Rate of dehydration when given compounds are treated with conc. $\ce{H2SO4}$ is


According to me the answer should be $R > Q > S > P$ but my textbook solution gives the answer as R > Q > P > S (Sadly, there is no explanation).

Dehydration of secondary & tertiary alcohols in the presence of conc. $\ce{H2SO4}$ proceed through E1 mechanism whereas most primary alcohols proceed through E2 mechanism.

In the molecular R, the carbocation formed after the alcohol group leaves the molecule after protonation is a tertiary carbocation and not just that, it is resonance stabilized as well.

In the molecule Q, you get a secondary carbocation and this is resonance stabilized as well.

In molecule S, you get a secondary carbocation but there is no resonance stabilization.

In molecule P, the dehydration goes via E2 mechanism and there is a bulky cyclohexene group attached to the $\alpha$-carbon which should be contributing to steric hindrance.

If the question did not have molecule P, then its pretty much obvious that the answer is R > Q > S but with molecule P, I am not quite sure where to place it.

I thought that P will be least reactive since it proceeds via E2 mechanism and there is one bulky group attached to the carbon.

So how do I come up with a concrete answer to justify why molecule P should react faster than molecule S but slower than molecule Q.

  • 2
    $\begingroup$ Note that P can also form an allylic carbocation after protonation of OH group and subsequent loss of water. $\endgroup$ Apr 1, 2016 at 19:21
  • $\begingroup$ P gets dehydrated via E2 mechanism? Carbocation isn't formed. Now its even more confusing, if it goes through E1 mechanism it will form an allylic carbocation. How do I now decide if it would undergo E1 or E2 elimination? $\endgroup$
    – Yashas
    Apr 2, 2016 at 4:49

1 Answer 1


The rate of dehydration is related to the ease of formation of the carbocation and the energy of the carbocation intermediate.

The ease of formation of carbocation is tertiary>secondary>primary. That is the ease of abstraction of $\ce{OH2+}$ COULD BE be R>Q/S>P.

Let's look at a comparison of the energy of the carbocation intermediates in all these cases.

A carbocation may be stabilized by resonance by a carbon-carbon double bond next to the ionized carbon. Such cations as allyl cation $\ce{CH2=CH–CH2+}$ and benzyl cation $\ce{C6H5–CH2+}$ are more stable than most other carbocations. Molecules that can form allyl or benzyl carbocations are especially reactive. These carbocations where the C+ is adjacent to another carbon atom that has a double or triple bond have extra stability because of the overlap of the empty p orbital of the carbocation with the p orbitals of the $\pi$ bond. This overlap of the orbitals allows the charge to be shared between multiple atoms – delocalization of the charge - and, therefore, stabilizes the carbocation.

enter image description here

Let's look at them individually,

  • P - allylic cation, primary alcohol
  • Q - allylic cation, secondary alcohol
  • R - allylic cation, tertiary alcohol
  • S - NOT an allylic cation, secondary alcohol

$\ce{R}$ is clearly the winner.

Now we may conclude that the charge dispersal is better in resonance than it is in hyperconjugation (explanation)

Here are the 3D diagrams of the P carbocation. Hydrogen cation doesn't look like it has to face significant steric hindrance to perform protonation of the alcohol.enter image description here

enter image description here


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