Extension of an embedding to a field extension by a transcendental element

Let $K$ be a subfield of a field $K'$ and suppose we have an embedding $\phi:K \rightarrow L$ of $K$ into an algebraically close field $L$. Let $x \in K'$. If $x$ is algebraic over $K$, then we can extend $\phi$ to $K(x)$. What if $x$ is transcendental over $K$? Can we then extend $\phi$ to $K(x)$?

Thanks.

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Note: this is an edited version of a previous incomplete (and wrong....) answer. While I was editing. Gerry wrote his answer giving the very same counterexample. Sorry for the mishap ... :)

If there is a $z\in L$ transcendental over $\phi(K)$ you can extend $\phi$ by letting $x\mapsto z$.

But it is possible that such a $z$ doesn't exist: for instance $K=\Bbb Q$ embeds in $L=\overline{\Bbb Q}$ but there's no way to extend this to an embedding of $K^\prime=\Bbb Q(\pi)$ into $L$.

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But the way I read the question, $L$ may not be a transcendental extension of $\phi(K)$ –  Lubin Aug 12 '12 at 13:03
How do we know that such a $z$ exists? –  Manos Aug 12 '12 at 13:05
Oops !! I'll edit ..... –  Andrea Mori Aug 12 '12 at 13:06
I suppose the answer is “never, in case $L$ is an algebraic closure of $K$”. –  Lubin Aug 12 '12 at 13:12

As Andrea writes, yes, if there's a transcendental in $L$. As Lubin hints, no, if not. E.g., if $K$ is the rationals, and $L$ is the algebraic numbers, and $x=\pi$, it can't be done. Or if $K$ is the reals, and $L$ is the complex numbers, and $x$ is an indeterminate.

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