Edexcel Jan 2013 Paper 4 Q19 (with explained solutions) |

Skip to content

Edexcel Jan 2013 Paper 4 Q19

Answers available below

SECTION C Answer ALL the questions. Write your answers in the spaces provided. 19 An organic compound, X, was analyzed in a laboratory. (a) Compound X was found to have the following percentage composition by mass: carbon, C = 54.5% hydrogen, H = 9.1% oxygen, O = 36.4% (i) Use these data to calculate the empirical formula of compound X, showing your working. (2) 24 *P41214A02428*<br />
(ii) The mass spectrum of X is shown below. Relative abundance 100 80 60 40 20 15 88 30 45 60 75 90 m / z Use your answer to (a)(i), and the mass spectrum of X, to show that the molecular formula of compound X is C4H8O2. (b) The infrared spectrum of X has a broad peak at approximately 3500 cm1 and a sharp peak at approximately 1700 cm1. Identify the bond responsible for the peak at 3500 cm1 1700 cm1 (2) (2) *P41214A02528* 25 Turn over<br />

Show answer

SECTION C Question Number 19(a)(i) Acceptable Answers C : H : O Mole ratio / mol 54.5 : 9.1 : 36.4 12 1 16 (1) = 4.5417 : 9.1 : 2.275 = 1.996 : 4 : 1 = 2 : 4 : 1 C2H4O (1) Correct empirical formula of C2H4O, with or without working, scores (2) Reject Mark Mark Mark Question Number 19(a)(ii) First mark: Acceptable Answers Any mention of 44 or of doubling C2H4O (1) Second mark: Any mention of 88 in the context of the mass spectrum eg mentions molecular ion / M+ / heaviest peak / peak furthest to the right / annotation at 88 on the mass spectrum itself / highest m value z Reject 88 obtained just by adding up the relative atomic masses in C4H8O2 scores (0) for 2nd scoring point Acceptable Answers (1) Reject OH / OH CO / C=O / CO (Peak at 3500 cm1) OH (1) Allow OH (Peak at 1700 cm-1) C=O (1) Penalise extra extension bond on an otherwise correct answer once only (eg OH and C=O scores (1)) IGNORE any names for the bonds suggested even if incorrect Question Number 19(b)<br />
Mark Acceptable Answers Reject X is an aldehyde scores (0) for this scoring point / X is not a ketone scores (0) for this scoring point 2 so) First mark: (X is neutral) so not a (carboxylic) acid (1) IGNORE X doesnt have a charge as it is neutral / X is not an alkali / X is not a base Second mark: (X does not react with Tollens) so is not an aldehyde / is a ketone (1) Third mark: (X reacts with H+ / Cr2O7 is an alcohol /contains an OH (group) / contains ROH / contains hydroxyl (group) (1) IGNORE not an acid if this is deduced solely from the H+ / Cr2O7 Fourth mark: a primary or a secondary (alcohol) both needed OR (X is) not tertiary (alcohol) (1) Mark each point separately NOTE: X is a primary or a secondary alcohol scores both the third and fourth marks ALLOW Correct formulae for the functional groups, instead of their names 2 information Question Number 19(c)(i)<br />
Question Number 19(c)(ii) Acceptable Answers Reject (primary or secondary) alcohol and ketone NOTE BOTH names are required here Just hydroxyl for alcohol and/or C=O /carbonyl for ketone/ Mark<br />
Question Number 19(d) Acceptable Answers Reject MARKING ADVICE Check answer for the suggested structure of X. If the correct structure is shown Mark OH Mark answer according to the following. However if no structure for X is shown or an incorrect structure for X is proposed, mark answer according to COMMENTS scheme below MARKS CAN BE AWARDED FROM SUITABLY ANNOTATED FORMULAE FOR X. First mark: Four different H / hydrogen / proton environments (1) Any five from following seven points: Either Application of the (n+1) rule to peak J (which is a quartet / splits into four) or application of the (n+1) rule peak M (which is a doublet / splits into two) (1) Any mention to explain no splitting for peak L as there is no H is attached to the adjacent carbon (1) Peak L (CH3) next to C=O (1) Peak M (CH3) next to CH (1) Peak K OH (1) Peak J (CH) next to CH3 (1) Any one correctvalue quoted within0.2 of the following chemical shifts: 1.4(M) or 2.2 (L) or 3.7(K) or 4.2 (J) (ppm) (1) Just four different chemical environments If any incorrect chemical shift OR A RANGE of chemical shifts is quoted, this scoring point is not available<br />
(Total 18 marks) Total marks 90 Final mark (Compound X is) CH3CH(OH)COCH3 NO other compound allowed. ACCEPT any unambiguous formula, e.g. displayed formula Peak J OH Peak M Peak K Peak L ACCEPT 3-hydroxybutan-2-one (1)<br />
COMMENT Strategy for marking answers with an incorrect structure for X or where no structure is suggested for X. The maximum mark in such cases is FOUR OUT OF SEVEN or TWO OUT OF SEVEN IF NO STRUCTURE DRAWN (as second and fourth marks are not available) Scoring points: First mark: States four different H / hydrogen / proton environments Second mark: Structure drawn for X has exactly 4 hydrogen environments Third mark: Peak K is due to OH Fourth Mark: Providing the structure drawn for X would produce ONE of these splits. Any ONE of Application of the (n+1) rule to peak to explain a peak which is a quartet splits into four or Application of the (n+1) rule to peak to explain a peak which is a doublet / splits into two or Application of the (n+1) rule to peak to explain a peak which is a singlet due to a CH3 next to C=O SEE NEXT PAGE FOR MAXIMUM MARKS AVAILABLE FOR SOME LIKELY INCORRECT STRUCTURES FOR X<br />
C H Max 4 Max 4 Max 3 Max 3 Max 3 Max 2 Max 3 C H C O Max 2 Max 2<br />

Question 15 16 17 18 19

About us | Need Help? | Feedback

Proudly powered by WordPress | Theme: Sydney by aThemes.