here i have just a few more examples these are all nucleophilic conditions followed by an acidic workup second step but in this case i've included some stereochemistry to think about and the thing to keep in mind is this nucleophilic attack is essentially an sn2 step and when nucleophiles add they add opposite the side of the leaving group so if the epoxides a leaving group the nucleophile adds opposite it so in our first reaction we have nacn that's n a plus and c and minus is the nucleophile look at your two epoxide carbons and the one on the right is less substituted because it has a methyl and if you want to draw it in we have a hydrogen okay now in this case the epoxide is drawn in the point so what i'm going to do is add the nucleophile in the plane but from the side opposite from underneath so we break that open let's draw what we get on the left side it's not a chiral center so we don't really care about the stereochemistry there on the right i'm going to draw that cyano group pointing straight down because the epoxide leaving group bond was pointed up so this will do opposite which is down and then i have the methyl out and that methyl kind of gets pushed upwards but it's still out and if you want to draw it in the hydrogen is back now this of course in step two gets protonated so we protonate the oxygen and i'm going to leave the hydrogen implied so there's the final product in the second example we still have the epoxide on a chain but in this case the epoxide was shown with dashes and wedges so here the epoxide is on a wedge and my nucleophile is nh2 minus now if you look at this we have the two carbons of the epoxide and the one on the left is less substituted so we're going to add the nucleophile right here the one on the right if you just kind of think three dimensionally if this epoxide is pointed out towards you we have this fennel down this methyl group i'm going to put it toward the back so you don't have to do this but i think it might make it a little more clear just to put that on a dash to help us keep track of everything it might seem counterintuitive but it actually does not matter if you keep the fennel in the plane and put the methyl back or keep the methyl in the plane and put the fennel back and follow through with the stereochemistry in both cases you'll get the exact same molecule and if you don't believe me try it and assign the rns configurations so i just decided to put the methyl back so now i'm going to take this nh2 minus and add it to the epoxide carbon and open that up and draw my chain and on the right side i have this methyl that i put back the o ends up as o minus and then it gets protonated to boh and then on the left carbon this now has the nh2 group but it came in from the side opposite the epoxide so since the epoxide is out i want to put the nh2 going back so there's the product for that reaction one last example here we're using sodium hydroxide so that means o h minus is the nucleophile in this case both sides of the epoxide are equally substituted and the molecule is symmetric so it doesn't matter which side i add the oh minus two i'm just going to add it right here and break that open but remember this adds from the side opposite your leaving group so i'm going to put a wedge o h here and then the new o h that added let's draw it in blue just to differentiate it it's going to come in from the back right here so this is actually a way to get an anti-diol