Hydrolysis of Nitriles

This page looks at the hydrolysis of nitriles under either acidic or alkaline conditions.

When nitriles are hydrolysed you can think of them reacting with water in two stages - first to produce an amide, and then the ammonium salt of a carboxylic acid.

Acidic hydrolysis of nitriles

• The hydronium ion is a strong enough acid to react with the p bond of the cyano group

•  The cyano group has a partially negative nitrogen atom due to the higher electronegativity of the nitrogen atom. The resonance dipole of the cyano group also contributes to the negative character of the carbon atom.

• This is one of the resonance forms of this cation and, in this form, the carbon has a positive charge and is a strong Lewis acid.

• Water is a good enough nucleophile to attack strong Lewis acids such as carbocations.

• N-hydrogen imines are unstable bases and will readily react with the protonated hydroxyl group (which is a strong acid) to form the resonance stabilized carbocation.

• The water will again attack the positive carbon atom, which is a strong Lewis acid.

•  This product is a combination of a base (the amino group) and a strong acid (the protonated alcohol group) and will readily undergo an acid base reaction…shown as an intramolecular reaction here…it can also be an intermolecular reaction.

• This reaction produced a protonated amine which can act like a leaving group. An electron pair from the hydroxyl group can form an oxygen-carbon double bond and break the carbon-nitrogen bond. This will yield a resonance stabilized cation.

• This step of the reaction is a rapid acid base reaction of ammonia with the protonated carboxylic acid, which is a strong acid. This reaction yields the carboxylic acid and the non-nucleophilic ammonium ion.

Alkaline hydrolysis of nitriles

• The cyano group has an electropositive carbon atom due to the higher electronegativity of the nitrogen atom. The resonance dipole of the cyano group also contributes to the electropositive character of the carbon atom.

• The hydroxide anion is a strong enough base to react with the electropositive carbon atom of the cyano group. A double barbed arrow should be drawn from the hydroxy groups electron pair to the carbon atom and a second double barbed arrow should be drawn showing the p electrons moving to the nitrogen atom.

• The imine anion is a strong base and will easily pull a hydrogen atom from a water molecule in the solution.

• The imine product will undergo tautomerization to form an amide. This is favored by the bond energies of the reactant versus the product.

• The amide carbonyl is electropositive and will be attacked by hydroxide anion. Use a double barbed arrow to show the movement of the electron pair from the hydroxide anion to the carbonyl carbon atom. Show another double barbed arrow to indicate the movement of the p electrons to the oxygen atom.

• The product of this reaction can reform the carbonyl and eliminate either a hydroxide anion or an amide anion. If the hydroxide ion is the leaving group, the amide is reformed. The amide anion is a much stronger base and is a very poor leaving group making this equilibrium unfavorable. A small amount can leave and this leads to the next step of the mechanism.

• This step of the mechanism drives the reaction. The acid-base reaction between the carboxylic acid and the amide anion is extremely favorable and fast, which helps drive the previous equilibrium to completion.

 

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~~forgiveness if there are mistakes~~