Is Context Free Language closed under perfect shuffle? Note that this is not shuffle but perfect shuffle, defined as follows:

Let $w = a_{1}a_{2} \ldots a_{n}$ and $x = b_{1}b_{2} \ldots b_{n}$ be two strings of the same length. Then the perfect shuffle of $w$ and $x$ is defined as $a_{1}b_{1}a_{2}b_{2} \ldots a_{n}b_{n}$.

So the question is: are context free languages closed under perfect shuffle?
Nope. Shuffle $\{ a^n b^{2n} \mid n\ge1\}$ with  $\{ a^{2n} b^n \mid n\ge1\}$ .