hey guys, professor, Dave here. I know you're probably having a hard time with your SN2, SN1
E2, E1 mechanisms, especially deciding which one is going to happen
well don't worry, I'm here with some basic tips that will help you decide
when we're learning these first few mechanisms a
very common substrate is a simple haloalkane
so we want to understand that the
different substitutions that are possible, or degrees of substitution
of each haloalkane
are gonna tell us which mechanisms can or cannot happen
so for example we look at this tertiary haloalkane, we have to understand that
SN2 is not going to be possible because there is simply too much
steric hindrance for a nucleophile to approach that tertiary center
so steric hindrance means there's a lot of alkyl groups, that a lot of electron density
that is going to repel any incoming nucleophile that is negatively charged
so with a tertiary haloalkane we can rule out SN2
that's not going to occur, however SN1 is a very good candidate
because we know that the more substituted that a carbocation is
the more stable it is due to the hyperconjugation from the neighboring alkyl groups
so if this bromine were to leave it would leave a tertiary
carbocation which is stable so SN1 is a very good candidate
therefore E1 is also good candidate and E2
is also possible. so basically with tertiary me know no SN2.
with a primary haloalkane
SN2 now is very possible, there's not very much steric hindrance so it's going to
be very
possible for a nucleophile to approach
this carbon in this direction and so
SN2 is going to be possible, E2 is also possible
however SN1 and E1 are very unlikely to happen
because if this a bromine is to leave
that would leave a primary carbocation
a primary carbocation is very unstable
because it does not get to enjoy very much hyperconjugation from
neighboring alkyl groups
so because the primary carbocation is very unlikely to form
we can more or less rule out E1 and SN1
for a primary haloalkane there are some exceptions like an allylic system
but at this stage we can pretty much just take it for granted that a primary
carbocation is very unlikely to form.
with a secondary haloalkane there's not that much information that
we can get
if there's a lot of beta branching or some steric hindrance more towards the sides
that might be a factor in repelling a nucleophile that's going to do SN2
but basically with a secondary haloalkane
we can't really gather that much information about what mechanism is
going to occur
just by looking at the substrate, but for tertiary and primary
we are able to rule out some possibilities
the next thing we want to look at is the strength of the nucleophile
and leaving group so the first thing we want to understand is that nucleophilicity
parallels basicity, this means that there's a correlation between
how much a group is a able to coordinate to a proton
and how well group is able to coordinate to a molecule, so something that's a very
strong base
seeks to form a bond to a proton
but it could just as well act as a nucleophile and
and coordinate to a molecule so for example hydroxide
is a much stronger base
than water, it has a formal negative charge, it's a strong base. a water molecule
is not very basic, in fact it's neutral. so that means that if let's say we keep
it real simple, we're just trying to decide between
SN1 and SN2, we'll take a look at the eliminations later
if we look at this substrate and we look at what would happen whether hydroxide
or water were to attack
now the hydroxide is a very strong base so it's a very strong nucleophile
so it is capable of an SN2 reaction, so there's the SN2 product
however a water molecule is not strong enough to do SN2, it's not a strong nucleophile
so but what could happen is that if the bromine leaves, leaving
a carbocation intermediate
that is a very unstable situation
so water molecule would be a strong enough nucleophile
to coordinate to a carbocation and then deprotonate
so basically a water molecule
is capable of behaving as a nucleophile in an SN1 reaction.
recall that if we do an SN1 we're gonna get
a racemic mixture, so there's the two enantiomers
so that's the main takeaway, we want understand
that nucleophilicity parallels basicity. now that the halides are going to be
very common nucleophiles
as well as leaving groups so we want to understand how those behave
in a polar aprotic solvent, meaning the solvent does not have any
protons that are available for dipole-dipole interactions
it is going to be the case that the that fluorine or
I should say fluoride, let's make these all
the anions
the fluoride is going to be the strongest nucleophile because
it is the strongest base of the halide ions
but the reason for that is that
as you go down the periodic table, you go down the halogens
the ionic radius is getting larger as we go
and if we have a larger ion then that becomes more polarizable
so what that means is that its a larger volume with which that
ion can diffuse and spread apart, spread out
its negative charge so because the
negative charge on the fluoride is highly localized on a very small
ion, that makes it a very strong nucleophile
all the halide ions are good nucleophiles but in a polar aprotic
solvent it's going to be fluoride
that's the best. now if you're looking at a polar protic
solvent this trend is gonna reverse because
by the same token if the fluoride is much more basic
that means that it is gonna spend the most time in solvent interacting with any
available protons
let's say the protons on a water molecule, so we can see here
water is a polar protic solvent because these protons are available for
ion-dipole interactions, so this is an ion-dipole interaction
and there are several that the fluoride is able to participate in
because it has a very localized charge so because the fluoride ion is spending
much more time
interacting with molecules of solvent that is going to hinder its ability to act
as a nucleophile, whereas is it still will be able to
but in this case the iodide
is going to be the best because its charge is
more spread out, more diffuse, it's more polarizable
so it is going to be spending less time participating in interactions with molecules
of solvent which will make it
the best nucleophile of those.
