Answer:
ethyl 3-ethoxy-3-hydroxy-2-isopropyl-5-methyl hexanoate
Explanation:
In this case, we have a very strong base (sodium ethoxide). Therefore, this compound will remove a hydrogen from ethyl 3-methyl butanoate generating a carbanion.
This carbanion, can attack another ethyl 3-methyl butanoate molecule on the carbonyl group generating a new C-C bond and producing a negative charge in the oxygen.
Then the ethanol can protonate the molecule generating an "OH" group and the ethoxide.
See figure 1
I hope it helps!
