can we define a tensor structure on $K(\operatorname{Proj}\text{-}R)$ to make it tensor triangulated category Let $K(\operatorname{Proj} R \bmod)$ be the homotopy category of projective R-mod. I was wondering is it possible to equip $K(\operatorname{Proj} R \bmod)$ in order to make a tensor triangulated category?
 A: A homotopy between two maps $f,g:C\to D$ of chain complexes can be seen as a map $H: C\otimes I\to D$, where $I$ is $...\to 0\to R\to R\oplus R\to 0$. While tensor products don't generically admit diagonals, there is a diagonal map $\Delta: I\to I\otimes I$, perhaps seen most easily by applying the cellular chain complex functor to the inclusion of the topological interval into the square.${}^1$ 
Thus given homotopies $H$ and $H':C'\otimes I\to D'$, we get a homotopy 
$$HH':C\otimes C'\otimes I\to^{C\otimes C'\otimes\Delta} C\otimes C'\otimes I\otimes I\to^{H\otimes H'}D\otimes D'$$
In other words, the tensor product preserves the homotopy relation between maps of chain complexes, and so descends to a tensor product on the homotopy category $K(R)$ of arbitrary complexes of $R$-modules. 
This does give $K(R)$ a tensor triangulated category structure. It's symmetric, additive, and respects suspension because the same was true on $C(R)$. It's also exact, since the mapping cones in $K(R)$ are defined in terms of direct sums and suspensions. 
So we see there's no problem at all at the level of $K$, whether you restrict to projectives or not. Furthermore, since $K^b(R-\mathrm{proj})\cong D^b(R-\mathrm{proj})\cong D^b(R)$, this same construction gives the tt-structure on the derived category of $R$. The only issue to be careful about is that the tensor products on $D^b(R)$ and $K^b(R)$ do not agree for non-projective complexes, since the tensor product I described above does not respect quasi-isomorphisms. 
${}^1$ Note that while there are big issues with the homology of a product of spaces, there are no such issues with the chain complex of a product of spaces. $C_*^{\mathrm{cell}}$ sends products to tensor products precisely.
