Answer:
(a) \[C{{H}_{3}}-\underset{1-Butene}{\mathop{C{{H}_{2}}-CH}}\,=C{{H}_{2}}+HBr\xrightarrow{Mark.addition}\]\[C{{H}_{3}}-C{{H}_{2}}-\underset{I(major)}{\mathop{\underset{\begin{smallmatrix}
| \\
Br
\end{smallmatrix}}{\mathop{\overset{*}{\mathop{CH}}\,}}\,}}\,-C{{H}_{3}}+\underset{C(minor)}{\mathop{C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}C{{H}_{2}}Br}}\,\]
Since \[I\] contains a chiral
carbon, it exist in two enantiomers(A and B) which are mirror images of each
other.
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