
Methane cannot be prepared by:
A. Wurtz reaction
B. Decarboxylation
C. Kolbe's reaction
D. Sabatier-senderens reaction
Answer
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Hint: For solving, this question, we need to understand the concept of alkanes. We know that alkanes are saturated hydrocarbons which are made of carbon and hydrogen atoms. It has a single bond between carbon-carbon and carbon-hydrogen. Methane, ethane, propane, butane, etc are all examples of alkanes.
Complete step by step answer:
We know that Wurtz reaction is a coupling reaction in which inorganic group polymers, whereby two alkyl halides are reacted with sodium metal in dry ether solution to form a higher alkane. We can write the reaction occurring in Wurtz reaction as,
$RX + 2Na + XR \to R - R + 2NaX$
In the above reaction, methane is not formed. Therefore, the option A is correct.
We must remember that the removal of carbon dioxide from carbonyl groups is called decarboxylation. We can write an example for this reaction as,
$C{H_3}COONa + NaOH\xrightarrow{{CaO}}C{H_4} + N{a_2}C{O_3}$
Methane is formed in the above reaction. Therefore, the option B is incorrect.
We have to remember that in Kolbe’s reaction, a carboxylation chemical reaction occurs that proceeds by heating sodium with carbon dioxide under pressure. Then, the product is treated with sulfuric acid. The final product is an aromatic hydroxy acid which is also known as salicylic acid. We can write the chemical equation for this reaction as,
$2RCOOK + 2{H_2}O \to R - R + 2C{O_2} + 2KOH + {H_2}$
In the above reaction, methane is not formed. Therefore, the option C is correct.
We must remember that the Sabatier Senderens reaction involves the reaction of hydrogen with carbon dioxide at elevated temperatures and pressures in the presence of a nickel catalyst to produce methane and water.
${C_n}{H_{2n}} + {H_2} \to {C_n}{H_{2n + 2}} + heat$
In the above reaction methane is not formed as a product. Therefore, the option D is correct.
From the definition we know that, Kolbe and Sabatier - Senderens reaction both form R-R bonds as a product so methane cannot be formed by these methods. From which we can conclude that methane cannot be prepared by Wurtz reaction, Kolbe’s reaction, and Sabatier - Senderens reaction.
So, the correct answer is Option B.
Note: We must know that the methane has a molecular formula of $C{H_4}$ in which one carbon atom is surrounded by a four hydrogen atom to form a tetrahedron and having sp3 hybridisation. Methane is used in various fields due to its abundance on earth and gaseous state at normal conditions.
In the oven, water heater, automobiles methane is used as fuel. In stream generators methane is used as a fuel to produce electricity.
Complete step by step answer:
We know that Wurtz reaction is a coupling reaction in which inorganic group polymers, whereby two alkyl halides are reacted with sodium metal in dry ether solution to form a higher alkane. We can write the reaction occurring in Wurtz reaction as,
$RX + 2Na + XR \to R - R + 2NaX$
In the above reaction, methane is not formed. Therefore, the option A is correct.
We must remember that the removal of carbon dioxide from carbonyl groups is called decarboxylation. We can write an example for this reaction as,
$C{H_3}COONa + NaOH\xrightarrow{{CaO}}C{H_4} + N{a_2}C{O_3}$
Methane is formed in the above reaction. Therefore, the option B is incorrect.
We have to remember that in Kolbe’s reaction, a carboxylation chemical reaction occurs that proceeds by heating sodium with carbon dioxide under pressure. Then, the product is treated with sulfuric acid. The final product is an aromatic hydroxy acid which is also known as salicylic acid. We can write the chemical equation for this reaction as,
$2RCOOK + 2{H_2}O \to R - R + 2C{O_2} + 2KOH + {H_2}$
In the above reaction, methane is not formed. Therefore, the option C is correct.
We must remember that the Sabatier Senderens reaction involves the reaction of hydrogen with carbon dioxide at elevated temperatures and pressures in the presence of a nickel catalyst to produce methane and water.
${C_n}{H_{2n}} + {H_2} \to {C_n}{H_{2n + 2}} + heat$
In the above reaction methane is not formed as a product. Therefore, the option D is correct.
From the definition we know that, Kolbe and Sabatier - Senderens reaction both form R-R bonds as a product so methane cannot be formed by these methods. From which we can conclude that methane cannot be prepared by Wurtz reaction, Kolbe’s reaction, and Sabatier - Senderens reaction.
So, the correct answer is Option B.
Note: We must know that the methane has a molecular formula of $C{H_4}$ in which one carbon atom is surrounded by a four hydrogen atom to form a tetrahedron and having sp3 hybridisation. Methane is used in various fields due to its abundance on earth and gaseous state at normal conditions.
In the oven, water heater, automobiles methane is used as fuel. In stream generators methane is used as a fuel to produce electricity.
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