Question

The major product formed in the reaction is $$ \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{~N}_{2}^{+} \mathrm{Cl}^{-}+\mathrm{H}_{3} \mathrm{PO}_{2}+\mathrm{H}_{2} \mathrm{O} \longrightarrow $$ (A) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OH}\) (B) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{Cl}\) (C) \(\mathrm{C}_{6} \mathrm{H}_{6}\) (D) \(\mathrm{C}_{6} \mathrm{H}_{5}-\mathrm{C}_{6} \mathrm{H}_{5}\)

Short answer

The major product formed in the reaction is \(\mathrm{C}_{6} \mathrm{H}_{6}\) (option C).

Step-by-step solution

Write the balanced reaction.

For this step, we need to write the balanced chemical reaction involving the reactants and the potential products.

Identify the type of reaction.

Determine the type of reaction occurring based on the reactants and the possible products.

Determine the major product.

Analyze the reaction and identify the most likely major product, taking into consideration the likely reaction pathway and properties of the reactants.

Verify the choice with the given options.

Compare the determined major product with the given options (A), (B), (C), and (D) to choose the correct answer. Solution:

Write the balanced reaction.

The balanced chemical reaction is not provided directly. So, we look at the potential products in the given options and write balanced reactions involving these: (A) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{N}_{2}^{+} \mathrm{Cl}^- + \mathrm{H}_{3} \mathrm{PO}_{2} + \mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OH} + \cdots\) (B) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{N}_{2}^{+} \mathrm{Cl}^- + \mathrm{H}_{3} \mathrm{PO}_{2} + \mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{Cl} + \cdots\) (C) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{N}_{2}^{+} \mathrm{Cl}^- + \mathrm{H}_{3} \mathrm{PO}_{2} + \mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{C}_{6} \mathrm{H}_{6} + \cdots\) (D) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{N}_{2}^{+} \mathrm{Cl}^- + \mathrm{H}_{3} \mathrm{PO}_{2} + \mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{C}_{6} \mathrm{H}_{5} - \mathrm{C}_{6} \mathrm{H}_{5} + \cdots\)

Identify the type of reaction.

The reaction involves anilinium chloride, which comes from aniline. Hence, it might be a reduction reaction, as hypophosphorous acid (\(\mathrm{H}_{3} \mathrm{PO}_{2}\)) acts as a reducing agent.

Determine the major product.

In a reduction reaction involving anilinium chloride, the nitrogen will lose its positive charge and amine group will be substituted by hydrogen. So, the major product will be the one where the \(\mathrm{N}_{2}\) group is replaced by hydrogen.

Verify the choice with the given options.

The product with the \(\mathrm{N}_{2}\) group replaced by hydrogen is phenylhydrazine (\(\mathrm{C}_{6}\mathrm{H}_{6}\), benzene). This corresponds to the option (C). Thus, the major product formed in the reaction is \(\mathrm{C}_{6} \mathrm{H}_{6}\). Therefore, the correct answer is (C).

Key concepts

Reduction Reactions

Reduction reactions are a fundamental class of chemical reactions where a molecule, atom, or ion gains electrons. This process often involves the removal of oxygen or the addition of hydrogen. In many cases, the molecule being reduced, also known as the oxidizing agent, experiences a decrease in its oxidation state.
In the context of aromatic hydrocarbons involving diazonium salts, like \[\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{~N}_{2}^{+} \mathrm{Cl}^{-}\], reduction reactions can transform complex groups into simpler ones. For example, reducing diazonium salts can lead to the formation of hydrocarbons or other derivatives as seen with benzene formation in our exercise. The role of the reducing agent is crucial as it provides the electrons for reduction. Hypophosphorous acid (\(\mathrm{H}_3\mathrm{PO}_2\)) is an excellent reducing agent for such transformations. This acid donates electrons, facilitating the replacement of the diazonium group (\(\mathrm{N}_{2}^{+}\)) with hydrogen, resulting in benzene (\(\mathrm{C}_{6} \mathrm{H}_{6}\)).

Balanced Chemical Equations

A balanced chemical equation represents the conservation of mass. It shows the same number of atoms for each element in the reactants and the products. Balancing equations requires adjusting coefficients to achieve this equilibrium, ensuring that no atoms are lost or created in the reaction process.
For the chemical reaction given in the exercise: \[\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{~N}_{2}^{+} \mathrm{Cl}^{-} + \mathrm{H}_{3} \mathrm{PO}_{2} + \mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{C}_{6} \mathrm{H}_{6} + \cdots\],understanding how to balance the equation helps us ensure that every molecule and ion involved is accounted for. Elements not fully represented in the simplified reaction, like phosphorus from \(\mathrm{H}_{3}\mathrm{PO}_{2}\), must be considered along with balanced charges to provide the complete equation. Balancing reactions is key to predicting yields and understanding reaction stoichiometry.

Major Product Identification

Identifying the major product in a chemical reaction involves understanding the reaction mechanism and knowing the stability and reactivity of intermediate compounds.
In the case of our exercise, diazonium salts often undergo substitution reactions, where the diazo group \(\mathrm{N}_{2}^{+}\) is readily replaced by other groups. For aromatic hydrocarbons, especially with a strong reducing agent like \(\mathrm{H}_3\mathrm{PO}_2\), the diazo group can be reduced to form a simple hydrocarbon. Here, our exercise predicts the major product as benzene (\(\mathrm{C}_{6} \mathrm{H}_{6}\)), which results from replacing the \(\mathrm{N}_{2}^{+}\) group with a hydrogen atom.
Understanding the final structure helps answer which product is most likely favored, especially when multiple possible pathways exist. Predictions should always be verified against provided choices, ensuring accurate interpretation.

Diazonium Salt Chemistry

Diazonium salts play a vital role in organic chemistry, particularly in synthetic transformations. These salts contain a functional group represented by \(\mathrm{N}_{2}^{+}\), which can be easily modified into other functional groups, playing a crucial role in substitution reactions.
In \\(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{~N}_{2}^{+}\), the diazonium group is highly reactive, making it an ideal starting point for creating various derived products. Reactivity of diazonium salts often leads to replacement of the diazo group with nucleophiles or through reduction as seen in our exercise where benzene is formed by substituting \(\mathrm{N}_{2}\) with hydrogen. This versatility makes diazonium salts ubiquitous in organic synthesis, allowing for diverse transformations in the creation of aromatic compounds.
Knowing the chemistry of diazonium salts is essential for understanding broader applications, including dye creation and other complex aromatic systems.