Problem 28 - A Complete Resource Book in Chemistry for JEE Main

Question

When ethyl bromide and \(\mathrm{n}\)-propyl bromide is allowed to react with sodium, in ether, they form (a) mixture of four alkanes (b) mixture of three alkanes (c) mixture of two alkanes (d) single alkane

Step-by-Step Solution

Answer

(b) Mixture of three alkanes

Step 1: Understand the Reaction Conditions

This situation refers to the Wurtz reaction, where haloalkanes react with sodium metal in an ether solvent to form higher alkanes.

Step 2: Determine the Reactants

The reactants are ethyl bromide ( ext{C}_2 ext{H}_5 ext{Br}) and ext{n}-propyl bromide ( ext{C}_3 ext{H}_7 ext{Br}).

Step 3: Predict the Products

Ethyl bromide and ext{n}-propyl bromide can combine to produce different alkanes: ethyl groups can pair with each other, propyl groups can pair with each other, and ethyl can also combine with propyl.

Step 4: List Possible Alkanes

The potential alkanes formed are: butane ( ext{C}_4 ext{H}_{10} from ethyl groups), hexane ( ext{C}_6 ext{H}_{14} from propyl groups), and pentane ( ext{C}_5 ext{H}_{12} from the combination of ethyl and propyl).

Step 5: Count the Alkanes

There are three distinct alkanes possible: butane, pentane, and hexane.

Key Concepts

Understanding Haloalkanes

Haloalkanes, also known as alkyl halides, are a class of organic compounds that contain a halogen atom (such as chlorine, bromine, or iodine) bonded to an alkyl group. These compounds are often the starting materials for various chemical reactions, including the Wurtz reaction. The presence of the halogen atom makes haloalkanes reactive, as they can readily undergo nucleophilic substitution or elimination reactions.

• The carbon-halogen bond in haloalkanes is polar, with the carbon atom having a partial positive charge and the halogen having a partial negative charge.
• This polarity makes the carbon atom susceptible to attacks by nucleophiles, leading to the breaking of the carbon-halogen bond and the formation of new compounds.
• Haloalkanes can be classified based on the type of carbon atom bonded to the halogen: primary (one carbon bonded), secondary (two carbons bonded), or tertiary (three carbons bonded).

Understanding the nature of haloalkanes is crucial when predicting the outcomes of their reactions, such as those involving sodium in ether.

The Role of Sodium in Ether

In the Wurtz reaction, sodium acts as a reducing agent when combined with haloalkanes in the presence of an ether solvent. Ether is not just any solvent; it plays a key role in ensuring the reaction proceeds smoothly.

• Sodium metal reacts with the halogen of the haloalkane, forming sodium halide and generating free radicals of the alkyl groups.
• These free radicals are highly reactive and can quickly combine to form a variety of alkanes.
• Ether serves as an inert solvent that stabilizes the reaction environment and prevents unwanted side reactions.

The choice of ether as a solvent is particularly important because it does not react with either the sodium metal or the haloalkanes, thus providing a stable medium for the reaction to occur.

Formation of Higher Alkanes

When ethyl bromide and -propyl bromide react with sodium in ether, a mixture of higher alkanes is formed. This process showcases the ability of the Wurtz reaction to construct longer carbon chains.

• The alkyl radicals generated can couple in various combinations, leading to new alkanes with different carbon chain lengths.
• With ethyl and -propyl bromide, the possible alkanes are butane ( C_4H_{10}), pentane ( C_5H_{12}), and hexane ( C_6H_{14}).
• Butane is formed from the coupling of two ethyl groups, hexane from two -propyl groups, and pentane from the combination of an ethyl and a -propyl group.

Thus, the Wurtz reaction is valuable for synthesizing higher alkanes from simpler haloalkanes, demonstrating the power of organic synthesis techniques.