Take benzene and two reactions you already know, nitration and bromination. Nitrate first and then brominate, and the product is 1-bromo-3-nitrobenzene. Brominate first and then nitrate, and the product is 1-bromo-4-nitrobenzene, a constitutional isomer of the first. The reagents are the same both times; only the order changed.
By this point in the chapter you have every other piece of a synthesis. Thinking backwards finds the bonds a target needs. Making carbon–carbon bonds lists the ten or so reactions that build a skeleton. Functional group interconversion turns one group into another, and Protecting groups showed how to hide a group that would get in the way. What remains is the order of the steps, and that is where most routes that look right on paper fail.
Plan backwards, then write forwards
Suppose your target is a tertiary alcohol, and your backward analysis cuts it into a ketone and a Grignard reagent. Suppose, too, that the ketone carries an OH group at its other end. On the retrosynthetic tree, the branching diagram of disconnections you drew while working backwards, every arrow looks legitimate. Write the steps forward, one after another, and the problem shows at once: the Grignard reagent meets the OH before it reaches the ketone.
That is why retrosynthesis and the forward route are two separate jobs. Retrosynthesis is a search: you try disconnections, and dead ends are expected. The forward route is a claim: every step has its reagents and its product, and you have checked that each product survives the step after it.
So finish by writing the forward route out in full, even when the backward analysis felt convincing. A mistake in order is invisible in a retrosynthetic tree and obvious in a forward list.
Four questions to ask of every step
Ask these four questions of each step in the forward route. Each one catches a different kind of mistake.
- Does the reagent have anything else to attack? Look at every functional group in the molecule, not only the one you are aiming at. LiAlH₄ meant for a ketone will also reduce an ester elsewhere in the same molecule. This is the question that finds a missing protecting group.
- Does the product of this step survive the next one? Say one step makes an aldehyde with a hydrogen on its α carbon, and the next step runs in aqueous base. The aldehyde reacts with itself in an aldol reaction, so it is gone before the next step can use it.
- Is the regiochemistry controlled? Regiochemistry is which of two possible positions reacts. Wherever two positions could react, name the reagent that chooses between them. Water adding to propene is an example: acid-catalyzed hydration puts the OH on C2, and hydroboration–oxidation puts it on C1.
- Is the stereochemistry controlled? If the target has defined stereochemistry, some step has to set it, and you should be able to say which one. The section on stereochemistry below shows how to find it.
Ordering on an aromatic ring
Go back to the two routes at the top of this page. In route A, benzene is nitrated and then brominated. In route B, the same two steps run the other way round.
Whichever group goes on first is already on the ring when the second electrophile arrives, and it decides where that electrophile lands. NO₂ directs meta, so route A gives the meta product. Br directs ortho and para, so route B gives mostly the para product. Ortho/meta/para directing effects worked through this same pair.
When a question asks for a benzene ring with two groups on it, choosing the order is usually the whole question.
A group already on the ring can also stop a later step from running at all. Friedel–Crafts alkylation and acylation fail on a ring that carries a meta-directing deactivator: NO₂, C≡N, a carbonyl group, SO₃H or CF₃. They also fail on a ring that carries an NH₂ group, for a different reason. The nitrogen’s lone pair binds the Lewis acid, AlCl₃. The nitrogen then carries a positive charge, which deactivates the ring, so adding more catalyst does not help. A Friedel–Crafts step therefore goes early, before any of these groups is on the ring.
The same thinking decides when to turn a nitro group into an amine. An NO₂ group comes through nitration, halogenation and sulfonation unchanged, and it directs every later electrophile meta. An NH₂ group causes trouble in each of those steps. It activates the ring so strongly that bromination does not stop at one bromine. Strong acid protonates it to –NH₃⁺, which directs meta instead of ortho, para. And, as above, it binds the Lewis acid.
So carry the nitrogen through the ring chemistry as NO₂, and reduce it to NH₂ at the end. The reduction uses H₂ over a metal catalyst, or a metal in acid such as Fe/HCl or Sn/HCl. H₂ over a catalyst also reduces any C=C or C≡C in the molecule, while a metal in acid leaves them alone.
One more aromatic rule applies when you attach a carbon chain. Friedel–Crafts alkylation goes through a carbocation, and a carbocation can rearrange. Acylation goes through an acylium ion, which is resonance-stabilized and does not. Electrophilic aromatic substitution showed the result with 1-chloropropane, and 1-bromopropane behaves the same way: with AlCl₃, benzene gives mostly isopropylbenzene. To attach a straight chain, acylate first. Then remove the C=O with a Clemmensen reduction (Zn(Hg), HCl) or a Wolff–Kishner reduction (H₂NNH₂, KOH, heat).
