it is question which needs little bit thinking.
now follow me what i m explaining.
Products.... "D"=> CH3-CH(=O) , "D"=> CH3-CH(=O) , "E"=> CH(=O)-CH(=O) .....2 get back A we have to remove the =O &join with double bonds.... The structure is... "A" => CH3-CH=CH-CH=CH-CH3 (note the positions of red carbons..)
So on reduction first one double bond will break & the the other giving u foll. structures..
"B" =>CH3-CH(H)-CH(H)-CH=CH-CH3 & "C"=>CH3-CH(H)-CH(H)-CH(H)-CH(H)-CH3 (blue H are added ones)
Last part on oxidation of "B" double bond breaks & Changed to -COOH group....so u get ...
"F" => CH3-CH(H)-CH(H)-C(=O)(OH) along with CH3-C(=O)(OH).
its a good question.
now rate me!!!!!!!!!!!!!