A) Propene
B) 1-butene
C) 2-butene
D) 2-pentene
Correct Answer: C
Solution :
Key Idea: Anti Markownikoffs addition is shown when unsymmetrical reagent reacts with unsymmetrical double or triple bond containing compound. \[\underset{(a)}{\mathop{\underset{propene}{\mathop{C{{H}_{3}}-CH=C{{H}_{2}}}}\,}}\,\,\,\underset{(b)}{\mathop{\underset{butene-1}{\mathop{C{{H}_{3}}-C{{H}_{2}}-CH=C{{H}_{2}}}}\,}}\,\] \[\underset{(c)}{\mathop{\underset{butene-2}{\mathop{C{{H}_{3}}-CH=CH-C{{H}_{3}}}}\,}}\,\] \[\underset{pentene-2}{\mathop{C{{H}_{3}}-C{{H}_{2}}-CH=CH-C{{H}_{3}}}}\,\] \[\because \] Choice (c) butene-2 is symmetrical alkene \[\therefore \] It does not show Anti Markownikoff addition.
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