Methane cannot be prepared by :

Wurtz reaction

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Soda Lime process

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Kolbe's reaction

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Sabatier-senderens reaction

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Solution

The correct option is D Sabatier-senderens reaction
Wurtz reaction: It is a coupling reaction thus, two alkyl halide will combine to form higher alkane. Hence, the smallest alkane that can be formed is ethane.
$R X + 2 N a + X R \to R - R + 2 N a X$

Kolbe's electrolysis: Here, the salt of carboxylic acid will undergo decarboxylation to give alkanes.
$C H_{4}$ can't be prepared by this method. The smallest alkane that can be formed is ethane ( $C_{2} H_{6}$ )
$2 R C O O K + 2 H_{2} O electrolysis ⟶ R - R + 2 C O_{2} + 2 K O H + H_{2}$

Sabatier senderens reaction: Here, unsaturated hydrocarbon is reduced to alkane. The minimum carbon chain of unsaturated hydrocarbon is two i.e., Ethene. Hence, it reduced to form Ethane.
$C_{n} H_{2 n} + H_{2} N i \to C_{n} H_{2 n + 2}$

Soda Lime reaction: Here, carboxylic acid will undergo decarboxylation to form alkane containing one C atom less than the carboxylic acid.
Thus, acetic acid will form methane by Soda Lime reaction.
$C H_{3} C O O H N a O H ⟶ C H_{3} C O O N a + H_{2} O$
$C H_{3} C O O N a + N a O H C a O - - \to Δ C H_{4} + N a_{2} C O_{3}$ .

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