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Here's the question

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enter image description here

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 $(c)$?

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  • $\begingroup$ yes, but could you please elaborate a little more $\endgroup$ – Prakhar Sep 23 '16 at 16:39
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The answer is (b). The initial step of the reaction should take energy. Otherwise the reactant molecule itself is unstable. The very initial portion of the curve in (c) is downward-sloped, which doesn't make sense. If the reacting alkene could relax to a different conformation without any activation barrier, then it would do so. But the fact that the reactant is a stable, isolable molecule with a well-defined chemical structure indicates that no such barrier-less relaxation is possible.

Thus, (b) is a better answer than (c).

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  • $\begingroup$ no, the answer is c $\endgroup$ – Prakhar Sep 23 '16 at 16:39
  • $\begingroup$ Not according to me it isn't. $\endgroup$ – Curt F. Sep 23 '16 at 16:41
  • $\begingroup$ well, the answer key could be wrong, but your answer doesnt really convince me much $\endgroup$ – Prakhar Sep 23 '16 at 16:43
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    $\begingroup$ I would agree with you on the initial downward bit, which makes 0 sense. However, (b) also doesn't make sense to me, with a carbocation intermediate that is more stable than the starting alkene. Even (c) says that the tertiary carbocation is lower energy than the alkene. I would prefer to believe that carbocations are high-energy intermediates, and that if you added, say, perchloric acid to the double bond, and let the system reach equilibrium, you would not obtain a salt of the carbocation with perchlorate counteranion, because that is what is implied. I don't like any of these answers. $\endgroup$ – orthocresol Sep 23 '16 at 17:03
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    $\begingroup$ As Curt said and as I said c doesn't make sense either. You are basically choosing the least bad answer, and between b and c, which answer is least bad is really quite subjective. I would argue for this. i.stack.imgur.com/SbKkb.png $\endgroup$ – orthocresol Sep 23 '16 at 17:34

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