Catalytic hydrogenation

Stir cyclohexene with a little powdered palladium under hydrogen gas. It takes up one molecule of H₂ and becomes cyclohexane. One hydrogen atom has joined each carbon of the old C=C. This is catalytic hydrogenation: adding H₂ across a multiple bond with the help of a metal. You met it in Alkene structure as a way to measure alkene stability; this section shows how it works.

Gaining bonds to hydrogen is a reduction, the opposite of the oxidation in the last section, where the carbons gained bonds to oxygen.

Why a metal, and not a reagent in solution? Reducing a carbon usually means handing it a hydride, H⁻. A hydride and an alkene's pi bond are both nucleophiles, so an isolated C=C gives the hydride nothing electrophilic to attack. The hydride reagents that do reduce C=O are taught later, in Oxidation & Reduction. A carbon–carbon multiple bond needs a different route: bring the alkene and the hydrogen together on a metal surface and let them combine there.

The basic reaction

Three metals are the usual choices. Palladium on carbon (Pd/C) is palladium dusted onto powdered charcoal. Platinum is used as the metal or as its oxide, PtO₂. Raney nickel is a spongy, finely divided nickel. With any of them, the two hydrogen atoms of H₂ add across a C=C, and the alkene becomes an alkane.

The metal is a catalyst: it speeds the reaction up and comes out unchanged. Without it, hydrogen gas and an alkene sit together in a flask at room temperature and nothing happens. That is not because the reaction is uphill. The alkane is lower in energy than the alkene and H₂ it came from, so the reaction releases heat. The problem is the barrier: the H–H bond has to break first, and without a surface to help, that costs too much.

How the metal delivers the hydrogens

The metal surface does the work in stages. H₂ molecules adsorb onto the metal, which means they stick to its surface, and each one splits into two H atoms bound to the metal. The alkene adsorbs too. It lies flat, with its pi bond facing the metal. One H atom then moves across to one carbon. For a moment the other carbon is still bonded to the metal. A second H atom then moves to that carbon, and the finished alkane, which no longer sticks, drifts away.

A flat alkene has two faces, the side above its plane and the side below. Both hydrogens come up from the metal, so both reach the face lying on it, and the other face cannot be reached. Hydrogenation is therefore a syn addition: the two new C–H bonds form on the same face.

A ring makes this visible. Hydrogenate 1,2-dimethylcyclohexene over platinum, and both hydrogens go in on the face that was touching the metal. That leaves both methyls on the other face, so the product is cis-1,2-dimethylcyclohexane.

Predict syn addition. Real hydrogenations are a little less tidy. The first hydrogen transfer can run backwards: the metal can take back an H, sometimes from a different carbon, before the second H arrives. Each round trip gives the double bond a chance to shift along the chain or to flip between cis and trans. Palladium does this most: the same 1,2-dimethylcyclohexene over Pd gives a real share of the trans product. Platinum stays much closer to clean syn addition, which is why the example above uses it. Partial hydrogenation of vegetable oils turns some cis double bonds into trans ones in the same way, and that is where trans fats come from (see Biomolecules).

Syn fixes the arrangement, not the mirror image

Now give the two ring carbons different groups. 1-Ethyl-2-methylcyclohexene can lie on the metal with either face down. Whichever face is down, both hydrogens arrive on it, so every molecule comes out cis. But the two faces give mirror images: one product has both new hydrogens on hashes, and the other has both on wedges.

The catalyst has no preference between the two faces, so half the molecules go each way. The product is one diastereomer, the cis one, formed as a racemic mixture of its two enantiomers.

This is what "syn" does and does not settle. It fixes the relative stereochemistry: which new group is on the same side as which. It does not fix the absolute stereochemistry, which enantiomer you get. The same holds for an open-chain alkene of known geometry (E or Z) that gains two stereocenters. Syn addition gives one diastereomer, as a racemate unless that diastereomer is meso (see the next paragraph). This is not special to hydrogenation. An achiral starting material and an achiral reagent cannot favor one mirror image over the other.

Sometimes both faces give the same molecule. cis-1,2-Dimethylcyclohexane, from 1,2-dimethylcyclohexene, has a mirror plane, so it is meso and has no enantiomer. There is nothing to pair up, and the reaction gives a single product. Open chains do the same: (Z)-3,4-dimethylhex-3-ene gives only meso-3,4-dimethylhexane, while its E isomer gives the racemic pair.

