Say a route has built the carbon chain of butan-1-ol, but the target is butanoic acid. The skeleton is already right, and only the group on the end carbon has to change. Changing one functional group into another like this is a functional group interconversion. The reactions in Making carbon–carbon bonds build the skeleton; interconversions do most of the rest.
Each interconversion is usually easy on its own. Even so, routes often go wrong at this stage, because there are so many of them. They are easier to keep straight as one map than as a long list, and the map is the oxidation ladder from Oxidation & Reduction.
Two kinds of move
Start with ethanol. Oxidize it to ethanal, and the carbon that carries the OH trades a C–H bond for a second bond to oxygen. That is an oxidation, so it needs an oxidant such as PCC. Turn ethanol into bromoethane instead, and the same carbon trades its bond to oxygen for a bond to bromine. Its bonds to hydrogen stay as they were, so nothing is oxidized or reduced. PBr₃ does the job, and it is not an oxidant or a reductant.
Every interconversion is one of those two kinds:
- Up or down the ladder. The carbon gains bonds to O, N or halogen and loses bonds to H, or the reverse. The rungs run alkane, alcohol, aldehyde or ketone, carboxylic acid. These moves need an oxidizing or a reducing agent.
- Sideways along a rung. One bond to O, N or halogen is traded for another, as in alcohol to alkyl halide, or acid to ester to amide. These moves are substitutions, and they need no redox reagent.
Naming the kind of move tells you which class of reagent to reach for before you recall a single reagent name. It also catches the two commonest planning errors: trying to reach a carboxylic acid from an alcohol by a substitution, and trying to turn an ester into an amide with a reducing agent.
Where the other groups sit on the ladder
The ladder was built from four groups: alkane, alcohol, aldehyde and carboxylic acid. Most other groups can be placed with one count: the number of bonds from the carbon to O, N or halogen, with a double bond counting as two and a triple bond as three. One bond puts a group on the alcohol rung, two on the aldehyde and ketone rung, and three on the acid rung. Alkenes and alkynes have no such bonds, so they need a different test, which comes below.
The alcohol rung. An alkyl halide, an ether and an amine each have one bond from the carbon to a halogen, an oxygen or a nitrogen. So each sits beside the alcohol, and interconverting them is substitution.
The alkene belongs on the alcohol rung too. It is placed by a convention: count over the two carbons of the double bond together, and ask what adding water does to them. Adding water to ethene gives ethanol. One carbon gains a C–H bond and the other gains a C–O bond, so the change is neither an oxidation nor a reduction, as Oxidation levels in organic chemistry showed for hydration. Removing water reverses it. So alcohol, alkyl halide and alkene form a triangle, and you can move between any two of them without an oxidant or a reductant.
The aldehyde and ketone rung. An acetal carbon has two single bonds to oxygen. An imine carbon has a double bond to nitrogen, which counts as two. An alkyne belongs here for the same reason an alkene belongs on the alcohol rung: hydrating propyne gives acetone (propan-2-one), and no oxidant is involved.
The carboxylic acid rung. An ester, an amide, an acyl chloride and an anhydride each have three bonds from the carbonyl carbon to O, N or Cl. A nitrile carbon has a triple bond to nitrogen, which also counts as three. So nitrile → carboxylic acid is a hydrolysis and needs no oxidant. A primary amine, RCH₂NH₂, has only one bond from that carbon to nitrogen and sits on the alcohol rung. So nitrile → primary amine drops two rungs, which is why it needs LiAlH₄.
The moves worth knowing cold
The three tables below collect interconversions from earlier chapters. The first holds the moves between rungs, the second the moves along one rung, and the third the moves that make or use an alkene, an alkyne or an epoxide.
