Edexcel International A Level Chemistry: Mass spectra and infrared spectroscopy (YCH11) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

This question set tests Topic 10D, Mass spectra and IR, of the Pearson Edexcel International Advanced Subsidiary/Advanced Level in Chemistry (XCH11/YCH11) specification, Issue 1, September 2017, namely outcomes 10.21 to 10.23 (Core Practical 8). It is assessed in the IAS written examination for Unit 2: Energetics, Group Chemistry, Halogenoalkanes and Alcohols. If stuck, use the Topic 10D study guide or the Topic 10D revision notes.

Data for this set. Use integer masses: H = 1, C = 12, O = 16, Cl = 35 or 37, Br = 79 or 81. Use these IR ranges (cm⁻¹): C–H alkane 2962–2853; =C–H alkene 3095–3010; C–H aldehyde 2900–2820 and 2775–2700; C=C 1669–1645; O–H alcohol 3550–3230 (broad); O–H carboxylic acid 3300–2500 (very broad); C=O aldehyde 1740–1720, ketone 1720–1700, carboxylic acid 1725–1700; C–Cl 800–600; C–Br 600–500; N–H amine 3500–3300.

Questions

1. This question is about the mass spectrum of 2-methylpropane, (CH₃)₃CH.

(a) State what is meant by the molecular ion. [1] (b) The base peak is at m/z 43. Write an equation for the fragmentation that forms this ion. [2]

2. State the bond and the wavenumber range of an IR absorption that shows the functional group in each compound.

(a) hex-1-ene [1] (b) propan-1-amine [1] (c) 1-chloropentane [1]

3. Predict three features of the IR spectrum of 4-hydroxypentan-2-one, CH₃COCH₂CH(OH)CH₃. Include one feature that is absent. [3]

4. A sample is one of pentan-1-ol, pentan-2-ol or pentan-3-ol. Its mass spectrum shows a small peak at m/z 88, the base peak at 59 and a peak at 29. There are no peaks at 31, 45 or 73.

(a) Identify the ion at m/z 59 and the radical lost when it forms. [2] (b) Identify the alcohol and explain why the other two are ruled out. [2]

5. In Core Practical 6, 2-methylpropan-2-ol is converted into 2-chloro-2-methylpropane with concentrated hydrochloric acid. The product’s mass spectrum shows peaks at m/z 92 and 94 (heights 3 : 1), peaks at 77 and 79 (heights 3 : 1), and the base peak at 57.

(a) Explain what the peaks at 92 and 94 show. [2] (b) Identify the ions at m/z 77 and 57. [2] (c) Describe how the IR spectrum of the product shows that no 2-methylpropan-2-ol remains, and give one absorption expected for the product. [2]

6. Suggest how each pair could be told apart using the method stated.

(a) Hex-1-ene and cyclohexane. Explain why the molecular ion peak does not help, and give an IR difference. [2] (b) Hexanal and hexan-2-one. Give one IR difference and one mass spectrum difference. [2] (c) Propan-1-amine and propan-1-ol, using IR only. [2]

7. A 2.90 g sample of compound R contains 1.80 g of carbon, 0.30 g of hydrogen and 0.80 g of oxygen. Its molecular ion is at m/z 58. Its IR spectrum has a broad absorption at 3400 cm⁻¹, a medium absorption at 1650 cm⁻¹ and a weak absorption at 3080 cm⁻¹. Nothing absorbs between 1680 and 1750 cm⁻¹. The mass spectrum also has peaks at m/z 41 and 31.

(a) Calculate the empirical formula of R. [2] (b) Deduce the molecular formula of R. [1] (c) Deduce the functional groups in R from the IR data. Include the evidence from the missing absorption. [3] (d) Suggest the structure and name of R, using the peaks at 41 and 31. [2]

8. Cyclohexanone is made from cyclohexene in three steps:

cyclohexene → bromocyclohexane → cyclohexanol → cyclohexanone

For each step, give one IR absorption that disappears and one that appears. [6]

9. 2-bromobutane, CH₃CHBrCH₂CH₃, reacts with potassium hydroxide. Under one set of conditions it forms product A; under another it forms product B.

