Pentan-1-ol, CH3CH2CH2CH2CH2OH, and pentan-2-ol, CH3CH(OH)CH2CH2CH3, are each oxidised separately using acidified potassium dichromate(VI), K2Cr2O7. (a) State the colour change observed at the dichromate ion when a successful oxidation of either alcohol occurs. [2] (b) A sample…

Answer: ((a)) The mixture changes from orange to green as the orange dichromate(VI) ion, Cr2O7^2-, is reduced to the green chromium(III) ion, Cr^3+. ((b)) CH3CH2CH2CH2CHO, pentanal; distilling the aldehyde off as it forms removes it from contact with the oxidising agent before it can be oxidised further to the carboxylic acid. ((c)) CH3COCH2CH2CH3, pentan-2-one; it cannot be oxidised further, because the carbonyl carbon of a ketone is bonded to two other carbon atoms and has no hydrogen atom left that can be removed by the oxidising agent.. Part (a): The colour change of the oxidising agent Acidified potassium dichromate(VI) contains the dichromate(VI) ion, Cr2O7^2-, in which chromium is in the +6 oxidation state; this ion is orange. When it successfully oxidises an alcohol,…

AS Chemistry 9701: Oxidising Alcohols to Aldehydes and Ketones

Pentan-1-ol, $\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{CH}_2\text{OH}$ , and pentan-2-ol, $\text{CH}_3\text{CH(OH)}\text{CH}_2\text{CH}_2\text{CH}_3$ , are each oxidised separately using acidified potassium dichromate(VI), $\text{K}_2\text{Cr}_2\text{O}_7$ .

(a) State the colour change observed at the dichromate ion when a successful oxidation of either alcohol occurs. [2]

(b) A sample of pentan-1-ol is heated with a limited amount of acidified potassium dichromate(VI), and the organic product is distilled off from the reaction mixture as soon as it forms. Give the structural formula and name of this organic product, and explain why distilling it off as it forms prevents further oxidation. [3]

(c) A separate sample of pentan-2-ol is heated under reflux with an excess of acidified potassium dichromate(VI). Give the structural formula and name of the organic product formed, and state whether this product can be oxidised further by the dichromate ion, explaining your answer. [3]

Part (a): The colour change of the oxidising agent

Acidified potassium dichromate(VI) contains the dichromate(VI) ion, $\text{Cr}_2\text{O}_7^{2-}$ , in which chromium is in the $+6$ oxidation state; this ion is orange. When it successfully oxidises an alcohol, it is itself reduced to the chromium(III) ion, $\text{Cr}^{3+}$ , which is green. So a positive oxidation is seen as the mixture turning from orange to green.

Part (b): Distilling pentan-1-ol to stop at the aldehyde

Pentan-1-ol is a primary alcohol, so oxidation converts its $\text{-CH}_2\text{OH}$ group into a $\text{-CHO}$ group: $\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{CH}_2\text{OH} + [\text{O}] \rightarrow \text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{CHO} + \text{H}_2\text{O}$ The product is pentanal, $\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{CHO}$ . An aldehyde can, in principle, be oxidised further to a carboxylic acid, so to isolate the aldehyde the mixture is heated so that pentanal (which has a lower boiling point than the alcohol) distils out of the flask as soon as it forms. Removing it immediately from the reaction mixture keeps it out of contact with the acidified dichromate(VI), preventing the second oxidation step from occurring.

Part (c): Refluxing pentan-2-ol to give a ketone

Pentan-2-ol is a secondary alcohol, so oxidation converts its $\text{CH(OH)}$ group into a $\text{C=O}$ group: $\text{CH}_3\text{CH(OH)}\text{CH}_2\text{CH}_2\text{CH}_3 + [\text{O}] \rightarrow \text{CH}_3\text{COCH}_2\text{CH}_2\text{CH}_3 + \text{H}_2\text{O}$ The product is pentan-2-one, $\text{CH}_3\text{COCH}_2\text{CH}_2\text{CH}_3$ . Unlike an aldehyde, this ketone cannot be oxidised any further: its carbonyl carbon is bonded to two carbon-containing groups (a $\text{CH}_3$ group and a propyl chain) rather than to a hydrogen atom, so there is no hydrogen on the carbonyl carbon that the oxidising agent could remove without breaking a carbon-carbon bond. Because no further reaction is possible, heating under reflux (which returns any vapour to the flask for continued reaction) simply gives the ketone as the final product. There is no need to distil it off immediately, since there is no over-oxidation to prevent.

Final answers

(a) Orange to green, as $\text{Cr}_2\text{O}_7^{2-}$ (orange) is reduced to $\text{Cr}^{3+}$ (green).
(b) Pentanal, $\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{CHO}$ . Distillation removes it before further oxidation to the carboxylic acid can occur.
(c) Pentan-2-one, $\text{CH}_3\text{COCH}_2\text{CH}_2\text{CH}_3$ . Cannot be oxidised further, as its carbonyl carbon has no hydrogen atom to remove.