OCR A-Level Chemistry Synthesis and analytical techniques (02), June 2017: Question 15

1 mark · Medium difficulty · Multiple Choice

Identify which of the given compounds are structural isomers of C6H12O2.

Practise this question

Question

Multiple-choice question 15 asks which of three compounds (1: hexanoic acid, 2: ethyl butanoate, 3: propyl propanoate) are structural isomers of C6H12O2. Four options A (1, 2 and 3), B (Only 1 and 2), C (Only 2 and 3), and D (Only 1) are provided along with a box for the student's answer.
Question text

15 Which compound(s) is a/are structural isomer(s) of C6H12O2?

1 hexanoic acid

2 ethyl butanoate

3 propyl propanoate

A 1, 2 and 3

B Only 1 and 2

C Only 2 and 3

D Only 1

Your answer

[1]

Mark scheme

Show the mark scheme Mark scheme indicating the correct answer for question 15 is A.

15 A 1

How to answer it

Structural Isomers of C₆H₁₂O₂

What this question tests

This question assesses your ability to determine and apply molecular formulas to different organic functional groups, specifically carboxylic acids and esters. It tests your structural visualization skills, naming conventions (nomenclature), and understanding of functional group isomerism.

Question 15

Exam Breakdown & Solutions

✅ Correct Answer

The correct option is A (1, 2 and 3).

All three listed compounds share the exact molecular formula C₆H₁₂O₂ and are structural isomers of each other.

💡 Key Knowledge

  • General Formula: Carboxylic acids and esters share the general molecular formula CₙH₂ₙO₂ (when non-cyclic with one double bond). For n = 6 , the formula is C₆H₁₂O₂ .
  • Hexanoic acid: A straight-chain carboxylic acid with 6 carbons ( CH₃CH₂CH₂CH₂CH₂COOH ).
  • Ethyl butanoate: An ester made from ethanol (2 carbons) and butanoic acid (4 carbons), totalling 6 carbons ( CH₃CH₂CH₂COOCH₂CH₃ ).
  • Propyl propanoate: An ester made from propan-1-ol (3 carbons) and propanoic acid (3 carbons), totalling 6 carbons ( CH₃CH₂COOCH₂CH₂CH₃ ).

🧠 Exam Technique

Do not waste time drawing out full displayed formulas for every option under high-pressure exam conditions. Instead, quickly deduce the total number of carbon atoms from the IUPAC name stems and prefixes:

  • Hex- = 6 carbons (Acid)
  • Ethyl (2) + butanoate (4) = 6 carbons (Ester)
  • Propyl (3) + propanoate (3) = 6 carbons (Ester)

❌ Common Errors

Students often incorrectly assume that esters and carboxylic acids cannot be isomers of one another because they belong to different homologous series. Remember that they exhibit functional group isomerism!

Mark Scheme Allocation: 1 mark for selecting option A. No working out required, but careful carbon-counting is essential.

Topics

Module 6: Organic chemistry and analysis · Module 4: Core organic chemistry · 6.1 Aromatic compounds, carbonyls and acids · 4.1 Basic concepts and hydrocarbons

Question and mark scheme from the OCR A-Level Chemistry examination, Synthesis and analytical techniques (02), June 2017. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.