📘 IB Understanding
Reduction reactions represent the reverse of oxidation pathways. You must know the reagents and products for the reduction of carbonyl compounds and nitrobenzene, and be able to combine reactions to design synthesis routes.
Reduction of Carbonyl Compounds
Aldehydes, ketones, and carboxylic acids can all be reduced back to alcohols. Because hydrogen is added to the molecule, we represent the reducing agent with the symbol [H] in balanced equations.
Reducing Agents
The choice of reducing agent depends on the strength required to reduce the specific functional group:
- Sodium borohydride (\(\text{NaBH}_4\)): A milder reducing agent. It is safe to use in aqueous or alcoholic solutions (like ethanol). It is strong enough to reduce aldehydes and ketones, but is not strong enough to reduce carboxylic acids.
- Lithium aluminium hydride (\(\text{LiAlH}_4\)): A much more powerful reducing agent. It reacts violently with water and alcohols, so it must be dissolved in a completely dry organic solvent like dry ether (ethoxyethane) at room temperature. The reaction is followed by the addition of a dilute acid (such as \(\text{H}_2\text{SO}_4\)) to release the alcohol. It is strong enough to reduce carboxylic acids, as well as aldehydes and ketones.
Summary of Carbonyl Reductions
| Starting Material | Reducing Agent | Product | Equation |
|---|---|---|---|
| Aldehyde (\(\text{R-CHO}\)) | \(\text{NaBH}_4\) or \(\text{LiAlH}_4\) | Primary alcohol | \(\text{R-CHO} + 2[\text{H}] \rightarrow \text{R-CH}_2\text{OH}\) |
| Ketone (\(\text{R-CO-R'}\)) | \(\text{NaBH}_4\) or \(\text{LiAlH}_4\) | Secondary alcohol | \(\text{R-CO-R'} + 2[\text{H}] \rightarrow \text{R-CH(OH)-R'}\) |
| Carboxylic Acid (\(\text{R-COOH}\)) | \(\text{LiAlH}_4\) (in dry ether) | Primary alcohol | \(\text{R-COOH} + 4[\text{H}] \rightarrow \text{R-CH}_2\text{OH} + \text{H}_2\text{O}\) |
Reduction of Nitrobenzene
Nitrobenzene (\(\text{C}_6\text{H}_5\text{NO}_2\)) can be reduced to form phenylamine (\(\text{C}_6\text{H}_5\text{NH}_2\), also known as aniline). Phenylamine is an important starting material in the manufacture of dyes and pharmaceuticals. This reduction is carried out in a two-stage process:
1Stage 1: Reflux with Tin and Concentrated Hydrochloric Acid
Nitrobenzene is heated under reflux with a mixture of tin (\(\text{Sn}\)) and concentrated \(\text{HCl}\). The tin acts as a reducing agent in the acidic environment. Because of the strongly acidic conditions, the product formed is the protonated phenylammonium ion (\(\text{C}_6\text{H}_5\text{NH}_3^+\)) rather than free phenylamine:
2Stage 2: Reaction with Sodium Hydroxide
Sodium hydroxide solution (\(\text{NaOH}\)) is added to the reaction mixture. The hydroxide ions (\(\text{OH}^-\)) remove a proton from the phenylammonium ion, liberating the free, insoluble phenylamine, which appears as oily droplets:
Organic Synthesis & Reaction Pathways
Organic synthesis involves designing a sequence of reactions to convert a starting material into a desired target molecule. In the IB exam, you may be asked to outline a multi-step pathway, stating the reagents and conditions for each step.
The Reaction Pathway Map
The diagram below summarises all the key reaction pathways you must know for both Standard Level and Higher Level organic chemistry:
Worked Example: Designing a Synthesis Route
💡 Practice Problem
Outline a reaction pathway to synthesize ethyl ethanoate starting from ethene. Include reagents and conditions for each step.
Step 1: Convert Ethene to Ethanol
- Reagents: Steam (\(\text{H}_2\text{O}\))
- Conditions: Phosphoric acid (\(\text{H}_3\text{PO}_4\)) catalyst, 300°C, 60 atm
- Equation: \(\text{CH}_2\text{=CH}_2(g) + \text{H}_2\text{O}(g) \rightarrow \text{CH}_3\text{CH}_2\text{OH}(l)\)
Step 2: Convert a portion of the Ethanol to Ethanoic Acid
- Reagents: Acidified potassium dichromate (\(\text{K}_2\text{Cr}_2\text{O}_7\text{ / H}_2\text{SO}_4\))
- Conditions: Heat under reflux
- Equation: \(\text{CH}_3\text{CH}_2\text{OH}(l) + 2[\text{O}] \rightarrow \text{CH}_3\text{COOH}(aq) + \text{H}_2\text{O}(l)\)
Step 3: React Ethanol with Ethanoic Acid to form the Ester
- Reagents: Concentrated sulfuric acid (\(\text{H}_2\text{SO}_4\)) catalyst
- Conditions: Heat gently under reflux
- Equation: \(\text{CH}_3\text{COOH}(aq) + \text{CH}_3\text{CH}_2\text{OH}(l) \rightleftharpoons \text{CH}_3\text{COOCH}_2\text{CH}_3(l) + \text{H}_2\text{O}(l)\)
⚠️ Common Exam Mistakes
- Using \(\text{NaBH}_4\) to reduce a carboxylic acid. It is not a strong enough reducing agent; you must specify \(\text{LiAlH}_4\) in dry ether.
- Forgetting to state that the reduction of nitrobenzene requires a second step (adding \(\text{NaOH}\)) to release the free phenylamine from the acidic reaction mixture.
- Using incorrect symbols: \([\text{H}]\) represents a reducing agent (hydrogen atoms), whereas \([\text{O}]\) represents an oxidising agent (oxygen atoms).
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