21.4 Chemistry of Acid Halides

Preparation and Reactions of Acid Halides

Acid halides are formed by replacing the $-OH$ group of a carboxylic acid with a halogen atom. This swap dramatically increases reactivity: the halogen is a far better leaving group than hydroxide, and its electron-withdrawing effect makes the carbonyl carbon strongly electrophilic. That combination is why acid halides sit at the top of the carboxylic acid derivative reactivity ladder and serve as starting points for synthesizing nearly every other derivative.

Preparation of Acid Halides

The general strategy is straightforward: treat a carboxylic acid with a reagent that swaps $-OH$ for a halide.

Acid chlorides are prepared using thionyl chloride ($SOCl_2$):

1. The carboxyl oxygen attacks the electrophilic sulfur of $SOCl_2$, displacing a chloride ion.
2. That chloride ion then attacks the carbonyl carbon, replacing the activated leaving group.
3. The byproducts are $SO_2$ and $HCl$, both gases that bubble out of the reaction mixture. This drives the equilibrium forward and makes purification easy.

Acid bromides are prepared using phosphorus tribromide ($PBr_3$):

1. The carboxyl oxygen attacks the electrophilic phosphorus of $PBr_3$, displacing a bromide ion.
2. The bromide ion attacks the carbonyl carbon, forming the acid bromide.
3. Byproducts are $HBr$ and phosphorous acid ($H_3PO_3$).

Acid chlorides are far more common in practice because $SOCl_2$ is inexpensive and the gaseous byproducts simplify workup.

Structure and Reactivity

An acid halide has an acyl group ($RCO-$) bonded directly to a halogen. Two features work together to make these compounds exceptionally reactive:

• Electrophilic carbonyl carbon. The electronegative halogen pulls electron density away from the already-polarized $C=O$, making the carbon a strong electrophile.
• Excellent leaving group. Halide ions (especially $Cl^-$) are stable and weakly basic, so they depart easily once a nucleophile attacks.

Because of this, acid halides react with a wide range of nucleophiles through the nucleophilic acyl substitution mechanism: a nucleophile attacks the carbonyl carbon, forming a tetrahedral intermediate, and the halide departs to restore the $C=O$.

Nucleophilic Acyl Substitution Reactions

Every reaction below follows the same core mechanism. A nucleophile attacks the electrophilic carbonyl carbon, a tetrahedral intermediate forms, and the halide leaves. What changes is the nucleophile, which determines the product.

• Water → Carboxylic acid. Water attacks the carbonyl carbon, displacing the halide. A proton transfer gives the carboxylic acid. This reaction (hydrolysis) is so favorable that acid halides fume in moist air.
• Carboxylate ion → Anhydride. A carboxylate ion ($RCOO^-$) attacks the carbonyl carbon, displacing the halide. This is the standard lab method for preparing anhydrides.
• Alcohol → Ester. An alcohol attacks the carbonyl carbon, displacing the halide. A base (often pyridine or triethylamine) is added to scavenge the $HX$ produced, preventing it from protonating the alcohol or causing side reactions.
• Amine → Amide. An amine ($RNH_2$ or $R_2NH$) attacks the carbonyl carbon, displacing the halide. Because $HX$ is generated, you typically need two equivalents of the amine: one to react, and one to neutralize the acid. Alternatively, a non-nucleophilic base can serve as the acid scavenger.

Reduction and Grignard Reactions

These reactions don't follow the simple acyl substitution pattern. Instead, they involve nucleophilic addition steps and produce alcohols.

Reduction with $LiAlH_4$ → Primary alcohol

1. A hydride ion ($H^-$) from $LiAlH_4$ attacks the carbonyl carbon, displacing the halide and forming an aldehyde intermediate.
2. A second hydride attacks the aldehyde carbonyl, giving an alkoxide.
3. Aqueous workup protonates the alkoxide to yield the primary alcohol.

The aldehyde intermediate is actually more reactive than the starting acid halide, so it reacts with the second equivalent of hydride immediately. You can't stop at the aldehyde stage with $LiAlH_4$. (To isolate the aldehyde, use a milder reducing agent like $DIBAL-H$ at low temperature or lithium tri-tert-butoxyaluminum hydride.)

Reaction with Grignard reagents ($RMgX$) → Tertiary alcohol

1. The alkyl group ($R^-$) of the Grignard reagent attacks the carbonyl carbon, displacing the halide and forming a ketone intermediate.
2. A second equivalent of the Grignard reagent attacks the ketone carbonyl, forming an alkoxide.
3. Aqueous workup protonates the alkoxide to yield a tertiary alcohol with two identical $R$ groups from the Grignard reagent.

Just like with $LiAlH_4$, the ketone intermediate is more electrophilic than the starting acid halide, so a second addition occurs readily. The product is a tertiary alcohol bearing two identical substituents from the Grignard reagent plus the original $R$ group from the acid halide.