Question

Which of the following alkane can be prepared in large amounts by Wurtz reaction? (a) \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CHCH}_{2} \mathrm{CH}\left(\mathrm{CH}_{3}\right)_{2}\) (b) CCCC(C)C CCC (c) \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CCH}_{2} \mathrm{CH}_{3}\) (d) \(\mathrm{CH}_{3} \mathrm{CH}\left(\mathrm{CH}_{3}\right) \mathrm{CH}_{2} \mathrm{CH}_{3}\)

Short answer

The alkane (d) can be prepared in large amounts by Wurtz reaction.

Step-by-step solution

Understanding Wurtz Reaction

Wurtz reaction involves the coupling of two alkyl halides in the presence of sodium metal to form a higher alkane. The reaction generally works best for forming symmetric alkanes, avoiding steric hindrance that could cause side reactions or low yield.

Analyzing the Compound Options

Let's list out the molecular structure of each given option: (a) \((\mathrm{CH}_3)_{2} \mathrm{CHCH}_2 \mathrm{CH}(CH_{3})_{2}\) indicates a branched alkane. (b) \((\mathrm{C}_{4} \mathrm{H}_{10})\) denotes 2-methylpentane, linear with one branch. (c) \((\mathrm{CH}_{3})_{3} \mathrm{CCH}_{2} \mathrm{CH}_{3}\) represents a molecule with three equivalent methyl groups and two more carbons, forming neopentane—requiring considerable branching in structure.(d) \((\mathrm{CH}_{3} \mathrm{CH}(\mathrm{CH}_{3}) \mathrm{CH}_{2} \mathrm{CH}_{3})\) is simply a linear chain, 2-methylbutane.

Evaluating Wurtz Suitability

Wurtz reaction is more effective for creating alkanes from simple precursors without branching or sterically hindrance complexities. Option (d) being the less branched alkane chain with only one methyl side chain is well-suited, allowing straightforward synthetic precursors from simpler haloalkanes.

Conclusion

Considering Wurtz reaction favors the formation of relatively unbranched alkanes, the best option is (d), \(\mathrm{CH}_{3} \mathrm{CH}(\mathrm{CH}_{3}) \mathrm{CH}_{2} \mathrm{CH}_{3}\). This structure is achievable by combining simple halo derivatives via Wurtz methodology, while others could introduce challenges due to branching.

Key concepts

Alkane Preparation

In organic chemistry, preparing alkanes is an essential task. Alkanes, being saturated hydrocarbons, consist only of single bonds, making them quite stable and significant in synthetic chemistry. There are several methods to prepare alkanes, but the Wurtz reaction is a popular choice due to its simplicity and direct approach.
This method is specifically useful for forming higher alkanes through the coupling of smaller alkyl halides using sodium metal.

• This reaction is advantageous for creating carbon-carbon bonds, which are crucial in constructing complex organic molecules.
• However, Wurtz reaction has limitations, particularly when dealing with highly branched alkanes due to possible side reactions.

Understanding alkane preparation helps in predicting outcomes when using Wurtz reaction and in finding suitable precursors for desired molecules.

Steric Hindrance

Steric hindrance significantly influences the efficiency of reactions like the Wurtz reaction. This phenomenon refers to the prevention of chemical reactions due to the size of groups within a molecule. When large groups are present, they can impede the approach and interaction of reagents, lowering the overall reaction yield.

This is especially relevant in the case of the Wurtz reaction:

• Severe steric hindrance occurs when bulky alkyl groups are involved, leading to decreased efficiency.
• Reactions involving simpler molecules or linear structures typically proceed with fewer issues.
• For example, a straightforward chain with minimal branching is often more reactive than a heavily branched alkane.

Therefore, minimizing steric hindrance is key when employing the Wurtz reaction for alkane preparation.

Symmetric Alkanes

Symmetric alkanes are those where the molecule has a reflective or rotational symmetry concerning carbon atoms. In the Wurtz reaction, the formation of symmetric alkanes is quite efficient. This is mainly because the reaction involves combining two identical or similar alkyl halides, leading to products with definite symmetry.

Consider the advantages of forming symmetric alkanes in these reactions:

• The simplicity of dealing with identical reactants can reduce side reactions, increasing the yield.
• The predictability of the resulting structure simplifies the synthesis process.
• They often lead to a more straightforward product purification and analysis.

Hence, symmetric alkanes are typically targeted when planning a syntheses using the Wurtz mechanism.

Sodium Metal Coupling

The key to the Wurtz reaction is sodium metal coupling. This process involves the use of sodium metal to facilitate the joining of two alkyl halide molecules, effectively forming a new carbon-carbon bond.
The general reaction mechanism involves sodium's donation of electrons to the halides, creating radicals that pair up to form the desired alkane.

• This method is renowned for its straightforward nature and effectiveness in constructing longer carbon chains from smaller segments.
• Sodium's reactivity makes it capable of breaking strong carbon-halide bonds, a critical process in forming new molecules.
• However, care must be taken as sodium is highly reactive and needs controlled conditions for safe and effective use.

Mastering sodium metal coupling is crucial for successfully executing Wurtz reactions, especially in preparative organic chemistry.