Edexcel Jun 2015 (IAL) Paper 5 Q24 (with explained solutions) |

Skip to content

Edexcel Jun 2015 (IAL) Paper 5 Q24

Answers available below

SECTION C Answer ALL the questions. Write your answers in the spaces provided. 24 Oil of cloves Oil of cloves is a traditional remedy for toothache; it is a topical treatment, which means that it is applied externally. The main active ingredient of oil of cloves is eugenol, which also gives cloves their characteristic smell. The structure of eugenol is shown below. HC CH2 H2C OH CH3 eugenol (a) A student suggested that eugenol could be prepared by an electrophilic substitution of 2-methoxyphenol, which is produced in the gut of desert locusts. This compound is one of the main components of the pheromones that cause locust swarming. The structure of 2-methoxyphenol is shown below. 2-methoxyphenol OH CH3 (i) Use your knowledge of electrophilic substitution reactions to suggest the structure of an electrophile that might be used in this reaction. (1) 22 *P45074A02228*<br />
(ii) Suggest an organic compound that could be used to produce the electrophile that you have given in (a)(i). (iii) Write the mechanism for the electrophilic substitution to prepare eugenol from 2-methoxyphenol, using the electrophile that you have given in (a)(i). (1) (3) *P45074A02328* 23 Turn over<br />
(b) Eugenol has been used in the manufacture of vanillin, the compound responsible for the flavour of vanilla. The process involves the conversion of eugenol to its isomer isoeugenol, which is then oxidized by hydrogen peroxide with a vanadium(V) oxide catalyst. HC CH2 H2C CH CH3 HC KOH OH CH3 eugenol H2O2 / V2O5 OH CH3 isoeugenol H2C CH3 CH3 OH vanillin OH CH3 methyl vanillyl ketone (i) Isoeugenol exists as two stereoisomers, whereas eugenol has just one structure. State the type of stereoisomerism shown by isoeugenol and explain why it can show this type of stereoisomerism, whereas eugenol does not. (2)24 *P45074A02428*<br />
*(ii) Discuss whether or not high resolution proton nmr spectroscopy could be used to distinguish between eugenol and a single isomer of isoeugenol. You should consider the numbers of peaks in the spectra, their splitting patterns and the areas under the peaks. You are not expected to suggest chemical shift values. (4)(iii) Suggest, in outline, a mechanism by which vanadium(V) oxide acts as a catalyst for the reaction between hydrogen peroxide and isoeugenol to form vanillin. You are not expected to give equations for the mechanism. (2)*P45074A02528* 25 Turn over<br />
(iv) A student suggested that the infrared spectroscopy data on page 5 of the Data Booklet could be used to show whether the purified vanillin contains a trace of methyl vanillyl ketone. Explain the basis of this idea and suggest why it may not work in practice. (2)(c) The Zeisel method for determining the purity of vanillin involves estimating the percentage by mass of methoxy group (CH3O) in the sample and comparing this with the percentage in pure vanillin. In a sequence of reactions, each methoxy group produces iodine which is estimated by titration with a sodium thiosulfate solution of known concentration. The Zeisel sequence is Step 1 Step 2 Step 3 Step 4 Step 5 ROCH3 + HIROH + CH3I CH3I + Br2CH3Br + IBr IBr + 3H2O + 2Br2HIO3 + 5HBr+ 5I + 6H+3I2 + 3H2O 2 + 2I 2S4O6 IO3 I2 + 2S2O3 (i) Deduce the number of moles of thiosulfate ions that are equivalent to one mole of methoxy group. Explain your answer. (2)26 *P45074A02628*<br />
(ii) A sample of vanillin was found to have 20.09% by mass of methoxy group. Calculate the percentage purity of the vanillin. (3) (Total for Question 24 = 20 marks) TOTAL FOR SECTION C = 20 MARKS TOTAL FOR PAPER = 90 MARKS *P45074A02728* 27<br />

