
Ethylene from ethyl bromide is obtained by treating it with
(A) Hydrogen
(B) Alcoholic caustic potash
(C) Aqueous caustic potash
(D) Aqueous caustic soda
Answer
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Hint: Ethylene is an unsaturated highly reactive organic compound which undergoes a number of reactions like addition, substitution, oxidation etc. Alkyl halides are used as a starting material for the synthesis of alkenes.
Complete Step By Step Answer:
When alkyl halide is treated with alcoholic potash it will cause dehydrohalogenation of the molecule to form alkene with the same number of carbon atoms as present in initial alkyl halide.
Halide of bromine and iodine is generally used in this reaction to form alkene. Alcoholic potash use in this reaction is prepared by dissolving potassium hydroxide in ethanol.
Alcoholic potash used in the reaction has a tendency to dissociate to form hydronium ion and alkoxide ion with the molecular formula of $ \left( {R{O^ - }} \right) $ . This alkoxide further removes the hydrogen atom from alkyl halide.
$ KO{H_{\left( {al} \right)}} \to K{O^ - } + {H^ + } $
This reaction is known as dehydrohalogenation reaction because it is completed by removal of hydrogen atom as well as halide atom from alkyl halide to introduce a double bond.
Reaction is accompanied with removal of hydrogen atoms from the carbon atom adjacent to the carbon having halide group which is designated as $ \left( \beta \right) $ carbon. Therefore, this is also known as $ \beta - $ elimination reaction.
Chemical reaction of ethyl bromide with alcoholic potash will give-
$ {C_2}{H_5}Br + KO{H_{\left( {al} \right)}} \to {H_2}C = C{H_2} + KBr + {H_2}O $
$ \Rightarrow $ From the above discussion we conclude that alcoholic potash is used to form ethylene from ethyl bromide.
Option (B) is correct.
Note:
Halides used in this reaction vary in their reactivity to leave the molecule, the order of ease of dehydrohalogenation is $ \left( {I > Br > Cl} \right) $ . The order of reactivity of alkyl halide for reaction of $ \beta - $ elimination reaction is $ \left( {3 > 2 > 1} \right) $ . Ethylene shows geometrical isomerism due to presence of unsaturated bonds within the molecule.
Complete Step By Step Answer:
When alkyl halide is treated with alcoholic potash it will cause dehydrohalogenation of the molecule to form alkene with the same number of carbon atoms as present in initial alkyl halide.
Halide of bromine and iodine is generally used in this reaction to form alkene. Alcoholic potash use in this reaction is prepared by dissolving potassium hydroxide in ethanol.
Alcoholic potash used in the reaction has a tendency to dissociate to form hydronium ion and alkoxide ion with the molecular formula of $ \left( {R{O^ - }} \right) $ . This alkoxide further removes the hydrogen atom from alkyl halide.
$ KO{H_{\left( {al} \right)}} \to K{O^ - } + {H^ + } $
This reaction is known as dehydrohalogenation reaction because it is completed by removal of hydrogen atom as well as halide atom from alkyl halide to introduce a double bond.
Reaction is accompanied with removal of hydrogen atoms from the carbon atom adjacent to the carbon having halide group which is designated as $ \left( \beta \right) $ carbon. Therefore, this is also known as $ \beta - $ elimination reaction.
Chemical reaction of ethyl bromide with alcoholic potash will give-
$ {C_2}{H_5}Br + KO{H_{\left( {al} \right)}} \to {H_2}C = C{H_2} + KBr + {H_2}O $
$ \Rightarrow $ From the above discussion we conclude that alcoholic potash is used to form ethylene from ethyl bromide.
Option (B) is correct.
Note:
Halides used in this reaction vary in their reactivity to leave the molecule, the order of ease of dehydrohalogenation is $ \left( {I > Br > Cl} \right) $ . The order of reactivity of alkyl halide for reaction of $ \beta - $ elimination reaction is $ \left( {3 > 2 > 1} \right) $ . Ethylene shows geometrical isomerism due to presence of unsaturated bonds within the molecule.
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