Mix acetone with cyanide ion and the two join. The cyanide carbon bonds to the carbonyl carbon, and the C=O oxygen ends up as an OH. That is a nucleophilic addition: a nucleophile (an electron-pair donor, from Nucleophiles) adds to the carbon of a C=O, and the C=O becomes a C–O single bond.
This is the main reaction of aldehydes and ketones. Hydrates, cyanohydrins, acetals, imines and the Wittig reaction, later in this chapter, are all versions of it.
Grignard reagents came in the chapter before this one (Organometallic Bonding & Grignard Reagents). NaBH₄ and LiAlH₄ (Oxidation & Reduction) are taught in full later. This page uses all three as examples.
Two steps, one example
Step 1: the nucleophile attacks. Cyanide ion, ⁻C≡N, has a lone pair and a negative charge on its carbon. That carbon attacks the carbonyl carbon of acetone and forms a new C–C bond. At the same time the pi bond of the C=O (the second of its two bonds) breaks, and its two electrons move onto oxygen. Two curved arrows show this.
The pi electrons must move because carbon cannot hold five bonds. The new bond needs room, so one existing bond has to go. The pi bond is the weakest one, and oxygen is the more electronegative atom, so the pair lands on oxygen.
The oxygen now has three lone pairs and a negative charge. An oxygen anion bonded to carbon is called an alkoxide.
The carbon changes shape. Before the attack it carries three groups and is flat: sp², trigonal planar, with 120° angles (Hybridization). After the attack it carries four groups and is tetrahedral: sp³, with angles near 109.5°. The species with this new sp³ carbon and the O⁻ is the tetrahedral intermediate. Every reaction on this page passes through one.
Step 2: the alkoxide takes a proton. An alkoxide is a strong base. It takes a proton from the nearest acid. Here that acid is HCN, which is present alongside the cyanide. The new O–H bond forms from an oxygen lone pair. The electrons of the old H–C bond stay on carbon, so cyanide ion is given back.
The product has an OH and a CN on the same carbon. It is called a cyanohydrin, and Hydrates & cyanohydrins covers it in full.
Now the general rule. Replace cyanide with any negatively charged nucleophile, and the two steps stay the same. First the nucleophile adds to the carbonyl carbon and pushes the pi electrons onto oxygen. Then the alkoxide takes a proton. Only the group that ends up on the carbon changes.
Nucleophile first, proton last: this is the order under basic conditions. It works when the nucleophile is strong enough to attack a neutral C=O, which usually means it carries a negative charge.
Under acid: the proton goes on first
Water is a weak nucleophile. On its own it adds to acetone only slowly. Add a trace of acid and the addition becomes much faster. The acid is a catalyst: it speeds the reaction up and comes out unchanged at the end. The figure below follows it through four steps.
Step 1: protonate the oxygen. The carbonyl oxygen uses a lone pair to take a proton from H₃O⁺. The oxygen, not the carbon, is the basic site of a C=O (Aldehydes & ketones).
Step 2: one cation, two drawings. The cation that forms is a single species, but no single drawing shows it well, so it is drawn as two resonance contributors (Resonance). In the first, the C=O stays and the positive charge sits on oxygen; every atom has a full octet. In the second, the pi electrons sit on oxygen and the positive charge sits on carbon; that carbon has only six electrons and an empty p orbital.
The second drawing explains the speed-up. A carbon with a positive charge and only six electrons is far more electrophilic (electron-poor, and so attractive to a nucleophile) than a neutral carbonyl carbon. A cation like this one, with its positive charge shared between a carbon and the oxygen bonded to it, is called an oxocarbenium ion.
Step 3: the weak nucleophile adds. A water molecule attacks the carbon with a lone pair. The figure draws the attack on the second contributor, so one arrow is enough. Drawn on the first contributor, the same attack needs two arrows: water to carbon, and the pi bond up onto oxygen. Either drawing is correct, as long as the arrows match it. The Acetals section uses the two-arrow version.
The water arrived neutral, so the oxygen it adds with ends up with three bonds and a positive charge. The intermediate is a protonated alcohol, not an alkoxide.
Step 4: deprotonate. A second water molecule takes the extra proton. The product is the neutral hydrate: two OH groups on one carbon. That step also makes a new H₃O⁺, so the acid comes back out.
The general order under acid is protonate, add, deprotonate. Under base, the base turns the nucleophile into its charged form (HCN into −CN, for example), and that anion is strong enough to attack the neutral C=O on its own. Under acid, the proton makes the carbonyl carbon more reactive instead. Both paths reach the same tetrahedral carbon.
