Organic Synthesis:
Synthesis is the most creative thing we do in organic chemistry.
Nomenclature is a straightforward application of rules, most spectroscopic
interpretation, at this level, requires little in the way of judgment. Structure and
properties of compounds are matters of simple observation. Synthesis, though,
has some science and some art to it. There is rarely a ‘right’ answer. Most
synthetic problems have more than one solution, and the trick is to judge which of
these is likely to have the best chance of success. Even the most experienced
chemists develop routes, which work well on paper but fail miserably in the lab.
But, that having been said, there are some guidelines, which are helpful in
designing a synthesis. What follows is an attempt to lay out those guidelines with
a few examples.
The best approach to designing a synthesis of a molecule when we have a
choice of starting materials is to work our way back from the target to simpler
compounds. This is called retrosynthesis. We try to imagine what might have
been the last step in the synthesis. The starting material for this reaction becomes
our new target. What was the last step in its preparation? We go backwards like
this until we reach an acceptable starting material. Each of these backward steps
is called a transform, and for each transform, there is a corresponding reaction.
In the first part of the course, we don’t have very many reactions (and
thus, transforms) available to us. Let’s assume that we have only gotten to the
halogenation of alkanes, and the Corey-House reaction. We can therefore make
alkyl halides, and couple them together to make larger alkanes. Let’s try to make
some larger alkanes.
Say we were asked to make hexane from butane. How do we even start?
If you have been given a specific starting material and product, as in this case, try
to identify the carbon atoms of the starting material in the product.
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*
*
*
*
*
*
*
We can see that we can have two arrangements. In the first one, two carbons (an
ethyl group) have been attached to the end of the butane from which we started.
In the other one, two one carbon groups (methyl groups) have been attached, one
to each end. So we can have two approaches, one of which requires the formation
of two C-C bonds, one which requires installing only one such bond. Let’s look
at the one that requires the fewer bond-forming steps, on the grounds that we can
probably do it in a smaller number of steps.
How then can we make the target molecule? Look at what needs to be
accomplished. We have to attach a two-carbon fragment to a four-carbon
fragment. The only method available to us is the Corey-House reaction. Can we
carry it out? This reaction requires the use of two alkyl halides, as shown below.
CuCl
Li
2
2
Cl
Li
2
CuLi
Cl
Since we are required to start from butane, this doesn’t work. On the other hand,
we know we can make alkyl halides from alkanes, and we’ve just shown how to
make hexane from chlorobutane, so we can presumably get from butane to
hexane, not directly, but via chlorobutane. In reality, most synthetic problems are
like this. Intermediate compounds lie along the pathway between available
starting materials and targets.
Of course, this isn’t a synthesis you would be likely to undertake. First,
hexane is cheap and readily available. Second, chlorination of alkanes isn’t
completely selective. We need to consider the efficiency of each step along the
way in order to decide if a synthesis is reasonable. Consider the preparation of 2methylpentane from alkanes of four carbons or less. We can imagine the source
of the carbons as shown below
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*
*
*
*
*
(a)
*
(b)
*
*
*
*
(c)
Of the three options, the last involves forming 2 new C-C bonds and so is the least
attractive. Of the other two, which to choose? If we are limited to Corey-house and free
radical halogenation, as in the previous case, then we have the following two
retrosyntheses, and their corresponding syntheses.
Corey-House
X
halogenation
+
X
Br
1. Li
Br2
2. CuBr
Cl2
Cl
Cl
+
X
Cl2
Cl
+
+
X
Cl
1. Li
Cl2
Cl
2. CuBr
Now, which route is the better? Here we run into the problem of selectivity.
Halogenation of alkanes is a somewhat selective process. Hydrogens on more subsituted
carbons are more easily replaced than those on less substituted carbons. Chlorination is
less selective in this way than bromination. Thus, for the second route, chlorination of 2methylpropane yields both 2-chloro-2-methylpropane (which we don’t want) and chloro-
2-methylpropane (which we do want). The undesired product is produced in greater
yield, so we have a problem with this route. Now look at the first route. We take
advantage of bromine’s greater selectivity for the secondary position to make 2bromopropane. Chlorination gives us a mixture of products, the ratio of which can be
manipulated a bit by varying the temperature. At room temperature, the ratio is about 6:4
in favour of the undesired product. At higher temperatures, we can get more of the
desired chloropropane.
So, given the choice between these two routes, we can most likely get the better
yield of desired product using the first. We will find that, in general, branch points are
good paces to make disconnections. This follows the common observation that
disconnections that simplify more are often better than those which simplify less.
OH
O
H
C
C
C
C
OH
O
C
H
C
C
C
Often, as in this case, you have options regarding the fate of the starting
carbon atoms. Now, determine what has to happen to the starting material to get to the
product in each case. In the first route above, we see that the carbonyl group has to be
reduced to a hydroxyl, and then a methyl group has to be added to the same position. In
the second route, the hydroxyl has to be reduced, and a methyl group has to be added to
the other end of the chain. Which is preferred? This is where the art comes in. As a
general policy, it is best not to try to carry out specific reactions on unfunctionalized
positions. If you want a reaction to occur at a specific place, it is best if there is already a
functional group at that location. Furthermore, only a limited set of reactions will lead to
the desired outcome. So, at first glance, the best choice of the two is the first.
