પ્રકિયા $- (2): CH_3 -CH = CH -CH_3 \xrightarrow{KMn{{O}_{4}}/NaI{{O}_{4}}}(C)$ $2$ મોલ
નીપજ $(B)$ અને $(C)$ શું હશે ?
Reaction \((2) \, CH_3 - CH = CH - CH_3 \xrightarrow[Leumix\,\operatorname{Re}agent]{KMn{{O}_{4}}/NaI{{O}_{4}}}2C{{H}_{3}}COOH\)
$I.\,\,C{H_2} = CH - C{H_3} + \,{H_2}O\,\xrightarrow{{{H^ + }}}$
$II. \,\,C{H_3} - CHO\,\xrightarrow[{(ii)\,{H_2}O}]{{(i)\,C{H_3}MgI}}$
$III. \,\, C{H_2}O\,\xrightarrow[{(ii)\,{H_2}O}]{{(i)\,{C_2}{H_5}MgI}}$
$IV.\,\,C{H_2} = CH - C{H_3}\,\xrightarrow{{KMn{O_4}}}$