_6H_5CH_2I, features an iodine atom attached to a benzyl group, which includes a reactive methylene (-CH_2-) attached to the phenyl ring. This structure makes benzyl iodide more prone to undergo nucleophilic substitution reactions.

During the experiment, benzyl iodide readily loses its iodide ion when boiled with NaOH, a characteristic behavior that distinguishes it from phenyl iodide. The resultant solution in test tube B, containing these released iodide ions, reacts with the AgNO_3 to form a promptly recognizable yellow precipitate of AgI. Understanding the chemical nature and reactivity of benzyl iodide is integral for accurate identification and results interpretation in chemical experiments.