(a) Predict the m/z values and relative heights of the molecular ion peaks of 2-bromobutane, and explain your answer. [3] (b) A has its molecular ion at m/z 74 and a broad IR absorption at 3550–3230 cm⁻¹. Give the conditions used and name A. [2] (c) The mass spectrum of A has a large peak at m/z 45. Identify the ion and write an equation for its formation from the molecular ion. [2] (d) B has its molecular ion at m/z 56 and no O–H absorption. Give one IR absorption expected for B. [1]

10. Four colourless liquids, W, X, Y and Z, are pentan-1-ol, pentanal, pentan-2-one and 1-chloropentane, in no particular order.

Test	W	X	Y	Z
Warm with acidified potassium dichromate(VI)	turns green	turns green	stays orange	stays orange
Warm with Benedict’s solution	no change	red precipitate	no change	no change
Warm with aqueous silver nitrate in ethanol	no change	no change	no change	white precipitate forms slowly

(a) Identify W, X, Y and Z, giving a reason for each. [4] (b) State two features of the IR spectrum of Y that confirm it is a ketone. [2] (c) Give a test, with its result, that shows W contains an –OH group. [1]

Answers

1. (a) The ion formed when a whole molecule loses one electron, M⁺ [1]. (b) (CH₃)₃CH⁺ → (CH₃)₂CH⁺ + CH₃• : the C₃H₇⁺ ion with its positive charge [1]; the uncharged methyl radical, with masses balancing, 58 = 43 + 15 [1]. Examiner insight: An equation that puts the charge on the methyl group, or leaves the fragment uncharged, usually loses credit, because only the positive ion is detected.

2. (a) C=C, 1669–1645 cm⁻¹ [1]. (b) N–H, 3500–3300 cm⁻¹ [1]. (c) C–Cl, 800–600 cm⁻¹ [1]. Examiner insight: “A peak at 1650” alone is weak; name the bond and quote the range from the data given.

3. Broad O–H (alcohol) at 3550–3230 cm⁻¹ [1]. Strong C=O (ketone) at 1720–1700 cm⁻¹ [1]. No aldehyde C–H band at 2775–2700 cm⁻¹ (or no very broad acid O–H at 3300–2500) [1]. Examiner insight: With two functional groups present, predict an absorption for each.

4. (a) 88 − 59 = 29, so C₂H₅• is lost [1]. The ion is CH₃CH₂CHOH⁺, C₃H₇O⁺ [1]. (b) Pentan-3-ol, with the OH on the middle carbon [1]. Pentan-1-ol would give CH₂OH⁺ at 31, and pentan-2-ol would give CH₃CHOH⁺ at 45 or lose CH₃• to give 73; pentan-3-ol has an ethyl group on each side of the C–OH carbon, so it loses only C₂H₅• [1]. Examiner insight: Rule out the other isomers using specific missing peaks; a name with no reason earns little.

5. (a) Two peaks two units apart in a 3 : 1 ratio show one chlorine atom, from ³⁵Cl and ³⁷Cl [1]. They are the molecular ions of C₄H₉Cl: 92 with ³⁵Cl and 94 with ³⁷Cl [1]. (b) m/z 77 is (CH₃)₂CCl⁺ (with ³⁵Cl), formed by loss of CH₃• [1]. m/z 57 is (CH₃)₃C⁺, formed by loss of Cl•; it is a single peak because it has no chlorine [1]. (c) No broad O–H absorption at 3550–3230 cm⁻¹ [1]. A C–Cl absorption at 800–600 cm⁻¹ [1]. Examiner insight: Quote the ratio and say “one chlorine atom”; “contains a halogen” is too vague for credit.

6. (a) Both are C₆H₁₂, so both have M⁺ at m/z 84 [1]. Hex-1-ene absorbs at 1669–1645 cm⁻¹ (C=C) and 3095–3010 cm⁻¹ (=C–H); cyclohexane does not [1]. (b) IR: hexanal has aldehyde C–H at 2775–2700 cm⁻¹ and hexan-2-one does not [1]. MS: hexanal can give CHO⁺ at m/z 29; hexan-2-one gives CH₃CO⁺ at 43 [1]. (c) Propan-1-amine: N–H absorption at 3500–3300 cm⁻¹ [1]. Propan-1-ol: broad O–H absorption at 3550–3230 cm⁻¹ [1]. Examiner insight: Describe both compounds in each pair; stating only what one shows gives incomplete evidence.

