Does birational imply D-equivalent? It is well-known that there are Calabi-Yau's who are not birational but are derived equivalent. However I am interested in seeing D-equivalence as a weakening of birationality. 

Q. If $X$ and $Y$ are smooth projective varieties and $X$ is birational to $Y$, does it follow that $D(X)\simeq D(Y)$?

My guess would be that no such result exists so far, but I cannot see a good obstruction to it happening. I know a lot of work is done to show that flips, flops and blowups are derived equivalent so it would mean that if the result above were true, the later would be easier to prove.
 A: The answer to the question as literally asked is "no". If $X$ is an surface and $Y$ is $X$ blown up at a point, then the rank of the Grothendieck $K_0$-group of $Y$ is one larger than that of $X$. Since $K_0$ is recoverable from the derived category, this shows that $\mathcal{D}(X) \not \cong \mathcal{D}(Y)$. 
This is why people ask not only that $X$ and $Y$ be birational, but impose conditions on the canonical class, such as that $X$ and $Y$ are Calabi-Yau, or the notion of $K$-equivalence in Kawamata.
A: This is conjectured to hold for varieties with trivial canonical bundle (Bondal-Orlov), and is known to be true for Calabi-Yau 3-folds (Bridgeland established this in his seminal paper https://arxiv.org/abs/math/0009053). In higher dimensions this is one of the basic problems in the field, and, to the best of my knowledge, very little progress has been made towards its solution.
Let me relate it to another question. A folklore conjecture is that derived invariant varieties have the same Hodge numbers (this is proven in dimension at most 3 by Popa and Schnell, an independent proof was given by Abuaf). However, birational varieties need not have the same Hodge numbers; the easiest counterexample is $X=\mathbf{P}^2$ and $Y=\mathbf{P}^1\times\mathbf{P}^1$. These are birational but cannot be derived equivalent (by Bondal-Orlov they would have to be isomorphic!).
