C2H5Br +KOH (alc) ->

C2H5Br + KOH(aq.)---->C2H5OH + KBr
C2H5Br + KOH(alc.)---->C2H4 + KBr + H2O
Hope it clears your doubt!
  • 40
The product obtained from the given reaction is ethene.
  • 2
C2h7 + 2na
  • 0
C2H5Oh+KBr
  • -2
C2H4+KBr+H2O
  • -2
C2H5Br +KOH = C2H5OH + KBr
It gives us ethanol and pottasium bromide.
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C2H5Br+KOH(Aq)?C2H5OH+ KBr
C2H5Br+KOH(Alk)?C2H4+KBr+ H2O
  • 1
C2H5Br+KOH
  • 0
C2H5Br + KOH(aq.)---->C2H5OH + KBr
C2H5Br + KOH(alc.)---->C2H+ KBr + H2O
  • 3
C2H5Br + KOH(aq.)---->C2H5OH + KBr
C2H5Br + KOH(alc.)---->C2H+ KBr + H2O  
  • 0
C2H5Br + KOH(aq.)---->C2H5OH + KBr
C2H5Br + KOH(alc.)---->C2H+ KBr + H2O    
  • 0
C2H5Br + KOH(aq.)---->C2H5OH + KBr
C2H5Br + KOH(alc.)---->C2H+ KBr + H2O                        
  • 2
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