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Organic chemistry
Course: Organic chemistry > Unit 9
Lesson 6: Other reactions and synthesisSynthesis of substituted benzene rings II
Synthesis practice problems. Created by Jay.
Want to join the conversation?
- I want to practice more retrosynthesis problems. Where can I go?(8 votes)
- I think making some synthesis problems yourself is almost as good as doing them. If you think to yourself, "I'm going to start with benzene and just keep reacting it until I can't anymore," you'll get a lot of practice thinking about how reactions can be applied to a molecule, and you'll also be thinking about these problems from a test maker's point of view, which can be helpful when you're taking tests yourself.(1 vote)
- if u try to reduce ur NO2 group, wouldnt it reduce the carboxilic acid 2? (reduce the bonds to the oxygens.) 1:36(5 votes)
- I think it's because carboxylic acids need really strong reducing agents (like LiAlH4) to reduce and iron with HCl isn't strong enough. Because Al is obviously less electronegative than Cl and lets go of its hydride more easily ('Course Cl+ is REALLY unstable)(2 votes)
- what does NBS stand for?(2 votes)
- atwhy can't the carboxylic acid be added through one step in a friedal-crafts? 3:11(3 votes)
- I think chloroformic acid is too unstable to be used in a Friedel-Crafts acylation, and there might be unfavorable side reactions from the carboxylic acid group. It's a lot safer to do the Friedel-Crafts alkylation (or acylation followed by reduction) then oxidize the alkyl group.(1 vote)
- Atwhy does he call Benzoic Acid as Benzilic/Benzylic Acid? 1:52(2 votes)
- He is saying Benzoic acid, it's just a bit difficult to hear.(2 votes)
- At, Could you explain the mechanism of this reaction? How do i turn this alkyl group into a carboxylic acid? 3:00(2 votes)
- So toluene (methylbenzene) can be oxidised straight to benzoic acid by using KMnO4 as well which is a more widely used method, if you want the mechanism for that see this:
http://iosrjournals.org/iosr-jac/papers/vol7-issue6/Version-1/D07611627.pdf
it's pretty involved though, so usually just better to remember that transformation!(2 votes)
- At :- Can i use Fridel Craft CH3CH2-Cl / AlCl3 to form Ar-CH2-CH3, and the continue with NBS to add Br on the Benzyl position, think that (CH-Br-Ch3) is meta directing because it will give the C (-)partial charge? And then finish with Br2/AlBr3. Or am i wrong? 4:59(2 votes)
- For the second synthesis why cannot we do a friedel craft alkylation then a NBS, hv, CCl4 to add the Br then do a bromination?(2 votes)
- Alkyl groups are ortho/para-directors, while carbonyl groups are meta-directors. If you added the alkyl group, you'd brominate in the ortho- and para-positions when you added bromine and ferric bromide.(1 vote)
- what does
I)NaHg in excess HCl
2)acid anhydride (CF3CO3H)
do to substituted benzene.(2 votes) - If the benzylic bromine substituent had been added to the benzene first, what kind of director would it be? I was thinking meta because the tri-fluoromethyl group in a previous video was described as a strong deactivator and meta directing. Could that have been done first and meta directed the Br?(2 votes)
- The trifluoromethyl group has three of the most electronegative atoms in the Periodic Table on the benzylic carbon. It is a powerful deactivator and meta director.
The 1-bromoethyl group has a mildly electron-withdrawing Br atom and a mildly electron donating methyl group on the benzylic carbon.
The group is probably a mild deactivator and meta director.(1 vote)
Video transcript
Once again, our goal
is to synthesize simple organic molecules
starting from benzene. And this time, our goal is to
make this molecule over here on the right. Once again, we're going to use
the concept of retro synthesis, so working backwards
and thinking about what could be an immediate
precursor to this molecule here. To do that, let's
analyze the groups that are attached to our ring. Right here we have
an amino group. And we know that's an
ortho/para director. We know that because of
the lone pair of electrons on this nitrogen right
next to our ring. And then over here, we have
a carboxylic acid functional group attached to our ring. And we know this carbonyl carbon
here is partially positive, which makes this
a meta director. Now, we haven't
covered any reactions that would install an amino
group meta to a carboxylic acid group. But we have talked about
how to put a nitro group meta to a carboxylic acid group. So let's turn that amino
group into a nitro group. So let's go ahead and draw
our benzene ring here. And we'll go ahead and put
our carboxylic acid down here. And instead of an amino group,
let's draw a nitro group, like that. So all we need to do
now is think about a way to reduce our nitro
group to our amino group. And there are many
ways to do it. One thing you could
use would be a metal like iron and some
hydrochloric acid. You could have used tin. You could have also used
hydrogen and a metal Catalyst like nickel. So there are many ways
to reduce a nitro group to an amino group
on a benzene ring. And now I know how to put
that nitro group on meta to my carboxylic acid. So we have a meta
director right here. And so if I take
that nitro group off, I can come up with the precursor
to this molecule, which should be just benzoic acid. So I'll go ahead and
draw benzoic acid here. And I have to think
about the reagents necessary for a nitration
reaction, which of course would be concentrated nitric acid
