Hydride Reductions

Hydride (H⁻) is the simplest nucleophile that can reduce a carbonyl. But free H⁻ is too reactive to handle; instead, chemists use borohydride or aluminum hydride reagents that deliver hydride in a controlled way. The two workhorses are NaBH₄ (mild) and LiAlH₄ (strong). Their selectivity difference is one of the most important practical distinctions in synthesis - and a recurring MCAT topic.

NaBH₄: The Mild Reducing Agent

Sodium borohydride (NaBH₄) reduces:

• Aldehydes → primary alcohols.
• Ketones → secondary alcohols.

NaBH₄ does NOT reduce:

• Esters, carboxylic acids, amides, nitriles (too unreactive for borohydride).
• Alkenes, alkynes.

Why selective? The B-H bond in NaBH₄ is only moderately polarized - the hydride is not strongly nucleophilic. It can attack highly electrophilic aldehydes and ketones but fails on less-reactive carbonyls (esters, amides) where the electron-donating OR or NR₂ groups reduce electrophilicity.

NaBH₄ tolerates water and alcohols as solvents (it is slow to react with them at 0°C to room temperature). Protic solvents are actually useful for the mechanism because they supply the proton needed to convert the alkoxide intermediate to the alcohol product.

LiAlH₄: The Strong Reducing Agent

Lithium aluminum hydride (LiAlH₄) reduces:

• Aldehydes → primary alcohols.
• Ketones → secondary alcohols.
• Carboxylic acids → primary alcohols.
• Esters → primary alcohols (+ alcohol byproduct from the OR group).
• Amides → amines (does NOT give alcohol; the amide nitrogen stays, gets a new C-H bond).
• Nitriles → primary amines.
• Alkyl halides → alkanes (rarely used for this).
• Epoxides → alcohols.

LiAlH₄ does NOT reduce:

• Alkenes (slowly, if at all).
• Benzene rings (no).

LiAlH₄ is a much stronger reducing agent because Al-H bonds are more polarized than B-H bonds (Al is less electronegative than B, so the H is more hydride-like).

LiAlH₄ is extremely sensitive to water, alcohols, and any acidic proton. It reacts violently with these, producing H₂ gas and potentially fires. It must be used in dry anhydrous ether or THF. Workup with aqueous acid (or a controlled quench with ethyl acetate) is required at the end.

The Mechanism

Both reagents deliver hydride to the carbonyl carbon via the same basic mechanism:

1. Hydride attacks the carbonyl carbon. The nucleophilic hydride pushes into the electrophilic C. The pi bond breaks, electrons flow to oxygen. Tetrahedral alkoxide intermediate forms.
2. Workup. The alkoxide is protonated by water or acid during workup, giving the alcohol.

For NaBH₄: the mechanism is usually drawn with direct hydride delivery from borohydride, with the borate byproduct handling subsequent equivalents. NaBH₄ can deliver up to four hydrides per molecule (4 H per B).

For LiAlH₄: similar, with AlH₃, AlH₂⁻, AlH⁻, and Al intermediates as the hydrides are delivered one at a time. LiAlH₄ also delivers 4 H per Al.

Stereochemistry

Both NaBH₄ and LiAlH₄ attack the carbonyl from whichever face is more accessible. For most substrates, this gives a roughly random approach and racemic products (or the more stable diastereomer for chiral substrates).

For stereospecific reductions, asymmetric reducing agents (CBS catalyst, Corey-Bakshi-Shibata; BINAL-H) or chiral ligand systems are used. These are rarely on the MCAT.

Practical Selectivity Table

Choosing the Right Reagent

• Need to reduce only aldehydes/ketones, leave esters/amides alone: NaBH₄.
• Need to reduce an ester to a primary alcohol: LiAlH₄.
• Need to reduce an amide to an amine: LiAlH₄.
• Need to reduce a nitrile to a primary amine: LiAlH₄.
• Need to reduce a ketone but protect a nearby ester: NaBH₄.

The MCAT almost always gives you the substrate and asks which product forms. If you recall the selectivity table, you win every time.