Reducing carbonyls

Stir butan-2-one with sodium borohydride, NaBH₄, in methanol, and the ketone becomes butan-2-ol. The C=O has turned into a CH–OH: one hydrogen has joined the carbon and one has joined the oxygen.

That is a reduction. The carbon has gained a bond to hydrogen and lost a bond to oxygen, so it has moved one rung down the ladder in Oxidation levels in organic chemistry. Oxidizing alcohols climbed that ladder; this section comes back down it.

Most carbonyl reductions use one of two reagents. Sodium borohydride, NaBH₄, is the mild one, and lithium aluminum hydride, LiAlH₄, is the strong one. Both deliver a hydride, a hydrogen that arrives with two electrons, written H⁻. You met both as examples in Nucleophilic addition. This section works out what each one reduces, and why.

Hydride in, proton on

The reduction of butan-2-one takes two steps. In step 1, a B–H bond of the borohydride ion, BH₄⁻, hands its hydrogen to the carbonyl carbon, and the bond's two electrons go with it. At the same moment the C=O pi bond breaks and its electrons move onto oxygen, because carbon cannot hold five bonds.

Step 1 leaves an alkoxide: an O⁻ on a carbon that is now tetrahedral. In step 2 the alkoxide takes a proton from a methanol molecule, and butan-2-ol forms.

LiAlH₄ gives up its hydride the same way, from an Al–H bond of the aluminohydride ion, AlH₄⁻. The arrows are identical.

Reducing a ketone usually makes a racemic mixture

Butan-2-ol has a stereocenter. Its carbinol carbon, the carbon that holds the OH, carries four different groups: OH, H, CH₃ and CH₂CH₃. Butan-2-one has no stereocenter, because its carbonyl carbon carries only three groups. So the reduction creates a stereocenter, and the question is which enantiomer forms.

Both form. The carbonyl carbon is trigonal planar, so the C=O and the two groups on the carbon lie in one plane, and that plane has two faces. A hydride that arrives from the front face leaves its H pointing toward you. One that arrives from the back face leaves it pointing away. The two products are mirror images: (S)-butan-2-ol from the front and (R)-butan-2-ol from the back.

Nothing in a flask of NaBH₄ and butan-2-one favors one face, so the two attacks happen equally often. The product is a racemic mixture, a 50:50 mix of two enantiomers, written (±)-butan-2-ol. It does not rotate plane-polarized light.

That gives a rule you can apply without drawing: an achiral starting material and an achiral reagent cannot give a single enantiomer. If a question asks for one enantiomer from a plain ketone reduction, the answer is that this method cannot give it. That needs a chiral reagent or an enzyme, which is beyond this course.

The heading says "usually" because of ketones whose two groups are the same. Acetone gives propan-2-ol, whose carbinol carbon carries two identical CH₃ groups, so there is no stereocenter and nothing to be racemic.

Why LiAlH₄ is so much stronger

Compare the two bonds that deliver the hydride. Boron's electronegativity is 2.04 and hydrogen's is 2.20, so a B–H bond is nearly nonpolar. Aluminum's is 1.61, so an Al–H bond leaves much more negative charge on the hydrogen. That hydrogen leaves as hydride far more readily, so LiAlH₄ is the much more reactive reagent.

The difference decides how each reagent is used. NaBH₄ reacts only slowly with methanol, ethanol or water. You can dissolve it in the alcohol and add the ketone, and the solvent supplies the proton in step 2.

LiAlH₄ reacts violently with water, and even with moisture in the air. A hydride is a strong base as well as a nucleophile, and it takes a proton from water to make hydrogen gas, H₂. So LiAlH₄ is used in a dry ether solvent, such as diethyl ether or THF (tetrahydrofuran, a cyclic ether).

With no proton source in the flask, the alkoxide from step 1 simply waits. Water or dilute acid goes in afterward, in a separate step called the aqueous workup, and protonates it. The reagent line shows both steps: 1. LiAlH₄, ether 2. H₃O⁺. An answer that writes only "LiAlH₄" over the arrow is incomplete, because without the workup there is no alcohol yet. For NaBH₄, either NaBH₄, CH₃OH or 1. NaBH₄ 2. H₃O⁺ is correct.

What each reagent reduces

Read the NaBH₄ column first. It reduces aldehydes and ketones, and of the other carbonyl compounds only the acid chloride. LiAlH₄ reduces every carbonyl compound in the table.

The split follows how electron-poor each carbonyl carbon is. In an ester, the oxygen of the OR group donates electron density into the C=O, so the carbonyl carbon is less positive than a ketone's (Nucleophilic addition). The nitrogen of an amide donates even more strongly, because nitrogen is less electronegative than oxygen and gives up its lone pair more easily. NaBH₄ is only reactive enough for the most electron-poor carbonyl carbons. LiAlH₄ is reactive enough for all of them.

