https://www.dropbox.com/s/wvvsxce2hkjff2j/20160927_200958.jpg?dl=0
why first step higher Ea? both species are not same charged, no
repulsion. step 1 involves breaking of a pi bond and forming a
bond.
step 2 involves breaking of a sigma bond
https://www.dropbox.com/s/9x3h7qzb0k6upvd/20160927_201057-1.jpg?dl=0
Your points are valid, but at the same time, bear in mind that the
Ea for the 1st step is somewhat higher than you would expect, due
to electrostatic repulsions between the partial negatively charged
O atoms between both reactants. The Ea for the 2nd step to cleave
the C-O bond is somewhat lower than you would expect, because of
electrostatic repulsions between the (partial and formal)
negatively charged O atoms in the unstable intermediate.
Finally, For MCQs, always choose the best answer. Options A
& D can be ruled out for obvious reasons. Between C and B (both
of which are not an accurate portrayal of the energy magnitudes
involved), B is more likely, because in option C, the products are
so thermodynamically unstable, that you would unlikely get much
alkaline hydrolysis of the ester (despite the non-reversible step
3), and yet you should already be familiar with the fact that
alkaline hydrolysis of esters occurs readily (ie. thermodynamically
feasible) upon heating.