Now what? For a relatively simple transformation like this one, it is a good idea
to see if you can accomplish the synthesis in one step. In this case, is there a reaction,
which does the following:
1. Form a carbon-carbon bond.
2. Form that bond to a carbonyl carbon.
3. Result in the conversion of that carbonyl group to a secondary hydroxyl
group?
At this point, a good grasp of synthetic reactions is important. You need to be
able to look at that list of criteria and come up with a yes or no answer to the
question “Can I do this in one step?” To help with this process, find reactions,
which satisfy one criterion. Then eliminate those, which do not satisfy the next
requirement, and finally the last. In this case, it’s probably a good idea to look at
the rather limited number of C-C bond forming reactions that we know. These
could be:
1. Corey-House: makes C-C bond, needs alkyl halide, makes alkane
2. Grignard reaction with a carbonyl: makes C-C bond, requires a
carbonyl in the starting material, makes an alcohol.
3. Wittig reaction: makes C-C bond, requires a carbonyl, makes an
alkene (which could be converted to alcohol later)
4. Grignard reaction with epoxide: makes C-C bond, requires epoxide,
gives alcohol, adds 2 carbons to the chain.
Of these options, only 2 and 3 look promising, and, of them, 2 is simpler. So one
route to the desired product might be the following:
O- MgBr+
OH
H+, water
O
+
CH3MgBr
CH3
H
The methyl group has been written out in full to make it obvious. The same result
could be obtained using option 3, the Wittig reaction:
CH3
OH
O
H
+
acid
R3PCH2
water
In practice, the Grignard reagent is more convenient, and the synthesis is
accomplished all in one pot, the addition of acid and water occurring during the
workup. The second route requires isolation of the alkene, probably by
distillation (boiling point about 30o C). This last point illustrates that, once a
viable set of reactions has been found, it is still necessary to consider the
efficiency, convenience and cost of the process (at least in a real lab situation).
Now try the following conversions, which can be accomplished by essentially one
step processes (that is, without isolation of intermediates).
CH3
O
(a)
O
H
(b)
O
(c)
MgBr
OH
(d)
O
Most synthesis problems which we see in the lab are not quite so straight
forward as the one above. Usually, we have a target compound, but no specific
starting material. Instead, we try to come up with a synthesis from “readily
available materials”. What do we mean by that? It may simply mean, “what’s
around the lab”, but for our purposes, we’ll say that it means materials of about
four or fewer carbons. All isomers of most kind of molecules in this size range
are cheap and easy to find. This includes the alcohols, ethers, acids, alkyl halides,
aldehydes, ketones, alkenes and alkynes. We usually also include benzene and its
simple derivatives, such as toluene, phenol and benzoic acid.
For example, imagine we had to prepare 3-hexanol from alcohols of four carbons
or less. Here is a sample retrosynthesis, and the corresponding reaction sequence.
Retrosynthesis:
a
+
MgBr
O
OH
b
c
OH
Br
d
HO
Reaction sequence:
OH
OH
+
PBr3
PCC
b
Mg
d
Br
c
O
MgBr
a
1.
O
2.acid, water
OH
There are a number of things to point out here. First, notice that for each
transform (a, b, c, d) there is a corresponding reaction. The last reaction matches
with the first transform, the second last reaction with the second transform, and so
forth. Second, details like reaction conditions are not usually written into
transforms. In fact, many people don’t even put in the actual compounds. Rather,
they just have skeletons indicating the electrophilic and nucleophilic sites, and try
to infer reactions from there. Although that’s the current convention, for this
document, we’ll use explicit compounds, at least for a while.
How was that retrosynthesis done? Notice that the restriction on starting
materials requires that at least one C-C bond be made. It could be more than one,
but at least one is required. So, of all the reactions which are available to the
organic chemist, we can be sure that we need one of the relatively small group of
C-C bond forming reactions. Second, the product must contain an alcohol group.
That may have survived from the starting material, or been produced along the
way. One good tip is to try to find disconnections that simplify the molecule a lot,
especailly disconnections at branch points. In this case, a disconnection at the
carbon bearing the OH is a good idea. Now, what reactions form a C-C bond and
result in an alcohol at that point? The reaction of a Grignard with an aldehyde
will give a C-C bond and a secondary OH group. Other reactions might work,
too. A Wittig reaction on the aldehyde would make the bond, but leave you with
an alkene, which could be hydrated to the alcohol. This means an extra reaction
step, and it isn’t clear which carbon will get the OH group. A Corey-House
reaction will make the C-C bond, but leave you with an alkane. Look at each C-C
bond forming reaction and consider the consequences. Having chosen the
Grignard transform, we then have two new targets: the aldehyde and the Grignard.
The aldehyde can be made from an alcohol. The Grignard reagent requires an
alkyl halide, which then becomes a target. Fortunately, it is easy to make an alkyl
halide from an alcohol, and there we are at a set of acceptable starting materials.