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Organic chemistry
Course: Organic chemistry > Unit 9
Lesson 6: Other reactions and synthesisReactions at the benzylic position
Free radical bromination, nucleophilic substitution, oxidation. Created by Jay.
Want to join the conversation?
- Do you have any videos on organometallics (excluding grinards) like the suzuki, stille, ring opening, cross coupling etc?(21 votes)
- At, I understand the losing a Hydrogen on the benzylic position can be resonance stabilized, while on any other carbon on the akyl group can't. But why not just losing a Hydrogen on the ring itself, as it can be resonance stabilized in the exact same way? 3:39(7 votes)
- I think the difference is that the ring has aromatic stabilization, while the benzylic position does not. This means that removing a benzylic hydrogen results in a smaller energy gain (and thus requires less energy) that removing a ring hydrogen, since the latter destroys the aromaticity of the ring.(2 votes)
- Why does the hydrogen turn into a radical at? 2:43(3 votes)
- This is a free radical reaction. NBS stands for N-bromosuccinimide. In the initiating step, the NBS loses the N-bromo atom. leaving behind a succinimidyl radical (S·):
NBS → S· + ·Br
It is the S· that removes the magenta hydrogen to form succinimide (SH):
S· + C6H5CH2CH2H3 → SH + C6H5CH(·)CH2CH3
Then C6H5CH(·)CH2CH3 + NBS → C6H5CHBrCH2CH3 + S·
and so on.(9 votes)
- Do you have any videos explaining the free radical substitution reaction with NBS on allylic positions?(4 votes)
- Jay says atthat substitutions at the benzylic position can be either SN1 or SN2 but the german Wikipedia article on nucleophilic Substitution states that SN2 reactions can never occur there. 4:20
On an Exercise i did with pretty much the same reagents fromI also thought that it would be SN2 but the solution said SN1. 5:35
So i guess Jay is wrong, but why can't SN2 occur here? Because even the primary carbokation is stabilized enough by the Benzene ring?(2 votes)- 1° benzylic halides typically react via an SN2 pathway,.
2° and 3° benzylic halides typically react via an SN1 pathway, via the resonance stabilised carbocation.(3 votes)
- Is there a way to add Chlorine to the benzylic position instead of bromine?(2 votes)
- Since Chlorine is a halogen it can be added in the same way. It would just have a different rate of reaction due to the difference in electronegativity.(2 votes)
- why doesn't and electrophilic aromatic substitution take place in this case ?(1 vote)
- Because you are using a nucleophile. Electrophilic substitution requires an electrophile.(2 votes)
- At aboutwhen substitution occurs at primary carbon via SN2, why it has to be the benzylic carbon? It doesn't seem like there's a radical that would need to be resonance stabilized, since the mechanism is concerted. 5:20(1 vote)
- Doesn't KMnO4-KOH also give benzoic acid as product on reaction with any type alkyl benzene?(1 vote)
- It does for all alkyl groups with a benzylic hydrogen. t-butyl groups do not react.(1 vote)
- Dear Khan Academy Community,
At, how would the reaction be affected if instead of Sodium Hydroxide we used something much more prominent like sunlight? 4:40
So, Benzyl bromide + Sunlight = What? A reaction does take place, right?(1 vote)
Video transcript
Reactions that occur at
the benzylic position are very important for
synthesis problems. So let's look at a few. We'll start with the
free radical bromination of alkyl benzenes. And so here is my alkyl
benzene, so a benzene ring, and I have an alkyl
group attached to that. So this is a carbon. And on that carbon, this
is the benzylic position. So this is will be a
benzylic hydrogen attached to that carbon, which is
necessary for this reaction. And then there are two other
things attached to that carbon here. So if we add some
NBS and some heat-- and this is a free
radical mechanism, so usually you add
something to initiate this, like peroxide--
you can see that we get a bromine added on
to the benzylic position in the place of
that hydrogen there. So let's look at
a quick reaction. So if I took
propylbenzene-- so I'm going to go ahead and
draw out propylbenzene-- so three carbons
coming off like that. And to it, I had some
NBS and some heat, and you could do this
in a solvent like carbon tetrachloride and
some peroxide here, add in some peroxide to
start your radical mechanism. And we have three carbons here. So there's one,
two, three carbons for a propylbenzene
on our alkyl group. But the only one that is
at the benzylic position is, of course, this
one right here. So that's where we're going
to end up adding our bromine. So when I draw the
product, I go ahead and I still have all
three of those carbons, but the bromine adds exclusively
to the benzylic position right here like that. And if I look at the
hydrogens that I have attached to this carbon--
so we started off with two hydrogens
attached to that carbon, and we'd left with only
one of those hydrogens. So let me just go
ahead and highlight the hydrogen in magenta here. So if we go back up here to
this general reaction, so this benzylic
hydrogen, I could say that it's either
one of these two. I'm going to go ahead and say
that it's this one right here. And so the other hydrogen
is attached to that carbon, and it is still left
for our final product. So the bromine took the place of
one of those benzylic hydrogens there. Now reason why the
reaction occurs only at the benzylic position is
because in this mechanism you form a radical. So let me just go ahead and
draw what the radical would look like for this
reaction here. So a radical for the
reaction of propylbenzene, so we have our benzene ring. And we already lost the
hydrogen in magenta, so now we're left with
just the hydrogen in blue here like that. And we're going to have
an unpaired electron right here on this carbon. And because this
radical, this electron, is right next to
this benzene ring, we can draw some
resonance structures. So I could take
this one electron and show it moving into here
with my fish-hook arrow. And then from this
pi bond, I could take one of those electrons
and move it into here. And the other one could
come off onto this carbon. So for a resonance structure
