# Rearrangement of 8a-ethyl-1,3,4,8a-tetrahydronaphthalen-4a(2H)-ol [closed]

really sorry for that previous post. I was actually half-asleep when I posted that q, hence the q was unclear. I am again reposting that question

The answer is given as A. Can anyone please explain HOW ??

• The last option is methylbenzene. – Aakhyat Singh Jun 21 '18 at 13:34
• Cation rearrangement should be expected here, yet the answer (C) is strange to me; I'd assume (A). – Ivan Neretin Jun 21 '18 at 14:37
• Sorry Ivan Neretin, typo error-ans is A.-but HOW ??? – Aakhyat Singh Jun 21 '18 at 15:45
• @Aakhyat Singh: If you are still interested, for reliable mechanism for this type of rearrangement, see my answer to this question. – Mathew Mahindaratne May 28 '19 at 18:33

Answer will be option A

First of all, name the carbon atom in the option A with $\ce{-Et}$ group attached to it as x, the $\ce{C}$ atom in the question with $\ce{-OH}$ attached to it as y, and the $\ce{C}$ atom in the Question with $\ce{-Et}$ attached to it as z

What will happen is, the $\ce{-OH}$ will get protonated forming $\ce{-OH2+}$. Concentrated acidic medium will facilitate formation of $\ce{H2O}$ by braking of bond between $\ce{C}$ and $\ce{O}$ in $\ce{C-OH2+}$ This will form a carbocation at the position y which will due to resonance reach the position x. Now, the $\ce{-Et}$ will do alkyl shift from z to x.

Now, a $\ce{C-H}$ bond at x will break and a $\ce{C=C}$ will form between the $\ce{C}$ atoms at x and z, thus forming an aromatic compound as major product i-e option A

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Formation of $\ce{C+}$ is facilitated because of $\ce{conc. H2SO4 + heat}$ because $\ce{(conc. H2SO4 + heat)}$ acts as an oxidising agent, hence oxidising the $\ce{C}$ atom (or you may also think like, formation of $\ce{H2O}$ in a concentrate system will liberate heat thus its formation is favourable)

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Alkyl shift from z to x is favourable as it drives the reaction towards formation of a much stable aromatic compound