it's professor Dave let's talk about halohydrins
so one addition reaction that we've learned is dihalogenation, and we saw how we can do an anti
addition where we put two halogens like two bromines on to an alkene, and they
will add an anti fashion and so let's look at a slight modification of this
let's say we've got this alkene and now we're looking at this alkene edge-on,
just for clarity of stereochemistry and let's say that we've got bromine so
we've got bromine let's go ahead and do the first step just like it was
dihalogenation. so we know that this pi bond can interact with bromine in such a
way that one of these bromine atoms will now be on the molecule. we'll kick that
off there and we'll make this bromonium ion intermediate so this is no different
than dihalogenation, it's the same first step. now the key difference here
is that we're running this not in aprotic right non-polar aprotic
solution we're running this in water so if we're gonna run this in water the
second step is actually gonna go a little bit of a different way. so we know
that when we do dihalogenation what was gonna happen next is that the
bromide ion that we generated right here was gonna go ahead and do sn2, pop open
the bromonium ion and we were gonna get our dihalogenated product. but the
thing is if we're running this in in water, running this in aqueous
conditions, it is always the case that if a molecule of solvent is able to do
chemistry that is going to do chemistry far before any other any kind of reagent
in solution because there's so much more solvent so just kinetically speaking if
we have water floating around water is going to go ahead and react and do sn2
because this bromonium ion intermediate is quite reactive where this bromine
doesn't really like to have a plus charge so whatever's around in solution
if it can do something it's gonna do it, so let's have water go ahead and attack
water is gonna do sn2 just the way the bromide ion would have if we were in inert
solvent and so waters gonna pop that open. I've omitted the deprotonation step
we know that this would have OH2 plus, something would grab the proton we just
saved a step but we've got this here and this is a halohydrin right so what is
a halohydrin, well the name kind of speaks for itself we've got a halogen
and we've got a hydroxyl and they are on adjacent carbons so a halogen and a
hydroxyl when they're on adjacent carbons that is a functional group called a halo
hydrin, and so this is the way that we're gonna make a halohydrin. once again it
is pretty much the same thing as dihalogenation just pay attention that we
are running this in aqueous conditions and so before bromide that we have
generated has a chance to go and do the SN2 in fact water is gonna go and do
that first because there's just so much water lying around.
it's aqueous solvent right we've got a molecule of solvent it's gonna do it. so
before we check out some other things about halohydrin formation let's just
look at one thing that halohydrins can do. and so let's say we have this cyclic
halohydrin and we have the the halogen and the hydroxyl trans to one another on
the ring, so we've got the dash and the wedge. if we if we put this in aqueous
base, aqueous base, so hydroxide let's say what this can do is this can actually
grab this proton right here and then these electrons can go ahead and do an
SN2 and kick off that bromine and we're gonna get an epoxide. so we do want to
be aware that this is one thing that halohydrins can do. but the thing is
this, in order for this oxygen atom to behave as a nucleophile and displace
this bromine we must have these groups trans to one another okay so on a linear
molecule they're gonna be able to rotate and find the right orientation but on a
cyclic molecule like this, this must be the orientation and the reason is that
this is an SN2 reaction right this is an SN2 just like any other we know that SN2
proceeds via a backside attack and so this bromine projecting that way
hydroxyl up this way that is in the correct orientation for this oxygen atom
to be able to access the LUMO and go ahead and do the SN2. so we will get the
epoxide, on the other hand over here if we have the cis halohydrin so we
do have the bromine and the hydroxyl on the same side of the molecule, if we use
the exact same conditions we're not going to get the epoxide because the
thing is that the hydroxyl is sticking out this way it is not going to be able
to come from behind and attack this this carbon in the correct orientation and so
this is actually not going to react, will not get the epoxide so that's the
definition of a halohydrin that is the mechanism to get a halohydrin and one
thing that a halohydrin can do. before we move on there is actually one more
thing to discuss regarding halohydrin formation there's actually some
regiochemistry that we want to be aware of and so that's going to go like this.
we know if we have this molecule right here you've got this alkene let's say we
just do a regular dihalogenation remember this is going to be a regular
dihalogenation because of this inert solvent right CCl4 or something similar
is not going to interfere with the reaction right we're going to be able to
get our dihalogenated product, solvent molecules will not interfere and so we
are going to get actually a mixture right because we we know that it's going
to be anti addition so we know that the bromines are going to be trans to one
another on the product, so we could get this but we also could get the other
situation right we could have this methyl on the wedge now and we could
have this bromine going back and then this bromine on the wedge over here so
not to confuse that let's just put that right there so these are the two this is
we're going to get a mixture of these two stereoisomers.
it is an anti addition so the bromines are trans to one another in either case
so that's just a regular dihalogenation but now remember what we
said if we do this but we run this in aqueous conditions we're gonna get our
halohydrin but let's take a look let's take a look at one very important thing.
let's take a look at the bromonium intermediate now right if we're gonna if
we're gonna look at things mechanistically let's say we are looking
at this
so we have our bromonium ion intermediate and now this is not very
happy this is not a happy situation because that bromine has a positive
charge and so this is actually gonna behave a little bit similarly to the
case remember when we were looking at epoxides and we were running
reactions in acidic conditions we had a specific regiochemistry where any
incoming nucleophile was going to go to the more substituted carbon and the
reason is this in order to try to alleviate that positive charge we are
actually going to start to slightly break one of these bonds and the thing
is that the bond that is more able to dissipate slightly is going to be the
one that is attached to the more substituted carbon because the
developing the developing partial positive charge on that intermediate is
going to be more more stabilized by the additional alkyl group surrounding and
so it is the case that this bond is becoming slightly weaker than this bond
and that means that this carbon ends up being more electrophilic. there is a
stronger partial plus right here this is more partially positive than this carbon
and so for that reason for precisely that reason water is going to
preferentially attack that carbon. so we may not immediately have have guessed
that we may think that that any incoming nucleophile is going to go to the less
substituted site but this is an electronic reason this is not a steric
reason, we're looking at electronics here and we're saying that because
the enhanced partial positivity at this site, water is going to attack here
instead and that is what's going to happen there and so again this is very
similar to when we were looking at protonated epoxides we saw in that case
too and for the same reasons that a nucleophile was going to attack the more
substituted carbon to pop open that epoxide and so we're
gonna get it's pretty much the same as here right we can have we can or that's
actually going to get pushed up right that methyl will get pushed up and we've
got the hydroxyl there and this will remain as a wedge bond there and so this
again they're still going to be trans like any halohydrin but we want to make
sure that we understand the regio chemistry here, that the water is going
to attack specifically this carbon and not this one in order to pop open that
bromonium ion intermediate and so that's a little bit more information about halohydrin formation.