CBSE Class 12 > Class 12 Videos > Sn1 vs Sn2: Summary
Sn1 vs Sn2 Summary Video Lecture - Class 12
FAQs on Sn1 vs Sn2: Summary
| 1. What is the difference between Sn1 and Sn2 reactions? |  |
Ans. Sn1 and Sn2 are both types of nucleophilic substitution reactions. The main difference between them lies in the reaction mechanism. Sn1 reactions proceed through a two-step mechanism involving the formation of a carbocation intermediate, while Sn2 reactions occur via a single-step mechanism with simultaneous bond formation and bond breaking.
| 2. Which factors determine whether a reaction follows an Sn1 or Sn2 mechanism? | |
Ans. Several factors influence whether a reaction proceeds via an Sn1 or Sn2 mechanism. The nature of the substrate, solvent, nucleophile, and leaving group all play a role. Sn1 reactions are favored by tertiary substrates, polar protic solvents, and weak nucleophiles. On the other hand, Sn2 reactions are preferred for primary substrates, aprotic solvents, and strong nucleophiles.
| 3. Why are Sn1 reactions typically slower than Sn2 reactions? | |
Ans. Sn1 reactions are generally slower than Sn2 reactions due to the involvement of a carbocation intermediate. The formation of the carbocation requires the breaking of a strong bond, which is a slow step. In contrast, Sn2 reactions occur in a single step, where the nucleophile directly attacks the substrate, leading to faster reaction rates.
| 4. How does the stereochemistry differ in Sn1 and Sn2 reactions? | |
Ans. Sn1 reactions often result in racemization or the formation of a mixture of enantiomers. This is because the nucleophile can attack the carbocation from either side, leading to the formation of both R and S enantiomers. In contrast, Sn2 reactions proceed with inversion of stereochemistry, resulting in the formation of a single enantiomer.
| 5. Can both Sn1 and Sn2 reactions occur simultaneously in a reaction? | |
Ans. It is highly unlikely for Sn1 and Sn2 reactions to occur simultaneously in a reaction. The reaction conditions, such as the type of substrate, solvent, and nucleophile, strongly favor either an Sn1 or Sn2 mechanism. However, in some cases, a reaction may exhibit characteristics of both mechanisms, known as a hybrid mechanism, where the reaction proceeds through a combination of Sn1 and Sn2 steps.
Text Transcript from Video
So here's a table
that summarizes
SN1 versus SN2 reactions.
The first thing you
should do is to look
at the structure of
the alkyl halides.
If it's a primary
alkyl halide, it's
always going to proceed
via an SN2 mechanism.
SN2 mechanisms require decreased
steric hindrance in order
for the nucleophile
to function properly.
So primary alkyl halides are
the least sterically hindered.
A primary alkyl halide not
react via an SN1 mechanism,
because an SN1 mechanism
produces carbyl cations,
and a primary carbyl cation
is not very stable at all.
So for all these reasons,
primary alkyl halides
will be SN2.
Secondary alkyl halides
are a little bit harder
to figure out, because it
could be either SN2 or SN1.
So to break that tie,
you have to also look
at your nucleophile
and your solvent.
If you have a strong nucleophile
and a polar aprotic solvent,
in the last video, we saw
polar aprotic solvents
favor SN2 mechanisms.
If you have a relatively
weak nucleophile
and a polar protic
solvent, that's
going to favor an SN1
mechanism, because
the polar protic solvent will
stabilize the resulting carbyl
cation.
Tertiary alkyl
halides are always
going to be SN1, because
tertiary alkyl halides produce
tertiary carbyl cations,
which are very stable.
SN2 mechanisms
wouldn't really happen
for tertiary alkyl
halides, because there
would be too much
steric hindrance.
Too much stuff in the way.
All right, so let's
do a bunch of examples
and see how to use this little
chart that we've created here.
So I start with this
as my alkyl halides,
and I'm going to react that
alkyl halide with water.
