Question

Which of the following alkanes cannot be synthesised by the Wurtz reaction in good yield? (1) \(\left(\mathrm{CII}_{3}\right)_{2} \mathrm{CII} \mathrm{CII}_{2} \mathrm{CII}\left(\mathrm{CII}_{3}\right)_{2}\) (2) \(\left(\mathrm{CII}_{3}\right)_{2} \mathrm{CII} \mathrm{CII}_{2} \mathrm{CII}_{2} \mathrm{CII}\left(\mathrm{CII}_{3}\right)_{2}\) (3) \(\mathrm{CII}_{3} \mathrm{CII}_{2} \mathrm{C}\left(\mathrm{CII}_{3}\right)_{2} \mathrm{CII}_{2} \mathrm{CII}_{3}\) (4) \(\left(\mathrm{CII}_{3}\right)_{3} \mathrm{C} \mathrm{CII}_{2} \mathrm{CII}_{2} \mathrm{CII}_{3}\)

Short answer

Alkanes (3) and (4) cannot be synthesized in good yield by the Wurtz reaction.

Step-by-step solution

- Understand the Wurtz reaction

The Wurtz reaction involves the coupling of two alkyl halides in the presence of sodium metal to form a larger alkane. The general reaction is: \[ 2 R-\text{X} + 2 \text{Na} \rightarrow R-R + 2 \text{NaX} \] where \(R\text{X}\) represents an alkyl halide.

- Analyze the structure of each alkane

We need to determine if each given alkane can be formed by coupling two identical alkyl halides. The structures given are: (1) \( \left(\mathrm{CII}_{3}\right)_{2} \mathrm{CII} \mathrm{CII}_{2} \mathrm{CII} \left(\mathrm{CII}_{3}\right)_{2} \) (2) \( \left(\mathrm{CII}_{3}\right)_{2} \mathrm{CII} \mathrm{CII}_{2} \mathrm{CII}_{2} \mathrm{CII} \left(\mathrm{CII}_{3}\right)_{2} \) (3) \( \mathrm{CII}_{3} \mathrm{CII}_{2} \mathrm{C}\left(\mathrm{CII}_{3}\right)_{2} \mathrm{CII}_{2} \mathrm{CII}_{3} \) (4) \( \left(\mathrm{CII}_{3}\right)_{3} \mathrm{C} \mathrm{CII}_{2} \mathrm{CII}_{2} \mathrm{CII}_{3} \). We need to check if each of these alkanes could be synthesized by the Wurtz reaction by analyzing their structure.

- Check for Symmetry in the Alkanes

For a molecule to be synthesized effectively by the Wurtz reaction, it should ideally be symmetrical. This is because the Wurtz reaction combines two alkyl groups. Let's check the symmetry of each alkane: (1) \( \left(\mathrm{CII}_{3}\right)_{2} \mathrm{CII} \mathrm{CII}_{2} \mathrm{CII} \left(\mathrm{CII}_{3}\right)_{2} \): symmetrical (2) \( \left(\mathrm{CII}_{3}\right)_{2} \mathrm{CII} \mathrm{CII}_{2} \mathrm{CII}_{2} \mathrm{CII} \left(\mathrm{CII}_{3}\right)_{2} \): symmetrical (3) \( \mathrm{CII}_{3} \mathrm{CII}_{2} \mathrm{C}\left(\mathrm{CII}_{3}\right)_{2} \mathrm{CII}_{2} \mathrm{CII}_{3} \): not symmetrical (4) \( \left(\mathrm{CII}_{3}\right)_{3} \mathrm{C} \mathrm{CII}_{2} \mathrm{CII}_{2} \mathrm{CII}_{3} \): not symmetrical.

- Determine the Synthesizable Alkanes

Symmetrical alkanes are more likely to be synthesized effectively by the Wurtz reaction. Based on the above analysis, alkanes (1) and (2) are symmetrical and can be synthesized in good yield by the Wurtz reaction.

- Identify the Unsynthesizable Alkane

Given that unsymmetrical alkanes are less likely to be synthesized in good yield by the Wurtz reaction, alkanes (3) and (4) cannot be synthesized effectively by this method. Therefore, options (3) and (4) are the alkanes that cannot be synthesized by the Wurtz reaction in good yield.

Key concepts

Wurtz reaction mechanism

The Wurtz reaction is a key method in organic chemistry for synthesizing higher alkanes. This reaction involves the coupling of two alkyl halides with the aid of sodium metal. When two alkyl groups come together, a new carbon-carbon bond forms. The general equation is:

\[ 2 R-\text{X} + 2 \text{Na} \rightarrow R-R + 2 \text{NaX} \]

In the context of this reaction, \(R\text{X}\) represents an alkyl halide. Sodium metal helps to initiate the formation of the bond between two alkyl radicals, producing the desired larger alkane and a byproduct of sodium halide (\(NaX\)).
Understanding this mechanism can help in predicting which types of alkanes can or cannot be efficiently synthesized through this reaction.

Symmetrical alkanes

Alkanes synthesized via the Wurtz reaction ideally begin as symmetrical molecules. Symmetrical alkanes have identical alkyl chains on both sides, leading to better coupling during the reaction.

Consider the comparison:

- For a symmetrical alkane like \( \text{CH}_3-\text{CH}_2-\text{CH}_2-\text{CH}_2-\text{CH}_3 \), the two alkyl halides used would be similar, leading to high yield synthesis.
- On the other hand, an asymmetrical alkane like \( \text{CH}_3-\text{CH}_2-\text{C}\text{H}\text{(CH}_3)_{2}-\text{CH}_2-\text{CH}_3 \) would have less efficient coupling reactions.

The effectiveness of the Wurtz reaction largely depends on the symmetry of the initial compounds. Symmetry ensures that the two alkyl halides can pair without complications, resulting in higher yields of the desired alkane.

Alkyl halides coupling

Coupling two alkyl halides is at the heart of the Wurtz reaction. This process requires sodium metal and produces a new carbon-carbon bond, lengthening the alkane chain.

Key notes on the coupling process:

- The reaction begins with sodium metal donating electrons to the alkyl halides, reducing them to alkyl radicals and creating sodium halide as a byproduct.
- The two alkyl radicals then combine to form a longer alkane.
- This process works best with alkyl halides that are identical or extremely similar, leading to symmetrical alkanes.

For instance, coupling two molecules of chloromethane (\( \text{CH}_3-\text{Cl} \)) through the Wurtz reaction yields ethane (\( \text{CH}_3-\text{CH}_3 \)). Such coupling scenarios are straightforward and usually highly efficient.

However, if the alkyl halides differ significantly, the yield drops. Non-symmetrical combinations can lead to a mix of unwanted byproducts and a less efficient reaction, making certain alkanes unsuitable for synthesis by this method.