Question

Methane cannot be prepared by :

• A. Wurtz reaction
• B. Soda Lime process
• C. Kolbe's reaction
• D. Sabatier-senderens reaction

Answer 1

Answer: (A), (C), (D)

The correct options are
A Wurtz reaction
C Kolbe's reaction
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.
$RX+2Na+XR\to R-R+2NaX$

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 ($C_2{H}_6$)
$2RCOOK+2H_{2}O\stackrel{\text{electrolysis}}{⟶}R-R+2C{O}_2+2KOH+H_2$

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.
$C_n{H}_{2n}+H_2\stackrel{Ni}{\to }{C}_n{H}_{2n+2}$

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.
$C{H}_{3}COOH\stackrel{NaOH}{⟶}C{H}_{3}COONa+H_{2}O$
$C{H}_{3}COONa+NaOH\underset{\Delta }{\overset{CaO}{\to }}C{H}_4+N{a}_{2}C{O}_3$.