Question:
In which of the following reactions, ethyl chloride is obtained in good yield?
In which of the following reactions, ethyl chloride is obtained in good yield?
Updated On: Jul 28, 2022
- Ethane $+ Cl _{2} \ce{->[{ hv }]}$
- Ethane $+ Cl _{2}($ excess $) \ce{->[{\text { hv}}]} $
- Ethene $+ Cl _{2} \ce{->[{\text { in dark }}]}$
- Benzene $+ Cl _{2} \longrightarrow$
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The Correct Option is A
Solution and Explanation
Ethane and limited quantity of chlorine reacts in presence of sunlight to form ethyl chloride. Whereas in presence of excess chlorine, hexa-chloro ethane (artificial camphor) is obtained.
$C _{2} H _{6}+ Cl _{2} {->[{ hv }]} C _{2} H _{5} Cl + HCl$
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Concepts Used:
Preparation of Haloalkanes
- Preparation by Alcohols: By using alcohol, we can very easily prepare Haloalkane as when R-OH is reacted with a suitable reagent it will form R-X. The following reagents will help in the formation of the reactions:
- Halogen Acids (HX)
- Thionyl chloride (SOCl2)
- Phosphorous halides (PX5 or PX3)
- Preparation by Free Radical Halogenation: The formation of alkyl bromides and alkyl chloride is very much possible with the help of free radical halogenation, but as these radicals are highly reactive and non-specific in nature, they can result in the formation of a mixture of products. Though, it's not a preferred method for the preparation of haloalkanes. For example, free radical chlorination can result in the formation of a number of haloalkanes, which in turn makes it difficult to set apart a single product.
- Preparation by Alkenes: An electrophilic addition reaction can be used to transform an alkene into a haloalkane as alkene will react with HX to form R-X.
Read More: Haloalkanes and Haloarenes