Up and down the ladder: each needs an oxidant or a reductant
| From | To | Reagent |
|---|---|---|
| Alcohol (1°) | Aldehyde | PCC, Swern or DMP |
| Alcohol (1°) | Carboxylic acid | Jones (aqueous CrO₃) |
| Alcohol (2°) | Ketone | Any of the above oxidants |
| Ketone/aldehyde | Alcohol | NaBH₄ |
| Ketone/aldehyde | Alkane (CH₂) | Clemmensen or Wolff–Kishner |
| Ester | 1° Alcohol | LiAlH₄ |
| Amide | Amine | LiAlH₄ |
| Nitrile | 1° Amine | LiAlH₄ |
| Alkene | Alkane | H₂, Pd/C |
| Alkyne | cis Alkene | H₂, Lindlar |
| Alkyne | trans Alkene | Na, NH₃(l) |
| Nitroarene | Aniline | H₂ and a catalyst, or Fe/HCl or Sn/HCl (the nitrogen, not the carbon, is reduced) |
| Ketone/aldehyde | Amine | R′NH₂ (or R′₂NH), NaBH₃CN — reductive amination |
Sideways along one rung: no redox reagent
| From | To | Reagent |
|---|---|---|
| Alcohol (1°/2°) | Alkyl halide | PBr₃, SOCl₂, or HX |
| Alcohol | Ether | NaH, then R′X (Williamson) |
| Alkyl halide | Nitrile | NaCN (SN2), which also makes a new C–C bond: the chain gains a carbon, and that carbon arrives at the acid level |
| Nitrile | Carboxylic acid | H₃O⁺, heat |
| Carboxylic acid | Ester | Alcohol, H⁺ (Fischer) |
| Carboxylic acid | Acyl chloride | SOCl₂ |
| Acyl chloride | Ester or amide | Alcohol, or amine |
The Wittig reaction turns a C=O into a C=C, but it is not in these tables. It makes a new C–C bond, and reactions that make a C–C bond fall outside this ladder bookkeeping, so the Wittig stays on the list in Making carbon–carbon bonds. The NaCN row above and the acetylide row below also make a C–C bond. They are kept here because each one also puts a new functional group in place, and each is marked.
Alkenes, alkynes and epoxides
| From | To | Reagent |
|---|---|---|
| Alkene | Alcohol (Markovnikov) | H₃O⁺, or oxymercuration |
| Alkene | Alcohol (anti-Markovnikov) | 1. BH₃ 2. H₂O₂, HO⁻ |
| Alkene | Diol (syn) | OsO₄ / NMO |
| Alkene | Diol (anti) | 1. mCPBA 2. H₃O⁺ |
| Alkene | Two carbonyls | 1. O₃ 2. Me₂S |
| Alcohol | Alkene | Conc. H₂SO₄, heat (E1) |
| Alkyl halide | Alkene | Strong base (E2); a bulky one such as t-BuO⁻ gives the less substituted alkene |
| Alkyl halide | Alkyne | Acetylide ion (SN2), which also makes a new C–C bond |
| Terminal alkyne | Methyl ketone | H₂O, H₂SO₄, HgSO₄ — an unsymmetrical internal alkyne gives a mixture |
| Terminal alkyne | Aldehyde | 1. R₂BH 2. H₂O₂, HO⁻ |
| Epoxide | β-substituted alcohol | Nu⁻ — it attacks the less hindered carbon |
Reagent pairs with opposite outcomes
Hydrogenate but-2-yne over Lindlar's catalyst and you get cis-but-2-ene. Reduce it with sodium in liquid ammonia and you get trans-but-2-ene. The starting material is the same, and only the reagent differs.
The fourth pair, Markovnikov against anti-Markovnikov, is drawn in the worked example at the end of this page.
Which reductant spares which group
Suppose a molecule carries both a ketone and an ester, and only the ketone should be reduced. NaBH₄ does exactly that. When a molecule has more than one reducible group, your choice among these three reagents, all from Reducing carbonyls, decides which ones are reduced.
| Reagent | Reduces | Leaves alone |
|---|---|---|
| NaBH₄ | Aldehydes and ketones (and acid chlorides); an aldehyde faster than a ketone | Esters, acids, amides, nitriles and nitro groups |
| LiAlH₄ | Aldehydes, ketones, esters, acids, amides and nitriles | Isolated C=C and C≡C bonds |
| H₂ over Pd/C at one atmosphere | Alkenes, alkynes and nitro groups | Ketones and esters |
Choose LiAlH₄ when every carbonyl group should be reduced. Knowing what each reagent spares tells you whether a group elsewhere in the molecule will survive a step. When it will not, the group has to be shielded first, which is the subject of Protecting groups.
