Skip to content
CIE Jun 2014 v3 Paper 4 Q7
Answers available below
Show answer
Question
1
2
3
4
5
6
7
8
![16 The combination of mass spectroscopy and NMR spectroscopy provides a powerful method of analysis for organic compounds. (a) The mass spectrum of a compound G contains M and M+1 peaks in the ratio of their heights of 74 : 2.5. Use these data to calculate the number of carbon atoms present in G. Show your working. PMT (b) The NMR spectrum of compound G is shown. [2] 12 10 6/ ppm (i) Use the Data Booklet and your knowledge of NMR spectroscopy to identify the type of proton responsible for each of the three absorptions. type of proton/ ppm 1.1 2.2 11.8 (ii) The addition of D2O causes one of these absorptions to disappear. Explain why this happens and state which absorption is affected.UCLES 2014 9701/43/M/J/14<br />
(iii) Draw the structural formula of G. 17 PMT (c) Several structural isomers of G exist. (i) Draw the structural formula of an isomer of G with only two absorptions in its NMR spectrum. [6] (ii) Use the Data Booklet to suggest where these absorptions would occur./ ppm peak 1 2 [3] [Total: 11]UCLES 2014 9701/43/M/J/14 [Turn over<br />](https://learnah.org/wp-content/uploads/2018/08/June_2014_v3_QP_-_Paper_4_CIE_Chemistry_A-level_7-0.jpg)
![Page 7 Mark Scheme Syllabus Paper GCE A LEVELMay/June 2014 9701 43 (a) PMT feature level of bonding formation of -helix secondary formation of disulfide bonds tertiary formation of ionic bonds tertiary linking amino acids primary (b) block letter name Deoxyribose Cytosine Phosphate Thymine (c) (i) H/hydrogen (bonds between bases) (ii) Bonds are weak and so require relatively little energy to break / are easily broken (d) (sugar, J) (base, M) DNA deoxyribose RNA ribose thymine / T uracil / U (a) Expression: n = or equivalent 100 2.5 1.1 74 n = 3.1 hence G has three carbon atoms (b) (i) ( 1.1) RCH3 or RCH2R or methyl or CH3 [3] 4[1] [1] [1] [1] [Total: 10] [1] [1] ( 2.2) (R)CH2CO(R) or CH3CO(R) ( 11.8) (R)COOH or (R)CONH(R) 3[1]Cambridge International Examinations 2014<br />
PMT [1] [1] [1] [1] [1] [1] [1] [1] [1] [1] [1] [1] [Total: 11] [1] [1] [1] [2] (b) (i) At point Q because the hydrocarbon tails region is hydrophobic/non-polar/ form van der Waals only or can dissolve in the fat-soluble area (ii) They all contain polar or hydrogen-bonding (groups) (c) (i) range 110 to 110 9 7 m (ii) (higher frequency radiation could) cause tissue/cell damage or mutation or harmful to cellsCambridge International Examinations 2014 [1] [1] [1] [1] [Total: 9] Page 8 Mark Scheme Syllabus Paper GCE A LEVELMay/June 2014 9701 43 (ii) The (OH) peak at11.8 (disappears) because of (O)H-D exchange or equation showing this (e.g. R-OH + D2OR-OD + HOD) (iii) CH3CH2CO2H (c) (i) OH OH CH CH or or O O or H3C H3C or (ii) If methyl ethanoate:2.02.13.34.0 Or if 1, 3-dioxolane:3.34.03.35.0 Or if 1, 2-dioxolane:0.91.43.34.0 Or if dihydroxycyclopropane:0.91.40.56.0 (a) (i) Amide or ester or peptide (ii) Hydrolysis (iii) Drug B (iv) two ester and one amide groups circled<br />](https://learnah.org/wp-content/uploads/2018/08/June_2014_v3_MS_-_Paper_4_CIE_Chemistry_A-level_7-0.jpg)