This sequence returns in the next sections. Acetal formation runs it twice, with a loss of water in between. Imine formation runs it and then loses water. Ester hydrolysis runs it on a carbon that carries a leaving group.
Protonate or collapse
Acetone's carbonyl carbon carries two CH₃ groups. Neither can leave, because CH₃⁻ would be a very poor leaving group. So acetone's tetrahedral intermediate has one option: its O⁻ takes a proton.
Acetyl chloride, CH₃COCl, carries a chlorine on its carbonyl carbon. Cl⁻ is a good leaving group (Leaving groups). So its tetrahedral intermediate has a second option: it can collapse. A lone pair on the O⁻ moves back down to re-form the C=O, and the C–Cl bond breaks, releasing Cl⁻.
Hydride reduction: NaBH₄ and LiAlH₄
Sodium borohydride, NaBH₄, turns propanal into propan-1-ol. The borohydride ion, BH₄⁻, is a boron atom bonded to four hydrogens, with a negative charge. Each B–H bond holds two electrons.
In step 1, one hydrogen moves to the carbonyl carbon and takes the B–H bond's electron pair with it. A hydrogen that arrives with two electrons acts as a hydride, H⁻. So the first arrow starts on the B–H bond, not on a free H⁻ ion.
The reaction runs in methanol. In step 2 the alkoxide takes a proton from a methanol molecule, and the alcohol forms in the flask.
Adding hydrogen to a C=O this way is a reduction. Reducing an aldehyde gives a primary alcohol: the carbon holding the OH is bonded to one other carbon. Reducing a ketone gives a secondary alcohol, where that carbon is bonded to two (Functional groups).
NaBH₄ is mild and selective. It reduces aldehydes and ketones, and it leaves esters, amides and carboxylic acids alone. It is mild enough to use in methanol or even water.
Lithium aluminum hydride, LiAlH₄, delivers hydride from an Al–H bond, with the same arrows. It is far more reactive and much less selective. It reduces almost every carbonyl: esters and carboxylic acids to primary alcohols, and amides to amines. It also reduces nitriles to amines.
LiAlH₄ reacts violently with water, so it is used in dry ether or THF (tetrahydrofuran, a cyclic ether used as a solvent). With no proton source in the flask, the alkoxide waits there. Water or dilute acid goes in afterwards, in a separate step called the aqueous workup. The reagent line shows it: 1. LiAlH₄, ether 2. H₃O⁺.
For NaBH₄ you will see two reagent lines: NaBH₄, CH₃OH, or 1. NaBH₄ 2. H₃O⁺. They describe the same reaction. The methanol supplies the proton as the reaction runs. The water or dilute acid added at the end breaks up the boron by-products and hands over any proton still needed. What is wrong is a hydride over an arrow with no proton source at all.
A molecule contains a ketone and an ester. You want only the ketone reduced. Use NaBH₄: it reduces the ketone and leaves the ester alone. If you want both groups reduced, use LiAlH₄.
Grignard and organolithium addition: building carbon skeletons
Methylmagnesium bromide, CH₃MgBr, adds to propanal and gives butan-2-ol. The product has one more carbon than the aldehyde.
CH₃MgBr is a Grignard reagent, RMgX, from Grignard reagents. Its C–Mg bond is polarized with δ− on carbon, so that carbon acts like a carbanion (a carbon with a lone pair and a negative charge): a strong carbon nucleophile. Organolithium reagents, RLi, behave the same way and are even more reactive; they come later in the course.
In step 1, the arrow starts on the C–Mg bond and ends on the carbonyl carbon. The bond that forms is a new carbon–carbon bond. That is the reason to use the reaction. In butan-2-ol that carbon is a new stereocenter; the flat carbonyl is attacked equally from both faces, so the product is racemic. With acetylide alkylation from Alkynes and the enolates of Enolate Chemistry, it is one of the few reliable ways to make a carbon skeleton bigger.
As with LiAlH₄, the proton comes in a separate aqueous workup. The magnesium alkoxide sits in the dry flask until dilute acid is added. The reagent line is 1. CH₃MgBr, dry ether 2. H₃O⁺.
Which class of alcohol forms? Count carbons at the carbonyl carbon. Propanal's carbonyl carbon carries one carbon group, the ethyl. The Grignard adds a second. So the OH carbon of butan-2-ol is bonded to two carbons, and the alcohol is secondary. Hydride on the same aldehyde adds no carbon, so propan-1-ol is primary.
