OCR A-Level Chemistry AS Depth in chemistry (02), June 2025: Question 6

✅ Correct Answer

• Peak P (m/z = 43): CH₃CO⁺ (or drawn displayed/skeletal structure with a positive charge).
• Peak Q (m/z = 72): CH₃COCH₂CH₃⁺ or [CH₃COCH₂CH₃]⁺ (the molecular ion).

📐 Fragment Mass Breakdown

1. Total Mᵣ of A: (4 × 12.0) + (8 × 1.0) + (1 × 16.0) = 72. Peak Q at m/z = 72 is the unfragmented molecular ion [M]⁺.
2. Peak P (m/z = 43): Loss of an ethyl radical, •CH₂CH₃ (mass 29): 72 − 29 = 43. This leaves the acylium ion CH₃–C≡O⁺ or CH₃CO⁺ .

🧠 Exam Technique

Always show a positive charge (+) on your fragments. Mass spectrometers only detect positively charged ions! Charges can be placed outside square brackets or directly on the fragment (e.g., CH₃CO⁺ ).

❌ Common Errors

• Forgetting the positive charge: Writing neutral formulas like CH₃CO loses the mark immediately.
• Incorrect fragment for m/z = 43: Suggesting propyl C₃H₇⁺ is chemically unreasonable for butanone, because forming C₃H₇⁺ requires rearranging the carbon skeleton rather than simple cleavage of the C–C bond next to the C=O group.

✅ Correct Answer

The peak at m/z = 73 is an M+1 peak caused by the presence of a naturally occurring carbon-13 (¹³C) atom in a small proportion of the molecules.

💡 Key Knowledge

• Naturally occurring carbon consists of approximately 98.9% ¹²C and 1.1% ¹³C.
• In any organic molecule, there is a small probability that one of the carbon atoms is a ¹³C isotope, producing an [M+1]⁺ peak at 1 unit higher than the molecular ion peak.

❌ Common Errors

• Attributing the peak to protonation ( [M + H]⁺ ). In standard electron ionisation (EI) mass spectrometry, molecules lose an electron; they are not protonated.
• Saying "isotopes" without specifying Carbon-13 or ¹³C.

✅ Correct Answer

It causes the covalent bonds to vibrate more (by absorbing energy, undergoing stretching and bending motions).

❌ Common Errors

DO NOT ALLOW: "Causes bonds to break" or "breaks bonds". Infrared radiation only has enough energy to excite vibrational energy levels; it does not cause homolytic or heterolytic bond fission (which requires high-energy UV or electron bombardment).

✅ Correct Answer

Two functional groups present:

• Alcohol (explained by: broad O–H absorption at 3200–3600 cm⁻¹)
• Alkene (explained by: sharp C=C absorption at 1620–1680 cm⁻¹)

💡 Data Sheet Matching

• O–H (alcohol): Very broad absorption band centred around 3350 cm⁻¹ (range: 3200–3600 cm⁻¹ ).
• C=C (alkene): Sharp, medium intensity absorption peak at roughly 1650 cm⁻¹ (range: 1620–1680 cm⁻¹ ).
• Absence of C=O: Note that the strong ketone peak at ~1715 cm⁻¹ is missing, confirming Compound B is no longer a carbonyl compound!

🧠 Exam Technique

• Name the functional group, not just the bond: Write alcohol (not just "O–H") and alkene (not just "C=C").
• Quote precise values or standard ranges: Always state both the specific bond and the wavenumber range from your Data Sheet (e.g., "O–H at 3200–3600 cm⁻¹").
• Note: Compound B has the molecular formula C₄H₈O. An isomer with one C=C and one O–H (an alkenol, such as but-3-en-1-ol) perfectly matches this formula!

❌ Common Errors

• Writing "hydroxyl" instead of alcohol when asked for the functional group name.
• Confusing the broad alcohol O–H peak (3200–3600 cm⁻¹) with a carboxylic acid O–H peak (2500–3300 cm⁻¹). Carboxylic acid O–H is extremely broad and overlaps the C–H region, and would require a C=O peak at ~1700 cm⁻¹.
• Quoting an observed wavenumber without naming the bond responsible (e.g., stating only "peak at 3350 cm⁻¹"). Both bond and wavenumber are required.