What happens when 1-bromopropane is heated with alcoholic caustic potash?
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When 1 bromopropane is heated with alcoholic KOH, the bromine with the alpha H (hydrogen from the carbon beside the carbon bonded with functional group) is removed and thus.. Forms 1 propene.
Ch3Ch2Ch2Br ---AL. KOH--> Ch3Ch=Ch2
Ch3Ch2Ch2Br ---AL. KOH--> Ch3Ch=Ch2
Answered by
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When 1-bromopropane is heated with alcoholic caustic potash, propene is formed.
How does a reaction take place?
- At the point when 1 bromopropane is warmed with alcoholic KOH, the bromine with the alpha H (hydrogen from the carbon alongside the carbon fortified with useful gathering) is eliminated and subsequently forms 1 propene.
- A solid base is given i.e alcoholic potash, subsequently, this is an Electrophilic reaction.
- Alcoholic potash (KOH) is a dehydrohalogenation specialist; it responds with an alkyl halide and alkene structures as a result.
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