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A-LevelChemistryHydroxy compounds9701/23 · May/June 2022 · Q59 Marks

5 (a) But-2-ene reacts with KMnO₄ to form organic product, Y. Y does not react with Na2CO3. A gas is produced when an excess of Na is added to Y. (i) Describe the conditions for the KMnO₄ used in the reaction to form Y from but-2-ene. (ii) 24.0 cm³ of gas is produced when an excess of Na is added to 0.001 mol of Y, when measured under room conditions. Assume that 1 mol of gas occupies 24.0 dm³ under room conditions. Deduce a possible structure of Y. Explain your answer. (b) Z contains three types of atom: carbon, hydrogen and a halogen. The mass spectrum of Z is recorded. Fig. 5.1 shows a section of the mass spectrum at m/e greater than 63. The fragment at m/e = 64 is the molecular ion peak. [Figure 5.1] (i) Deduce the number of carbon atoms present in a molecule of Z using Fig. 5.1. Show your working. (ii) Deduce which halogen is present in Z using Fig. 5.1. Explain your answer. (iii) There are also peaks at m/e = 29 and m/e = 49. Suggest the formulae of these fragments. Deduce the name of Z. m/e = 29 m/e = 49 name of Z

  1. (a)(i)Describe the conditions for the KMnO₄ used in the reaction to form Y from but-2-ene.[1]
  2. (a)(ii)24.0 cm³ of gas is produced when an excess of Na is added to 0.001 mol of Y, when measured under room conditions. Assume that 1 mol of gas occupies 24.0 dm³ under room conditions. Deduce a possible structure of Y. Explain your answer.[3]
  3. (b)(i)Deduce the number of carbon atoms present in a molecule of Z using Fig. 5.1. Show your working.[1]
  4. (b)(ii)Deduce which halogen is present in Z using Fig. 5.1. Explain your answer.[1]
  5. (b)(iii)There are also peaks at m/e = 29 and m/e = 49. Suggest the formulae of these fragments. Deduce the name of Z. m/e = 29 m/e = 49 name of Z[3]

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Mark scheme

Marking points from the official mark scheme (May/June 2022). The mark or mark type is shown in brackets.

  1. [1] (a)(i) cold + dilute
  2. [3] (a)(ii) M1 unbranched 4C structure AND any number of –OH in any position; M2 0.001 mol H2 made from 0.001 mol Y; AND; R-OH + Na → RONa + ½H2 OR use of 1 OH (group) → ½H2; M3 CH3CH(OH)CH(OH)CH3
  3. [1] (b)(i) 100 x 2.2 / 1.1 x 100 = 2
  4. [1] (b)(ii) chlorine / CI; AND; peak at M+2 represents the molecular ion with 37-Cl; (rather than 35–Cl as relative abundance of (peaks) M: M+2 is 100:33 / 3:1); OR; relative abundance of (peaks) M: M+2 is 100:33.3 / 3:1 (so peak at M+2 contains 37–Cl)
  5. [3] (b)(iii) m/e = 29: C2H5⁺; m/e = 49: CH2Cl⁺; name of Z: chloroethane

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About this question

Topic
Hydroxy compounds, A-Level Chemistry (syllabus 9701). Structured question, 9 marks.
Source
A-Level Chemistry, May/June 2022 Paper 2 Variant 3, question 5.

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