Going around rather than through
Some conversions have no direct reagent, or none that gives the product you want. They take two steps instead. Four of these detours come up often.
Putting the alkene where you want it
2-Methylbutan-2-ol can give two different alkenes. Heat it with concentrated H₂SO₄, and the reaction goes through a carbocation and gives mostly 2-methylbut-2-ene. That is the more substituted alkene, the one with more carbons attached to its double-bond carbons, and it is the product Zaitsev's rule predicts. With other alcohols the carbocation can also rearrange before it loses a proton, which gives you even less control.
To get the other alkene, make the alkyl halide first. HBr turns the alcohol into 2-bromo-2-methylbutane. An E2 elimination with a bulky base such as tert-butoxide, t-BuO⁻, then removes a hydrogen from one of the exposed CH₃ groups rather than from the more crowded CH₂, and gives mostly the less substituted alkene, 2-methylbut-1-ene.
A small base such as ethoxide would give mostly the Zaitsev alkene again. So on this route the base, not carbocation stability, decides where the double bond goes.
An anti-Markovnikov alkyl halide
For a bromide there is a one-step answer. HBr with a trace of peroxide adds to propene by a radical pathway and puts the Br on the end carbon, giving 1-bromopropane. Markovnikov / anti-Markovnikov explains why.
That peroxide effect works only with HBr. HCl and HI add the Markovnikov way whether a peroxide is present or not, so an anti-Markovnikov chloride or iodide takes two steps. Hydroboration–oxidation of propene gives the anti-Markovnikov alcohol, propan-1-ol, and then SOCl₂ turns it into 1-chloropropane.
An iodide comes from the same alcohol with phosphorus and iodine, a reagent pair the course has not used before. The two make PI₃ in the flask, and PI₃ replaces the OH just as PBr₃ does.
The two-step route also suits a bromide that needs a particular relative configuration, for example on a ring. In the radical route the carbon radical is nearly flat, so H–Br can reach either face. Hydroboration adds H and B to the same face, oxidation puts the OH exactly where the boron was, and PBr₃ then replaces the OH with inversion, so you can predict which diastereomer forms. Both routes still give a racemic product from an achiral alkene.
An aldehyde from an ester
Treat methyl butanoate with LiAlH₄ and you get butan-1-ol, not butanal. The aldehyde does form on the way, but it is more electrophilic than the ester it came from, so it takes a second hydride as soon as it appears.
There are two ways round this. One is DIBAL-H (diisobutylaluminum hydride), a milder hydride reagent. It stops at the aldehyde, but only when you use a single equivalent at −78 °C. The other is to reduce all the way to butan-1-ol with LiAlH₄ and then oxidize back up to butanal with PCC. The second route takes two steps and uses only reagents from the tables.
Moving a group along a chain
Sometimes a target is the starting material with its functional group on the neighboring carbon. Eliminate to the alkene, then add back with the opposite regiochemistry, which means that the group goes onto the other carbon of the double bond. This is the standard way to shift an OH from one carbon to the next, and it is almost always the intended answer to this kind of problem.
Target: propan-1-ol, from propan-2-ol.
Both compounds have three carbons and one OH, and both sit on the alcohol rung. Only the position of the OH differs, so the route needs no C–C chemistry and no redox reagent. The move is to eliminate, then add water back the other way round.
1. Conc. H₂SO₄, heat → propene.
2. BH₃, then H₂O₂/HO⁻ → propan-1-ol. Hydroboration is anti-Markovnikov, so it puts the OH on the less substituted carbon, C1.
Acid-catalyzed hydration in step 2 would put the OH straight back on C2.
What carries forward
Almost nothing on this page is new chemistry. Nearly every row is a reaction from an earlier chapter, sorted so you can look it up. What is new is a habit. Faced with a gap between two structures, ask whether the step moves up or down the ladder or sideways along a rung, and whether one reagent can make it or it needs two steps. Planning a multistep route puts that habit to work.