Show answer

Section C Question Number 24(a)(i) Acceptable Answer Reject Mark or ALLOW Positive charge on any part of the carbocation Structural / fully displayed / skeletal formulae Acceptable Answer Reject Mark Question Number 24(a)(ii) X = Cl / Br / I OR structural / fully displayed / skeletal formulae OR 3- chloro/bromo/iodo prop(-1-)ene No TE on incorrect electrophile in (a)(i) name without 3 WCH0501msc20150812.docx 11 July 2015<br />
Question Number 24 (a)(iii) Acceptable Answer TE on incorrect electrophile in (a)(i) If benzene used instead of substituted benzene OR If final product not 1,2,4 only MP1 & 2 scored Curly arrow from on / within the circle to positive C ALLOW Curly arrow from anywhere within the hexagon Arrow to any part of the electrophile including to the + charge (which can be anywhere on electrophile), OR Arrow to a point at least half the distance between ring and electrophile (1) Intermediate structure including charge with horseshoe covering at least 3 carbon atoms, and facing the tetrahedral carbon and with some part of the positive charge within the horseshoe. IGNORE substituent errors (incorrect position on ring or bond to substituent) at this marking point ALLOW dotted horseshoe (1) Curly arrow from CH bond to anywhere in the benzene ring reforming delocalized structure of a correct stable molecule. Ignore any involvement ofin the final step (1) AlCl4 Correct Kekul structures score full marks Mark 3 Reject Curly arrow on or outside the hexagon Partial bonds to H or CH3 except for dot and wedge in 3-D structure WCH0501msc20150812.docx 11 July 2015<br />
Question Number 24(b)(i) Stand alone marks Reject Mark Optical isomerism Acceptable Answer Geometric / E-Z / cis-trans isomerism (1) Because isoeugenol has (two) different groups attached to each of the carbon atoms of the double bond ALLOW Because eugenol has two hydrogen atoms on one of the carbon atoms in the C=C (1) IGNORE References to the barrier to free rotation about the C=C WCH0501msc20150812.docx 11 July 2015<br />
Reject Mark Question Number 24(b)*(ii) Acceptable Answer (1) If no other mark is scored both eugenol and isoeugenol have eight peaks scores 1 Candidates are only expected to interpret the spectra using knowledge of the (n+1) rule. EITHER The only (significant) difference is likely to be (in the peak areas / heights) due to the protons on the alkene chain This mark may be awarded if the use of the alkene chain is indicated but not stated Both will have three sets of peaks due to the three sets of protons on the alkene chain (1) The alkene chain will give two doublets and a quintet in both isomers (1) In isoeugenol the doublets will have different peak areas / heights under the peaks / peak heights in ratio 1:3 whereas in eugenol the doublets will be the same height (1) OR Eugenol has areas / heights in the ratio 2:1:2:1:1:1:1:3 and isoeugenol has peak areas / heights in the ratio 3:1:1:1:1:1:1:3 The alkene chain will give two doublets and a quintet in both isomers (1) In isoeugenol the doublets will have different peak areas / heights under the peaks / peak heights in ratio 1:3 whereas in eugenol the doublets will be the same height (1) OR The only (significant) difference likely to be in the splitting pattern of the peaks due to the protons on the alkene chain (1) In eugenol the protons at the end of the alkene chain are in different environments so eugenol will have four sets of peaks whereas isoeugenol will have three sets of peaks (1) (1) (1) WCH0501msc20150812.docx 11 July 2015<br />
24(b)*(ii) (cont) In eugenol the alkene chain will give three doublets and a quintet (1) In isoeugenol the alkene chain will give two doublets and a quintet (1) Question Number 24(b)(iii) V2O5 oxidizes isoeugenol / alkene Acceptable Answer Reject Just V2O5 oxidizes Mark substituent (to the aldehyde & ketone) (and V(V) is reduced to a lower oxidation state) OR Explanation in terms of isoeugenol reducing V2O5 (1) H2O2 oxidizes vanadium back to the +5 oxidation state (1) Mechanism with H2O2 oxidizing V2O5 as the first step scores max 1 If no other mark is scored vanadium(V) is reduced then oxidized scores 1 Ignore any reference to adsorption and desorption on the surface. Question Number 24(b)(iv) Vanillin has an aldehyde group, suggesting a Acceptable Answer peak in the range 1740-1720 (cm1) whereas methyl vanillyl ketone has a ketone group suggesting a peak in the range 1700-1680 (cm1) (The peaks occur at different wavenumbers so the ketone peak could be seen) (1) These are general ranges and might overlap in the particular spectra OR Vanillin is an aromatic aldehyde OR Concentration of the ketone might be too small for the peak to be observed (1) Reject Mark WCH0501msc20150812.docx 11 July 2015<br />
Question Number 24(c)(i) 6 (moles of S2O3 Acceptable Answer 2 per mole CH3O) (1) Stand alone mark In the sequence ROCH3CH3IIBrHIO33I26S2O3 2 (1) Question Number 24(c)(ii) Mr (vanillin) = 152 (1) Acceptable Answer EITHER % CH3O in pure vanillin = 100 x 31/152 = 20.3947% (1) % purity of the vanillin = 100 x 20.09 / 20.3947 = 98.5058% (1) OR 20.09% weighs 31 So 100% weighs 100 x 31/20.09 = 154.31 So apparent molar mass = 154.31 (1) Therefore % purity is 152 x 100/154.31 = 98.5058% (1) OR Apparent mass CH3O = 100 x 20.09/152 = 30.5368 (1) Therefore % purity is 100 x 30.5368 /31 = 98.5058% (1) Correct answer with no working scores 3 IGNORE SF except 1 SF Mark Reject Partial sequences Reject Mark Total for Question 24 = 20 marks Total for Section C = 20 marks WCH0501msc20150812.docx 11 July 2015<br />

Question 21 22 23 24 25

About us | Need Help? | Feedback

Proudly powered by WordPress | Theme: Sydney by aThemes.