Question:
The reaction condition leading to the best yield of C$_2H_5 Cl$ are-
The reaction condition leading to the best yield of C$_2H_5 Cl$ are-
Updated On: Jun 14, 2022
- $C_2H_6(excess)+CI_2 \, \, \, \, \underrightarrow{UV \, light}$
- $C_2H_6+CI_2 (excess) \, \, \, \xrightarrow[room \, temp]{dark}$
- $C_2H_6+CI_2(excess) \, \, \, \, \underrightarrow{UV \, light}$
- $C_2H_6+CI_2 \, \, \, \, \underrightarrow{UV \, light}$
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The Correct Option is A
Solution and Explanation
During chlorination of alkane, if exebss of alkane is treated with Cl$_2$(g) in presence of light or heat, chance of mono-chlorination predominate.
$C_2H_6(excess)+CI_2 \, \, \, \, \underrightarrow{UV \, light} \, CH_3CH_2CI+HCI$
$C_2H_6(excess)+CI_2 \, \, \, \, \underrightarrow{UV \, light} \, CH_3CH_2CI+HCI$
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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