White gets only 2 stones and some ko threats...
Herman: After capturing these two stones, white has only one eye, so will have to make a second one...
So here, leads to disaster, and should be at , allowing white to capture four stones for a second eye.
But can black do better?
How about this, for example?
AVAVT: oh sorry, I forgot the lack of eye, it was really hasty of mine >_<
How about this, then:
White can play at a to make 2 eyes, if Black play a White can play b for a ko:
Herman: Yes, now because of and , the position of the two marked stones makes sense. I'm not sure whether this is the full solution, maybe white can live without ko?
Dave: This is an interesting ko position aside of tsume-go considerations. When B starts the ko at , White should play as the first internal ko threat. If Black answers at a, then after winning the ko at b White can capture nine additional stones with c for extra profit. This increases the burden on Black. On the other hand, if Black answers with d, White ends up with many additional internal ko threats: a (Black captures two stone with e), throw in at a (Black captures with ), cut at f (Black plays g), capture two stones at c (Black recaptures), White atari at the marked Black stone (if Black wins the ko on the left instead of connecting, Black can not connect at a after White captures at c so the ko continues).
Dave: My first thought is that this is the type of problem where White cuts at instead of . Black's shortage of liberties means that and lead no where. Now Black can not atari at a and White can live with b or c. If Black plays c instead of , White peeps at a and descends at d after Black connects at b. Black can not play now due to shortage of liberties.
Unfortunately this all reverts to the main line when Black answers with instead of connecting at :-(