I'm going to heat up
the reaction like that.
So the first thing you do--
structure of your alkyl halide.
This carbon is the one that's
bonded to that bromine.
It's connected to
one other carbon.
Therefore, this is a
primary alkyl halide.
Primary alkyl halides
are always SN2.
So this is going to proceed
via an SN2 mechanism, which
we know is a
concerted mechanism.
What is the nucleophile?
Well, water's the
nucleophile, in this reaction.
Water is also the
solvent, making
this a solvolysis reaction.
And that might confuse some
people, because they say,
oh, wait this is a
relatively weak nucleophile,
and it's polar protic.
So why is this proceeding
via an SN2 mechanism?
The structure of your
alkyl halide always wins.
So this is a polar protic
solvent, which does favor SN1,
but since this is a primary
alkyl halide, it's only
going to react via
an SN2 mechanism,
although this reaction
does occur rather slowly.
So we know the first
step is our nucleophile
is going to attack our
partially positive carbon here.
At the same time,
these electrons
kick off onto our leaving group.
So what will that give us?
This is all concerted
SN2 mechanism.
So the water molecule is
going to add the same time
that the bromine leaves.
So our water molecule has now
added to our original molecule
here.
And let's see, it's going
to have an extra lone pair
of electrons, giving it a
positive 1 formal charge,
like that.
So what else is going to happen?
Well, another water molecule
is going to come along,
and instead of functioning
as a nucleophile, this time
it's going to
function as a base.
It needs to get rid of that plus
1 formal charge on the oxygen.
So this lone pair of electrons
is going to take that proton.
These electrons are going to
kick in here on this oxygen.
And so we would end up with
ethanol as our product.
Two lone pairs of electrons
on that oxygen, and a hydrogen
here.
So we're lucky
that we don't have
to worry about stereochemistry
for this reaction.
There are no chirality
centers created.
So ethanol is the only
product for this reaction.
So the approach is first,
look at your alkyl halide.
Figure out what
type of mechanism
you're going to use, and
then go through the problem
with the mechanism,
thinking about possibilities
for stereochemistry.
And then you'll finally
end up with your product.
So there's a lot of work
involved with these.
But if you go through
these problems
with those steps in mind,
It'll make them much easier.
Let's look at another example.
So if we have this as our
alkyl halide-- so a bromine
right here.
Notice stereochemistry
is involved,
and we're going to react
that alkyl halide with sodium
cyanide, so Na plus CN minus.
And our solvent will be DMF.
So we look at the structure
of our alkyl halide first.
So this is the carbon that
connected to our halogen.
That carbon is connected
to two other carbons,
making this a
secondary alkyl halide.
We know that secondary
alkyl halides can proceed
via SN1 or SN2
mechanisms, so we need
to look at the nucleophile
and the solvent.
The nucleophile
is cyanide anion,
which happens to be a
relatively strong nucleophile.
Our solvent is DMF, and
we saw in the last video
that's a polar aprotic solvent.
So, polar aprotic solvent,
strong nucleophile,
is an SN2 mechanism.
So this is going to proceed
via an SN2 mechanism.
So that means as a
nucleophile attacks,
our halogen is going to leave.
So as this negatively
charged carbon
is going to attack this carbon,
that happens at the same time
that the bromine leaves.
And the bromine is on
one side of the molecule.
The cyanide has attacked
from the opposite side
of the molecule.
So the steric hindrance
part to an SN2 mechanism.
So when you draw
your product-- when
you show the cyanide
substituting for the halogens
there-- you're going
to show your cyanide
going away from you in space.
Because it has to attack from
the opposite side of where
the bromine is leaving.
So you get inversion
of configuration
via your SN2 mechanisms.
So this will be the only
product for this reaction.
Let's look at another one.
Let's actually start with
the same alkyl halide.
OK, so here's our
same alkyl halide.
