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.