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enter image description here

When the above compound is heated with (conc.) $\ce{H2SO4}$ the $\ce{OH}$ will probably be protonated and it will leave, producing a carbocation. After that possibly there will be a hydride shift. Now, after this step what could be the possibilities. I was thinking of ring expansion (which could potentially form phenanthrene) but again I doubt that a negative charge on a $sp^2$ carbon of the benzene ring would attack the positive charge after C-C bond cleavage. Or another possibility is formation of a double bond to form a structure like:

enter image description here

Which is the correct (major) product formed?

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  • $\begingroup$ Ring expansion is possible and we get an aromatic compound I suppose which is quite stable.... $\endgroup$ – Suraj S May 7 '17 at 5:46
  • $\begingroup$ @SurajS For ring expansion we need to form a negative charge on the benzene ring which has to potentially attack the formed carbocation due to leaving of the $\ce{OH2+}$ group. Is that feasible? $\endgroup$ – user38977 May 7 '17 at 5:49
  • $\begingroup$ I think this process favors thermodynamics, Phenanthrene might be the major product... $\endgroup$ – Suraj S May 7 '17 at 5:50
  • $\begingroup$ Sorry I don't get why - ve charge on the benzene has to be formed... Isn't it th 3° carbon making way for the rearrangement so that 1° carbocation stabilizes to 2°? $\endgroup$ – Suraj S May 7 '17 at 5:59
  • $\begingroup$ Which "$1°$ carbocation stabilizes to $2^o$" are you talking about? I can't understand. Can you draw it and show? @SurajS $\endgroup$ – user38977 May 7 '17 at 6:02
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What I feel is that forming phenanthrene would be possible because of its high stability due to resonance. Initially, $\ce{-OH2}$ leaves out forming a 1°carbocation. Now the $\ce{\beta C-\gamma C}$ cleaves so as to make way for the better 2°carbocation which is also thermodynamically favourable. enter image description here

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