Reduction of Organic Compounds
Quick Notes
- Reduction = gain of hydrogen / loss of oxygen / decrease in oxidation state
- Carboxylic acids are reduced to aldehydes, then to primary alcohols
- Ketones are reduced to secondary alcohols
- Hydride ions (H⁻) are responsible for adding hydrogen in reduction
Full Notes
What Is Reduction in Organic Chemistry?
Organic reduction typically involves:
- Addition of a carbon–hydrogen bond
- Removal of a carbon–oxygen bond
- A decrease in the oxidation state of the carbon atom
Carbonyl compounds (like aldehydes, ketones, and carboxylic acids) can be reduced using reducing agents that act as a source of hydride (H⁻) ions.
Role of Hydride Ions (H⁻)
Hydride ions (H⁻) deliver hydrogen to the carbonyl group (C=O), converting it into an –OH group.
Hydrogen atoms (including the H⁻ from the reducing agent) are represented as [H] in the balanced equation.
Note: Names of specific reducing agents (e.g. LiAlH₄ or NaBH₄) are not required for your IB exams.
Reduction of Carboxylic Acids
Carboxylic acids can be reduced in two steps:
For Example: Ethanoic acid reduction
The reduction of ethanoic acid to ethanol occurs in two steps and can be written using [H] to represent the reducing agent:
- CH₃COOH + 2[H] → CH₃CHO + H₂O (ethanoic acid → ethanal)
- CH₃CHO + 2[H] → CH₃CH₂OH (ethanal → ethanol)
Overall: CH₃COOH + 4[H] → CH₃CH₂OH + H₂O
This shows the full reduction of a carboxylic acid to a primary alcohol, via the aldehyde intermediate, using a strong reducing agent like LiAlH₄.
Reduction of Ketones
Ketones can be reduced to secondary alcohols:
For Example: Propanone to propan-2-ol
CH₃COCH₃ + [H] → CH₃CH(OH)CH₃
Summary
- Reduction adds hydrogen, removes oxygen, or lowers oxidation state.
- Carboxylic acids reduce stepwise to aldehydes then to primary alcohols.
- Ketones reduce directly to secondary alcohols.
- Hydride ions deliver hydrogen to carbonyl compounds.
- Oxidation states can track changes and confirm the order of oxidation.
Structure 3.1 — Linked Course Question
How can oxidation states be used to show that the following molecules are given in increasing order of oxidation: CH₄, CH₃OH, HCHO, HCOOH, CO₂?
We can use the oxidation state of the carbon atom in each molecule to show how oxidised it is. Oxidation is defined as an increase in oxidation number, which corresponds to the carbon forming more bonds to oxygen or fewer bonds to hydrogen.
Let’s calculate the oxidation number of carbon in each compound:
- CH₄ (methane): H is +1; total of +4, so C must be –4
- CH₃OH (methanol): 3 H (+1) and 1 OH group (O = –2, H = +1) → total = +4 – 2 = +2, so C must be –2
- HCHO (methanal): 1 H (+1), 1 =O (–2), and 1 –H (+1) = total +2 – 2 = 0, so C is 0
- HCOOH (methanoic acid): 1 –H (+1), 1 =O (–2), and 1 –OH (O = –2, H = +1) → total = +2 – 4 = –2, so C must be +2
- CO₂ (carbon dioxide): 2 O atoms, each –2 = –4, so C must be +4
Conclusion: The oxidation number of carbon increases across the series:
CH₄ (–4) < CH₃OH (–2) < HCHO (0) < HCOOH (+2) < CO₂ (+4)
This confirms that the molecules are listed in increasing order of oxidation, as the carbon atom is progressively more oxidised.