Question

Match the following

List I	List II
A. Methane	1. Sabatier-Senderens reaction of propylene
B. Ethane	2. Wurtz reaction of methyl iodide
C. Propane	3. Decarboxylation of sodium acetate
D. Butane	4. Sodium butanoate at Kolbe electrolysis
	5. Sodium propanoate at Kolbe electrolysis

The correct match is:
A. A-2 B-3 C-1 D-5
B. A-4 B-2 C-1 D-5
C. A-3 B-2 C-1 D-5
D. A-2 B-3 C-1 D-4

Answer 1

Hint: To match the two lists we need to start with any list and compare it with another list. The reactions given in list II have general form. But we need to use the compounds given along with it to get the required output.

Complete step by step answer:
To match both the columns we can check any list one by one and then find the corresponding correct match. Matching through list 1 would be difficult so we will know about list 2 and then find its match in list 1.
First is Sabatier-Senderens reaction to propylene. It is a reaction that involves the reaction of hydrogen with carbon dioxide at elevated temperatures and pressures in the presence of nickel catalysts to produce methane and water. If we increase the number of carbons on the left side then its number will increase on the right side also. Its general equation is as given below.
${C_n}{H_{2n}} + {H_2} \to {C_n}{H_{2n + 2}} + heat$
According to the above equation, if propylene is the reactant then propene will be formed as a product. Therefore, C-1.
Second is the Wurtz reaction. It is a coupling reaction in which two alkyl halides are reacted with sodium metal in dry ether solution to form a higher alkane. Its general form is as given below.
$RX + 2Na + XR \to R - R + 2NaX$
Now, according to the above equation, if one carbon atom compound is used as reactant then higher alkane is formed. So, if methyl is used then ethane would be formed. Therefore, B-2.
Third is decarboxylation of sodium acetate. It is a chemical reaction that removes a carboxyl group and releases carbon dioxide. It is mainly removal of carbon atoms from a carbon chain. The term decarboxylation means replacement of a carboxyl group with a hydrogen atom. Its general form is as given as below.
$RC{O_2}H \to RH + C{O_2}$
When sodium acetate is heated with soda lime, methane and sodium carbonate will be produced by decarboxylation. It is one of the general methods to prepare alkanes. Therefore, A-3.
Kolbe electrolysis or Kolbe reaction is an organic reaction named after Hermann Kolbe. This reaction is a decarboxylative dimerization of two carboxylic acids. If a mixture of two different carboxylates are used, all combinations of them are generally seen as the organic product structures. General reaction is as given below.
$2RCOOK + 2{H_2}O \to R - R + 2C{O_2} + 2KOH + {H_2}$
In sodium propanoate by Kolbe electrolysis reaction at anode is given as,
$2{C_2}{H_5}CO{O^ \bullet } \to 2{C_2}{H_5}^ \bullet + 2C{O_2}$
$2{C_2}{H_5}^ \bullet \to {C_4}{H_{10}}$
Therefore, D-5

So, the correct answer is option C. A-3 B-2 C-1 D-5

Note:
Sabatier-Senderens reaction involves the reaction of hydrogen with carbon dioxide at elevated temperatures and pressures in the presence of nickel catalysts to produce methane and water. Wurtz reaction is a coupling reaction in which two alkyl halides are reacted with sodium metal in dry ether solution to form a higher alkane. Decarboxylation is a chemical reaction that removes a carboxyl group and releases carbon dioxide. Kolbe electrolysis reaction is a decarboxylative dimerization of two carboxylic acids.