
When ethyl bromide is heated with sodium in dry ether solvent, the alkane obtained is:
A. Butane
B. Propane
C. Ethane
D. A mixture of the above three
Answer
559.2k+ views
Hint: A reaction in which alkyl halide reacts with metallic sodium metal in presence of dry ether solvent to form symmetrical alkanes containing double the number of carbon atoms present in the alkyl halide and the name of the reaction is known as Wurtz reaction. It is used for the preparation of higher alkanes.
Complete step by step answer:
When two moles of ethyl bromide react with two moles of sodium in presence of dry or anhydrous ether it forms butane as a product. The above reaction is written below:
$2{C_2}{H_5} - Br + 2Na\xrightarrow{{dry\,ether}}{C_2}{H_5} - {C_2}{H_5} + 2NaBr$
(Butane)
It is an example of Wurtz reaction.
The obtained product butane is a symmetrical alkane with double the number of carbon atoms present in the ethyl bromide.
So, the correct answer is Option A .
Additional Information:
Uses of alkyl halides.
Alkyl halides find use in the preparation of alcohols, ethers, amines, esters etc.
$C{H_3}I$ which is also known as methyl iodide can be used as a methylating agent.
Alkyl halides are used for preparing the Grignard’s reagent.
Alkyl halides can be used as a solvent for relatively non-polar compounds.
Alkyl halides can be used as a starting material for the synthesis of a wide range of organic compounds.
Alkyl halides are used in fire extinguishers.
Note: Wurtz reaction generally fails with tertiary alkyl halides since under the basic conditions of the reaction, they prefer to undergo dehydrohalogenation to form alkenes.
When a mixture of two different alkyl halides are used, all the three possible alkanes are formed. The reaction involved in this process is written below:
$3{R^/} - X + 6Na + 3{R^{//}} - X\xrightarrow{{dry\,ether}}{R^/} - {R^/} + {R^{//}} - {R^{//}} + {R^/} - {R^{//}} + 6NaX$
That is why this is not a good method for the preparation of unsymmetrical alkanes.
Complete step by step answer:
When two moles of ethyl bromide react with two moles of sodium in presence of dry or anhydrous ether it forms butane as a product. The above reaction is written below:
$2{C_2}{H_5} - Br + 2Na\xrightarrow{{dry\,ether}}{C_2}{H_5} - {C_2}{H_5} + 2NaBr$
(Butane)
It is an example of Wurtz reaction.
The obtained product butane is a symmetrical alkane with double the number of carbon atoms present in the ethyl bromide.
So, the correct answer is Option A .
Additional Information:
Uses of alkyl halides.
Alkyl halides find use in the preparation of alcohols, ethers, amines, esters etc.
$C{H_3}I$ which is also known as methyl iodide can be used as a methylating agent.
Alkyl halides are used for preparing the Grignard’s reagent.
Alkyl halides can be used as a solvent for relatively non-polar compounds.
Alkyl halides can be used as a starting material for the synthesis of a wide range of organic compounds.
Alkyl halides are used in fire extinguishers.
Note: Wurtz reaction generally fails with tertiary alkyl halides since under the basic conditions of the reaction, they prefer to undergo dehydrohalogenation to form alkenes.
When a mixture of two different alkyl halides are used, all the three possible alkanes are formed. The reaction involved in this process is written below:
$3{R^/} - X + 6Na + 3{R^{//}} - X\xrightarrow{{dry\,ether}}{R^/} - {R^/} + {R^{//}} - {R^{//}} + {R^/} - {R^{//}} + 6NaX$
That is why this is not a good method for the preparation of unsymmetrical alkanes.
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