Question
Elimination of bromine from 2-bromobutane results in the formation of
A.
equimolar mixture of 1and 2-butene
B.
predominantly 2-butene
C.
predominantly 1-butene
D.
predominantly 2-butyne
Answer :
predominantly 2-butene
Solution :
\[\underset{\text{2-Bromobutane}}{\mathop{\overset{\begin{smallmatrix}
\,\,\,\,\,\,\,\,\,Br \\
\,\,\,\,\,\,\,|
\end{smallmatrix}}{\mathop{C{{H}_{3}}C{{H}_{2}}CHC{{H}_{3}}}}\,}}\,\xrightarrow{KOH\left( alc. \right)}\] $$\mathop {C{H_3}CH = CHC{H_3}}\limits_{{\text{ 2 - Butene }}\left( {{\text{80% }}} \right)} \,\, + $$ $$\mathop {C{H_3}C{H_2}CH = C{H_2}}\limits_{{\text{l - Butene}}\,\left( {20\% } \right)} $$
In elimination reaction of alkyl halide major product is obtained according to Saytzeff's rule, which states that when two alkenes may be formed, the alkene which is most substituted one predominates.