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Alcohols and Esters: Question 10

Syllabus 16.1

Structured AS 9 marks

2-Methylpropan-1-ol, (CH3)2CHCH2OH(\text{CH}_3)_2\text{CHCH}_2\text{OH}, can undergo the three separate reactions described below.

Reaction 1: 2-methylpropan-1-ol is heated under reflux with an excess of acidified potassium manganate(VII).

Reaction 2: 2-methylpropan-1-ol vapour is passed over a heated aluminium oxide catalyst.

Reaction 3: 2-methylpropan-1-ol is heated under reflux with ethanoic acid and a few drops of concentrated sulfuric acid.

(a) For Reaction 1, give the structural formula of the organic product and state the colour change observed. [3]

(b) For Reaction 2, name the type of reaction and give the structural formula of the alkene formed. [3]

(c) For Reaction 3, name the type of reaction and give the structural formula of the ester formed. [3]

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Worked solution

Classifying the alcohol

The carbon bearing the −OH-\text{OH} group in 2-methylpropan-1-ol, (CH3)2CHCH2OH(\text{CH}_3)_2\text{CHCH}_2\text{OH}, is attached to only one other carbon group, so this is a primary alcohol. It can be oxidised all the way to a carboxylic acid, and gives only a single dehydration product.

Part (a): Reaction 1. Full oxidation

Refluxing a primary alcohol with an excess of acidified potassium manganate(VII) oxidises it completely, via the aldehyde intermediate (which cannot escape under reflux), to the carboxylic acid:

(CH3)2CHCH2OH→(CH3)2CHCOOH(\text{CH}_3)_2\text{CHCH}_2\text{OH} \rightarrow (\text{CH}_3)_2\text{CHCOOH}

The product is 2-methylpropanoic acid, (CH3)2CHCOOH(\text{CH}_3)_2\text{CHCOOH}. As the purple manganate(VII) ion, MnO4−\text{MnO}_4^-, is reduced to the almost colourless manganese(II) ion, Mn2+\text{Mn}^{2+}, the solution changes colour from purple to colourless.

Part (b): Reaction 2. Dehydration

Passing the alcohol vapour over a heated Al2O3\text{Al}_2\text{O}_3 catalyst is an elimination reaction, removing the −OH-\text{OH} from C1 and a hydrogen atom from the adjacent carbon (C2) to form a double bond:

(CH3)2CHCH2OH→(CH3)2C=CH2+H2O(\text{CH}_3)_2\text{CHCH}_2\text{OH} \rightarrow (\text{CH}_3)_2\text{C}=\text{CH}_2 + \text{H}_2\text{O}

Since C2 (bearing the branching methyl groups) carries only one hydrogen available for elimination, only a single alkene, 2-methylpropene, (CH3)2C=CH2(\text{CH}_3)_2\text{C}=\text{CH}_2, is formed.

Part (c): Reaction 3, esterification

Refluxing the alcohol with ethanoic acid, using concentrated sulfuric acid as a catalyst, condenses the two together with loss of water. This is an esterification (condensation) reaction:

CH3COOH+(CH3)2CHCH2OH⇌CH3COOCH2CH(CH3)2+H2O\text{CH}_3\text{COOH} + (\text{CH}_3)_2\text{CHCH}_2\text{OH} \rightleftharpoons \text{CH}_3\text{COOCH}_2\text{CH(CH}_3)_2 + \text{H}_2\text{O}

The ester formed is 2-methylpropyl ethanoate, CH3COOCH2CH(CH3)2\text{CH}_3\text{COOCH}_2\text{CH(CH}_3)_2.

Final answers

  • (a) 2-methylpropanoic acid, (CH3)2CHCOOH(\text{CH}_3)_2\text{CHCOOH}; colour change purple → colourless.
  • (b) Elimination; 2-methylpropene, (CH3)2C=CH2(\text{CH}_3)_2\text{C}=\text{CH}_2.
  • (c) Esterification (condensation); 2-methylpropyl ethanoate, CH3COOCH2CH(CH3)2\text{CH}_3\text{COOCH}_2\text{CH(CH}_3)_2.