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Carboxylic Acids and Their Derivatives

Reduction and decarboxylation of carboxylic acids: reduction of carboxylic acids using LiAlH₄

  • 3 things to know
  • A2 Unit 4

Overview

Reducing carboxylic acids to primary alcohols with LiAlH₄ in dry ether, the equation using [H], why NaBH₄ cannot be used, and why the reduction cannot be stopped at the aldehyde.

What you need to know

Answer each one before you reveal it.

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Key term

Under what conditions is a reduction with LiAlH₄ carried out?

Key term

Why can NaBH₄ not be used in place of LiAlH₄ to reduce a carboxylic acid?

Key term

Can the reduction of a carboxylic acid be stopped at the aldehyde?

Questions from this subtopic

Each one with the model answer underneath.

  1. 1

    What is formed when a carboxylic acid is reduced by lithium tetrahydridoaluminate(III)?

    Model answer

    A primary alcohol is formed. The carboxyl group is reduced all the way to a CH₂OH group and a molecule of water is lost.

  2. 2

    Under what conditions is a reduction with LiAlH₄ carried out?

    Model answer

    The reagent is used in a dry ether solvent such as ethoxyethane at room temperature, with the apparatus kept free of moisture. The mixture is then treated with dilute acid in a separate work-up step to release the alcohol.

  3. 3

    Why must the solvent for a LiAlH₄ reduction be dry?

    Model answer

    Lithium tetrahydridoaluminate(III) reacts violently with water, giving hydrogen gas and destroying the reagent before it can reduce the organic compound. A dry ether dissolves the reagent without reacting with it.

  4. 4

    Write an equation, using [H], for the reduction of ethanoic acid.

    Model answer

    The equation is CH₃COOH + 4[H] → CH₃CH₂OH + H₂O. Four hydrogen atoms are needed because the reduction removes one of the oxygen atoms altogether.

  5. 5

    Name the product formed when propanoic acid is treated with LiAlH₄ in dry ether and the mixture is then acidified.

    Model answer

    Propan-1-ol is formed. The carboxyl carbon becomes the carbon bearing the hydroxyl group, so the number of carbon atoms is unchanged.

  6. 6

    Why can NaBH₄ not be used in place of LiAlH₄ to reduce a carboxylic acid?

    Model answer

    Sodium tetrahydridoborate(III) is a much milder reducing agent and can only reduce the strongly polarised carbonyl group of an aldehyde or ketone. The carbonyl carbon of a carboxylic acid is made less electron-deficient by the neighbouring hydroxyl oxygen, so it needs the far more powerful lithium tetrahydridoaluminate(III).

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