Let's look at reactions of epoxides in a few different conditions, so for reactions in
acid, we're typically going to see addition at the more substituted carbon be preferred,
so here we've got a tertiary carbon versus a primary, so we end up seeing that tertiary
carbon get our nucleophile methanol here and I say unhindered nucleophiles so something
up to about the bulkiness of an ethanol or isopropanol we're gonna see again this trend
hold, but if we got something up to a nucleophile bonded to say a t-butyl group, that's getting
to be a point where we wouldn't see as strong as a trend okay, but in general we're gonna
see this a whole substitution at the more substituted carbon is going to be preferred
and why that is, is we see a protonation of our epoxide from our acidic conditions and
so we've protonated our solvent, methanol here in our reaction, and so we'll see that
lone pair grab an extra proton, we are protonated and now all of a sudden this is a very strained
ring system and it tries to almost open up so it gets to the point where it stretches
this bond going towards the tertiary position, lengthens it out and puts a pretty significant
delta plus on the more stable substituted carbon, not quite a full carbocation and this
has to do with the ways really strained rings bond and their hybridization being a little
bit different than a normal sp3 center, but we create just enough room in this situation
for our nucleophile to come in, in the form of our solvent, attack this position, open
up the rest of the epoxide and now, we've got a protonated ether and we've got an alcohol
on the other side that opens up to the less substituted side so then we just take our
solvent molecule to finish out our bookkeeping with our arrows here, deprotonate and we've
got our products so to take this a second step, this is a work up step that would be
performed in lab to quench the excess acid once the reaction is over so we do our epoxide
opening ,we'd get to an equilibrium of our product and our protonated ethanol and then
we would get rid of that protonated ethanol by our weak base, the weak base prevents any
side reactions from occurring with other functional groups that may be present and sodium bicarbonate
would return our methanol and then break up into CO2 in water, so that's what that second
step is about, is isolation of the neutral at the end, okay alright so anyway let's look
at arene oxides and the dreaded NIH shift and its not dreaded it is just it's a mechanistic
curiosity that we need to look at here and so let's remember that a functional group,
an arene oxide, has two double bonds and it's almost a benzene ring, in fact this NIH shift,
in a protic solvent or acidic conditions, will occur by a protonation of our epoxide
and so we're going to form, let's remember we've got a meso compound here, the epoxides
are on the same side of a ring surface there so they are formed by syn additions and so
we need to remember that, now we've got a protonation there and at this point we have
a stable cation that we can form and so we go ahead and open this, let's show it occurring
to this way since we did show on our final product carbon having that oxygen so it opens
up and we see a stable allylic cation form and so we've got resonance that's occurring
throughout this system, we can resonate that positive charge and stabilize it so then what
occurs if we draw in our hydrogen's that we have okay now we've got we started with them
being back away from us and now one still back away from us, the others planar because
we have an sp2 cation, we can close this to a double bond and push, with neighboring group
assistance and chimeric assistance, to push that hydride over and accomplish the shift
and so what we end up with in the end is a protonated ketone and our two hydrogen's now,
the stereochemistry no longer matters because we've got two hydrogens on the same carbon
and what we've got is an enone, but normally a ketone's more stable, but here we tautomerization,
we can take our solvent molecule and deprotonate, kick back up to an alcohol and we form an
aromatic phenol, which is the most stable of all, and this is practically irreversible
at this point because of aromatic stability that we have in the ring so we'll see a significant
portion of phenol conversion in these acidic conditions.
So what if we have basic conditions, not for arene oxides, but for epoxides in general
like we saw in the first couple of examples there, we've got reactions going to be preferred
at the less substituted position, the most easily accessible, the lowest transition state
energy for that attack and so if we take a nucleophile such as a cyanide ion, we can
attack the less substituted position, kick up our alcohol and form directly here, an
alkoxide with our nucleophile on the other side, on the less substituted side, the alcohol
ends up on the more substituted side and then we add a little bit of weak acid at the end
so it doesn't interfere with other functional groups, well ammonium chloride, acetic acid
just fine here as well, and what we'll get is a protonation of our alcohol or alkoxide
and we get our final product and so that workup step again helps us to isolate the final product
and takes care of the O- exchange, brings the pH of our system up so we don't have any
loss to the aqueous layer as we're extracting away our byproducts of the reaction, okay
so attack at the less substituted positions always preferred, in these conditions we can
make a trans diol and so we'll see hydroxide is able to attack the backside of the epoxide
kicking up in an SN2 reaction, we end up with in one case where we attack the left-hand
side we end up with this intermediate that's then protonated by our workup step, again
with our acetic acid, this time we're just showing a couple of ways to do this, with
the workup step a nice weak acid and we get our trans diol, the other enantiomer is from
attack at this carbon, now we can also attack with organocuprates and in a similar way with
the basic conditions attack the less substituted side, the cuprate mechanism is coupling, we
won't show that, we can consider it as if we're conceptually as we're attacking this
position and kicking up to that alkoxide and so if we do that, we get our carbon backbone,
but now we have to think about protonating that, there's no other functional group present,
let's use a strong acid this time, like HCl, it's going to proton transfer to hydronium
to a hydronium ion in a water solvent and so we'll show it as coming off of water here
and so we get that protonation, no other functional group present, why not use a strong acid in
the workup step here, but we can attack with an organocuprate to couple carbons onto an
epoxide forming an alcohol at the more substituted position.