Problem 143 - A Complete Resource Book in Chemistry for JEE Main

Question

Match the following: List I List II (Reagent) \(\quad\) (Electrophiles) 1\. \(\mathrm{Cl}_{2}+\mathrm{AlCl}_{3}\) (i) \(\mathrm{NO}_{2}\) 2\. \(\mathrm{HNO}_{3}+\mathrm{H}_{2} \mathrm{SO}_{4}\) (ii) \(\mathrm{Cl}_{\cdots}\).. \(\mathrm{Cl} \ldots \mathrm{AlCl}_{3}\) or \(\mathrm{Cl}^{+}\) 3\. \(\mathrm{H}_{2} \mathrm{~S}_{2} \mathrm{O}_{7}\) (iii) \(\mathrm{SO}_{3} \mathrm{H}\) (or \(\left.\mathrm{H}_{2} \mathrm{SO}_{4}+\mathrm{SO}_{3}\right)\) 4\. \(\mathrm{Br}_{2}+\mathrm{Fe}\) (iv) \(\mathrm{SO}_{3}\) (v) Br... Br...FeBr \(_{3}\) or \(\mathrm{Br}^{+}\)

Step-by-Step Solution

Answer

1-(ii), 2-(i), 3-(iv), 4-(v)

Step 1: Analyze Reagent 1

The reagent is 1. \( \mathrm{Cl}_{2} + \mathrm{AlCl}_{3} \). This reaction creates an electrophile, typically chloride, \( \mathrm{Cl}^{+} \), because aluminum chloride, \( \mathrm{AlCl}_{3} \), acts as a Lewis acid and polarizes the \( \mathrm{Cl}_{2} \) molecule, pulling a \( \mathrm{Cl}^{-} \) away to stabilize the \( \mathrm{AlCl}_{3}^{-} \). The resulting electrophile is \( \mathrm{Cl}^{+} \), known in the list as option (ii).

Step 2: Analyze Reagent 2

The reagent is 2. \( \mathrm{HNO}_{3} + \mathrm{H}_{2} \mathrm{SO}_{4} \). This mixture generates a nitronium ion, \( \mathrm{NO}_{2}^{+} \). The sulfuric acid protonates the nitric acid, creating an effective electrophile, which is \( \mathrm{NO}_{2} \), known in the list as option (i).

Step 3: Analyze Reagent 3

The reagent is 3. \( \mathrm{H}_{2} \mathrm{~S}_{2} \mathrm{O}_{7} \). Also known as fuming sulfuric acid or oleum, this reagent predominantly yields \( \mathrm{SO}_3 \) as the electrophile, known in the list as option (iv).

Step 4: Analyze Reagent 4

The reagent is 4. \( \mathrm{Br}_{2} + \mathrm{Fe} \). This combination typically generates bromo-electrophiles like \( \mathrm{Br}^{+} \), similar to the action of iron halides. This matches option (v) in the list.

Step 5: Match Each Reagent with Electrophile

From the analysis:- \( \mathrm{Cl}_{2} + \mathrm{AlCl}_{3} \) matches (ii) \( \mathrm{Cl}^{+} \).- \( \mathrm{HNO}_{3} + \mathrm{H}_{2} \mathrm{SO}_{4} \) matches (i) \( \mathrm{NO}_{2} \).- \( \mathrm{H}_{2} \mathrm{~S}_{2} \mathrm{O}_{7} \) matches (iv) \( \mathrm{SO}_{3} \).- \( \mathrm{Br}_{2} + \mathrm{Fe} \) matches (v) \( \mathrm{Br}^{+} \).

Key Concepts

Electrophiles

Electrophiles are chemical species that are attracted to electrons. They seek out and react with nucleophiles, which are electron-rich species. This makes them very important in many chemical reactions, especially in organic chemistry. One common type of electrophilic reaction is the electrophilic substitution reaction, where an electrophile replaces another atom or group in a compound.

Some characteristics of electrophiles include:

• Having a positive charge or being neutral with an electron-deficient atom.
• Acting as electron acceptors in chemical reactions.

For instance, when chlorine ( Cl_2 ) is combined with aluminum chloride ( AlCl_3 ), an electrophile is generated in the form of chloronium ion ( Cl^+ ). Here, AlCl_3 acts as a Lewis acid by polarizing the Cl_2 molecule and creating the positively charged ion, Cl^+ .

Lewis Acid

A Lewis acid is a substance that can accept a pair of electrons to form a new chemical bond. Named after Gilbert N. Lewis, this concept is a broad definition compared to the more specific Brønsted-Lowry acids, which donate protons. Lewis acids are vital in the formation of electrophiles.

Some typical Lewis acids include:

• Metal cations such as Al^{3+} , which have an empty orbital, allowing them to accept electrons.
• Compounds like AlCl_3 , which reacts with Cl_2 to generate Cl^+ .

In the typical reaction Cl_2 + AlCl_3 , AlCl_3 serves as a Lewis acid. It helps to create an electrophile by drawing a Cl^- ion, thus facilitating the generation of Cl^+ , the chloronium ion.

Nitronium Ion

The nitronium ion ( NO_2^+ ) is a potent electrophile used in nitration reactions, especially in aromatic compounds. It is formed when nitric acid ( HNO_3 ) reacts with sulfuric acid ( H_2SO_4 ).

This is how the reaction typically works:

• The sulfuric acid protonates the nitric acid, facilitating the loss of a water molecule.
• This results in the creation of the NO_2^+ ion.

The nitronium ion is highly reactive and attacks the benzene ring in aromatic compounds, substituting a hydrogen atom for a nitro group ( NO_2 ). This type of reaction is widespread for introducing nitro groups into organic substances, making the nitronium ion crucial in organic synthesis.