question_answer

\[{{\left( {{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}} \right)}_{\text{2}}}\text{NH}>{{\text{C}}_{6}}{{\text{H}}_{\text{5}}}\text{NHC}{{\text{H}}_{\text{3}}}>{{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}\text{N}{{\text{H}}_{\text{2}}}>{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}\text{N}{{\text{H}}_{\text{2}}}\] \[{{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}\text{N}{{\text{H}}_{\text{2}}}>{{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}\text{NHC}{{\text{H}}_{\text{3}}}>{{\left( {{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}} \right)}_{\text{2}}}\text{NH}>{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}\text{N}{{\text{H}}_{\text{2}}}\]X and V respectively are

A) \[{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}\text{N}{{\text{H}}_{\text{2}}}>{{\left( {{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}} \right)}_{\text{2}}}\text{ NH }>{{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}\text{NHC}{{\text{H}}_{\text{3}}}>{{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}\text{N}{{\text{H}}_{\text{2}}}\]and \[{{\text{N}}_{\text{2}}}{{\text{O}}_{\text{5}}}\]

B) \[\text{2}.\text{4 }\times \text{l}{{0}^{-\text{4}}}\] and\[\text{mol }{{\text{L}}^{\text{-1}}}\]

C)  \[\text{mi}{{\text{n}}^{\text{-1}}}\]and\[{{N}_{2}}{{O}_{5}}\xrightarrow{{}}2N{{O}_{2}}+\frac{1}{2}{{O}_{2}}\]

D) \[\text{mol }{{\text{L}}^{\text{-1}}}\]

Correct Answer: C

Solution :

\[\text{C}{{\text{O}}_{\text{2}}}\] \[{{\text{O}}_{\text{2}}}\]      \[{{S}_{N}}1\] The main responsible factor here is organic peroxide which dissociates butene into free radicals and modifies the electrophilic addition to free radical addition which passes through free radicals as their intermediates. \[benzyl>allyl>{{1}^{\circ }}>{{2}^{\circ }}>{{3}^{\circ }}>Me\] \[Me>{{1}^{\circ }}>{{2}^{\circ }}>{{3}^{\circ }}>allyl>benzyl\] \[{{3}^{\circ }}>{{2}^{\circ }}>{{1}^{\circ }}>Me>allyl>benzyl\] The modified case of HBr in presence of peroxide is called Kharasch effect and considered as an example of anti-Markownikoff addition.