The general rule: count the carbon groups on the carbonyl carbon, then add one for a Grignard reagent (none for hydride). Formaldehyde, H₂C=O, has none, so a Grignard reagent gives a primary alcohol. Any other aldehyde has one, so it gives a secondary alcohol. A ketone has two, so it gives a tertiary alcohol.
An ester looks like an exception but is not. Its carbonyl carbon carries an OR group, which can leave. So after the first Grignard adds, the tetrahedral intermediate collapses and expels the alkoxide. That leaves a ketone in a flask that still holds Grignard reagent.
The ketone is more electrophilic than the ester it came from. The ester's OR group was donating electron density into the C=O, and that group has now left. So the ketone reacts faster than the ester did, and the reaction cannot be stopped halfway. The product is a tertiary alcohol with two identical new R groups. Carboxylic Acids & Derivatives is where that branch is taught in full.
Target: 2-phenylbutan-2-ol. The figure draws it: one carbon carrying an OH, a phenyl ring, a methyl and an ethyl.
Step 1: find the carbinol carbon. The carbinol carbon is the carbon that holds the OH. Everything in this analysis happens there.
Step 2: count its carbon groups. It has three, so the alcohol is tertiary. The carbonyl compound it came from already had two: it was a ketone.
Step 3: pick one C–C bond at that carbon and break it on paper. The three bonds are C–phenyl (a), C–ethyl (b) and C–methyl (c). Each one gives a valid route. Breaking a bond on paper to find starting materials is called a disconnection.
Step 4: read off the two pieces. The group cut away becomes the Grignard reagent. The rest, with C–OH rewritten as C=O, becomes the ketone. Below, Ph is short for phenyl, a benzene ring used as a group.
- Cut a (C–phenyl): PhMgBr + butan-2-one, CH₃COCH₂CH₃.
- Cut b (C–ethyl): CH₃CH₂MgBr + acetophenone, PhCOCH₃, a benzene ring carrying a C(=O)CH₃ group.
- Cut c (C–methyl): CH₃MgBr + propiophenone, PhCOCH₂CH₃, the same with an ethyl in place of the methyl.
All three routes work. A target alcohol usually offers several disconnections, and you choose by what you can buy. Check each one forwards before you commit: count the carbon groups on the carbinol carbon of the product it would give, and make sure there are three.
Nitrogen nucleophiles: where they go next
Acetone reacts with methylamine, CH₃NH₂, to give (CH₃)₂C=N–CH₃. With dimethylamine, (CH₃)₂NH, the product is CH₂=C(CH₃)–N(CH₃)₂ instead. Both reactions release water, but the new double bond lands in a different place.
Both amines start the same way. The nitrogen lone pair adds to the carbonyl carbon, giving a tetrahedral intermediate like every one above. Then that intermediate does something new: it loses water and forms a double bond again.
With a primary amine, RNH₂, the new double bond goes to nitrogen. The C=N product is an imine.
A secondary amine, R₂NH, uses its only N–H proton during the addition. After water leaves, the carbon–nitrogen double bond carries a positive charge on nitrogen (an iminium ion), and there is no N–H left to lose. So a proton leaves from the carbon next door instead. The new double bond sits between two carbons, next to the nitrogen: an enamine.
The mechanism of both, and the narrow pH window they need, belong to Imines and enamines. Here they matter only because they show the tetrahedral intermediate doing something other than taking a proton.
Other additions worth knowing
The cyanohydrin from the first example is useful because its nitrile can later be hydrolyzed to a carboxylic acid or reduced to an amine: a one-carbon extension with a handle on the end.
Water adds reversibly to give that hydrate, also called a gem-diol. Hydrates & cyanohydrins takes it up next.
The Wittig reaction, later in this chapter, uses a phosphorus ylide as the nucleophile. An ylide has a negative carbon bonded directly to a positive phosphorus. The ylide carbon adds to the C=O. Then the oxygen leaves bonded to the phosphorus, as triphenylphosphine oxide, Ph₃P=O, for the usual ylides. The product is an alkene, and its new C=C sits exactly where the C=O was. That control over where the double bond goes is why the Wittig reaction is one of the most used reactions in synthesis (The Wittig reaction).
What carries forward
The tetrahedral intermediate returns in Carboxylic Acids & Derivatives, where it collapses, and in Enolate Chemistry, where the nucleophile is an enolate carbon. Hydride and Grignard additions are the backbone of synthesis problems. The acid and base orders on this page apply to every carbonyl mechanism ahead.