• # question_answer 44)   The addition of $HBr$to 1-butene gives a mixture of products A, B and C.                 $\underset{(C)}{\mathop{C{{H}_{3}}-C{{H}_{2}}-C{{H}_{2}}-C{{H}_{2}}-Br}}\,$ The mixture consists of: (a) A and B as major and C as minor products      (b) B as major, A and C as minor products (c) B as minor, A and C as major products              (d) A and B as minor and C as major products

(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.