Question:
A mixture of ethyl iodide and n-propyl iodide is subjected to Wurtz reaction. The hydrocarbon which will not be formed is:
A mixture of ethyl iodide and n-propyl iodide is subjected to Wurtz reaction. The hydrocarbon which will not be formed is:
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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.
Updated On: Mar 18, 2026
- Butane
- Propane
- Pentane
- Hexane
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The Correct Option is B
Approach Solution - 1
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.
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Approach Solution -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)
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