Question

A mixture of ethyl iodide and n-propyl iodide is subjected to Wurtz reaction. The hydrocarbon which will not be formed is:

• A. Butane
• B. Propane
• C. Pentane
• D. Hexane

Hint:

The Wurtz reaction typically involves the coupling of two alkyl halides to form a higher alkane. The product depends on the type of alkyl halides used.

Answer 1

The Correct Option is B

The Wurtz reaction involves the coupling of alkyl halides in the presence of sodium metal to form higher alkanes. When ethyl iodide (C$_2$H$_5$I) and n-propyl iodide (C$_3$H$_7$I) are subjected to the Wurtz reaction, they can couple to form different hydrocarbons. 
Step 1: The reaction between two ethyl iodide molecules can form butane (C$_4$H$_{10}$), as shown by: \[ C_2H_5I + C_2H_5I \xrightarrow{{Na}} C_4H_{10}. \] The reaction between ethyl iodide and n-propyl iodide can form pentane (C$_5$H$_{12}$), as shown by: \[ C_2H_5I + C_3H_7I \xrightarrow{{Na}} C_5H_{12}. \] 
Step 2: However, propane (C$_3$H$_8$) cannot be formed in this reaction. This is because propane would require the coupling of two propyl iodide molecules, which is not possible here, as only ethyl and n-propyl iodides are involved. 
Step 3: Thus, the hydrocarbon that will not be formed is propane.

Answer 2

In the Wurtz reaction, alkyl halides react with sodium in dry ether to form higher alkanes through a coupling process. The general reaction is: \( 2R-X + 2Na \rightarrow R-R + 2NaX \), where \( R \) represents an alkyl group, and \( X \) is a halogen. In this case, we have ethyl iodide (\( C_2H_5I \)) and n-propyl iodide (\( C_3H_7I \)). Possible combinations from these two halides are:

• Ethyl + Ethyl: \[ C_2H_5-C_2H_5 = C_4H_{10} \] (Butane)
• Propyl + Propyl: \[ C_3H_7-C_3H_7 = C_6H_{14} \] (Hexane)
• Ethyl + Propyl: \[ C_2H_5-C_3H_7 = C_5H_{12} \] (Pentane)

Propane (\( C_3H_8 \)) cannot be formed because there is no suitable combination of these starting alkyl halides to produce it. Therefore, the hydrocarbon which will not be formed in this reaction is Propane.