This time, we're reacting
that with acetic acid.
So let's go ahead and draw
acetic acid in here like that.
All right, so once again, since
it's the same alkyl halide,
I know it's secondary.
So it's a secondary
alkyl halide.
This time, I have a
somewhat weak nucleophile.
My nucleophile would
be this oxygen here,
because it has lone pairs of
electrons on it like that.
But it doesn't really have
a negative 1 formal charge.
It's not really that
strong of a nucleophile.
It's also protic,
because this would
be the protic part
of that solvent.
So it's going to function as
our nucleophile and our solvent.
We go back up here.
We find polar protic solvent--
relatively weak nucleophile.
We'll proceed via
an SN1 mechanism.
We know the first step
in an SN1 mechanism
is formation of
our carbyl cation.
So these electrons
have to kick off
onto our bromine to
form our carbyl cation.
So this carbon
right here ends up
with a positive 1 formal charge.
That is a secondary
carbyl cation.
So secondary carbyl cations have
the possibility of rearranging,
but in this case,
there's nothing
that could happen
to rearrange to make
this a more stable
tertiary carbyl cation.
So it's going to stay as
a secondary carbyl cation.
But always think
about rearrangements
as being possible
via an SN1 mechanism.
So we have a secondary
carbyl cation.
This is where our nucleophile
is going to attack.
So, lone pair of electrons is
going to attack right here.
And I'm going to go ahead and
just jump to the products here.
So we're going to lose a
proton after that attacks,
and we're going to
get our products.
And we're going to
add on the oxygen.
And since this is
an SN1 mechanism,
the oxygen going to attack
from either above or below
this planar SP2 hybridized
carbyl cation here.
So you're going to get a mixture
of inversion of configuration
and retention of configuration,
as we saw in our SN1 video.
So when this oxygen
attacks, it could
attack from above or below.
If it attacks from
this side, the oxygen's
going to add on to here.
And then the rest the
molecule would look like that.
So I already took the
proton off in the mechanism,
just to save time.
So that's one possible product.
The other possible product--
instead of the oxygen
adding on as a wedge, it
could add on as a dash.
It attacks from
the opposite side
of your planar carbyl cation.
So these are your two
possible products.
So if I look at them, this one
over here-- this is a wedge.
This is retention of
absolute configuration,
and over here it's a dash.
This is inversion of
absolute configuration.
Remember, inversion is slightly
preferred over retention.
Now, these two happen to be
enantiomers of each other.
The absolute configuration
is different at this carbon.
So chirality centers are
important to think about.
If you're starting
with a chiral reactant,
and you end up with
a chiral product,
it's very important
to think about that.
Right up here, too,
for this one as well.
This is a chirality
center, so you
have to specify stereochemistry.
So stereochemistry, of
course, makes these problems
even harder than
they already are.
So secondary alkyl halides.
We did one example where it was
an SN2 mechanism with inversion
of configuration only,
and we did one example
where it was SN1,
in which case we
got both inversion and
the retention product
after our nucleophile attacked.
All right, let's do
one more example.
So let's look at this
alkyl halide like that.
So I'll put and iodine here.
We're going to react this alkyl
halide with methanol, so CH3OH.
Two lone pairs of electrons.
So what kind of an alkyl
halide is this guy?
So this carbon is
attached to 1, 2,
3 other carbons,
so it's tertiary.
So a tertiary alkyl halide.
I go back up to my
charts, and I say,
tertiary alkyl halides
are going to be SN1.
So I think to myself,
what is the SN1 mechanism?
The first thing that
happens is formation
of your carbyl cation, right?
So these electrons
kick off onto here,
forming a tertiary
carbyl cation,
which we know are very stable.
And the next step is
nucleophilic attack, right?
So lone pair of
electrons on the methanol
is going to attack right here.
And that, of course-- now
we have to show our oxygen
is now bonded to that carbon.
There's a hydrogen
on that oxygen.
