Consider the reaction of 2-Bromobutane with alc. KOH and as given below
( is Potassium Tertiary Butoxide)
Identical
Chain isomers
Positional isomers
Not isomeric
is a bulky base and extracts least hindred
from the substrate giving the Hoffman's alkene while
will give the Saytzeff's product
(P) and (Q) are thus positional isomers .
Therefore, the correct answer is Option (3).
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