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 \(A\) 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? (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

Step-by-Step Solution

Answer

B was \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{I}\).

Step 1: Understand the Reaction with NaOH

When phenyl iodide (\(\mathrm{C}_6\mathrm{H}_5\mathrm{I}\)) and benzyl iodide (\(\mathrm{C}_6\mathrm{H}_5\mathrm{CH_2}{I}\)) are boiled with \(\mathrm{NaOH}\) solution, the iodine atoms are replaced by \(\mathrm{OH}\) groups, forming the corresponding alcohols.

Step 2: Analyze Reaction with AgNO3

When the iodo compounds are reacted with \(\mathrm{AgNO}_3\), iodide ions (\(\mathrm{I}^-\)) will form a precipitate. The color of the precipitate helps identify the iodide: phenyl iodide forms a white one, while benzyl iodide forms a yellow one.

Step 3: Evaluate the Observations

Solution B gave a yellow precipitate, indicating the formation of silver iodide (AgI) specifically associated with benzyl iodide due to the immediate reaction with \(\mathrm{AgNO_3}\).

Step 4: Determine the Compound in Solution B

The yellow precipitate suggests that \(\mathrm{B}\) is \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{I}\) as it gives a yellow \(\mathrm{AgI}\) precipitate in aqueous solution due to iodide ions.

Key Concepts

Iodo Compounds

Iodo compounds are a fascinating subset of organic compounds that contain an iodine atom bonded to a carbon atom. These compounds, such as phenyl iodide \((\mathrm{C}_6\mathrm{H}_5\mathrm{I})\) and benzyl iodide \((\mathrm{C}_6\mathrm{H}_5\mathrm{CH}_2\mathrm{I})\), have interesting properties and reactions due to the nature of the iodine atom. Iodine, being a larger and more polarizable atom compared to other halogens, is more prone to undergoing nucleophilic substitution reactions. This makes iodo compounds very reactive in organic chemistry.

Here's an interesting aspect: when these compounds react with bases like \(\mathrm{NaOH}\), the iodine atom can be replaced by an \(\mathrm{OH}\) group. This transformation gives rise to alcohols, which are versatile in further organic synthesis.

Moreover, the iodine atom influences the physical properties of the compound, such as its boiling point and solubility. These properties make iodo compounds manageable when used in chemical experiments or reactions. Overall, their reactivity and properties make iodo compounds significant in the synthesis of various organic molecules.

Precipitation Reaction

A precipitation reaction is a fascinating chemical process where two soluble substances react in a solution to form an insoluble product, known as a precipitate. In the context of our original exercise, when an iodo compound reacts with \(\mathrm{AgNO}_3\), a classic precipitation reaction occurs.

This reaction is characterized by the formation of a solid precipitate—silver iodide \((\mathrm{AgI})\)—from iodide ions \((\mathrm{I}^-)]\) and silver ions \((\mathrm{Ag}^+)\). The color of the precipitate is key in identifying the type of iodo compound present. Notably, phenyl iodide forms a white precipitate, whereas benzyl iodide results in a yellow one. This color difference is a crucial diagnostic tool in chemistry.

Understanding precipitation reactions is vital for students because they illustrate how ions interact in a solution. It's a fundamental concept in analytical chemistry used to identify and separate compounds. The sensitivity of this reaction can be utilized to detect and quantify the presence of iodide ions, thereby elucidating the identity of mystery solutions.

Organic Reaction Mechanism

Organic reaction mechanisms delve into the 'how' and 'why' of chemical reactions. They offer deeper insights into the step-by-step transformation of reactants to products, providing a detailed map of electron shifts. When we examine the transformation of iodo compounds, like phenyl iodide and benzyl iodide reacting with \(\mathrm{NaOH}\), we observe a nucleophilic substitution mechanism.

In this process, the hydroxide ion \((\mathrm{OH}^-)\) acts as a nucleophile, attacking the carbon-iodine bond. This results in the displacement of the iodine atom and formation of alcohol.

The organic reaction mechanism also illuminates the intricate interplay of steric and electronic effects. For example, the benzyl group in benzyl iodide stabilizes the transition state due to resonance, making the reaction more favorable.

• Steric Effects: The substituents around the reactive sites can hinder the approach of nucleophiles.
• Electronic Effects: The electron-withdrawing or donating nature of substituents affects the reactivity of the molecule.

Learning these mechanisms is crucial for chemists to design and predict reactions. Every organic reaction holds a story of molecular transformation shaped by its mechanism, bringing chemistry to life!