Most chapters so far have run forwards: you had a starting material and a reagent, and you worked out the product. Synthesis runs the other way. You are given the product, called the target, and asked how to make it. There are far too many possible routes to find one by trying reactions and seeing where they lead.
So you work backwards. You ask which simpler molecules could make the target, then ask the same question of each of those, and stop when you reach molecules you can buy. This method is called retrosynthesis. You have already read an aldol product and a Diels–Alder ring backwards; this chapter turns that habit into a method.
The notation, and why it is different
Take butan-2-ol as a target. A Grignard reagent adds to an aldehyde to give a secondary alcohol (Grignard reagents). So ethylmagnesium bromide and ethanal would make butan-2-ol.
Written backwards, that plan reads: butan-2-ol ⇒ ethylmagnesium bromide + ethanal. The double-lined open arrow, ⇒, means “could be made from”. It is deliberately not the ordinary reaction arrow, because it points backwards in time, from a product to what it came from. You met it in The Diels–Alder reaction.
That backwards step broke one bond on paper: the bond from the ethyl group to the carbinol carbon, which is the carbon that carries the OH. Breaking a bond this way is called a disconnection, and it is marked with a squiggle across the bond. Draw the squiggle every time: many wrong answers cut a bond that the named reaction does not make. A disconnection is a thought, not a reaction. Nothing in a flask breaks the bond you crossed out; you are asking which two pieces, joined by a reaction you know, would give this bond.
There is a second move you can make backwards, and it cuts nothing. A functional group interconversion, or FGI, turns one group into another while the carbon skeleton stays the same: “this ketone could be made from this alkyne”, or “this amine could be made from this nitrile”. You write FGI above the ⇒ arrow.
An FGI makes the target no smaller, so it can feel like no progress. Its value is that the new group often reveals a bond you know how to make. Most real analyses alternate between the two moves: an FGI to expose a pattern, then a disconnection, then perhaps another FGI. The second worked example below runs exactly this way.
Synthons and their synthetic equivalents
Look again at the two pieces in the butan-2-ol figure. The ethyl piece carries a minus charge, because it is the one that attacks. The carbonyl carbon of ethanal carries a δ+, because it is the one that gets attacked. Idealized fragments like these, usually charged, are called synthons.
You cannot buy an ethyl anion. The reagent that behaves like one is the Grignard reagent CH₃CH₂MgBr, and it is called the synthetic equivalent of that synthon. The synthon is the idea; the synthetic equivalent is what you actually order. A carbonyl piece needs no stand-in: ethanal, with δ+ on its carbonyl carbon, is both the synthon and the compound you order. The table lists the common pairs.
| Synthon | Synthetic equivalent |
|---|---|
| R⁻ (carbanion) | RMgBr, RLi |
| R–C≡C⁻ | terminal alkyne + strong base (NaNH₂) |
| ⁻CN | NaCN |
| ⁻CH₂–CO–R (enolate) | the ketone + base |
| R⁺ (carbocation) | alkyl halide, or alcohol + acid |
| R–CO⁺ (acylium) | acyl chloride + AlCl₃ |
The first four synthons in the table attack, so they are nucleophiles. The last two get attacked, so they are electrophiles. A disconnection for a polar reaction always gives one of each, because in a polar reaction a nucleophile attacks an electrophile. The Diels–Alder, further down, is the exception: it joins two neutral molecules, so its pieces carry no charge.
Where to cut
Some disconnections are more useful than others. These are the ones to try first:
- Next to a functional group. The bonds next to a functional group are the ones you have reactions for. A C–C bond to a carbinol carbon suggests a Grignard. In a carbonyl compound, the carbon next to the C=O carbon is the α carbon and the one after it is the β carbon. When the β carbon carries an OH, the bond between the α and β carbons suggests an aldol.
- At a branch point. A branch is where two pieces were joined, and cutting there usually splits the target into two simpler pieces of similar size.
- Across a six-membered ring that contains a C=C. That is the Diels–Alder pattern, and one step turns the ring into two open-chain pieces.
- Through the middle of a symmetric target. The two halves are then the same, so you only have to plan one of them.
Now look at 4-hydroxy-4-methylpentan-2-one, drawn below. Its OH sits on the β carbon, two carbons along from the C=O carbon. An aldol makes exactly that arrangement, so the pattern tells you both which bond to cut and which reaction will make that bond.
A structural pattern that signals a particular disconnection is called a retron. Learning to see retrons is most of what makes an experienced chemist fast at this. Six are worth knowing now, and the figure after the list draws one example of each. To say which bond to cut, number the carbons, starting with the one that carries the functional group: the C=O carbon or the carbinol carbon is 1. These numbers are only for counting; they are not IUPAC locants.
- An alcohol is the retron for a Grignard. Cut a C–C bond to the carbinol carbon, the 1–2 bond. Pentan-3-ol comes from ethylmagnesium bromide and propanal.
- A β-hydroxy carbonyl, with the OH on carbon 3, is the retron for an aldol. Cut the 2–3 bond, between the α and β carbons. 3-Hydroxybutanal comes from two molecules of ethanal. A β-hydroxy carbonyl is also an alcohol, so a Grignard cut looks possible. But whichever C–C bond to the carbinol carbon you cut, one piece still carries the other C=O, and the Grignard reagent would attack it too. Make the aldol cut.
- An α,β-unsaturated carbonyl, with a C=C between carbons 2 and 3, is the retron for an aldol condensation: an aldol followed by loss of water. Cut the C=C. But-2-enal also comes from two molecules of ethanal.
