
Haloalkane in the presence of alcoholic KOH undergoes:
A.elimination
B.polymerization
C.dimerisation
D.substitution
Answer
448.3k+ views
Hint: We know that the type of solvent is one of the factors that determines whether we get elimination reaction or substitution reaction. The fraction of water to ethanol in the solvent matters where water promotes nucleophilic substitution reactions and alcohol encourages elimination reaction.
Complete step by step answer:
We must know that the Haloalkane in the presence of alcoholic \[KOH\] undergoes elimination reactions. For example:
\[C{H_3}C{H_2}Br{\text{ }} + {\text{ }}KOH{\text{ }}\left( {alc.} \right){\text{ }} \Rightarrow {\text{ }}{H_2}C = C{H_2} + {\text{ }}KBr{\text{ }} + {\text{ }}{H_2}O\]
Alcoholic potassium hydroxide, \[KOH\] solution that functions as solvent gives alkoxide ions that act as a strong base. This base abstracts \[\beta \]-Hydrogen atom from saturated substrate - alkyl halide.
Abstracted \[\beta \]-Hydrogen atom is then transferred to the alkyl part to form an alkane and simultaneously a molecule of \[HCl\] is eliminated.
The basicity of hydroxide ions is considerably lower than the basicity of alkoxide ions. Hydroxide ion is significantly getting hydrated in aqueous solution. Hence, hydroxide ions cannot abstract \[\beta \]-hydrogen atoms of alkyl chloride to eliminate \[HCl\] and form alkene.
Other than the solvent, there are many factors that decide whether the method will undergo elimination reaction or substitution reaction. These factors include, type of alkyl halide, temperature, concentration of \[KOH\] solvent, etc.
To favor elimination reactions, one must use higher temperature, pure ethanol as solvent and concentrated \[KOH\] solution.
Hence, option A is the correct option.
Note:
We generally use alcoholic \[KOH\] to form Alkene from Alkyl Halides whereas aqueous \[KOH\] is used to form alcohols from Alkyl Halides.
When we use aqueous potassium hydroxide, or \[KOH\] as a solvent, then it gives hydroxide ions \[\left( {O{H^ - }} \right)\] that acts as a nucleophile and attacks the alpha Carbon atom of the substrate of alkyl halide. This process gives alcohol as the product, thereby undergoing Substitution reaction.
Complete step by step answer:
We must know that the Haloalkane in the presence of alcoholic \[KOH\] undergoes elimination reactions. For example:
\[C{H_3}C{H_2}Br{\text{ }} + {\text{ }}KOH{\text{ }}\left( {alc.} \right){\text{ }} \Rightarrow {\text{ }}{H_2}C = C{H_2} + {\text{ }}KBr{\text{ }} + {\text{ }}{H_2}O\]
Alcoholic potassium hydroxide, \[KOH\] solution that functions as solvent gives alkoxide ions that act as a strong base. This base abstracts \[\beta \]-Hydrogen atom from saturated substrate - alkyl halide.
Abstracted \[\beta \]-Hydrogen atom is then transferred to the alkyl part to form an alkane and simultaneously a molecule of \[HCl\] is eliminated.
The basicity of hydroxide ions is considerably lower than the basicity of alkoxide ions. Hydroxide ion is significantly getting hydrated in aqueous solution. Hence, hydroxide ions cannot abstract \[\beta \]-hydrogen atoms of alkyl chloride to eliminate \[HCl\] and form alkene.
Other than the solvent, there are many factors that decide whether the method will undergo elimination reaction or substitution reaction. These factors include, type of alkyl halide, temperature, concentration of \[KOH\] solvent, etc.
To favor elimination reactions, one must use higher temperature, pure ethanol as solvent and concentrated \[KOH\] solution.
Hence, option A is the correct option.
Note:
We generally use alcoholic \[KOH\] to form Alkene from Alkyl Halides whereas aqueous \[KOH\] is used to form alcohols from Alkyl Halides.
When we use aqueous potassium hydroxide, or \[KOH\] as a solvent, then it gives hydroxide ions \[\left( {O{H^ - }} \right)\] that acts as a nucleophile and attacks the alpha Carbon atom of the substrate of alkyl halide. This process gives alcohol as the product, thereby undergoing Substitution reaction.
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