Operation on two CFLs which results in a non-CFL Suppose we have two context-free Languages $L_{1}$ and $L_{2}$, and we form a new language $L_{3}=L_{1} \circ L_{2} = \{\alpha\beta | \alpha \in L_{1},\beta \in L_{2},|\alpha|=|\beta|\}$. Is it possible that $L_{3}$ is not context-free?
Let's denote your operation by $\diamond$, so the question is, if $L_1$ and $L_2$ are CFLs, is $L_1\diamond L_2$ always a CFL? The answer is no.

Let $L_1=L_2=\{0^n1^n\mid n\ge 0\}$. This is, of course, a CFL. However, 
$$
L_1\diamond L_2=\{0^n1^n0^n1^n\mid n\ge 0\}
$$
and it's not too hard to show that this language is not context-free.