Provide a mechanism for the following reaction. Include all intermediates, resonance structures, charges, and electron pushing arrows to obtain full credit.

4,4‐dimethoxy‐2,3,4,5,6,7‐hexahydro‐1H‐indene and sulfuric acid

I’m stuck with this problem. I got as far as step 2 of the mechanism. I know the double bond acts as a nucleophile and attacks one of the hydrogen on sulfuric acid thus creating its conjugated base $\ce{HSO4-}$ leaving a carbocation on carbon 1.

In previous problems I have done the carbocation will rearrange to form a more stable cation. I’m told it’s done by either a hydride shift (which I understand) or an alkyl shift. I’m having a hard time grasping the concept of a six-membered ring going into a five-membered ring. And essentially understanding why there is a shift.

  • $\begingroup$ #1) In an analogy: Are you able to number non-H atoms in your starting material in a pattern like a shoelace (this time with definitive start and end), and may you place this on the product? Don't feel bad if you need to draw such numbering patterns for you, at the beginning you probably need multiple attempts to find your way through. #2) You probably recognize $\ce{H2SO4}$ as an acid, capable to donate protons. What could be sites in the organic molecule accepting protons -- even if only in an equilibrium? #3) Sketch how these new molecule could serve as starting materials. $\endgroup$ – Buttonwood Sep 24 '19 at 20:41
  • $\begingroup$ #4) Probably there are multiple pathways found; which one is more reasonable, than others? (It is not saying that the reaction definitively works this or that way, but finding what makes more sense.) This site won't provide you right on the sleeves the answer. But if you document what you already tried to tackle the question, it is easier to recognize where you need help. $\endgroup$ – Buttonwood Sep 24 '19 at 20:43
  • $\begingroup$ Thank you for you feedback. I was not clear with my question. $\endgroup$ – Lromero2 Sep 24 '19 at 22:45
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    $\begingroup$ I seriously question the product in this reaction. The carbon bearing two methoxy groups and one hydroxyl group is the intermediate in ester hydrolysis. Orthoesters are acceptable but not this 2/3-orthoester. $\endgroup$ – user55119 Sep 25 '19 at 23:17
  • $\begingroup$ @user55119 You are right, the targeted compound (the «2/3-orthoester») is unknown to Elsevier's Reaxys database; regardless if searched as drawn here, or allowing further than methox-substition as only $\ce{-C(OMe)2OH}$. $\endgroup$ – Buttonwood Sep 27 '19 at 21:20

The reaction of the ketal 1 with aqueous H2SO4 is more likely to follow the pathway 1 --> 2 --> 3 to form α,β-unsaturated ketone 3. This is a typical ketal hydrolysis in acid medium. For the proposed reaction pathway 1 --> 4 --> 5 --> 6, species 6 is unstable. Indeed, if it were formed as an intermediate, it would afford ester 7a and ultimately the carboxylic acid 7b. Is there a possibility that the actual reactant doesn't bear methoxy groups but methyl groups? If this is the case, then the reaction 8 --> 9 is more reasonable.

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  • Many organic reactions' steps may be described by equilibria. In other words, at least on microscopic scale, both «forward» and «backward reaction» is possible. So a homework for you is to become fluent in this reversibility. As an example relevant here: enter image description here

  • I'm not aware about an better approach than to learn a handfull of rearrangement reactions till you recognize «typical pattern», e.g. molecular substrates which are more likely than others to undergo this type of chemistry. Here, the formation of the product bears similarity to the Pinacol rearrarangement. As already identified by yourself, an intermediate carbocation is needed to trigger the cascade like, enter image description here

    and to memorize larger alkyly and phenyl groups are more likely to move, than methyl groups (e.g., Beckmann rearrangement). In early exercises it helped me to recall that each single dash not only represents to carbon atoms connected with each other, but equally a pair of two electrons forming a bond between these two atoms in questions. Depending on teacher and school, a drawing like the following may be permissible until creating enough «muscle memory»: enter image description here

    Probably soon you will learn that drawing a bent arrow anyway symbolizes the movement of two electrons (contrasting to the fish hooks to symbolize moves of single electrons) which renders drawing to dots for the two electrons in question frequently redundant.

  • The formation of six- and five-membered carbocyles is thermodynamically favourable. If a six-membered ring interconverts into a five-membered carbocycle might be of lesser importance for your current level of study, than to recognize patterns typical for this kind of reaction. It depends both on thermodynamics as well as kinetics, and over time you develop a reasoning when one, or the other favours the starting materials, or the products. In some way, it is like learning grammar of a foreign language (e.g., I speak, you speak, s/he speaks, we speak, you speak, they speak). As such, keep speaking Chemistry.

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