The acid chloride is the exception in the NaBH₄ column. Its chlorine donates very little and pulls electron density away, so its carbonyl carbon is even more electron-poor than a ketone's. Even NaBH₄ reduces it, and it goes all the way to the primary alcohol, for the reason the next section gives for esters.

Neither reagent reduces an isolated alkene or alkyne, meaning one that is not joined directly to a C=O. A hydride is a nucleophile, and a C=C has no electron-poor carbon for it to attack. Reducing a C=C is the job of Catalytic hydrogenation.

Two rows end in something other than an alcohol. LiAlH₄ turns an amide into an amine and a nitrile into a primary amine. Each has its own section below.

Esters take two hydrides

Ethyl propanoate and LiAlH₄, then workup, give propan-1-ol and ethanol. The old carbonyl carbon ends up with two new hydrogens, so the ester has taken two hydrides.

The first hydride adds exactly as it does to a ketone and gives a tetrahedral intermediate. With butan-2-one, that intermediate carries only carbon groups and an H, so all it can do is wait for a proton. The ester's intermediate also carries an ethoxy group, OCH₂CH₃, and that group can leave.

A lone pair on the O⁻ moves back down to re-form the C=O, and the bond to the ethoxy oxygen breaks. Ethoxide ion, CH₃CH₂O⁻, leaves, and propanal is left behind. Nucleophilic addition called this move a collapse.

Propanal is more electrophilic than the ester it came from, because the donating ethoxy group has gone. So a second hydride adds to it faster than the first one added to the ester, and the aldehyde is used up as fast as it forms. After workup the product is the primary alcohol, and the ethoxide becomes ethanol.

An acid chloride follows the same path. Its tetrahedral intermediate expels chloride, an even better leaving group than ethoxide, and the aldehyde that forms is reduced again. That is why the acid chloride row reads "1° alcohol" for both reagents.

Stopping at the aldehyde: DIBAL-H

Ethyl propanoate with one equivalent of DIBAL-H at −78 °C, followed by aqueous workup, gives propanal. DIBAL-H is diisobutylaluminum hydride, (i-Bu)₂AlH: an aluminum atom carrying one hydrogen and two isobutyl groups, (CH₃)₂CHCH₂–. One equivalent means one mole of reagent for each mole of ester.

DIBAL-H is neutral, and its aluminum has an empty orbital, so it is a Lewis acid (Lewis acids/bases). It first bonds to the carbonyl oxygen and then passes its hydride to the carbon. The oxygen of the tetrahedral intermediate stays bonded to aluminum.

At −78 °C that intermediate holds together and keeps its ethoxy group, so no aldehyde forms while DIBAL-H is still active. The workup destroys the leftover reagent and breaks the O–Al bond. Only then does the intermediate fall apart to propanal and ethanol, and by then there is no hydride left to reduce the aldehyde.

The cold matters. With two equivalents in a warmer flask, the intermediate collapses while hydride is still present, and DIBAL-H takes the ester all the way to the primary alcohol.

Carboxylic acids: the proton goes first

Propanoic acid with LiAlH₄, then workup, gives propan-1-ol. The first thing the reagent does, though, is act as a base. A hydride takes the acid's O–H proton, and hydrogen gas bubbles off.

That leaves the carboxylate ion, RCO₂⁻. Its negative charge makes its carbonyl carbon a poor target for a negative hydride, so it reacts far more slowly than an ester. LiAlH₄ still reduces it, and the workup gives the primary alcohol. The acid uses up one extra hydride on the way, so reducing an acid takes excess reagent.

NaBH₄ does not reduce a carboxylic acid, and workup gives the acid back.

Amides give amines

N,N-Dimethylacetamide, CH₃CON(CH₃)₂, is an amide: a carbonyl carbon bonded to a nitrogen. With LiAlH₄ and then workup it gives N,N-dimethylethanamine, CH₃CH₂N(CH₃)₂. The C=O has become a CH₂, and the nitrogen is still attached. An amide reduced by LiAlH₄ gives an amine, not an alcohol.

The first hydride adds as usual. The difference lies in which group leaves the tetrahedral intermediate. The nitrogen group, (CH₃)₂N⁻, would be a very strong base, much stronger than the ethoxide that leaves an ester: their conjugate acids have pKa values of about 38 and 16 (pKa). So it is a very poor leaving group (Leaving groups). The oxygen can leave instead. When the C–O bond breaks, its electron pair goes to the oxygen. That oxygen is bonded to aluminum, so it leaves as an O–Al group, a far weaker base than (CH₃)₂N⁻.

So the nitrogen lone pair moves in to form a C=N bond, and the C–O bond breaks. The result is an iminium ion, a C=N with a positive charge on the nitrogen, like the one in Imines and enamines. A second hydride adds to its carbon, the C=N pi electrons move onto nitrogen, and the amine is formed.

