Problem 38 - Arihant Objective Approach to Organic Chemistry for IIT JEE and NEET all Medical Entrances Part 4

Question

A compound on treatment with NaOH followed by addition of \(\mathrm{AgNO}_{3}\) produces white precipitate at room temperature, the precipitate is soluble in \(\mathrm{NH}_{4} \mathrm{OH}\). The compound is identified as: [AFMC 1997] (a) vinyl chloride (b) benzyl chloride (c) chlorobenzene (d) ethyl bromide

Step-by-Step Solution

Answer

Benzyl chloride (b) forms a white precipitate with AgNO₃, soluble in NH₄OH.

Step 1: Recall the reagent behavior

The compound treated with NaOH followed by AgNO₃ produces a white precipitate. This behavior indicates the presence of a halide ion (Cl⁻ or Br⁻). AgNO₃ reacts with halide ions to form a precipitate (AgCl - white, AgBr - pale yellow).

Step 2: Consider the solubility in NH₄OH

From the problem, the precipitate is soluble in NH₄OH. AgCl is soluble in excess NH₄OH, while AgBr is not. This indicates that the precipitate is AgCl.

Step 3: Identify the possible halide

Given the halide identified in Step 2 is chloride, evaluate the compounds given: vinyl chloride (CH=CHCl), benzyl chloride (C₆H₅CH₂Cl), chlorobenzene (C₆H₅Cl), and ethyl bromide (C₂H₅Br). Compounds with 'chloride' are potential candidates because they contain Cl⁻.

Step 4: Analyze compound reactivity and structure

Chlorobenzene is unreactive to this treatment due to the stability of the C-Cl bond in aromatic compounds. Vinyl chloride and benzyl chloride can release Cl⁻ ions because they are more reactive than chlorobenzene. Ethyl bromide would produce a bromine precipitate, not chlorine.

Step 5: Determine the final compound

Based on reactivity, vinyl chloride gives a stable vinyl cation and does not form precipitate with Ag⁺ readily. Benzyl chloride, however, forms a reactive benzyl cation, readily releasing Cl⁻ when treated with NaOH, matching the criteria.

Key Concepts

Nucleophilic Substitution

Nucleophilic substitution is a fundamental reaction in organic chemistry where a nucleophile replaces a leaving group in a molecule. Let's break this down a bit more: a nucleophile is a chemical species that donates an electron pair to form a chemical bond. These nucleophiles can be ions like OH⁻ or neutral molecules like NH₃. The leaving group is typically a halogen, such as a chloride (Cl⁻) or bromide (Br⁻) ion. Let's think of this process like a square dance, where partners (atoms) switch around, creating new combinations (molecular structures) but keeping the ballroom (stoichiometry) the same."

There are two primary types of nucleophilic substitution reactions in organic chemistry:

• SN1 reactions: The `S` stands for substitution, `N` for nucleophilic, and `1` indicates a unimolecular reaction. This means the rate of reaction depends solely on the concentration of the substrate. In SN1 reactions, a carbocation intermediate is formed and then attacked by the nucleophile.


• SN2 reactions: Similar to SN1, but bimolecular. Here, the nucleophile attacks the substrate simultaneously as the leaving group departs, resulting in a one-step, concerted process.

For the problem mentioned, benzyl chloride undergoes an SN1 reaction. Benzyl chloride is excellent for SN1 due to the stability of the resulting benzyl cation, making it reactive when treated with NaOH, leading to its transformation.

Halide Identification

Determining which halide ion is present in a compound can often involve simple precipitation reactions. In the case of the exercise, after treating a compound with NaOH followed by AgNO₃, a white precipitate indicated the presence of a chloride ion. But how did we find out it was chloride and not bromide? Let's dive in!

Silver nitrate (AgNO₃) is used in this identification due to its reaction with halide ions. When AgNO₃ is added to a solution containing Cl⁻, it forms silver chloride (AgCl), which shows up as a white precipitate. Similarly, silver bromide (AgBr) forms a pale yellow precipitate. However, a clue lies in the behavior of these silver halides in ammonium hydroxide (NH₄OH):

• Silver chloride (AgCl): Soluble in excess NH₄OH, which dissolves the precipitate.


• Silver bromide (AgBr): Not soluble in excess NH₄OH.

Thus, in this case, the solubility of the precipitate in NH₄OH helped confirm that the halide formed was indeed silver chloride, indicating chloride ions were present.

Reaction Mechanism Analysis

Understanding the detailed steps of a reaction—its mechanism—is vital in predicting the behavior of organic compounds. Analyzing these mechanisms involves examining the structural changes, energy barriers, and intermediates formed during the reaction.

In the given scenario, detailed step-by-step analysis led us to identify that benzyl chloride was the responsive compound in question. Let's break down the approach:

• Structural analysis: Observing the given compounds, chlorobenzene is known for its strong C-Cl bond due to resonance stabilization and does not react under conditions similar to NaOH and AgNO₃ treatment.


• Reactivity comparison: Vinyl chloride has stability in its vinyl cation, thus less likely to precipitate with Ag⁺. However, benzyl chloride forms a reactive benzyl cation when treated with NaOH, liberating Cl⁻ ions readily.


• Solubility tests: The resulting silver chloride from benzyl chloride's NaOH and AgNO₃ treatment aligns with the white precipitate description that dissolves in NH₄OH.

Through this rigorous mechanism analysis, we demystified which chemical interactions took place, allowing us to identify the compound confidently as benzyl chloride. This approach is a cornerstone in synthetic and analytical chemistry, enabling chemists to predict and manipulate chemical behaviors effectively.