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I have recently come across this organic chemistry problem. enter image description here I thought it would undergo simple Diels–Alder reaction enter image description here But it didn't turn out to be. The product to me looks like a nucleophilic attack of (I) on (II). Can anybody say me why exactly this abnormality seen and what is the mechanism of this reaction?

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    $\begingroup$ You're not going to achieve a s-cis conformation for a [4+2] cycloaddition. The dithiide protecting group is way too bulky to be in plane with the vinyl ether. This is a conjugate addition into the enone, followed by aromatization of the ring, and addition of the phenol moiety into the oxocarbenium. $\endgroup$ – Zhe Oct 3 '17 at 17:52
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    $\begingroup$ I thought about the bulkiness of the dithiane group but I couldn't get any further ideas. Can you please put up the solution. $\endgroup$ – Suraj S Oct 3 '17 at 17:58
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    $\begingroup$ You should have enough hints from my first comment. Please think about the requirements for a [4+2] cycloaddition and the implications thereof. Build models if necessary. $\endgroup$ – Zhe Oct 3 '17 at 18:11
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    $\begingroup$ I suggest editing the title. As it stands now, it is misleading. See, there is a thing called inverse electronic demand D.-A., and that's an abnormal Diels-Alder reaction for sure. Yours is not a Diels-Alder at all. $\endgroup$ – Ivan Neretin Oct 3 '17 at 18:33
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    $\begingroup$ @IvanNeretin an inverse electron demand DA is not abnormal. Diene + dienophile leading to a cyclohexene derivative is a DA reaction, doesn't matter what's the mechanism. $\endgroup$ – DSVA Oct 4 '17 at 10:08
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Firstly, Why doesn't Diels-Alder reaction happen here?

On first sight the reaction seem completely feasible as both the diene and the dienophile are good species that could undergo Diels-Alder, but no. It could be proven by just analyzing the products (the actual one and the expected one).

  • Resonance. The property that literally defines organic stability. The DA product has not got even a trace of the stabilization that the actual product receives. The actual product gets to experience extended resonance due to the benzene ring that is formed.
  • The second factor being that the actual product has got TWO (yes...two) aromatic rings fused together, each having 6 electron resonance.
  • Steric hindrance. The DA product experiences Steric hindrance while the reaction happens due to the bulky dithiane group. While the other path is free from this.
  • Strain. Consider the carbon that is holding the dithiane group in both the products. In the DA product the carbon is $sp_3$ while the other product has $sp_2$ (perfect 120 degrees required, and 120 degrees received). In fact, all carbons in the actual product are $sp_2$, other than the ones in the dithiane group obviously.

All of these factors encourage the reaction to proceed in the second path which leads to an amazing yield of 62%.

P.S. I kinda feel stupid after typing "actual product" like 10 times.


Now, the MECHANISM (ohh yeahh!)

Let's start with para-benzoquinone. It is an $\alpha-\beta ketone$. Ergo, the $\beta$ hydrogen is acidic. So, lets prepare it for the incoming nucleophile.

enter image description here

umm...something's missing. Yeah!

enter image description here

OK. Looks better.

Now that we are done with this, we will get to the other reactant. Observe that the methoxy group on the $sp_2$ makes it a good nucleophile. But there is a problem, the dithiane group is too bulky to be drawn again and again. Therefore,

enter image description here

Properties of 'G' :

*bulky, must be kept away

*observing the product, where it is intact, we can say that it is useless for the reaction.

Coming back to the reaction, this nucleophile attacks our first molecule.

enter image description here

Here the (*) marked carbon is activated due to the methoxy group, hence the attack is at that site.

Now the carbocation can stabilized using proton transfer onto the oxygen.

enter image description here

The encouraging factor for this is the achievement of an aromatic ring in the molecule.

Observe the (*) marked carbon in the above image. We see that it is strained due to the thiane group present on + the methoxy group. Therefore the methoxy could abstract the proton from the adjacent carbon and exit as methanol.

enter image description here

Yet again, another encouraging factor for this exit is the formation of another aromatic ring.

Replacing the 'G" with it's true form we have...(drum roll)

enter image description here

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    $\begingroup$ 1) Resonance is only an argument if all potential products are in full thermodynamic equilibrium and can convert back and forth which should not be the case here. 2) I am really not sure whether you should count the benzofurane system as two aromatic rings; it’s not like there are two benzene systems being generated. 3) Angle strain shouldn’t play any significant role in any of the products or intermediates. 4) I believe your mechanistic steps to be … let’s say suboptimal. $\endgroup$ – Jan Oct 16 at 15:52
  • $\begingroup$ 1) Resonance stabilisation could be used as an arguement between two reaction paths (right?). As the product will be more thermodynamically stable hence easier to be formed. 2) Benzofurane system can be considered as two separate aromatic rings due to the Sacrificial Aromatic Character displayed by them (as seen between Napthalene and Benzene). 3) I strongly believe that angle strain should play a role. If the system is too strained there would be a higher chance of reversion of the product into the parent molecules. 4) Could you please be more specific about the steps you find "suboptimal". $\endgroup$ – Still_a_kid Oct 17 at 14:26
  • $\begingroup$ 1) Only if there is indeed a conflict of pathways. In OP’s example, the DA pathway fails long before the question of resonance or not can even be considered, it simply does not happen so it is not useful to compare product stabilities. 2) Eh, more like 1.5 aromatic systems but that doesn’t have much bearing on the answer. 3) There is no intermediate in this pathway (or the impossible DA pathway) where any carbon atom is susceptible to any angle strain so it is irrelevant to this question. 4) The cyclisation is most certainly not concerted, … $\endgroup$ – Jan Oct 17 at 14:50
  • $\begingroup$ The proton transfer will most likely happen before cyclisation, and methanol elimination will probably follow a different order of steps to name just the most striking. $\endgroup$ – Jan Oct 17 at 14:51
  • $\begingroup$ All of your comments make so much sense!! Anyways, I think there is a serious defect in the way things are happening in the mechanism I proposed. Thanks for pointing it out (still_a_kid...got to learn a lot :) ). $\endgroup$ – Still_a_kid Oct 17 at 15:45

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