Reduction Methods: Choosing a Reagent That Reduces Only What You Want

The two hydride reagents

Note also that neither reduces an isolated alkene. Carbon–carbon double bonds are not electrophilic enough for hydride attack, which is why catalytic hydrogenation is needed for them.

Catalytic hydrogenation

Hydrogen with a metal catalyst reduces alkenes, alkynes, nitro groups and aromatic rings under forcing conditions. Addition of hydrogen occurs from the same face of the molecule, so the stereochemistry is syn.

Deliberately poisoning a catalyst to reduce its activity is the standard way to stop at the alkene stage. Choosing between the three rows to obtain a required geometry is a routine question.

Other reductions worth knowing

• Clemmensen reduces a carbonyl all the way to a methylene group under acidic conditions.
• Wolff–Kishner does the same under basic conditions.
• Rosenmund reduces an acid chloride to an aldehyde using a partially poisoned catalyst, stopping before the alcohol.
• Dissolving metal reduction gives trans alkenes from alkynes and reduces aromatic rings under Birch conditions.

The Clemmensen and Wolff–Kishner pair is a favourite question because they achieve the same transformation under opposite conditions. The choice is made by what else the molecule contains: an acid-sensitive group rules out Clemmensen, a base-sensitive one rules out Wolff–Kishner. Stating that reasoning is the expected answer.

Why the two hydrides differ

The aluminium–hydrogen bond is longer and weaker than the boron–hydrogen bond, and aluminium is less electronegative than boron. Both factors make the hydride in lithium aluminium hydride far more loosely held and therefore a much stronger nucleophile.

That single difference explains the whole selectivity table. A ketone is electrophilic enough for either reagent. An ester is less electrophilic, because the ester oxygen donates electron density into the carbonyl, so only the stronger donor reaches it. A carboxylic acid is worse still, since the reagent is first consumed deprotonating the acidic hydrogen.

The same reasoning predicts the solvent restriction. A stronger hydride donor is also a stronger base, so lithium aluminium hydride reacts violently with any protic solvent, while borohydride is mild enough to be used in alcohols and even in water.

A decision procedure

1. Identify what must be reduced and to what level.
2. List everything else in the molecule that a reducing agent might attack.
3. Choose the mildest reagent that achieves the target — borohydride before aluminium hydride, poisoned catalyst before ordinary.
4. Check the conditions against any acid- or base-sensitive groups.
5. If nothing is selective enough, protect the vulnerable group first.

Step three is the general principle of the whole topic: use the weakest reagent that will do the job, because every increase in strength is an increase in what else gets attacked.

Frequently asked questions

Why does sodium borohydride not reduce esters?

Because it is a much weaker hydride donor, and an ester carbonyl is less electrophilic than a ketone due to donation from the ester oxygen. The combination puts it out of reach.

Why must lithium aluminium hydride be used in anhydrous conditions?

Because it reacts violently with water and with any acidic hydrogen, destroying the reagent and releasing hydrogen gas.

How do I reduce a ketone in the presence of an ester?

Use sodium borohydride, which reduces the ketone and leaves the ester untouched.

Why can neither hydride reduce an isolated alkene?

Because a carbon–carbon double bond is not sufficiently electrophilic to be attacked by hydride. It requires catalytic hydrogenation or dissolving metal conditions.