OCR A-Level Chemistry Synthesis and analytical techniques (02), June 2018: Question 9

✅ Correct Answer & Breakdown

The correct option is B ( CH₃CHOHCN ).

Looking at the provided IR spectrum:

• There is a broad absorption peak around 3200 - 3400 cm⁻¹ , corresponding to an O–H stretch (alcohol).
• There is a sharp, medium absorption peak around 2200 - 2250 cm⁻¹ , corresponding to a C≡N stretch (nitrile).
• There is no very broad absorption over 2500 - 3300 cm⁻¹ which would indicate a carboxylic acid O–H, ruling out C.
• There is no strong carbonyl (C=O) stretch around 1630 - 1820 cm⁻¹ , ruling out C and D.

💡 Key Knowledge (Data Booklet Values)

• O–H (alcohol): Broad peak at 3200 – 3600 cm⁻¹
• C≡N (nitrile): Sharp peak at 2220 – 2260 cm⁻¹
• C=O (carbonyl): Strong, sharp peak at 1630 – 1820 cm⁻¹
• O–H (carboxylic acid): Very broad peak at 2500 – 3300 cm⁻¹

🧠 Exam Technique

• Process of elimination: Scan the high-wavenumber region first (>1500 cm⁻¹). Identify key diagnostic peaks before looking at the options.
• Check for the sharp nitrile peak near 2200 cm⁻¹ —it is isolated and very distinct, making it an excellent starting anchor for spotting molecules containing CN groups.
• Cross-reference functional groups: Molecule B is the only option containing both an alcohol (O–H) and a nitrile (C≡N).

❌ Common Errors

• Confusing O-H types: Mistaking the alcohol O–H peak for a carboxylic acid O–H (which overlaps heavily with C–H regions and appears much broader).
• Ignoring the 2200 cm⁻¹ region: Overlooking the sharp nitrile peak and guessing an alcohol like ethanol (A), which lacks the 2200 cm⁻¹ spike entirely.
• Assuming C=O is present: Looking for a carbonyl peak when none exists, leading students to incorrectly select carboxylic acids (C) or amides (D).