Which of the following alkanes cannot be made in good yield by Wurtz reaction?

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1. **Principle of Wurtz Reaction:** The Wurtz reaction involves the coupling of two alkyl halide molecules with sodium metal in dry ether to form higher alkanes ($2RX + 2Na \rightarrow R-R + 2NaX$). This method is suitable for preparing **symmetrical alkanes** containing an **even number** of carbon atoms .
2. **Analysis of Options:**
   - **n-Butane ($C_4H_{10}$):** Even number of carbons (4). Symmetrical ($C_2H_5-C_2H_5$). Can be prepared from bromoethane.
   - **n-Hexane ($C_6H_{14}$):** Even number of carbons (6). Symmetrical ($C_3H_7-C_3H_7$). Can be prepared from 1-chloropropane.
   - **2,3-Dimethylbutane ($C_6H_{14}$):** Even number of carbons (6). Symmetrical dimer of isopropyl chloride. Can be prepared effectively.
   - **n-Heptane ($C_7H_{16}$):** Odd number of carbons (7). It is unsymmetrical. To prepare it, one would need two different alkyl halides (e.g., propyl halide and butyl halide). This results in a mixture of three products (propane, octane, and heptane) due to self-coupling and cross-coupling, making separation difficult and the yield of n-heptane poor .
3. **Conclusion:** Alkanes with an odd number of carbon atoms, like n-Heptane, cannot be prepared in good yield using the Wurtz reaction.

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