How does aluminium react with bases to form aluminates? An example of reaction: 

$$\ce{2Al + 2KOH + 6H2O-&gt;2 K[Al(OH)4] + 3H2 ^}$$

Aluminium is not ionic, how then does it attract the $\ce{OH-}$ groups to bond with them into the complex ion $\ce{[Al(OH)4]-}$?

I started re-reading non-organic chemistry at a slightly deeper level recently after going through the basics of organic chemistry. This reaction stumped me. I've re-read some stuff about aluminium and complex ions, but can't seem to understand it yet. 

There's no positive charge on $\ce{Al}$, yet it reacts with hydroxide groups floating around. Where will the electron from a non-ionic aluminium atom go? 

I cannot 'imagine' the reaction, visualize its mechanism, and this makes it hard to remember it.



P.S. Aluminium hydroxide seems to be mysterious in its structure. I quote Chemguide:


  The chemistry textbooks that I have to hand aren't too clear about the structure of aluminium hydroxide as far as the degree of covalent character is concerened, and a web search (until I got totally bored with it!) didn't throw up any reliable chemistry sites which discussed it. Several geology sites describe the mineral gibbsite (a naturally occurring form of aluminium hydroxide) in terms of ions, but whether it actually contains ions or whether this is just a simplification as a convenient way of talking about and drawing a complicated structure, I don't know.
There's no positive charge on Al, yet it reacts with hydroxide groups floating around.


Your observation is correct. In the beginning, aluminium in its elemental state has the oxidation state 0. Apparently, that changes in the course of the reaction:
$$\ce{Al &lt;=&gt; Al^3+ + 3e-}$$


  Where will the electron from a non-ionic aluminium atom go?


Again, your observation is correct. If one species is oxidized, another one must get reduced. In this case, protons from the water are reduced to form hydrogen gas:

$$\ce{ 2H2O + 2e- &lt;=&gt; 2OH- + H2 ^}$$

If you combine both half reactions and balance the stoichiometry, you end up with the initial equation for the whole redox reaction.