I haven't learned the claissen condensation yet so take this with a grain of salt. In the step before R40- is kicked out by the shift of pi electrons, the molecule already has a negative charge on the other oxygen. So perhaps there is not as high an energetic barrier to forming the R4O- leaving group for that reason. It's just trading a one strong base for another strong base.
I would not call an alkoxide a strong base. When I think of strong bases, I think of KOH, lithium diisopropyl amide, and the like.
For SN2, wouldn't an alkoxide ion be an equivalently bad leaving group as hydroxide?
No. However, it is not necessarily a great leaving group either. Only under circumstances do you have an alkoxide acting as a leaving group (good thing too, can't have ethers falling apart willy-nilly
). You'll notice that in this case, with the claisen condensation, the alkoxide that is being kicked out is already actually in solution. The reaction is run in a mixture of the alcohol and the corresponding sodium salt.
Also is it just me or does that mechanism use "R4" in 2 different ways: once as a base in the beginning of the mechanism, and again as an alkyl group in the second ester molecule? That confused me at first.
It does. This is because, they want you to notice that whatever group is going to be acting as the leaving group they also want to use as a base. This is because there is the possibility that the base can instead act like a nucleophile and attack the carbonyl and exchange the ester alkoxide out. However, if we use the same alkyl chain then we don't really care if that happens because the structure is unchanged.