Aldehydes and ketones both contain the carbonyl functional group (\(\text{C=O}\)). However, their differing chemical environments lead to distinct behaviors when treated with oxidizing and reducing agents. This lesson covers how to oxidize aldehydes, how to reduce both types of carbonyls using sodium tetrahydridoborate(III) (\(\text{NaBH}_4\)), and how to carry out chemical tests to distinguish them in the laboratory.
🔑 Key Principle
Aldehydes are easily oxidized to carboxylic acids using acidified potassium dichromate(VI) or mild reagents like Tollens' and Fehling's. Ketones resist oxidation because breaking a C-C bond is energetically unfavorable. Both aldehydes and ketones are reduced to primary and secondary alcohols, respectively, by \(\text{NaBH}_4\) via a nucleophilic addition mechanism.
1. Oxidation of Aldehydes and Ketones
Aldehydes have a hydrogen atom attached to the carbonyl carbon atom (\(\text{R-CHO}\)). This hydrogen atom makes them highly susceptible to oxidation. When heated under reflux with an oxidizing agent, aldehydes are converted into carboxylic acids:
\[ \text{RCHO} + \text{[O]} \rightarrow \text{RCOOH} \]
The standard laboratory reagent for this oxidation is acidified potassium dichromate(VI), \(\text{K}_2\text{Cr}_2\text{O}_7 / \text{H}_2\text{SO}_4\). During the reaction, the orange dichromate(VI) ions (\(\text{Cr}_2\text{O}_7^{2-}\)) are reduced to green chromium(III) ions (\(\text{Cr}^{3+}\)).
Ketones have two alkyl groups attached to the carbonyl carbon (\(\text{R-CO-R}'\)). Because there is no hydrogen atom directly bonded to the carbonyl carbon, ketones cannot be oxidized without breaking a strong carbon-carbon single bond. Consequently, ketones resist oxidation under standard laboratory conditions. The mixture remains orange when heated with acidified potassium dichromate(VI).
2. Chemical Tests to Distinguish Carbonyls
The difference in ease of oxidation is exploited to distinguish between aldehydes and ketones in qualitative tests using mild oxidizing agents:
An aqueous solution of silver nitrate in excess ammonia, containing the complex ion \( \text{[Ag(NH}_3\text{)}_2\text{]}^+ \). It acts as a mild oxidizing agent to detect aldehydes.
An alkaline solution containing copper(II) ions complexed with tartrate ions. It appears dark blue and is used to oxidize aldehydes to carboxylic acids.
Test 1: Tollens' Reagent (The Silver Mirror Test)
When Tollens' reagent is warmed with an aldehyde, the aldehyde is oxidized to a carboxylate ion, and the silver(I) complex ions are reduced to metallic silver. The silver deposits on the inner wall of the test tube, forming a reflective silver mirror:
\[ \text{Ag(NH}_3\text{)}_2^+(\text{aq}) + \text{e}^- \rightarrow \text{Ag(s)} + 2\text{NH}_3(\text{aq}) \]
Ketones do not react, and the solution remains clear and colorless.
Test 2: Fehling's Solution
When Fehling's solution is warmed with an aldehyde, the copper(II) complex ions oxidize the aldehyde and are reduced to copper(I) oxide. The color changes from a clear blue solution to a brick-red precipitate:
\[ 2\text{Cu}^{2+}(\text{aq}) + 2\text{OH}^-(\text{aq}) + 2\text{e}^- \rightarrow \text{Cu}_2\text{O(s)} + \text{H}_2\text{O(l)} \]
Ketones do not react, and the solution remains clear blue.
3. Reduction of Aldehydes and Ketones
Aldehydes and ketones can be reduced back to alcohols using sodium tetrahydridoborate(III), \(\text{NaBH}_4\), in aqueous solution. In these equations, the reducing agent is represented by the symbol [H].
- Aldehydes are reduced to primary alcohols:
\[ \text{RCHO} + 2\text{[H]} \rightarrow \text{RCH}_2\text{OH} \] - Ketones are reduced to secondary alcohols:
\[ \text{RCOR}' + 2\text{[H]} \rightarrow \text{RCH(OH)R}' \]
Mechanism: Nucleophilic Addition (Reduction)
The reduction occurs via a nucleophilic addition mechanism. The tetrahydridoborate ion (\(\text{BH}_4^-\)) acts as a source of hydride ions (\(\text{H}^-\)). The hydride ion has a lone pair of electrons and acts as the nucleophile.
The steps of the mechanism are as follows:
- Attack on the carbonyl carbon: The hydride ion (\(\text{H}^-\)) attacks the electron-deficient carbonyl carbon (\(\text{C}^{\delta+}\)). Simultaneously, the polar \(\text{C=O}\) \(\pi\) bond breaks, with both electrons shifting to the oxygen atom. This forms a negatively charged alkoxide intermediate.
- Protonation: The lone pair of electrons on the negatively charged oxygen atom of the intermediate attacks a hydrogen atom in a water molecule (or a hydrogen ion from water), protonating the intermediate to form the final alcohol and regenerating a hydroxide ion (\(\text{OH}^-\)).
When drawing this mechanism:
1. Always label the partial charges (\(\delta+\) and \(\delta-\)) on the polar carbon and oxygen of the \(\text{C=O}\) bond.
2. Ensure the first arrow starts from the lone pair or negative charge of the hydride ion (\(\text{H}^-\)), pointing to the carbonyl carbon.
3. Ensure the second arrow starts from the \(\text{C=O}\) double bond itself and points to the oxygen atom.
4. Do not omit the intermediate alkoxide ion; the oxygen must show a negative charge (\(\text{O}^-\)) and a lone pair of electrons for the second step.
Describe two chemical tests that could be used to distinguish between them, including observations for each.
Solution:
Test 1: Tollens' Reagent (The Silver Mirror Test)
- Add a few drops of Tollens' reagent to a sample of each compound in separate test tubes and warm gently in a water bath.
- Observation with pentanal (aldehyde): A shiny silver mirror forms on the inner wall of the test tube.
- Observation with pentan-2-one (ketone): No change; the solution remains clear and colorless.
Test 2: Fehling's Solution Test
- Add a few drops of blue Fehling's solution to a sample of each compound in separate test tubes and warm in a water bath.
- Observation with pentanal (aldehyde): The blue solution turns green and then forms a brick-red precipitate.
- Observation with pentan-2-one (ketone): No change; the solution remains blue.
Draw the structure of the product and name it.
Step 1: Write the balanced chemical equation.
Since propanone is a ketone, it will be reduced to a secondary alcohol. Two hydrogen atoms are added (one to the carbonyl carbon, and one to the carbonyl oxygen):
\[ \text{CH}_3\text{COCH}_3 + 2\text{[H]} \rightarrow \text{CH}_3\text{CH(OH)CH}_3 \]
Step 2: Draw the product structure and name it.
The product has a three-carbon chain with a hydroxyl group (\(\text{-OH}\)) on carbon-2:
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H - C - C - C - H
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H H H
Product Name: Propan-2-ol (a secondary alcohol).