here, I can draw my ring, and I can show these
pi bonds over here. Now I can show a
double bond into here. There's still a hydrogen
attached at this point. And then I have my
other carbons like that. And then there is an
electron out on this carbon. So the benzylic radical
is resonance stabilized. And if you formed a radical
at the other two carbons on the rings-- not on the
ring, on the akyl group right here-- these are not
resonance stabilized, because they're not
right next to the ring. And so that's just a
little bit of insight as to why the reaction occurs
only at the benzylic position. It's because of resonance
stabilization of the benzylic radical. So let's look at another
type of reaction. And this, of course, would
be a substitution reaction of a benzylic halide. So I put a bromine,
here but you can imagine a different halogen. And if you add a
nucleophile in a solvent, depending on the classification
of your alkyl halide portion of the molecule and
what solvent you use, the mechanism could
be either SN1 or SN2. But of course, the end result
would be for your nucleophile to substitute in
for your halogen at the benzylic position. So if we were to look at
a reaction here-- so let me just go ahead and show one. So let's say we started
with benzyl bromide here. So this is our benzyl
bromide molecule. And to benzyl bromide, we
add some sodium hydroxide, so I have Na plus and OH minus. I'm going to go ahead
and draw my lone pairs on my hydroxide
anion, which is going to function as a nucleophile. So this is our nucleophile. And if I look at the
structure of this alkyl halide over here-- so this carbon
that's bonded to my bromine is bonded to one other
carbon, so that's primary. And if we're talking about
a primary alkyl halide, decreased steric hindrance, and
so your mechanism would be SN2. So if you think about an SN2
type mechanism for this, which is a concerted mechanism,
your nucleophile is going to attack this
carbon at the same time that these electrons kick
off onto your bromine to form the bromide anion as
a relatively stable leaving group. And so this is a
concerted mechanism. Everything happens at the same
time in an SN2 type mechanism. And we end up with an
OH replacing our Br. So we form benzyl alcohol
for this reaction. So that's SN2. If you thought about an SN1
type reaction, let's go ahead and think about that. If I started with a tertiary
alkyl halide-- so something like this-- so if I
were to analyze that, put my lone pairs of
electrons on this bromine, the carbon that is
attached to my bromine-- is attached to one, two,
and three other carbons. So therefore, it's tertiary. So you could think about
an SN1 type of mechanism. And we know in an
SN1 type mechanism-- let me go ahead and
just write that here. So in an SN1 type
mechanism, your first step is dissociation. So you're going to get these
electrons in here kicking off onto your halogen
onto your bromine to form your bromide anion. And you would, therefore,
be taking a bond away from this carbon here
at the benzylic position. So let me go ahead and draw
the resulting carbocation. So I have my ring. I have these two methyl groups
coming off of that carbon. And that is where I have
a plus 1 formal charge. So I have a benzylic
carbocation here, which is resonance stabilized. So I could think about these
electrons moving into here, and I could go ahead and
draw a resonance structure. So for this resonance
structure, these pi electrons are over here. Now, there's a double bond
between those two carbons. And let me show
those electrons here. So the electrons in blue
have moved out to here to form a pi bond, took a
bond away from this carbon. So that carbon gets our
plus 1 formal charge. And you could, of
course, continue. But the point is that the
presence of our benzene ring allows for resonance
stabilization of our benzylic
carbocation, which means that the
substitution is going to occur at the
benzylic position, because of that resonance
stabilization here. So let's look at
one more reaction. And this is oxidation
of alkyl benzenes. And so here, I have
my alkyl benzene. So once again, carbon with a
benzylic hydrogen and then two other things attached
to my carbon here. If I added some sodium
dichromate source of protons, so something like sulfuric
acid, and I heat up my reaction, I can oxidize my
alkyl side chain to a carboxylic acid
functional group over here. So this would, of course, be
the benzoic acid molecule. You could also do this
oxidation with something like permanganate and
some heat as well. So that's another possibility. Let's go ahead
and do an example. Let's start with-- let's
do butylbenzene this time. So if I have butylbenzene,
so four carbons, for my alkyl group here. And instead of rewriting
all that stuff, I'm just going to put
some ditto marks here. So we add some sodium
dichromate source of proton, sulfuric
acid, and some heat. We're going to oxidize
our side chain. And it doesn't really
matter the length of the alkyl group you have. In this case we
have four carbons. We're still going to
get a carboxylic acid. The reaction is still going to
occur at the benzylic position. And we get a
carboxylic acid here. Again, a complicated mechanism,
so I won't go into it. But again the length of
carbons that you have, it doesn't matter. So you could add
on a bunch more, and you would still get
this as your products. What about if you try to do
this reaction with something like tert-Butylbenzene. So instead of
butylbenzene, now we have tert-Butylbenzene
like that. And we went ahead and added
sodium dichromate and sulfuric acid and heat again. We heated it up. We wouldn't get any
kind of reaction. So no reaction this time. And that's because we don't have
any benzylic hydrogens present. So we back here to
our generic reaction. There's a benzylic hydrogen. If I go down to here,
here's a methyl group, here's a methyl group,
here's a methyl group. And then this is a carbon. So we don't have any
benzylic hydrogens. There are no hydrogens
attached to this carbon here, which is necessary
for the reaction to occur. And so we see no
reaction in that case. So just keep these
reactions in mind when you're doing synthesis
problems, because you will see that they
are very useful.