Chemistry, asked by skant4803, 1 year ago

Elimination of bromine from 2-bromobutane results in the formation of :-
a) equimolar mixture of 1- and 2-butene
b) predominantly 2-butene
c) predominantly 1-butene
d) predominantly 2-butyne

Answers

Answered by piya8991
1

Answer:

c)..predominantly 1 butene

I guess....hope it helps

Answered by kynakhanna6
1

Answer: Option b)

Explanation: Reaction of alkyl halide with KOH (alc) elimination reaction take place and produces alkenes.

CH3-CH2-Br + KOH (alc.)  =  H2C=CH2 + KBr + H2O

The reaction will give us Saytzeff's product predominantly which is 2 Butene.

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