Question

An organic compound (x) was heated with \(\mathrm{NaOH}\) solution and then acidified with dilute \(\mathrm{HNO}_{3} \cdot\) Addition of silver nitrate solution gave a white precipitate completely soluble in ammonium hydroxide. The compound (x) is (a) \(\mathrm{p}\) -chlorotoluene (b) vinyl chloride (c) \(\mathrm{p}\) - dichlorobenzene (d) benzyl chloride

Short answer

Answer: (d) benzyl chloride

Step-by-step solution

Identify the functional groups in each compound

For each given compound, we need to identify the functional groups that can react with \(\mathrm{NaOH}\) and silver nitrate. Here are the given compounds and their functional groups: - (a) \(\mathrm{p}\) -chlorotoluene (aromatic and halogen, no reactive functional group) - (b) vinyl chloride (alkene and halogen, no reactive functional group) - (c) \(\mathrm{p}\) - dichlorobenzene (aromatic and halogen, no reactive functional group) - (d) benzyl chloride (benzylic halide and halogen, reactive functional group)

Analyze the reactions expected for each functional group

Now, let's analyze the expected reactions for each functional group and see if they fit the given reaction conditions: - (a) Aromatic and halogen groups will not react with \(\mathrm{NaOH}\) and form a soluble complex with silver nitrate after acidification. - (b) Alkene and halogen groups will not react with \(\mathrm{NaOH}\) and form a soluble complex with silver nitrate after acidification. - (c) Aromatic and halogen groups will not react with \(\mathrm{NaOH}\) and form a soluble complex with silver nitrate after acidification. - (d) The benzylic halide in benzyl chloride can react with \(\mathrm{NaOH}\) to produce a carboxylic acid. After acidification and the addition of silver nitrate, carboxylic acids tend to form a white precipitate that is soluble in ammonium hydroxide.

Deduce the compound based on the reaction analysis

Based on the reaction analysis, we can conclude that compound (d) benzyl chloride is the compound (x) that fits the given conditions of the problem. The white precipitate observed after treating with silver nitrate is silver carboxylate, which is soluble in ammonium hydroxide. So, the correct answer is: (d) benzyl chloride

Key concepts

Functional Groups in Organic Compounds

Understanding functional groups is crucial in organic chemistry because they determine how a molecule will react in chemical processes.
Functional groups are specific groups of atoms within molecules that have characteristic properties.
In the context of the exercise, we're examining the functional groups of four compounds to predict their reactivity:

• Aromatic and halogen groups: Found in compounds like p-chlorotoluene and p-dichlorobenzene. These groups are generally unreactive with \(\mathrm{NaOH}\) and silver nitrate.
• Alkene and halogen groups: Present in vinyl chloride. Like aromatic and halogen groups, these don't tend to participate in reactions with these reagents under normal conditions.
• Benzylic halide group: Found in benzyl chloride. This group is reactive because the benzylic position stabilizes the transition state during chemical reactions, making this compound interesting in this scenario.

Functional groups help us predict the kinds of reactions a molecule can participate in, assisting in identifying and working with organic compounds effectively.

Reactions with NaOH and Silver Nitrate

When dealing with organic compounds, reactions with \(\mathrm{NaOH}\) and silver nitrate can reveal important information about the substance.
Here's how this applies to the compounds in the exercise:

• NaOH (Sodium Hydroxide) Reaction: Often used to saponify or attack ester groups, but in the context of this exercise, it reacts with benzylic halides such as benzyl chloride, forming carboxylic acids. This reaction is a key indicator of active functional groups capable of undergoing substitution reactions.
• Silver Nitrate Reaction: Often used to test for halide ions. In aqueous solutions, silver nitrate (\(\mathrm{AgNO_3}\)) reacts with halides to form insoluble silver halides. The formation of a white precipitate of silver carboxylate is indicative of an initial NaOH reaction leading to a carboxylate ion, which reacts with silver ions.

Knowing these types of reactions helps in predicting which reactions are possible and what the products might be, thus facilitating the identification of the compound in question.

Identification of Organic Compounds

Identifying organic compounds involves analyzing their reactions and predicting their behavior under specific conditions.
The procedure outlined in the exercise centers on observing how compound (x) reacts with chemical reagents:

• Reactivity Analysis: Reactions with \(\mathrm{NaOH}\) and \(\mathrm{AgNO_3}\) can provide clues about compound structure. In the exercise, only benzyl chloride reacted to form a white precipitate, suggesting a reactive group presence.
• White Precipitate Formation: This indicates the presence of silver halide or silver carboxylate. These precipitates are typically insoluble except in specific reagents like ammonium hydroxide, which dissolves the white precipitate, helping identify the compound.
• Conclusion Drawing: From these observations, benzyl chloride fits the bill. Its reactivity with NaOH followed by precipitation in silver nitrate points to its identity as the compound containing a benzylic halide, confirmed by the solubility in ammonium hydroxide.

This methodical approach highlights how observing reactions can lead to the accurate identification of unknown organic compounds.