Why $∀x(P(x)→Q)$ is equivalent to $∃xP(x)→Q$ Given $x$ occurs free in $P$. If $x$ does not occur free in $Q$, then $∀x(P(x)→Q)$ is semantically equivalent with $∃xP(x)→Q$. How to understand this statement. And also, need an example to show that the formula will not necessarily be equivalent if $x$ does occur free in $Q$.
I guess what struggles me is I'm not quite understand what does "free" mean in this case. But it would be better to have answer for the above question. 

Let $\mathcal M$ a structure whatever, with domain $D$. 
Two cases :
(i) $∃xP(x)$ is False.
Thus $∃xP(x) \to Q$ is True.
But to say that $∃xP(x)$ is False in $\mathcal M$ means that for $a \in D$ whatever, we have that $P(x)[a/x]$ is False, and thus $(P(x) \to Q)[a/x]$ is True in $\mathcal M$.
And this means in turn that $\mathcal M \vDash ∀x(P(x) \to Q)$.
(ii) $∃xP(x)$ is True, i.e. there is an object $a \in D$ such that $P(x)[a/x]$ is True.
Two sub-cases :
(a) $Q$ is False. Thus $∃xP(x) \to Q$ is False.
But $(P(x) \to Q)[a/x]$ is False for the $a \in D$ above, and thus $∀x(P(x) \to Q)$ is False in $\mathcal M$.
(b) $Q$ is True. Thus $∃xP(x) \to Q$ is True.
But if $Q$ is True, also $(P(x) \to Q)[a/x]$ is True, for $a \in D$ whatever.
And this means that $∀x(P(x) \to Q)$ is True in $\mathcal M$.

Why the condition "$x$ not free in $Q$" is necessary ?
First of all we have to agree on the way to define what does it mean : $\mathcal M \vDash \varphi(x)$ with $x$ free.
One convention is to consider the $\text {Closure}$ $\forall x \varphi(x)$ of the formula, i.e. : 

$\mathcal M \vDash \varphi(x) \text { iff } \mathcal M \vDash \text {Cl}(\varphi(x))$.

Consider now a simple interpretation with domain $D = \{ 0,1 \}$ and interpret both $P(x)$ and $Q(x)$ as $(x=0)$.
The LH formula is : $\forall x \ ((x=0) \to (x=0))$ while the RH formula is : $\exists x \ (x=0) \to (x=0)$.
