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
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Answer:
c)..predominantly 1 butene
I guess....hope it helps
Answered by
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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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