so leaving group ability is basically the opposite side of the coin
so its precisely for the same reason
that hydroxide is a very very strong nucleophile
that's the same reason why it's a terrible leaving group so
hydroxide is very basic, it's unstable
it seeks to coordinate to something so it should follow that if it's such a
good nucleophile
then it's not really going to be in a rush to leave so
that's why a hydroxyl group is going to be a bad
leaving group because it's so basic. however if we tack on another proton
another proton makes that a water molecule
water is a very good leaving group because a water molecule
will be very stable once it has left so
that's why the halides as well are very good leaving groups
because bromide is perfectly happy on it's own, it's got a full octet
so that's not gonna be a problem and we can look at the same rules
with which we examined the nucleophilicity of the halides
to examine which ones are going to be the best leaving group
because we can see depending on whether it's a polar protic or aprotic solvent
we will know how well
it is going to interact with the protons in solvent
but those are all going to the pretty good leaving groups.
a third thing that we want to look at is the steric hindrance associated with the
nucleophile
so let's take a look at these, the hydroxide and some alkoxides
so these all have O minus as
they're trying to behave as a nucleophile, so we have roughly the same
basicity. there's a small
variation but it's all O minus so there's not much difference in the
basicity here so
technically the nucleophilic strength is roughly the same
however look at what's happening as we are
adding more and more carbons branching
this is increasing the steric hindrance in the nucleophile
and there is a point where a
nucleophile will no longer be able to approach a substrate
because of all the steric hindrance and so hydroxide being very small
is able to approach even a
secondary system with the beta branching, itdoesn't really matter
it's quite small
however a tert-butoxide is the classic
E2 promoter because it is simply to sterically hindered
to approach a substrate and substitute so
the more sterically hindered the base
the less likely it is to substitute and the more likely it is to eliminate.
the last thing we want to look at his temperature
so let's take a system like this, we've got a secondary haloalkane, we've got
hydroxide. hydroxide is very strong, you're probably looking at
SN2 or E2. we can look at temperature to choose which one is going to occur
so let's take two hypothetical temperatures, let's call
0 a cold temperature and 100 is gonna be a hot temperature
so it as it turns out cold temperatures
favor substitution so at 0 it's much more likely to get the SN2 product, and
hot temperatures favor
elimination so at 100 we're much more likely to get
one of the two possible elimination products. now the reason for this
lies in the Gibbs free energy equation. I hope we will remember this from gen-chem.
so Delta G equals Delta H minus T Delta S
so Delta G is the term that is Gibbs free energy and refers to the spontaneity
of any given reaction so a negative value for Delta G
means it's going to be a spontaneous reaction, positive will be non-spontaneous
Delta H is a change in enthalpy so that's basically energy, the energy
that's associated with the
with a change in the reaction and then T is temperature.
Delta S is a change in entropy or disorder so
one thing we need to understand is that an elimination reaction is more
entropically favorable
than a substitution reaction, and here's why. we have a substrate
and we have a nucleophile, we have the product and the leaving group
so that's two molecules or atoms generating
two things, so two things two things. however with an elimination
so we have a the same set up at the start here
but look what happens at the end, we have the product ,we have a water molecule
and a bromide, so two things become three things
more disorder, that's entropically favorable.
now the thing about the equation is that
the change in entropy is being multiplied
by temperature, so that means that at very high temperatures
the entropic favorability or unfavorability
becomes a greater factor in determining the spontaneity of the reaction
so let's say we have a very entropically favorable situation like an elimination
at a high temperature this whole T Delta S term becomes more significant
and it becomes more likely that Delta G is gonna end up having a negative value
so that's why at high temperatures elimination is going to be favored
it can be tricky to choose which mechanism is occurring but as long as we
look at the substrate and the nucleophile we ought to be able to eliminate
some options
so let's put it all together and look at some examples
if we're gonna look at something like this
the first thing we can do is look at the substrate and see if we can eliminate
any of the possible mechanisms
so this is a primary haloalkane so immediately we can get rid of SN1 and E1,
we're not going to form that primary carbocation and
SN1 and E1 both require a carbocation intermediate to occur
so we're gonna get rid of those options, we know it's going to be a SN2 or E2.
