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I was thinking of dehydrating this alcohol:
3-methyl-1-phenylbutan-2-ol

Its structure is:

enter image description here

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?

Loong
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Dusty_Wanderer
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    The double bond in the first product is in conjugation with the aromatic system which is energetically favoured – Waylander Jun 19 '21 at 07:41
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    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.. – Safdar Faisal Jun 19 '21 at 08:10
  • isnt zaitzev violated then ? I have learnt that hoffmann product is major only in the following cases : (i) bulky base (ii) poor leaving group (iii) gamma carbon is neo carbon (iv) thermal elimination recations – Dusty_Wanderer Jun 19 '21 at 08:30
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    @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. – Nisarg Bhavsar Jun 19 '21 at 09:44
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    Zaitsev's rule is useful when there are no other significant factors involved. Conjugation to an adjacent aromatic system is a significant factor. – Waylander Jun 19 '21 at 10:30
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    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. – TRC Jun 19 '21 at 13:31
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    Also, if I'm not wrong, dehydration of secondary alcohols proceeds by carbocation intermediate - in that case the initial carbocation would rearrange ot give the stable benzyl cation, not the other way (again, because resonance in benzyl cation gives more stability than hyperconjugation in the others). After rearrangement, you have little room for any other product except the first one. – TRC Jun 19 '21 at 13:33
  • @Waylander Are you sure? But the t-butyl cation is more stable than benzyl cation. See : https://chemistry.stackexchange.com/a/86692/121235 – An_Elephant Aug 24 '22 at 09:49
  • @omjoglekar The second product do not come from 2° cation but come from stable 3° - t(butyl) cation ( after a 1,2-hydride shift) awhich is more stable than benzylic cation. I suspect that your answer key is correct. – An_Elephant Aug 24 '22 at 09:52
  • @An_Elephant SLIGHTLYy more stable. They will be in equilibrium which allows the more energetically favourable alkene to form. – Waylander Aug 24 '22 at 10:12
  • @Waylander Similarly like alkenes, won't "more energetically favourable" compounds form always like alkene and the stability of these cations is useless? – An_Elephant Aug 24 '22 at 14:05
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    @An_Elephant Not necessarily. I this case the conjugation with the aromatic system is the most important factor. In alkyl systems the cation stability will be the major factor. – Waylander Aug 24 '22 at 17:07

1 Answers1

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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.

Safdar Faisal
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  • 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 : https://chemistry.stackexchange.com/a/86692/121235 – An_Elephant Aug 24 '22 at 09:54