There's also a methyl group.
There's still a lone
pair of electrons,
and that gives this oxygen
a plus 1 a formal charge.
So in the last step
with this mechanism,
another methanol
molecule comes along,
and this time
functions as a base.
So it's going to
take this proton,
and these electrons
in here are going
to kick off onto your oxygen,
and finally, we are done.
So we'll go ahead
and draw the product.
So oxygen and a methyl
group is your product here.
So this is an SN1 mechanism
with a stable tertiary carbyl
cation, and then there's
also an acid-base reaction
to finish this one off.
I look at my product,
and I think to myself,
do I have any chirality centers?
And I don't, for this reaction.
So that means I
don't have to worry
about specifying any
kind of stereochemistry.
This will be my only product.
So this is how to do SN1
versus SN2 reactions.
First look at the structure
of your alkyl halide,
and then think about, do I
need to look at my nucleophile
and solvent, and figure out
what type of mechanism you have.
And then you run through
your whole mechanism
and figure out about
stereochemistry,
and then you finally end
up with your product.
So there's a lot involved
with SN1 versus SN2 reactions.
If you follow these
steps, I think
it should make it a
lot easier for you.
that summarizes
SN1 versus SN2 reactions.
The first thing you
should do is to look
at the structure of
the alkyl halides.
If it's a primary
alkyl halide, it's
always going to proceed
via an SN2 mechanism.
SN2 mechanisms require decreased
steric hindrance in order
for the nucleophile
to function properly.
So primary alkyl halides are
the least sterically hindered.
A primary alkyl halide not
react via an SN1 mechanism,
because an SN1 mechanism
produces carbyl cations,
and a primary carbyl cation
is not very stable at all.
So for all these reasons,
primary alkyl halides
will be SN2.
Secondary alkyl halides
are a little bit harder
to figure out, because it
could be either SN2 or SN1.
So to break that tie,
you have to also look
at your nucleophile
and your solvent.
If you have a strong nucleophile
and a polar aprotic solvent,
in the last video, we saw
polar aprotic solvents
favor SN2 mechanisms.
If you have a relatively
weak nucleophile
and a polar protic
solvent, that's
going to favor an SN1
mechanism, because
the polar protic solvent will
stabilize the resulting carbyl
cation.
Tertiary alkyl
halides are always
going to be SN1, because
tertiary alkyl halides produce
tertiary carbyl cations,
which are very stable.
SN2 mechanisms
wouldn't really happen
for tertiary alkyl
halides, because there
would be too much
steric hindrance.
Too much stuff in the way.
All right, so let's
do a bunch of examples
and see how to use this little
chart that we've created here.
So I start with this
as my alkyl halides,
and I'm going to react that
alkyl halide with water.
I'm going to heat up
the reaction like that.
So the first thing you do--
structure of your alkyl halide.
This carbon is the one that's
bonded to that bromine.
It's connected to
one other carbon.
Therefore, this is a
primary alkyl halide.
Primary alkyl halides
are always SN2.
So this is going to proceed
via an SN2 mechanism, which
we know is a
concerted mechanism.
What is the nucleophile?
Well, water's the
nucleophile, in this reaction.
Water is also the
solvent, making
this a solvolysis reaction.
And that might confuse some
people, because they say,
oh, wait this is a
relatively weak nucleophile,
and it's polar protic.
So why is this proceeding
via an SN2 mechanism?
The structure of your
alkyl halide always wins.
So this is a polar protic
solvent, which does favor SN1,
but since this is a primary
alkyl halide, it's only
going to react via
an SN2 mechanism,
although this reaction
does occur rather slowly.
So we know the first
step is our nucleophile
is going to attack our
partially positive carbon here.
At the same time,
these electrons
kick off onto our leaving group.
So what will that give us?
This is all concerted
SN2 mechanism.
So the water molecule is
going to add the same time
that the bromine leaves.