and concentrated sulfuric acid. And so once again,
because my carbonyl here on my carboxylic acid
is meta directing, the nitro group ends up
in the meta position. So now I have been benzoic acid. And I need to make a
benzoic acid from benzene. And if you think about
one of the reactions we did in the benzylic
position video, we were able to
create a benzoic acid molecule from an alkylbenzene. And so the immediate
precursor to this molecule must be some sort
of alkylbenzene, so a benzene ring with
an alkyl group on it. So just to make
things easy, I'll just make it a methyl group
here, so a methyl group like that, so toluene as
our immediate precursor to benzoic acid. To oxidize the
alkyl side chain, I need to add something like
sodium dichromate, so Na2Cr2O7. A source of protons, so
something like sulfuric acid, and some heat. And so this is oxidation of
an alkyl benzene, like that. You could have also
done this reaction with permanganate
and heat as well. And you could have chosen
any number of carbons on your alkyl side chain. So I just chose one
to make it easy. So this carbon, when
this group gets oxidized, would turn into that
carbon right here. Now we would need to convert
benzene into toluene. And so we could do that using
a Friedel-Crafts alkylation reaction. And so we need one carbon. So for our alkyl
chloride, we would need a one-carbon alkyl
chloride, so CH3 Cl. And our catalyst for
Friedel-Crafts alkylation would be aluminum
chloride, like that. So our synthesis is complete. Let's go ahead and run through
the reactions very quickly. And let's make sure that
everything makes sense. So we start with benzene. And we do a
Fridel-Crafts alkylation to put a methyl group onto our
benzene ring to form toluene. Next, we oxidized that
alkyl side chain with sodium dichromate to convert it
into a carboxylic acid, which is now a meta director. So when you nitrate your
ring, the nitro group gets put on in a position
meta to your carboxylic acid. Finally, you reduce
your nitro group to turn it into an amino
group, and you are done. So let's do one more
synthesis problem here. And so we can see
that, this time, our target molecule is
over here on the right. And immediately, I see a bromine
at the benzylic position, the carbon right next
to our benzene ring. And so I know a
reaction that will put a bromine in the
benzylic position. If you remember, that's the free
radical bromination reaction. And so I can go ahead
and draw the precursor to that molecule, which
would be a benzene ring. This bromine over
here is on our ring. And we would have just an
ethyl group now, like that. And so I would need
some NBS, some heat. I could use carbon tetrachloride
as a solvent, a peroxide to initiate the free
radical mechanism. And this is a free radical
bromination reaction, which puts a bromine at
the benzylic position. All right, trying to figure out
a precursor to this molecule. Let's analyze what sort of
groups we have on our ring. So this ethyl group
is an alkyl group. And remember, that's
an ortho/para director. This bromine here, of course,
is also an ortho/para director because of lone pairs
of electrons on it. So this is an
ortho/para director. We have two ortho/para
directors on our ring, but those ortho/para directors
are meta to each other. So we need to think about a
way to turn one of ortho/para directors into a meta director. And a good way of
doing that would be to turn our alkyl
group into a group that has a carbonyl on
it, so an acyl group. So thinking about the
precursor molecule here, I draw my benzene ring. I'm going to leave
my bromine here. And instead of an
alkyl group, we're going to have an acyl group
now, because our acyl group is a meta director. So this carbon right here
is partially positive. So we need to think
about how to reduce an acyl group to an alkyl group. And one way of
doing that would be to do a Clemmensen reduction. So you take amalgamated zinc,
so we take some zinc here, so an amalgam with
mercury, and also some HCL. And so that will reduce our
acyl group to and alkyl group, like that. Now that we have an
acyl group on there, which is a meta director,
that would allow us to install the bromine meta
to that acyl group. So when I think about the
precursor to this molecule, I just take off that bromine. And so I have my ring with
an acyl group on that ring. And once again, since this
is a meta director here, all I need to do is a
bromination reaction. So once again, I would
need some bromine. And I would need our
catalyst, so FeBr3. And that reaction will work. Finally, I have to go from
benzene to this molecule. And once again, I can do that
with a Friedel-Crafts acylation reaction. I would need two carbons to
do my Friedel-Crafts acylation reaction. So from my acyl
chloride, I need to make sure I have two carbons. And once again,
my catalyst would be something like
aluminum chloride for this Friedel-Crafts
acylation. And so this would be
the complete synthesis. Let's once again run
through each step starting with benzene. We do a Friedel-Crafts acylation
to install an acyl group on our ring, like that,
which is a meta director. So if we follow that
with a bromination, the bromine is installed
meta to our acyl group. Our acyl group is then
reduced, using a zinc amalgam, and converted into an
ethyl group right here. And then finally, we do a
free radical bromination to put a bromine at
the benzylic position. And we are done. So think retro synthesis. And then also
double-check and make sure your synthesis makes sense.