# Reaction coordinate

Here's the question

For a reaction to take place the net energy should go down, snd thus option $(d)$ can be discarded.

We are now left with options $(a),(b),(c)$

$3^{\circ}$ carbocation is more stable than a $2^{\circ}$ carbocation therefore the $3^{\text{rd}}$ dip in the graph should be lower than the $2^{\text{nd}}$ dip as more stable means lower energy

Thus now we are left with options $(b)$ and $(c)$

How do you decide between $(b)$ and $(c)$?

• yes, but could you please elaborate a little more – Prakhar Sep 23 '16 at 16:39