likewise if we look at the strength at the base, that's a strong base
a strong nucleophile, it doesn't have to wait for a carbocation to occur
it can bully around the substrate and make a reaction happen
so by that reasoning again we're looking at SN2
or E2, so we've just gotten rid of half of the options
and now we're down to SN2 and E2 just from looking at the substrate
and the base so between SN2 and E2
we might have to look at maybe a temperature. let's say we're doing this
at a hundred degrees
that is going to cause us to favor E2 because hot
temperatures favor elimination so number one
we're gonna go with that as our
E2 product. so number two
same setup right, but both by looking at the substrate and the base
both those things are telling us that we're looking at SN2 or E2 again
now looking at the cold temperature this time maybe we're gonna favor
SN2, so number 2
we're looking at that as our most likely product
SN2. so again we're looking at the substrate, we're looking at the strength of the base
and we're looking at a temperature, that's gonna allow us to
figure out what's going on. so let's look at this one now, we have a secondary
substrate so
with the secondary haloalkane we're not really gonna be able to figure out too
much about which reaction
or which mechanism is going to occur but we're looking at water
as the nucleophile so water is a weak base
so it is a poor nucleophile, that means we can get rid of SN2 and E2 as possibilities
because it's not a strong enough nucleophile to do either of those
reactions, either of those mechanisms, so we know we're gonna get
SN1 or E1.
between the two we can look at the temperature again
this time let's say zero so that means we're gonna favor
SN1 because colder temperatures favor substitution so
we know that we're gonna be looking at this
and that is our
SN1 product. okay so once again
we're just looking at the base and we're seeing that that is weak
that is a weak nucleophile and a weak nucleophile will not be able to do SN2
or E2 so we're getting rid of those possibilities
and then we look to the temperature. so how about this, we've got the same substrate
now we've got that tert-butoxide so if we remember from before
tert-butoxide is the classic E2 promoter. now the reason for this is
it is strong, it is an alkoxide, it
has a formal negative charge, so alkoxides are strong bases
so a strong base is a strong nucleophile
so it's going to be able to do SN2 or E2 it does not need to wait for a carbocation
it's gonna make something happen, it sees the substrate and it is
either gonna try to coordinate
or is gonna do an acid-base reaction. because it's a strong base
now the thing is that, remember before we were talking about steric hindrance
of the nucleophile, there is too much steric hindrance
associated with this nucleophile for it to coordinate
it cannot do SN2 so this
can only do E2 so
there is our E2 product
so remember, tert-butoxide
always gonna go E2. so now let's look at a tertiary
haloalkane. so with a tertiary we can remember
we're not going to be able to do SN2 so we're
we're getting rid of that possibility
the thing is hydroxide is strong, hydroxide is a strong base so it's a strong
nucleophile so it's able to do
SN2, E2, it doesn't need to wait for a carbocation
to form so SN1 and E1 aren't going to happen, that's just going to take too long
hydroxide finds a substrate in solution it's gonna bully it around, it's going to do something
but we just said it can't do SN2
so the only possibility is therefore E2, we don't even have to look to a temperature
we're seeing that this is gonna want to either substitute or eliminate
it cannot substitute because of the steric hindrance associated with the substrate
this time so that is going to give us the E2 product
so number five, we added a carbon this time so there's that
so there is our E2 product
incapable of doing SN2. now same substrate
now we're looking at methanol, so methanol
is weak, there is no formal charge there
it is comparable to water in terms of basicity, just has that
extra methyl group instead of a hydrogen but
it's completely covalent, there's no
its neutral, there's no negative charge so that is a weak
base so it's a poor nucleophile so it is not strong enough to you SN2
or E2 so it's gonna be SN1 or E1, we're going to look at the temperature
100 degrees, once again that favors elimination so
but it is weak so it's gonna do E1, so it has to wait
for that bromine to leave, once that bromine leaves and we have a carbocation
then those
beta protons become much much more acidic such that even a weak base like
methanol
is going to be able to extract one. so let's say number six
is our
E1 product
so here's a few examples and and we're able to see
how each mechanism could be promoted in certain situations depending
on the substitution of the substrate
strength of the base or nucleophile
and the temperature, so as long as you just think critically about
any example you're given you can always look to something like this
and begin to get rid of something, whatever's going on
and narrow it down to the answer.
thanks for watching, guys. subscribe to my channel for more tutorials
and as always feel free to email me with questions