Edexcel Jun 2013 (R) Paper 1 Q24 (with explained solutions) |

Skip to content

Edexcel Jun 2013 (R) Paper 1 Q24

Answers available below

24 Alkenes are unsaturated hydrocarbons. They are used in the industrial production of many organic compounds. (a) Add structural formulae to the flowchart below to show the organic product formed in each addition reaction of 2-methylpropene. (4) H2(g) with nickel catalyst Br2(l) C C CH3 CH3 HBr(g) mixture of isomers *P42982A01924* 19 Turn over<br />
(b) Suggest a mechanism for the reaction of 2-methylpropene with bromine, Br2(l). Include curly arrows. (c) Ethene, C2H4, was prepared from ethanol, C2H5OH, by the following reaction C2H5OH o C2H4 + H2O A chemist reacted 9.2 g of ethanol, C2H5OH, and obtained 4.2 g of ethene. Calculate the percentage yield of ethene in the reaction. (3) (2) (Total for Question 24 = 9 marks) 20 *P42982A02024*<br />

Show answer

Question Number 24(a) Acceptable Answers Reject Mark (1) for each correct product ALLOW correct displayed / skeletal / semi-skeletal / structural / semi-structural formula in each case ALLOW any order of symbols after or before each carbon ALLOW brackets or no brackets around Br/ CH3 for example CH2BrCH3CBrCH3 6CH0101R 1306<br />
Question Number 24(b) Acceptable Answers H3C H3C Br Br H3C H3C Br (:) Br (1) for both arrows (1) for carbocation (1) for arrow H3C H3C Br Br First mark Double-headed arrow from alkene must start from somewhere on C=C bond Partial charge on Br2 molecule must be correct if shown Second mark is for either correct primary or secondary carbocation and is a standalone mark Third mark Double-headed arrow from bromide ion can start from the minus sign, a lone pair on Br , or from the Br and can go to the C or the + sign on the intermediate The negative charge must be present on the bromide ion The final product, if shown, must be correct to gain third mark Mechanisms with other electrophiles (e.g. HBr, BrOH) can score 2nd and 3rd marks Mark Reject Single-headed arrow Bromine / bromide free radicals Single-headed arrow (Penalise again) 6CH0101R 1306<br />
Mark Question Number 24(c) Acceptable Answers Reject (0) for 4.2 x 100% 9.2 First mark is for calculating the theoretical maximum mass of ethene from 9.2 g ethanol:- (46 g C2H5OH gives 28 g C2H4 so 9.2 g C2H5OH gives maximum mass of) 5.6 g C2H4 (1) Second mark is for calculating the percentage yield from candidates theoretical maximum mass:- (4.2/5.6 x 100% =) 75 (%) IGNORE s.f. except 1 s.f. OR First mark Amount of ethene = 4.2/28 = 0.15 (mol) and amount of ethanol = 9.2/46 = 0.20 (mol) (1) Second mark % yield = 0.15/0.20 = 75 % (1) NOTE Correct answer with no working scores (2) % yield TE on candidates theoretical mass / moles only if % yield 100% If molar masses are reversed, award one mark for 27.8% Total for Question 24 = 9 marks 6CH0101R 1306<br />

Question 18 19 20 21 22 23 24 25

About us | Need Help? | Feedback

Proudly powered by WordPress | Theme: Sydney by aThemes.