Target: 1-bromo-4-propylbenzene. From: benzene and any reagent with three carbons or fewer.
Count and disconnect. The target has nine carbons: six in the ring and three in the chain, with a bromine para to the chain. Two bonds have to be made, ring to chain and ring to bromine.
Choose the chain reaction. An alkyl group directs ortho, para, and so does bromine, so either order puts the two groups para to each other. The order is decided by something else. Alkylation with 1-bromopropane would put on an isopropyl group, as the paragraph above showed, so the chain has to go on by acylation.
Route.
1. Propanoyl chloride, AlCl₃ → propiophenone. The three-carbon chain is now joined to the ring through its C=O carbon.
2. Zn(Hg), HCl (Clemmensen) → propylbenzene. The C=O becomes CH₂, and the straight chain survives.
3. Br₂, FeBr₃ → 1-bromo-4-propylbenzene. Propyl directs ortho, para, and the para product dominates because the chain crowds the ortho positions.
Why not brominate first? That order works too, but it is slower. Bromine deactivates the ring, so the acylation that follows it runs more slowly. The bulky acylium ion still goes mostly para, so the product is the same; only the rate suffers. Acylating the unsubstituted ring first is the better order, and that reason is what an exam question on this route usually asks for.
Ordering around a Grignard reagent
Protecting groups opened with a Grignard reagent and an alcohol in the same molecule. The Grignard carbon takes the O–H proton faster than it adds to any carbonyl, and the reagent is destroyed. An N–H, an S–H or a CO₂H does the same. Adding a Grignard reagent to a molecule that contains one of these is the most common error in student routes.
There are two ways out for an alcohol. You can protect it as a silyl ether, or you can order the route so that the O–H is not yet there when the Grignard reagent arrives. For an N–H, an S–H or a CO₂H, this course has no mask that a Grignard tolerates, so reordering is the way out. Reordering adds no steps, so try it first. The same idea covers any sensitive group: if a group cannot survive later conditions, install it after them.
A second ordering rule concerns SN2 steps. Cyanide and acetylide ions make C–C bonds by SN2, and an SN2 slows sharply as the carbon gets more crowded. It fails on a tertiary carbon, and an acetylide, which is also a strong base, gives mostly elimination on a secondary one. So if a route both branches a carbon and runs an SN2 on it, run the SN2 first, while the carbon is still primary.
Target: 2-methylhexan-2-ol, CH₃CH₂CH₂CH₂C(CH₃)₂OH. From: butan-1-ol and any reagent with two carbons or fewer.
Count. The target has seven carbons and butan-1-ol has four, so three carbons must be added. The target is a tertiary alcohol, and a tertiary alcohol is the Grignard retron, the pattern that points to a Grignard disconnection next to the OH carbon.
Disconnect. Two of the three groups on the carbinol carbon, the carbon that carries the OH, are methyls. Cut both at once. That leaves two CH₃⁻ synthons, supplied by CH₃MgBr, and the acyl piece, supplied by pentanoyl chloride. A Grignard reagent adds twice to an acid chloride or an ester, giving a tertiary alcohol with two identical groups. That double addition is usually a nuisance. Here the two groups really are identical, so it is exactly the reaction you want.
Close the gap. Pentanoyl chloride has five carbons, and butan-1-ol has four. So the route must add one carbon, reach the carboxylic acid oxidation level, and then add the two methyls. Making a five-carbon acid from butan-1-ol is the example worked in Making carbon–carbon bonds; this route picks it up from there.
1. PBr₃ → 1-bromobutane. The OH becomes Br with no change in oxidation level, and the carbon stays primary.
2. NaCN in DMSO → pentanenitrile. This is the first new C–C bond, made by SN2 on a primary halide.
3. H₃O⁺, heat → pentanoic acid; then SOCl₂ → pentanoyl chloride. Nitrile, acid and acid chloride all sit at the acid oxidation level, so neither step needs an oxidant. Making the acid chloride also keeps step 4 safe: a free CO₂H would destroy the Grignard reagent.
4. 2 CH₃MgBr, then H₃O⁺ → 2-methylhexan-2-ol.
Ask the four questions. Does any reagent have something else to attack? No. The Grignard reagent is safe because nothing in the molecule carries an O–H when it arrives; the only OH in the target is made by the Grignard step itself. Does each product survive the next step? Yes. Is the regiochemistry controlled? Nothing here could react at two positions. Is the stereochemistry controlled? The target has no stereocenter, so there is nothing to set. If it had one, this route would make it as a racemic mixture, because every reagent in it is achiral.
Count again at the end. The carbon count runs 4, 4, 5, 5, 5, 7. It changes only at the two C–C steps, the cyanide in step 2 and the double methylation in step 4. Two of the four steps make the skeleton. The other two only change which functional group is present.