7. (a) Moles: C 1.80/12 = 0.150; H 0.30/1 = 0.30; O 0.80/16 = 0.050 [1]. Divide by 0.050: 3 : 6 : 1, so the empirical formula is C₃H₆O [1]. (b) Empirical formula mass = 36 + 6 + 16 = 58, which equals M⁺, so the molecular formula is C₃H₆O [1]. (c) Broad 3400 cm⁻¹ is in the alcohol O–H range [1]. 1650 cm⁻¹ is C=C, supported by =C–H at 3080 cm⁻¹ [1]. No absorption at 1680–1750 cm⁻¹ means no C=O, so R is not propanal or propanone [1]. (d) CH₂=CHCH₂OH, prop-2-en-1-ol [1]. m/z 31 is CH₂OH⁺; m/z 41 is CH₂=CHCH₂⁺ (C₃H₅⁺), from loss of OH•, 58 − 17 = 41 [1]. Examiner insight: Show the division step and ratios in part (a); a bare formula gives no evidence of method.

8. Step 1: C=C at 1669–1645 cm⁻¹ (or =C–H at 3095–3010) disappears [1]; C–Br at 600–500 cm⁻¹ appears [1]. Step 2: C–Br at 600–500 cm⁻¹ disappears [1]; broad O–H at 3550–3230 cm⁻¹ appears [1]. Step 3: O–H at 3550–3230 cm⁻¹ disappears [1]; C=O (ketone) at 1720–1700 cm⁻¹ appears [1]. Examiner insight: Quote a range for every absorption; a bond name without a wavenumber is usually not enough.

9. (a) m/z 136 and 138 [1], of about equal height [1]. The formula is C₄H₉Br, and bromine is roughly half ⁷⁹Br and half ⁸¹Br, giving C₄H₉⁷⁹Br⁺ and C₄H₉⁸¹Br⁺ [1]. (b) Aqueous potassium hydroxide, heated [1]. A is butan-2-ol [1]. (c) CH₃CHOH⁺ (C₂H₅O⁺) [1]. CH₃CH(OH)CH₂CH₃⁺ → CH₃CHOH⁺ + C₂H₅• [1]. (d) C=C at 1669–1645 cm⁻¹ (or =C–H at 3095–3010 cm⁻¹) [1]. Examiner insight: In (b), “KOH” alone does not show which reaction happened; say aqueous for substitution, as ethanolic gives the alkene.

10. (a) W is pentan-1-ol: oxidised by dichromate but not an aldehyde [1]. X is pentanal: oxidised and gives a red precipitate with Benedict’s [1]. Y is pentan-2-one: not oxidised and has no halogen [1]. Z is 1-chloropentane: hydrolysis releases chloride ions, which give white silver chloride [1]. (b) Strong C=O at 1720–1700 cm⁻¹ [1]. No aldehyde C–H at 2775–2700 cm⁻¹ and no broad O–H [1]. (c) Add PCl₅: misty fumes of hydrogen chloride [1]. Examiner insight: Each identification needs the observation that supports it; a list of names with no reasons gains little.

Where marks are usually lost

• Calling the base peak the molecular ion.
• Leaving the charge off fragment ions, or putting it on the radical.
• Quoting a wavenumber with no bond, or a bond with no wavenumber.
• Calling the 3300–2500 cm⁻¹ acid band an alcohol O–H.
• Identifying a carbonyl as an aldehyde without the 2775–2700 cm⁻¹ evidence.
• Missing the 3 : 1 (Cl) or 1 : 1 (Br) pattern, or swapping them.
• Writing “KOH” without stating aqueous or ethanolic.

Next steps

• Revisit weak areas in the revision notes and study guide.
• Book a free trial class with a tutor.
• Browse the Edexcel IAL Chemistry course page and the outcome checklist.
• Spot gaps with all free 10-minute diagnostics.

Official syllabus

Pearson Edexcel International Advanced Subsidiary/Advanced Level in Chemistry (XCH11/YCH11) specification, Issue 1, September 2017 (Pearson Education Limited). Mass spectra and IR is part 10D of Topic 10, outcomes 10.21 to 10.23, in Unit 2.