hey it's professor Dave let's talk about exhaustive methylation.
so we're very familiar with elimination reactions and we know that to do an elimination we
need a good leaving group so very common leaving groups we've seen thus far are
things like halogens that sort of stuff but what if we have an amine what if we
have an amine where we want to remove an amino group so we haven't really seen a
technique yet that would allow us to do this but let's look at one now so we're
familiar with maybe some of the chemistry of amines and one thing that
we're able to do with amines is an alkylation such as a methylation so this
is pretty straightforward we can just react with methyl iodide and we're gonna
attack here just just a regular sn2 reaction so we're gonna grab that methyl
displace the iodide ion right so now this methyl is on there it's attached to
this nitrogen there the nitrogen has picked up another bond so it's got a
formal positive charge so we'll just lose a proton one of those two protons
is gonna is going to go and we're gonna be here we're gonna be left here and
we've gone from the primary amine to the secondary amine so that was a
methylation we performed a methylation but this this nitrogen still has a lone
pair right there's still a lone pair on this nitrogen and so there's no reason
why we can't go ahead and methylate again so let's do the same thing we'll
attack there and we'll kick this off so this time we'll take the deprotonation
as rendered we don't always have to draw every every proton transfer so we've
gone here to the tertiary amine right we've put another methyl group on there
but now there's still a lone pair right once we lose that last proton we've
still got that lone pair on nitrogen so let's methylate yet again right this
is why we call this exhaustive methylation because we're methylating
many times as many as many times as we can and so let's go ahead and get
another do another methylation get another methyl group there just place
the iodide now we've got the quaternary ammonium salt right iodide would be
acting as the counterion we didn't we didn't draw it in there but
so we've got we've got our quaternary situation there's no more lone pair on
nitrogen no more methylations can be done cannot
accommodate any more bonds that's the maximum we're gonna go so at this point
we react with a mixture of silver oxide and water and then don't forget we also
have the iodide ions right every time we do the methylation we're displacing an
iodide ion so they're our iodide ions in solution and what this mixture does is
it reacts to produce silver iodide and then the hydroxide ion so now by doing
that we've created a basic solution and this base we know that hydroxide can do
elimination reactions and so what ends up happening is because of the high the
high degree of steric hindrance associated with this group right here
we've got three methyl groups sticking out in different directions
it's actually going to prefer Hofmann elimination which is atypical
we're used to hydroxide giving us the Zaitsev elimination product but
here it's actually gonna go Hofmann and all we're gonna do is we're just going
to grab that proton from this side go right here and we're gonna kick off this
entire group so remember earlier we said we wanted to get rid of this amino group
this was not a good leaving group right if this left it would be NH2 -
extremely basic not not gonna leave very bad leaving group but what we've done is
by methylating it we've forced it to have a formal positive charge and that
makes it a great leaving group just the way a hydroxide is a bad leaving group
but then if you protonate it you've got a water molecule with a positive charge
on the oxygen then that's gonna be a great leaving group so the same thing
here so we're gonna kick that off and we get our Hofmann elimination product and
then we've got our our trimethylamine side product as well remember when this
gets kicked off we've got nitrogen connected to three to three methyl
groups and then we get that lone pair back as well so this is a Hofmann
elimination by exhaustive methylation of a means it is a great way to get rid of
an amino group on a substrate when you want to get rid of it for whatever
reason in your synthetic pathway and we just want to remember that when we
eliminate with the base when we eliminate this group
with the base we are gonna go Hofmann we're gonna get the more sterically
available proton and we're gonna generate the less substituted alkene
product which is the Hofmann product.