So our water molecule has now
added to our original molecule
here.
And let's see, it's going
to have an extra lone pair
of electrons, giving it a
positive 1 formal charge,
like that.
So what else is going to happen?
Well, another water molecule
is going to come along,
and instead of functioning
as a nucleophile, this time
it's going to
function as a base.
It needs to get rid of that plus
1 formal charge on the oxygen.
So this lone pair of electrons
is going to take that proton.
These electrons are going to
kick in here on this oxygen.
And so we would end up with
ethanol as our product.
Two lone pairs of electrons
on that oxygen, and a hydrogen
here.
So we're lucky
that we don't have
to worry about stereochemistry
for this reaction.
There are no chirality
centers created.
So ethanol is the only
product for this reaction.
So the approach is first,
look at your alkyl halide.
Figure out what
type of mechanism
you're going to use, and
then go through the problem
with the mechanism,
thinking about possibilities
for stereochemistry.
And then you'll finally
end up with your product.
So there's a lot of work
involved with these.
But if you go through
these problems
with those steps in mind,
It'll make them much easier.
Let's look at another example.
So if we have this as our
alkyl halide-- so a bromine
right here.
Notice stereochemistry
is involved,
and we're going to react
that alkyl halide with sodium
cyanide, so Na plus CN minus.
And our solvent will be DMF.
So we look at the structure
of our alkyl halide first.
So this is the carbon that
connected to our halogen.
That carbon is connected
to two other carbons,
making this a
secondary alkyl halide.
We know that secondary
alkyl halides can proceed
via SN1 or SN2
mechanisms, so we need
to look at the nucleophile
and the solvent.
The nucleophile
is cyanide anion,
which happens to be a
relatively strong nucleophile.
Our solvent is DMF, and
we saw in the last video
that's a polar aprotic solvent.
So, polar aprotic solvent,
strong nucleophile,
is an SN2 mechanism.
So this is going to proceed
via an SN2 mechanism.
So that means as a
nucleophile attacks,
our halogen is going to leave.
So as this negatively
charged carbon
is going to attack this carbon,
that happens at the same time
that the bromine leaves.
And the bromine is on
one side of the molecule.
The cyanide has attacked
from the opposite side
of the molecule.
So the steric hindrance
part to an SN2 mechanism.
So when you draw
your product-- when
you show the cyanide
substituting for the halogens
there-- you're going
to show your cyanide
going away from you in space.
Because it has to attack from
the opposite side of where
the bromine is leaving.
So you get inversion
of configuration
via your SN2 mechanisms.
So this will be the only
product for this reaction.
Let's look at another one.
Let's actually start with
the same alkyl halide.
OK, so here's our
same alkyl halide.
This time, we're reacting
that with acetic acid.
So let's go ahead and draw
acetic acid in here like that.
All right, so once again, since
it's the same alkyl halide,
I know it's secondary.
So it's a secondary
alkyl halide.
This time, I have a
somewhat weak nucleophile.
My nucleophile would
be this oxygen here,
because it has lone pairs of
electrons on it like that.
But it doesn't really have
a negative 1 formal charge.
It's not really that
strong of a nucleophile.
It's also protic,
because this would
be the protic part
of that solvent.
So it's going to function as
our nucleophile and our solvent.
We go back up here.
We find polar protic solvent--
relatively weak nucleophile.
We'll proceed via
an SN1 mechanism.
We know the first step
in an SN1 mechanism
is formation of
our carbyl cation.
So these electrons
have to kick off
onto our bromine to
form our carbyl cation.
So this carbon
right here ends up
with a positive 1 formal charge.
That is a secondary
carbyl cation.
So secondary carbyl cations have
the possibility of rearranging,
but in this case,
there's nothing
that could happen
to rearrange to make
this a more stable
tertiary carbyl cation.
So it's going to stay as
a secondary carbyl cation.
But always think
about rearrangements
as being possible
via an SN1 mechanism.
