Ethyl iodide and $n -$ propyl iodide are allowed to undergo Wurtz reaction. The alkane which will not be obtained in this reaction is:

Butane

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Propane

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Pentane

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Hexane

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Solution

The correct option is B Propane
In Wurtz reaction, alkane is prepared by reducing alkyl halide using Na in dry ether. It is a coupling reaction where two alkyl group will combine to form higher alkanes.
$R - X + 2 N a + X - R_{1} d r y e t h e r ⟶ R - R_{1} + 2 N a X$ .
In given reaction, Ethyl iodide and n-propyl iodide are involved i.e., $R \neq R_{1}$
Therefore, we have 3 possibilities,
(i) $R$ will couple with $R$ ( $R$ = Ethyl)
Ethyl is two carbon chain so it react with another ethyl to form four carbon chain, Butane.
$C H_{3} C H_{2} - I + 2 N a + I - C H_{2} C H_{3} d r y e t h e r ⟶ C H_{3} C H_{2} C H_{2} C H_{3} + 2 N a I$ .
(ii) $R_{1}$ will couple with $R_{1}$ ( $R_{1}$ = n-Propyl)
n-propyl is three carbon chain so it react with another n-propyl to form six carbon chain, Hexane.
$C H_{3} C H_{2} C H_{2} - I + 2 N a + I - C H_{2} C H_{2} C H_{3} d r y e t h e r ⟶ C H_{3} C H_{2} C H_{2} C H_{2} C H_{2} C H_{3} + 2 N a I$ .
(iii) $R$ will couple with $R_{1}$
Ethyl and n-propyl will combine to form five carbon chain, Pentane.
$C H_{3} C H_{2} - I + 2 N a + I - C H_{2} C H_{2} C H_{3} d r y e t h e r ⟶ C H_{3} C H_{2} C H_{2} C H_{2} C H_{3} + 2 N a I$ .

Thus, propane is not prepared with this combination of alkyl halide in Wurtz reaction.

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