AQA A-Level Chemistry Paper 3, 2023: Question 2

8 marks · Hard difficulty · State/Explain/Describe

Isomerism and Dehydration of Alcohols This question involves drawing displayed and skeletal formulas of isomers of pentan-2-ol, identifying functional group isomers, depicting alcohols that resist dehydration, completing a dehydration reaction mechanism with curly arrows, naming the mechanism, and drawing the structure of an isomeric product.

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Question text

02 This question is about isomerism and the dehydration of alcohols.

Pentan-2-ol has the molecular formula C5H12O

02.1 Draw the displayed formula of an unbranched position isomer of pentan-2-ol that can

be dehydrated to form a single alkene.

[1 mark]

02.2 Draw the skeletal formula of a chain isomer of pentan-2-ol that can be dehydrated to

form a mixture of alkenes.

[1 mark]

02.3 Draw the structure of an unbranched functional group isomer of pentan-2-ol.

[1 mark]

02.4 Another isomer of pentan-2-ol is an alcohol that is not dehydrated when heated with

concentrated sulfuric acid.

Draw the structure of this isomer.

7 [1 mark]

02.5 An incomplete mechanism for the dehydration of a compound is shown.

Complete the mechanism for this reaction by drawing two curly arrows on the

intermediate.

Name the mechanism for this reaction.

[3 marks]

02.6 An isomer of the final product can also form in the reaction in Question 02.5.

Draw the structure of this isomer.

[1 mark]

Mark scheme

Show the mark scheme

Question Answers Additional comments/Guidelines Mark

2.1 Displayed formula of pentan-1-ol NOT pentan-3-ol

NOT –OH

2.2 Skeletal formula of 3-methylbutan-2-ol or 2-methylbutan-2-ol IGNORE numbers on C atoms

IGNORE ‘dots’ at junctions

IGNORE other non-skeletal structures

IGNORE skeletal structure of pentan-2-ol 1

NOT other incorrect skeletal structures

NOT O–H

NOT if bond clearly to H of OH

2.3 one of these compounds Any structural representation of correct

compound

– A-LEVEL CHEMISTRY – –

2.4 Any structural representation of correct

compound 17

2.5

M1/M2 list principle for additional arrows on any

structure

M1 loss of H2O: arrow from C–O bond to O

M1 NOT if arrow to +

M2 loss of H+: arrow from correct C–H bond to correct C–C bond

M3 IGNORE acid-catalysed / dehydration

M3 elimination

NOT nucleophilic / addition / electrophilic

2.6 Any structural representation of correct

compound

If skeletal CH2 not needed 1

Allow rings in place of C6H5

How to answer it

Step-by-Step Guide: Isomerism and Dehydration of Alcohols

Question (a): Displayed Formula of an Unbranched Position Isomer

Step 1: Identify an Unbranched Position Isomer

The question asks for an unbranched alcohol that can dehydrate to form a single alkene. The correct answer is pentan-1-ol .

Answer: Pentan-1-ol (ensure the OH group is attached to the terminal carbon atom).

Teacher Tip: Always ensure the displayed formula shows all bonds explicitly, as missing bonds may lose marks.

Question (b): Skeletal Formula of a Chain Isomer

Step 1: Identify a Chain Isomer

Possible chain isomers are 3-methylbutan-2-ol or 2-methylbutan-2-ol .

Answer: Either skeletal formula is acceptable. Ensure to use zig-zag lines and correctly place the -OH group.

Teacher Tip: Practice drawing skeletal formulas accurately as they are commonly tested. Avoid numbering carbons in skeletal structures unless specified.

Question (c): Functional Group Isomer

Step 1: Identify Functional Group Isomers

A functional group isomer of pentan-2-ol would be an ether. Examples include methoxybutane or ethoxypropane .

Answer: Any correct ether structure is acceptable.

Teacher Tip: Functional group isomers retain the same molecular formula but differ in functional groups (e.g., alcohol vs ether). Be prepared to convert between these forms.

Question (d): Alcohol That Is Not Dehydrated

Step 1: Identify Tertiary Alcohols

Tertiary alcohols do not dehydrate as they lack a β-hydrogen. The correct answer is 2-methylpropan-2-ol .

Answer: 2-methylpropan-2-ol (a tertiary alcohol).

Teacher Tip: Memorise the rule that tertiary alcohols do not undergo dehydration, as this is a common exam question.

Question (e): Dehydration Mechanism

Step 1: Complete the Mechanism

To complete the mechanism, draw the following:

A curly arrow from the C-O bond to the oxygen atom to show the loss of water.
A curly arrow from the adjacent C-H bond to the C=C bond to form the double bond.

The mechanism is elimination .

Teacher Tip: Ensure your curly arrows are precise and start from bonds or lone pairs. Avoid drawing arrows to or from charges incorrectly.

Question (f): Isomer of the Final Product

Step 1: Draw the Structural Isomer

The isomer of the final product would have the double bond in a different position. Ensure the structure reflects this correctly.

Answer: A positional isomer of the alkene product.

Teacher Tip: For positional isomers, move the double bond while maintaining the overall structure of the molecule.

Topics

Organic Chemistry · 3.3.1 Introduction to Organic Chemistry · 3.3.5 Alcohols

Question and mark scheme from the AQA A-Level Chemistry examination, Paper 3, 2023. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.