Why does Beryllium react to make neutrons? I've seen the reaction:
$$
^{9}_{4}\mbox{Be} + ^{4}_{2}\mbox{He}\, \rightarrow\, ^{12}_{\,\,6}\mbox{C} + \mbox{n}\,.
$$
But what I don't understand is why Carbon 13 is not made, or if there is actually some carbon 13 made?
 A: The first thing to understand in a reaction like $^{9}$Be($\alpha$,n)$^{12}$C is that, indeed, you have a $^{9}$Be and an $\alpha$ particle interacting (note that in the lab frame, with stationary $^{9}$Be you need ~300keV $\alpha$ particles to see much of anything, and at 2MeV $\alpha$s the cross section has increased by roughly 5 orders of magnitude). As you note, this leads to a $^{13}$C compound nucleus should they actually get together. This results in the emission of a neutron leaving $^{12}$C. So, you did read correctly that a $^{13}$C is involved in there somewhere. So why don't you get something like $^{9}$Be($\alpha$,nothing)$^{13}$C instead? Well, in the reaction you need to conserve energy and momentum, and this is easiest to see in the center-of-mass frame. Now one does find a few nuclear reactions like A(n,$\gamma$)B, but remember that the momentum of a photon is really small, so it really only works at precise energies.
OK, to the literature. In The Reaction of $^{9}$Be($\alpha$,n)$^{12}$C one finds in the introduction:

The reaction $^{9}$Be($\alpha$, n)$^{12}$C (Q = 5.704 MeV) is a convenient source of fast neutrons
for calibration purposes. The neutron group leading to the ground state of $^{12}$C
is well separated in energy from the neutron group leading to the first excited state
(at 4.43 MeV excitation) and the gamma ray from the decay of this state is the only
gamma ray present.

Well, right off the bat one sees that $^{9}$Be($\alpha$,$\gamma$)$^{13}$C is not observed. And a way to look at that is through nuclear energy level diagrams. For light nuclei, the TUNL Nuclear Data site is a good one. Looking at A=13, you get a plot that looks like:

The compound nucleus formed is not in the ground state (red box on the left). In fact it is more than 10MeV above the ground state. What can it do? Well, it looks around and sees the red box on the right - the $^{12}$C+n exit ramp. Now it has a way of keeping energy and momentum conserved, and that way has a large Q value of about 5.7MeV. Plus it has a choice of several neutron emission energies on top of that.
Now, besides being a fairly easy neutron source, this reaction leads to several other interesting bits. One is in the Determination of the $^{9}$Be($\alpha$,n)$^{12}$C reaction rate, which notes that in a supernova where a neutron- and $\alpha$-rich environment exits, the $^{4}$He($\alpha$n,$\gamma$)$^{9}$Be($\alpha$,n)$^{12}$C reaction chain 'dominates the creation of carbon' - so, we should be very happy that it exists, being carbon-based life forms.
Another one is Spectroscopy of $^{13}$C above the $\alpha$ threshold with $\alpha$ + $^{9}$Be reactions at low energies where the authors use the cross-sections of the elastic and inelastic $\alpha$ scattering as well as the $^{9}$Be($\alpha$,n)$^{12}$C reaction to try to tease out the nuclear structure of $^{13}$C. In particular, if you remember that $^{8}$Be quite rapidly and happily decays into 2 $\alpha$ particles (at a Q of some 17MeV), one starts wondering why 3 $\alpha$ particles seem quite happy to make a $^{12}$C nucleus rather than falling apart. Well, they do seem to like it, and there is some evidence that the 3 $\alpha$ particles join up and don't really want the spare neutron flying around in $^{13}$C. Pretty mean of them, but as the odd nucleon out it gets ejected from the group readily. Anyway, I'm rapidly going astray of the original question.
Bottom line: conserve both energy and momentum (always easier to see in the center-of-mass frame). Use available resources (like energy level diagrams at TUNL, or the Evaluated Nuclear Data Files (ENDF) or Evaluated Nuclear Structure Data Files (ENSDF) available at various mirrors. And, have fun playing with nuclear physics.
