IB CHEMISTRY

The Examiner's Assessment Bank

Structure 3.2 Exam Practice

Section A: High-Difficulty MCQs

1. What is the IUPAC name for (CH3)2CHCH(OH)CH2CH3(CH_3)_2CHCH(OH)CH_2CH_3?

A. 1,1-dimethylpropan-2-ol
B. 3-methylbutan-2-ol (Correct)
C. 2-methylbutan-3-ol
D. Pentan-3-ol

2. Which molecule exhibits optical isomerism?

A. 1-chlorobutane
B. 2-chlorobutane (Correct)
C. 1-chloropropane
D. 2-chloropropane

3. What is the IHD of Aspirin (C9H8O4C_9H_8O_4)?

A. 4
B. 5 (Correct)
C. 6
D. 7

Section B: Structured Data-Response

Q1: Isomerism & Properties [5 marks]

Consider the isomers of C4H10OC_4H_{10}O.

  1. Draw the structure of a tertiary alcohol isomer. [1]
  2. Compare the boiling point of this alcohol with its functional group isomer, ethoxyethane. Explain. [3]
  3. Identify which isomer of C4H9OHC_4H_9OH is optically active. [1]

Examiner's Markscheme

1. Structure of 2-methylpropan-2-ol.

2. The Alcohol has a higher boiling point. [1] Reason: Alcohols contain Hydrogen Bonding (strongest IMF). [1] Ethers only have Dipole-Dipole forces. [1]

3. Butan-2-ol. Asterisk on C2 (*).

Q2: Structural Elucidation (The Triangulation) [6 marks]

An unknown Compound X (C4H8O2C_4H_8O_2) presents the following data:

  • IR: Strong, sharp peak at 1740 cm⁻¹ and no broad absorption above 3000 cm⁻¹.
  • 1H NMR:
    • Signal A: 1.3 ppm, Triplet, Integration 3H
    • Signal B: 2.0 ppm, Singlet, Integration 3H
    • Signal C: 4.1 ppm, Quartet, Integration 2H

Deduce the structure of Compound X. Justify using all data.

Examiner's Markscheme

  • IHD: 1 (Consistent with C=O).
  • IR: Peak at 1740 (C=O) + Absence of O-H = Ester. [1]
  • NMR: Triplet (3H) + Quartet (2H) = Ethyl group. [1]
  • Singlet (3H) indicates Isolated methyl next to C=O. [1]
  • Signal C (4.1 ppm) confirms CHβ‚‚ is next to O (Ester link). [1]
  • Structure: Ethyl Ethanoate (CH3COOCH2CH3CH_3COOCH_2CH_3). [2]