Question

\(\Lambda\) mixture of cthyl iodide and n-propyl iodide is subjected to Wurtz reaction. The hydrocarbon which will not be formed is (1) butanc (2) propanc (3) pentane (4) hexane

Short answer

Propane (C3H8) will not be formed.

Step-by-step solution

- Understand the Wurtz Reaction

The Wurtz reaction involves the coupling of alkyl halides in the presence of sodium metal to form higher alkanes.

- Identify reactants

The reactants are ethyl iodide (C2H5I) and n-propyl iodide (C3H7I).

- Possible combinations

Combine each pair of reactants to determine the possible products: 1. Ethyl Iodide + Ethyl Iodide: C2H5-C2H5 (Butane, C4H10) 2. n-Propyl Iodide + n-Propyl Iodide: C3H7-C3H7 (Hexane, C6H14) 3. Ethyl Iodide + n-Propyl Iodide: C2H5-C3H7 (Pentane, C5H12)

- Identify the missing product

From step 3, the only hydrocarbon not formed from either reactant is Propane (C3H8).

- Final Answer

The hydrocarbon that will not be formed is Propane (C3H8).

Key concepts

Alkyl Halides

Alkyl halides are organic compounds that contain carbon, hydrogen, and halogen atoms. They are commonly used in organic synthesis due to their reactivity.
These compounds are characterized by a carbon-halogen bond, where the halogen can be fluorine, chlorine, bromine, or iodine.

In the Wurtz reaction, alkyl halides act as the reactants that form higher alkanes when combined with sodium metal.
The reactivity of alkyl halides increases with the size of the halogen atom, meaning that alkyl iodides are generally more reactive than alkyl chlorides or bromides in the Wurtz reaction. This makes them ideal candidates for this type of coupling reaction.

Coupling Reactions

Coupling reactions are a class of chemical reactions where two molecules are joined together with the aid of a metal catalyst.
In the context of the Wurtz reaction, sodium metal serves as the catalyst.

There are different types of coupling reactions, but in the Wurtz reaction specifically, two alkyl halides react to form a higher alkane.
The general mechanism involves the formation of a highly reactive radical or carbanion intermediate, which then couples with another alkyl halide molecule to form the final product.
This reaction is particularly useful for synthesizing symmetrical alkanes when identical alkyl halides are used.
However, when different alkyl halides are used, a mixture of products can be formed.

Hydrocarbon Synthesis

Hydrocarbon synthesis refers to methods used to produce hydrocarbons, which are compounds consisting only of carbon and hydrogen.
The Wurtz reaction is one such method for synthesizing higher alkanes from smaller alkyl halides.

This reaction is especially valuable because it allows for the construction of larger, more complex alkane molecules from simpler ones.
For example, in the given exercise, ethyl iodide and n-propyl iodide can combine to form butane, hexane, and pentane.
This gives chemists a versatile tool for building a wide variety of hydrocarbons, essential for both research and industrial applications.

Sodium Metal

Sodium metal plays a crucial role in the Wurtz reaction as the reducing agent and catalyst.
It provides the necessary electrons to break the carbon-halogen bonds in the alkyl halides, forming reactive intermediates that then couple to produce the desired hydrocarbon.

Handling sodium metal requires caution as it is highly reactive, especially when in contact with water.
During the Wurtz reaction, sodium must be used under controlled conditions typically in an inert solvent like dry ether.
The reactivity of sodium ensures the breakdown of the C-X bonds in alkyl halides, facilitating the formation of a variety of hydrocarbons efficiently.