Carboxylic Acids and Derivatives: Question 1

Worked solution

Two different outcomes from oxidising a primary alcohol

A primary alcohol can be oxidised to two different products depending on how the reaction is carried out:

• Distillation as the aldehyde forms removes it from the reaction mixture before the oxidising agent can react with it again, so the reaction stops at the aldehyde (partial oxidation).
• Refluxing with an excess of oxidising agent keeps everything in contact for long enough that the intermediate aldehyde is oxidised a second time, all the way to the carboxylic acid (full oxidation).

Why A is correct

Heating hexan-1-ol under reflux with an excess of acidified potassium dichromate(VI) gives full oxidation to hexanoic acid. Using $[\text{O}]$ to represent oxygen from the oxidising agent:

$\text{CH}_3(\text{CH}_2)_4\text{CH}_2\text{OH} + 2[\text{O}] \rightarrow \text{CH}_3(\text{CH}_2)_4\text{COOH} + \text{H}_2\text{O}$

As the reaction proceeds, the orange dichromate(VI) ions, $\text{Cr}_2\text{O}_7^{2-}$, are reduced to green $\text{Cr}^{3+}$ ions, giving a visible colour change that confirms oxidation is occurring. Reflux (heating with continuous condensation and return of vapour to the flask) ensures the volatile intermediate aldehyde, hexanal, is not lost before it can be oxidised further.

Why the other options are wrong

• B describes the method for preparing the aldehyde (hexanal) only: distilling the product off as it forms removes it before the second oxidation step can occur, so the carboxylic acid is never obtained.
• C is wrong because $\text{LiAlH}_4$ is a powerful reducing agent, not an oxidising agent. It could reduce a carboxylic acid to a primary alcohol (the reverse of what is wanted here), but it cannot oxidise an alcohol to an acid.
• D is wrong because aqueous sodium hydroxide has no oxidising ability at all; refluxing an alcohol with it brings about no reaction of this kind.

Final answer

A. Excess acidified potassium dichromate(VI) under reflux fully oxidises the primary alcohol to the carboxylic acid, whereas distillation (B) would stop at the aldehyde, $\text{LiAlH}_4$ (C) is a reducing agent, and NaOH(aq) (D) does not oxidise alcohols.