So we have a secondary
carbyl cation.
This is where our nucleophile
is going to attack.
So, lone pair of electrons is
going to attack right here.
And I'm going to go ahead and
just jump to the products here.
So we're going to lose a
proton after that attacks,
and we're going to
get our products.
And we're going to
add on the oxygen.
And since this is
an SN1 mechanism,
the oxygen going to attack
from either above or below
this planar SP2 hybridized
carbyl cation here.
So you're going to get a mixture
of inversion of configuration
and retention of configuration,
as we saw in our SN1 video.
So when this oxygen
attacks, it could
attack from above or below.
If it attacks from
this side, the oxygen's
going to add on to here.
And then the rest the
molecule would look like that.
So I already took the
proton off in the mechanism,
just to save time.
So that's one possible product.
The other possible product--
instead of the oxygen
adding on as a wedge, it
could add on as a dash.
It attacks from
the opposite side
of your planar carbyl cation.
So these are your two
possible products.
So if I look at them, this one
over here-- this is a wedge.
This is retention of
absolute configuration,
and over here it's a dash.
This is inversion of
absolute configuration.
Remember, inversion is slightly
preferred over retention.
Now, these two happen to be
enantiomers of each other.
The absolute configuration
is different at this carbon.
So chirality centers are
important to think about.
If you're starting
with a chiral reactant,
and you end up with
a chiral product,
it's very important
to think about that.
Right up here, too,
for this one as well.
This is a chirality
center, so you
have to specify stereochemistry.
So stereochemistry, of
course, makes these problems
even harder than
they already are.
So secondary alkyl halides.
We did one example where it was
an SN2 mechanism with inversion
of configuration only,
and we did one example
where it was SN1,
in which case we
got both inversion and
the retention product
after our nucleophile attacked.
All right, let's do
one more example.
So let's look at this
alkyl halide like that.
So I'll put and iodine here.
We're going to react this alkyl
halide with methanol, so CH3OH.
Two lone pairs of electrons.
So what kind of an alkyl
halide is this guy?
So this carbon is
attached to 1, 2,
3 other carbons,
so it's tertiary.
So a tertiary alkyl halide.
I go back up to my
charts, and I say,
tertiary alkyl halides
are going to be SN1.
So I think to myself,
what is the SN1 mechanism?
The first thing that
happens is formation
of your carbyl cation, right?
So these electrons
kick off onto here,
forming a tertiary
carbyl cation,
which we know are very stable.
And the next step is
nucleophilic attack, right?
So lone pair of
electrons on the methanol
is going to attack right here.
And that, of course-- now
we have to show our oxygen
is now bonded to that carbon.
There's a hydrogen
on that oxygen.
There's also a methyl group.
There's still a lone
pair of electrons,
and that gives this oxygen
a plus 1 a formal charge.
So in the last step
with this mechanism,
another methanol
molecule comes along,
and this time
functions as a base.
So it's going to
take this proton,
and these electrons
in here are going
to kick off onto your oxygen,
and finally, we are done.
So we'll go ahead
and draw the product.
So oxygen and a methyl
group is your product here.
So this is an SN1 mechanism
with a stable tertiary carbyl
cation, and then there's
also an acid-base reaction
to finish this one off.
I look at my product,
and I think to myself,
do I have any chirality centers?
And I don't, for this reaction.
So that means I
don't have to worry
about specifying any
kind of stereochemistry.
This will be my only product.
So this is how to do SN1
versus SN2 reactions.
First look at the structure
of your alkyl halide,
and then think about, do I
need to look at my nucleophile
and solvent, and figure out
what type of mechanism you have.
And then you run through
your whole mechanism
and figure out about
stereochemistry,
and then you finally end
up with your product.
So there's a lot involved
with SN1 versus SN2 reactions.
If you follow these
steps, I think
it should make it a
lot easier for you.
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Apr 20, 2026 Last updated
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