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In this reaction after the attack of lone pairs on $\ce{H+}$ ions, a stable $3^{°}$ carbocation is formed. But seeing the six membered ring and the double bonds already present, I can't help but think that there's some way of obtaining a benzene ring through rearrangements. Can someone suggest a mechanism for this?

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  • $\begingroup$ Be careful with your thinking. H+ ions do not attack. The lone pair on the alcohol attacks (or accepts ) H+. You should always think of mechanisms in terms of the movement of electrons. Have you attempted a mechanism? Can you show us what you have tried? $\endgroup$ – Michael Lautman May 28 at 13:43
  • $\begingroup$ Yes I phrased it wrong. The lone pair on alcohol attacks on the H+ ions to convert it into a good leaving group ie. water and the water leaves to form a carbocation as shown. $\endgroup$ – Sameer Thakur May 28 at 13:54
  • $\begingroup$ chemistry.stackexchange.com/questions/98588/… $\endgroup$ – Mithoron May 30 at 23:02
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I think Sameer Thakur was in right track when started to write the mechanism. But the path got lost at the end. I don't see reason to have a methide shift followed by a hydride shift and then proton abstraction. The 1,2-methide shift gives you a very stable $3^\circ$-carbocation, which is compatible with the initial $3^\circ$-carbocation given by elimination of water. Thus, I think the following mechanism is a very reliable one for gaining aromaticity:

Rearrangement to Aromatic

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  • $\begingroup$ You are right. The hydride shift is redundant in my solution. $\endgroup$ – Sameer Thakur May 28 at 19:03
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While there are examples of 1,2-methide shifts, this question has an uncanny resemblance to the dienone-phenol rearrangement whose mechanism was first elucidated by Woodward and Singh in 1950. Dienone 1 under acidic conditions undergoes rearrangement to phenol 6 and not, based on earlier speculation, to phenol 4. The direct 1,2-methide shift (2b --> 3) does not occur but rather the reaction proceeds through the spiro carbocation 5. continued

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Applying this approach to the carbocation 7 generated from the alcohol in this question, spiro carbocation 8 forms the predicted tetrahydronaphthalene 10. One way to distinguish between the two mechanisms is via a labeling experiment. All of the carbon label in 7 will retain its location as the red star in 10 if the 1,2-methyl shift mechanism applies. The spiro mechanism will partition the label ~50:50 between the two ring benzylic carbons. For related studies, see reference 2.

1) R. B. Woodward and T. Singh, J. Am. Chem. Soc., 1950, 72, 494.
2) A. J. Waring, J. H. Zaidi and J. W. Pilkington, J. Chem. Soc, Perkin Transactions I, 1981, 1454.

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It will not work the way the OP likely envisions. To make an aromatic ring you need to move the methyl group into the bottom ring, but all of these bottom positions are saturated. No place for the proposed rearranging group to land.

There is, however, another possibility if the system from a nonclassical carbocation. To make the nonclassical carbocation, the methyl group moves over the bridging bond but not all the way across so that the bond between the methyl group is delocalized between the two bridgehead carbons. In this structure, if it forms, the delocalized electrons are also conjugated across the bridge like an additional pi bonding pair so that the top ring (6 conjugated electrons) and the bridge structure (2 electrons) are both aromatic. Possibly other users more familiar with the system can indicate whether this structure forms, leading to a new question.

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  • $\begingroup$ Why do we need to move the methyl group in the lower ring? $\endgroup$ – Sameer Thakur May 28 at 14:47
  • $\begingroup$ Upper ring otherwise cannot be fully conjugated. $\endgroup$ – Oscar Lanzi May 28 at 14:51
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Okay I figured it out. First I drew the resonating structure for the carbocation and then performed a methyl shift followed by a hydride shift. Then the H+ ion leaves via the dehydration mechanism making the ring aromatic in the process.

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  • $\begingroup$ Your first two "resonance structures" are confusing. Have you migrated the methyl group (not resonance) or turned the molecule over? $\endgroup$ – Withnail May 28 at 15:29
  • $\begingroup$ You can get the second structure, just by resonance among the 4electron, 5 center pi system on the top ring. Then turn the molecule over. $\endgroup$ – Withnail May 28 at 15:34
  • $\begingroup$ Not sure about this. I'm still curious about the possibility of a nonclassical ion. $\endgroup$ – Oscar Lanzi May 28 at 15:47
  • $\begingroup$ @Withnail I moved the double bonds towards left through resonance and then turned the molecule over. It is confusing but it would proceed the same way even without turning it. $\endgroup$ – Sameer Thakur May 28 at 15:51
  • $\begingroup$ @OscarLanzi what do you mean by a nonclassical ion? $\endgroup$ – Sameer Thakur May 28 at 15:51

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