Why does a Diels–Alder reaction not occur here?
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.



umm...something's missing. Yeah!

 

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, 

 

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.



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.

 

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.



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)