What gets reduced, and in what order

Put a molecule under H₂ over Pd/C at one atmosphere, the pressure of a hydrogen balloon, and its C=C is reduced in minutes. The C=O of a ketone in the same flask is left alone. Catalytic hydrogenation does not reduce everything. Its preference for some groups over others is its chemoselectivity. Learn the list below as an order, from easiest to hardest:

• Readily reduced: alkenes; alkynes, all the way to alkanes; imines (C=N double bonds); nitro groups, –NO₂, which become amines, –NH₂; and azides, –N₃, which also become amines.
• Needs forcing conditions: ketones and aldehydes, which react only at higher pressure over a more active catalyst such as PtO₂. Over Pd/C under ordinary conditions they survive.
• Essentially untouched: esters, carboxylic acids and amides. A benzene ring needs severe conditions, high pressure and high temperature, so it survives an ordinary hydrogenation intact.

Hydride reagents such as NaBH₄ reverse this preference; see Oxidation & Reduction, and the worked example below.

Stopping an alkyne halfway, in either geometry

Take but-2-yne, CH₃–C≡C–CH₃. It is an alkyne: its two middle carbons share a triple bond, one sigma bond and two pi bonds, from sp carbons (see Hybridization), so the four carbons lie in a straight line. Alkynes, the next section, covers them in full. Two pi bonds mean hydrogenation can happen twice, and the useful chemistry here is stopping after the first.

Over ordinary Pd/C, but-2-yne takes up two molecules of H₂ and becomes butane: the reaction runs past the alkene. Two other sets of conditions stop at but-2-ene, and they give opposite geometries.

Lindlar catalyst is palladium on calcium carbonate that has been poisoned: partly deactivated on purpose. The poisons are lead and quinoline, a nitrogen compound that clings to the metal. The catalyst is active enough to reduce an alkyne and too weak to reduce the alkene that forms. It is still a surface reaction, so the addition is syn, and the product is the cis (Z) alkene: cis-but-2-ene from but-2-yne.

Sodium in liquid ammonia is a dissolving-metal reduction. Ammonia boils at −33 °C, so it is used as a cold liquid, and sodium dissolves in it. There is no surface. Instead the alkyne takes on an electron, then a proton, then a second electron and a second proton:

• Sodium gives the alkyne one electron. The result is a radical anion: one carbon carries an unpaired electron, and the other carries a lone pair and a negative charge.
• The negative carbon takes a proton (H⁺) from ammonia. That leaves a vinyl radical, a radical on a carbon of a C=C.
• A second sodium atom gives that carbon one more electron, making a vinyl anion: a lone pair and a negative charge on a carbon of a C=C.
• The vinyl anion takes a second proton from ammonia, and the alkene is finished.

The vinyl radical flips between its two bent shapes very fast, and it spends most of its time in the trans shape, where the two methyls are apart. The vinyl anion flips much more slowly, so it keeps the trans shape until ammonia protonates it. The proton goes where the lone pair was, and the product is the trans (E) alkene: trans-but-2-ene from but-2-yne.

Heats of hydrogenation, as a measuring tool

But-1-ene, cis-but-2-ene and trans-but-2-ene all take up H₂ to give the same product, butane. Each releases heat as it does so, and the three amounts differ:

Compare the three values rather than reading any one alone. All three end at the same molecule, so the one that gives out the least heat must have started lowest. trans-But-2-ene releases 2.7 kcal/mol (11 kJ/mol) less than but-1-ene, so it was 2.7 kcal/mol more stable to begin with. The 1.0 kcal/mol (4 kJ/mol) gap between cis and trans is the cost of two methyls crowding each other on the same side.

The order that falls out, trans-disubstituted > cis-disubstituted > monosubstituted, is the alkene stability order from Alkene structure, which lists the same values as ΔH° with a minus sign. This reaction is how that order was measured. The comparison only works between alkenes that give the same alkane. 2-Methylpropene gives 2-methylpropane rather than butane, so its 28.4 kcal/mol (119 kJ/mol) is not strictly on this scale.

The same measurement comes back later. Conjugation & Pericyclic Reactions uses it to show that buta-1,3-diene is about 15 kJ/mol (about 3.6 kcal/mol) more stable than two separate double bonds would be.

What carries forward

Two ideas carry forward. First, stereochemistry: hydrogenation is syn, and the Lindlar and dissolving-metal pair is the standard way to build an alkene of known geometry from an alkyne, which is a large part of why alkynes are useful in synthesis. Second, chemoselectivity: set the order above beside the hydride reagents of Oxidation & Reduction, and you can answer most questions of the form "which reagent, and what survives?"