Choosing reagents and conditions

Which reagent, and why that one?

A chemist rarely changes one group in a molecule that has only one group. A keto ester has a ketone and an ester. A flavour compound has a double bond and an alcohol. The task is to change one group and leave the others alone. So the question for each reagent is: what does it do, and what does it leave?

Oxidising agents and where they stop

Oxidation takes hydrogen away from the carbon that holds the OH. How far it goes depends on the reagent and the conditions.

• PCC in dichloromethane: a primary alcohol gives an aldehyde, a secondary alcohol gives a ketone, a tertiary alcohol gives nothing.
• Acidified $\text{K}_{2}\text{Cr}_{2}\text{O}_{7}$, product distilled out: the same as PCC: aldehyde, ketone, nothing.
• Acidified $\text{K}_{2}\text{Cr}_{2}\text{O}_{7}$, heated under reflux: a primary alcohol gives a carboxylic acid, a secondary alcohol gives a ketone, a tertiary alcohol gives nothing.
• $\text{KMnO}_{4}$, hot: a carboxylic acid, a ketone, and nothing (unless the conditions are forced).

Distil or reflux? The aldehyde boils lower than the alcohol, because it cannot hydrogen-bond to itself. Ethanal boils at $21\ ^\circ\text{C}$ and ethanol at $78\ ^\circ\text{C}$. Add the alcohol drop by drop to hot acidified dichromate and the ethanal distils off as it forms, before it is oxidised further. To get the acid, heat the alcohol with excess dichromate under reflux, so that nothing escapes and the aldehyde stays in the flask until it is oxidised.

The colour tells you the stage. Orange dichromate (Cr$^{6+}$) turns green as Cr$^{3+}$ forms. Purple permanganate in acid turns almost colourless as $\text{Mn}^{2+}$ forms.

Reducing agents and what they leave

• Aldehyde or ketone C=O: reduced by $\text{NaBH}_{4}$ in ethanol, by $\text{LiAlH}_{4}$ in dry ether, and by $\text{H}_{2}$ over Ni at pressure.
• Ester: reduced by $\text{LiAlH}_{4}$ only.
• Carboxylic acid: reduced by $\text{LiAlH}_{4}$ only.
• Isolated C=C: reduced by $\text{H}_{2}$ over Pd or Ni only. Neither hydride touches it.

So NaBH$_4$ can reduce the ketone of a keto ester and leave the ester. LiAlH$_4$ reduces both, and hydrogen over palladium reduces only a C=C double bond. LiAlH$_4$ reacts violently with water and alcohols, so it is used in dry ether, and the work-up with water comes after the reduction. NaBH$_4$ is gentle enough to be used in ethanol or water.

Counting electrons: dichromate and ethanol

Each carbon of ethanol that loses two hydrogens (to the aldehyde) gives up 2 electrons. Going on to the acid takes 4 electrons in all. One dichromate ion, $\text{Cr}_{2}\text{O}_{7}^{2-}$, takes 6 electrons, to make two $\text{Cr}^{3+}$.

The first needs one dichromate for three ethanol, and the second needs two. This is the chemistry of the old breath test, where orange dichromate turned green in proportion to the alcohol.

The line for the acid is twice as high as the line for the aldehyde.

A back titration

A sample of unknown ethanol content is added to a measured amount of dichromate, in excess. When the reaction is over, the dichromate left is measured with iron(II) of known concentration:

The dichromate that reacted with the sample is what was added, minus what was left. Then the ethanol follows from the 3 : 2 ratio of the acid equation.

Atom economy

The atom economy of a reaction is $\dfrac{M_{\text{wanted}}}{M_{\text{all reactants}}} \times 100$, using the balanced equation. A reaction that uses a heavy oxidising agent has a low atom economy, because most of the reactant mass ends in waste. Dehydrogenation of ethanol over copper, $\text{CH}_{3}\text{CH}_{2}\text{OH} \to \text{CH}_{3}\text{CHO} + \text{H}_{2}$, has a high one, because the only other product is hydrogen.

Practise with new numbers

Try it in Python

The program works out the dichromate for a sample of ethanol both ways, then the ethanol found by a back titration.

Common mistakes

• Using one ratio for both products. The ethanal needs one dichromate for three ethanol, and the acid needs two.
• Heating the alcohol under reflux and expecting an aldehyde. The aldehyde is oxidised before it can leave. Distil it out.
• Using NaBH$_4$ to reduce an ester or an acid. It does not. Use LiAlH$_4$.
• Adding water to LiAlH$_4$ before the reduction. It destroys the reagent.
• Forgetting the iron(II) ratio. One dichromate takes six Fe$^{2+}$.

Remember these

• PCC stops at the aldehyde. Hot dichromate under reflux goes on to the acid. Tertiary alcohols stay.
• NaBH$_4$ reduces only aldehydes and ketones. LiAlH$_4$ also reduces acids and esters. Hydrogen over Pd reduces C=C.
• Ethanal: 3 ethanol to 1 dichromate. Ethanoic acid: 3 ethanol to 2 dichromate. Dichromate to iron(II): 1 to 6.
• Atom economy compares the wanted product with all the reactants.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. Which reagent changes the ketone of a keto ester $\text{CH}_{3}\text{COCH}_{2}\text{CH}_{2}\text{COOCH}_{3}$ into an alcohol and leaves the ester?

2. Ethanol is added dropwise to hot acidified dichromate in a flask fitted for distillation. What is collected, and why that?

3. What colour change shows that dichromate has oxidised an alcohol?

4. How many moles of Fe$^{2+}$ react with 0.0010 mol of dichromate?

References

• A-Level and IB Chemistry texts on the oxidation of alcohols, with the usual distillation and reflux practicals.
• Morrison and Boyd, Organic Chemistry, the chapters on alcohols.
• OpenStax Chemistry 2e, Chapter 20, section 20.2 (the oxidation of alcohols).