Setting the stereochemistry
Add bromine to (E)-hex-3-ene, and the product is meso-3,4-dibromohexane. Add it to (Z)-hex-3-ene, and the product is the (3R,4R)/(3S,4S) pair instead. The alkene you start with decides which stereoisomer you get. The worked example at the end of this section draws both cases.
A reaction in which the stereoisomer you start with fixes the stereoisomer you get is stereospecific, as Addition reactions defined it. You have met several. Bromine adds anti, one atom to each face of the double bond, because bromide opens the bromonium ion from the side opposite the bridge. OsO₄ adds both OH groups syn, to the same face. Catalytic hydrogenation delivers both hydrogens from the metal surface to one face. The Diels–Alder reaction is stereospecific too. The figure shows the first three on ring alkenes.
Look at what each reaction decides and what it leaves open. It decides the relative configuration, how the two new stereocenters relate to each other: trans for the two bromines on the ring, cis for the two OH groups of the diol. It does not decide the absolute configuration, which of the two mirror-image molecules forms. The reagent is equally likely to approach either face of the flat alkene, so a chiral product forms as a racemic mixture, 50 : 50 of the two enantiomers. When the product is meso, as cis-1,2-dimethylcyclohexane is, both faces give the same compound.
The rule, then, is to name the step that fixes each stereocenter in the target. Often that is the step that makes the stereocenter. Sometimes it is an earlier step, as the next example shows. If a route gives the wrong relative configuration, go back to the step that fixed it and change that step.
Target: meso-3,4-dibromohexane. From: acetylene (ethyne) and bromoethane.
Count. The target has six carbons, and each starting material has two. So at least two C–C bonds have to be made.
Work backwards, one step at a time. Two bromines on neighboring carbons come from adding Br₂ to an alkene, so the compound one step back is hex-3-ene. An alkene of a chosen geometry comes from reducing an alkyne, so the compound before that is hex-3-yne. Hex-3-yne disconnects next to the triple bond into but-1-yne and bromoethane. But-1-yne disconnects the same way into acetylene and bromoethane, so hex-3-yne is really acetylene with an ethyl group added to each end.
Route.
1. NaNH₂, then CH₃CH₂Br → but-1-yne. The halide is primary, as the SN2 requires.
2. NaNH₂, then CH₃CH₂Br → hex-3-yne.
3. Na in liquid NH₃ → (E)-hex-3-ene, the trans alkene.
4. Br₂ → meso-3,4-dibromohexane.
Which step sets the stereochemistry? Step 3, not step 4. Bromine adds anti to a simple alkene like this one, so step 4 offers no choice. The choice is the geometry of the alkene it adds to. Anti addition to the trans alkene gives the meso compound. Anti addition to the cis alkene, made with H₂ over Lindlar catalyst, gives the (3R,4R) and (3S,4S) pair instead.
Check the order. Steps 1 and 2 have to come before step 3, because the acetylide chemistry needs the triple bond. Step 3 needs a reagent that stops at the alkene. Na in NH₃ and H₂ over Lindlar catalyst both stop there. H₂ over Pd/C does not: it reduces the alkyne all the way to hexane, which leaves nothing for the bromine to add to.
Common failures worth recognizing in your own work
Most failed routes break one of the rules above. Three more mistakes are worth checking for on their own.
An oxidation that overshoots. Butan-1-ol with Jones reagent (CrO₃ in aqueous acid) gives butanoic acid, not butanal, because the water lets the oxidation run on past the aldehyde. When a step needs the aldehyde, use an anhydrous oxidant such as PCC (Oxidizing alcohols).
A single enantiomer from achiral inputs. Suppose a route starts from achiral compounds, uses only achiral reagents in an achiral solvent, and claims one enantiomer of a chiral product. That route is wrong: as the stereochemistry section showed, the reagent is equally likely to approach either face, so the product is racemic.
A route that claims one enantiomer has to use one of four ways out: a chiral starting material, a chiral reagent, a chiral catalyst or enzyme, or a resolution, which separates the two enantiomers afterwards. Relative configuration needs none of these, as the section on stereochemistry showed.
Counting carbons wrong. Check at the end that the product of the forward route has the same molecular formula as the target. A product with one carbon too many usually means a one-carbon extension, with cyanide, CO₂ or formaldehyde, was used where none was needed.
What carries forward
This section draws on every chapter before it, and it shows what those chapters were for: a reaction you can only recite is not yet a reaction you can use. When a synthesis problem stalls, the useful move is rarely to recall a new reagent. Count the carbons, find the C–C bond that has to form, and ask which of the ten or so C–C reactions could have made it.