I was thinking of dehydrating this alcohol:

Its structure is:

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My concern is which of the following would it form:

enter image description here

The problem is that the first one comes from a stable benzyl secondary carbocation while the second one comes from a secondary carbocation but forms a more substituted alkene. Which one is preferred and why? (I feel that the second one should be preferred because of Saytzeff's (Zaitzev's) rule stating that more substituted alkene is formed)

Given answer was that the first one is formed (that too trans, which is I know why). But why the first one?

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    $\begingroup$ The double bond in the first product is in conjugation with the aromatic system which is energetically favoured $\endgroup$
    – Waylander
    Commented Jun 19, 2021 at 7:41
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    $\begingroup$ It's better to think of the Zaitsev's rule as the more stable carbocation is preferred. The thermodynamics of this reaction show that the stability of the final product depends on the stability of the double bond formed.. $\endgroup$ Commented Jun 19, 2021 at 8:10
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    $\begingroup$ @omjoglekar Zaitev rule isn't a law of compulsion. What happens, happens. Don't get bogged down it some rule of thumb fails, rather try learning the mechanism so that you can predict products with better accuracy. $\endgroup$ Commented Jun 19, 2021 at 9:44
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    $\begingroup$ Zaitsev's rule is useful when there are no other significant factors involved. Conjugation to an adjacent aromatic system is a significant factor. $\endgroup$
    – Waylander
    Commented Jun 19, 2021 at 10:30
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    $\begingroup$ Saytzeff rule's more substituted alkene is formed when all substituents are alkyl groups, so as to have more hyperconjugative structures and stabilize the double bond. In the general case, stability by resonance > stability by hyperconjugation. So a double bond resonating with the benzene ring is more stable. $\endgroup$
    – TRC
    Commented Jun 19, 2021 at 13:31

1 Answer 1


Zaitsev's rules are nothing but a set of rules that were used to explain experimental results that were observed. It was not the other way around.

The real rule you have to use here is the Hammond's Postulate. This states:

The transition state of a reaction resembles either the reactants or the products, to whichever it is closer in energy.

So, in this case, the transition state is closer to the final product and so resembles the product which is a double bond. Therefore, the more thermodynamically stable product is preferred.

In essence, even Zaitsev's rule is used to explain the same thing, the reason for a more substituted alkene being favored is because of the higher number of hyperconjugative structures that are made possible due to more groups attached.

Therefore in this case, the product formed will be 3-methyl-1-phenylbut-1-ene (E) or (Z), since the stereochemistry is unknown.

  • $\begingroup$ But the t-butyl carbocation formed after a 1,2-hydride shift is more stable than a benzylic carbocation formed after 1,2-hydride shift. See this : chemistry.stackexchange.com/a/86692/121235 $\endgroup$ Commented Aug 24, 2022 at 9:54

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