A nitrile has a carbon triple-bonded to nitrogen, R–C≡N. Propanenitrile, CH₃CH₂C≡N, gives propan-1-amine, CH₃CH₂CH₂NH₂, with LiAlH₄ and then workup. The nitrile carbon takes one hydride for each of its two pi bonds, and workup puts two H on the nitrogen.

Hydrogen gas and a metal catalyst

Cyclohexanone under H₂ at high pressure over platinum oxide, PtO₂, gives cyclohexanol. The two atoms of H₂ add across the C=O: one H lands on the carbon and one on the oxygen.

This is the reaction of Catalytic hydrogenation, applied to a C=O instead of a C=C. A C=O is much harder to hydrogenate than a C=C. Under H₂ at one atmosphere over Pd/C, a C=C is reduced and a plain (aliphatic) ketone survives. That ketone needs higher pressure and a more active catalyst, such as PtO₂ or Raney nickel. An ester survives ordinary hydrogenation conditions.

So H₂ over Pd/C and NaBH₄ complement each other. The first reduces a C=C and leaves a plain ketone alone, and the second does the reverse.

A ketone whose C=O is joined directly to a benzene ring is the exception. Over Pd/C, the C=O of 1-phenylpropan-1-one can be reduced past the alcohol, even all the way to CH₂, so do not count on a ketone like this surviving.

Choosing the weaker reagent

In a molecule with two reducible groups, selectivity usually comes from choosing the weaker reagent on purpose. NaBH₄ reduces a ketone and leaves an ester alone, while LiAlH₄ reduces both.

All the way to CH₂: Clemmensen and Wolff–Kishner

1-Phenylpropan-1-one is a ketone: a benzene ring joined to a C=O, with an ethyl group on the other side. Heat it with zinc amalgam (zinc alloyed with mercury), Zn(Hg), and concentrated HCl, and it becomes propylbenzene. The C=O has become a CH₂. This is the Clemmensen reduction, first met in Imines and enamines.

A hydride reagent takes a ketone down one rung, to an alcohol. The Clemmensen reduction takes it down two, because both bonds to oxygen are replaced by bonds to hydrogen.

The Wolff–Kishner reduction does the same job in base. Hydrazine, H₂N–NH₂, first condenses with the ketone to give a hydrazone, C=N–NH₂, as in Imines and enamines. Heating the hydrazone with KOH then removes both nitrogens as N₂ gas and leaves a CH₂.

The two reactions are learned as a pair because their conditions are opposite. The Clemmensen needs strong acid, and the Wolff–Kishner needs strong base and heat. Choose the one the rest of the molecule can survive.

An acetal, for example, falls apart in aqueous acid (Acetals & hemiacetals). A ketone in a molecule that also carries an acetal therefore needs the Wolff–Kishner. In the other direction, a bromine on a secondary carbon would be lost by E2 elimination in hot KOH (E2), which rules out the Wolff–Kishner for that molecule.

Neither reaction turns an ester or an amide into a CH₂. In the Wolff–Kishner, hydrazine adds to the C=O, and an ester's tetrahedral intermediate then collapses and expels its OR group, so no hydrazone forms. The Clemmensen also leaves esters and amides alone; learn that as a fact. So C=O to CH₂ is a reaction of aldehydes and ketones only.

The Wolff–Kishner mechanism starts from the hydrazone and runs in five steps:

• Step 1. Hydroxide takes a proton from the NH₂ group.
• Step 2. The new lone pair on that nitrogen forms an N=N bond. The C=N pi electrons swing onto the carbon and pick up a proton from water. The carbon now has its first new H.
• Step 3. Hydroxide takes the last N–H proton.
• Step 4. The N⁻ lone pair forms a third bond between the two nitrogens, and the C–N bond breaks with both its electrons staying on carbon. N₂ leaves as a gas. The carbon left behind is a carbanion: a carbon with a lone pair and a negative charge.
• Step 5. The carbanion takes a proton from water, and the second new H is in place.

Step 4 drives the reaction. N₂ is a very stable molecule, and as a gas it escapes the flask, so it cannot come back and react in reverse.

The Clemmensen reduction happens on the surface of the zinc. Its mechanism is not well understood and is not drawn with curved arrows, so learn its conditions and its result.

A preview. Aromatic Chemistry uses both reactions. It attaches a C=O group to a benzene ring and then reduces it to CH₂, which puts a straight alkyl chain on the ring without the rearrangement a direct alkylation can suffer.

What carries forward

Hydride reduction is the nucleophilic addition of Nucleophilic addition, with H⁻ as the nucleophile. The reagent table is the part to keep at hand. Most "which reagent" questions in synthesis come from it, and the amide-to-amine row is the one most often missed. Amines uses that row to make amines, and Synthesis & Retrosynthesis draws on every reduction on this page.