Catalytic hydrogenation — Oxidation & Reduction

Hydride reagents cannot reduce an isolated C=C, because a hydride is a nucleophile and an alkene has no electrophilic end to attack. Reducing a carbon–carbon multiple bond needs a different mechanism entirely: adsorb both the alkene and hydrogen gas onto a metal surface and let them combine there.

The basic reaction

H₂ with a metal catalyst — palladium on carbon (Pd/C), platinum, or Raney nickel — adds two hydrogens across a double bond to give an alkane. The catalyst is essential: hydrogen gas and an alkene in a flask together do nothing at room temperature, because breaking the H–H bond costs too much without a surface to do it on.

The mechanism explains the one stereochemical fact you need. Both hydrogens are delivered from the metal surface, which the alkene is lying against, so they can only arrive on the face that is touching it. Hydrogenation is therefore a syn addition: the two new C–H bonds form on the same face.

On a ring, syn addition is directly visible in the product. Hydrogenating a 1,2-disubstituted cyclohexene gives the cis product, because both hydrogens arrive on the same face of the ring. That is a stereochemical claim you can be examined on, and it comes straight from the surface mechanism.

What gets reduced, and in what order

Catalytic hydrogenation is powerful but not indiscriminate, and the ordering matters more than the list:

Readily reduced: alkenes, alkynes (all the way to alkanes), imines, nitro groups (to amines), and azides.
Requires forcing conditions: ketones and aldehydes need higher pressure and a more active catalyst such as PtO₂. Under ordinary conditions over Pd/C they survive.
Essentially untouched: esters, carboxylic acids and amides. Benzene rings need severe conditions — high pressure and temperature — which is why an aromatic ring survives an ordinary hydrogenation intact.

That ordering is what makes the reaction useful rather than merely strong. A molecule containing an alkene, a ketone and a benzene ring is hydrogenated at the alkene alone over Pd/C at one atmosphere — the exact complement of the hydride reagents from the last section, which reduce the ketone and leave the alkene.

Stopping an alkyne halfway, in either geometry

An alkyne over ordinary Pd/C is reduced past the alkene straight to the alkane. Two modified conditions stop it at the alkene, and they give opposite geometries — which makes the pair one of the most useful stereochemical tools in the course.

Lindlar catalyst — palladium on calcium carbonate, poisoned with quinoline and lead. Deliberately deactivated so it reduces an alkyne to an alkene and no further. It is still a surface reaction, so the addition is syn, and the product is the cis (Z) alkene.
Dissolving metal reduction — sodium in liquid ammonia. Not a surface reaction at all: it proceeds through radical anion intermediates that adopt the less hindered arrangement, so the two hydrogens end up on opposite sides and the product is the trans (E) alkene.

Learn these two as a pair with the geometry attached, because that is how they are examined: "make the cis alkene" means Lindlar and "make the trans alkene" means Na/NH₃. One alkyne, two conditions, two stereoisomers you can choose between.

Worked example — three reagents, one molecule

A molecule contains a disubstituted alkene, a ketone and a benzene ring. Give the product of each:

H₂, Pd/C, 1 atm: the alkene is reduced to an alkane. The ketone and the ring survive.

NaBH₄, then aqueous workup: the ketone becomes a secondary alcohol. The alkene and the ring survive.

H₂, PtO₂, high pressure: now the ketone is reduced too, and the alkene certainly is. The benzene ring still needs more than this.

The lesson is the one running through this whole chapter: the substrate does not decide the product on its own. The reagent does.

Heats of hydrogenation, as a measuring tool

The same internal alkyne, three sets of conditions. Lindlar and sodium in ammonia both stop at the alkene and hand you opposite geometries; ordinary Pd/C does not stop at the alkene at all.The split comes from where the hydrogens are delivered. Lindlar is a deliberately poisoned surface, and an alkene lying against a surface can only be reached from the face touching it, so both hydrogens arrive on that face and the product is cis. Sodium in ammonia never uses a surface: it adds an electron, then a proton, twice over, and the geometry is fixed at the vinyl anion formed by the second electron transfer, which is configurationally stable and sits with its two R groups apart. The radical before it inverts far too fast to decide anything. Same two hydrogens, same alkyne, opposite answers.

Because hydrogenating any alkene gives the same kind of product — an alkane — the heat released is a direct measure of how stable the alkene was to begin with. A more stable alkene starts lower down and releases less energy.

This is the measurement behind two claims made earlier in the course. It is how alkene stability was shown to rise with substitution, and it is how the conjugation chapter put a number — about 15 kJ/mol — on the extra stability of buta-1,3-diene against two isolated double bonds. The reaction is useful as a ruler as well as a transformation.

What carries forward

Two things. The syn stereochemistry is a genuine stereochemical handle, and the Lindlar/dissolving-metal pair is the standard way to build a defined alkene geometry from an alkyne — which is why alkynes are so useful as synthetic intermediates. And the chemoselectivity ordering, held next to the hydride table from the last section, is most of what you need to answer "which reagent, and what survives".