- A 1,3-dicarbonyl, two C=O carbons with one carbon between them, is the retron for a Claisen. In a β-keto ester, number from the ester C=O carbon and cut the 2–3 bond, which joins the CH₂ to the ketone C=O carbon. Ethyl 3-oxobutanoate comes from two molecules of ethyl acetate.
- A 1,5-dicarbonyl, two C=O carbons numbered 1 and 5, is the retron for a Michael addition. Cut the 2–3 bond, between the α carbon of one carbonyl and the β carbon of the other. 3-Acetylheptane-2,6-dione comes from pentane-2,4-dione and methyl vinyl ketone. The donor is a 1,3-dicarbonyl, so its enolate forms easily and adds cleanly (Michael and Robinson).
- A cyclohexene is the retron for a Diels–Alder. Number the ring from the C=C, so the alkene carbons are 1 and 2, and cut the 3–4 and 5–6 bonds. Cyclohex-3-ene-1-carbaldehyde comes from buta-1,3-diene and propenal.
Target: 2-phenylbutan-2-ol, PhC(CH₃)(OH)CH₂CH₃. Its carbinol carbon carries a phenyl, a methyl and an ethyl group, as the figure after this example shows.
Spot the retron. It is an alcohol with C–C bonds to its carbinol carbon, which is the Grignard retron.
Disconnect. Cut the bond to the ethyl group (cut c in the figure). The synthons are CH₃CH₂⁻ and acetophenone, PhCOCH₃, whose carbonyl carbon is the δ+ end.
Name the equivalents. They are CH₃CH₂MgBr and acetophenone. Both can be bought, so the route is one step.
Check forwards. CH₃CH₂MgBr attacks the carbonyl carbon of the ketone, and aqueous workup protonates the alkoxide to give 2-phenylbutan-2-ol. ✓
The carbinol carbon has three C–C bonds, so there were three possible cuts: to the ethyl, to the methyl or to the phenyl. All three are sound Grignard disconnections. Prefer the one whose starting materials are easiest to get. Having several valid answers is normal in synthesis, and a question that asks for “a” synthesis rather than “the” synthesis is telling you so.
Target: hexan-3-one, CH₃CH₂COCH₂CH₂CH₃. Permitted: acetylene and bromoethane.
Count first. The target has six carbons, acetylene has two and bromoethane has two. So at least two new C–C bonds must be made, and one disconnection will not be enough. The first job is to find a move that shows where those bonds can go.
Retro step 1: change the group first. Hydrating an alkyne gives a ketone (Alkynes), so an FGI can trace the ketone back to an alkyne. Write hexan-3-one ⇒ hex-3-yne, with FGI over the arrow. No C–C bond has been cut yet. What has changed is that the new target has a skeleton you know how to build.
Retro step 2: now disconnect. Hex-3-yne, CH₃CH₂C≡CCH₂CH₃, is an internal alkyne, and an internal alkyne is the retron for acetylide alkylation. The negative carbon of an acetylide ion attacks a primary alkyl halide and forms a new C–C bond (Organolithiums and acetylides), so cut the C4–C5 bond, between an alkyne carbon and the ethyl group beside it. The synthons are CH₃CH₂C≡C⁻ and CH₃CH₂⁺. Their equivalents are but-1-yne with NaNH₂, and bromoethane.
Retro step 3: cut again. But-1-yne, CH₃CH₂C≡CH, is one cut from acetylene. Cutting its C2–C3 bond gives HC≡C⁻ and CH₃CH₂⁺, which means acetylene with NaNH₂, and bromoethane a second time. Every piece is now on the permitted list, so the analysis stops.
Write it forwards.
1. Acetylene, with NaNH₂ in liquid NH₃ and then CH₃CH₂Br → but-1-yne. The halide is primary, as the SN2 requires.
2. NaNH₂, then CH₃CH₂Br → hex-3-yne.
3. H₂O, H₂SO₄, HgSO₄ → hexan-3-one.
Check the last step. Hydration can put the oxygen on either alkyne carbon. Hex-3-yne is symmetric, so both ways give the same ketone. An unsymmetrical internal alkyne would give a mixture of two ketones, and the route would be worthless.
Notice the order of the moves. The first retro step changed a group instead of cutting a bond, and that is what made the C–C disconnections visible. Suppose you had cut the ketone directly instead, at the C4–C5 bond, between its α carbon and the next carbon out. The synthons would be an enolate of butan-2-one and CH₃CH₂⁺, which is an enolate alkylation. That route needs butan-2-one, which is not on the permitted list. Butan-2-one also has α hydrogens on both sides of its C=O, so the base would have to remove a proton from the CH₃ side only (Kinetic & thermodynamic enolates).
Knowing when to stop
You keep disconnecting until every piece is something you can reasonably buy: simple alkyl halides, small alcohols, benzene and its common derivatives, acetylene, and simple ketones and esters. A question usually names the permitted starting materials, and that list is the real definition of “done”.
The second rule is about length. Every step loses some material and adds one more chance to fail, and the yields multiply. Ten steps at 80% each return 0.8¹⁰ of the material, about 11%. So fewer steps is usually better, and a short route with one slightly awkward step usually beats a longer, more elegant one.
Step count is only a guide, so compare two routes by multiplying out their yields. Eight steps at 70% each return 0.7⁸, about 6%. That shorter route delivers less than the ten-step route above, because its steps are worse.
What carries forward
Retrosynthesis is a way of using everything else in the course, not a new set of facts. The next four sections supply what it draws on: the short list of ways to make a carbon–carbon bond, how to turn one functional group into another, what to do when a reagent cannot tell two groups apart, and how to order the steps so that each one survives the ones after it.