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Edexcel Jun 2010 Paper 4 Q17
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![PMT SECTION B Answer ALL the questions. Write your answers in the spaces provided. 17 A bromoalkane has the molecular formula C4H9Br. The ionic equation for the hydrolysis of this compound with aqueous sodium hydroxide is shown below. C4H9Br + OHC4H9OH + Br (a) The rate of hydrolysis was investigated by mixing a large excess of the bromoalkane with aqueous sodium hydroxide, and measuring the time taken for all the hydroxide ions to be used up. This was carried out with different initial concentrations of the bromoalkane and the hydroxide ions. The results are shown in the table below. Experiment [C4H9Br] /mol dm3 [OH] /mol dm3 Time for OH to be used up/s Initial rate /mol dm3 s1 0.017 0.034 0.034 0.0012 0.0012 0.0020 42 21 35 2.9105 5.7105(i) Complete the missing value of the initial rate in the table. (ii) State the order of the reaction with respect to C4H9Br and to OH. Justify each answer by reference to the concentrations of both reactants. (1) (3) Order with respect to C4H9BrReasonOrder with respect to OHReason(iii) Deduce the rate equation for the reaction. (1) Rate =*N36506A0924* Turn over<br />
PMT (iv) Use the results for the first experiment in the table to calculate the rate constant and give its units. (2) Units(b) What evidence supports the theory that there is more than one step in the reaction mechanism? (1)(c) Write the mechanism for the hydrolysis of C4H9Br which is consistent with your rate equation. Show the structure of C4H9Br clearly in your mechanism. (3) 10 *N36506A01024*<br />
PMT *(d) Explain why primary and tertiary bromoalkanes are hydrolysed by different mechanisms. (2)(Total for Question 17 = 13 marks) *N36506A01124* 11 Turn over<br />](https://learnah.org/wp-content/uploads/2018/08/l_smalljpgJune_2010_QP_-_Unit_4_Edexcel_Chemistry_A-level_17-0.jpg)
![Section B Question Number 17 (a)(i) Question Number 17 (a)(ii) Acceptable Answers Reject Mark 5.7 x 10 5 /5.71 x 10 5 /5.714x 10 5 /0.000057 IGNORE SF except 1 (ie dont accept 6 x 10 5) Acceptable Answers Reject Mark C4H9Br: first order / 1 (1) (going from first to second experiment) rate doubles when concentration / number of moles doubles (and [OH] constant )/ rate and concentration increase in proportion (1) ALLOW time halves instead of rate doubles OH : zero order / 0 and (going from second to third expt) as increase in concentration does not affect rate (and [C4H9Br] constant ) (1) ALLOW doubling in concentration of OH instead of increase in concentration ALLOW time increases by the same factor as increase in hydroxide concentration (5/3) May refer to experiment number rather than concentrations Question Number 17 (a)(iii) Acceptable Answers Reject Mark Rate = k[C4H9Br] OR Rate = k[C4H9Br]1[OH]0 ALLOW k in lower or upper case Rate equation must be consistent with orders in (a)(ii) If no order is given for hydroxide in (ii) mark cannot be given<br />
Question Number 17 (a)(iv) Acceptable Answers Reject Mark k= 2.9 x 10-5 0.017 = 1.7 x 10-3 / 1.71 x 10-3 / 1.706 x 10-3 s-1 ALLOW k=1.68 x 10-3 (value obtained from experiment 2 or 3) value of k (1) units (1) stand alone mark ALLOW TE from (a)(iii) IGNORE SF except 1 Rate = k[C4H9Br]2 gives k= 0.10036 dm3 mol-1s-1 Rate = k[C4H9Br][OH-] gives k= 1.42 dm3 mol-1s-1 ALLOW k=1.39 dm3 mol-1s-1 (value obtained from experiment 2 or 3) Rate = k[C4H9Br][OH-]2 gives k= 1184.6 dm6 mol-2s-1 Rate = k[C4H9Br]2[OH-] gives k= 83.62 dm6 mol-2s-1 Question Number 17(b) Acceptable Answers Reject Mark [OH] is (in chemical equation but) not in rate equation / not in rate determining step (so is in a step other than rate determining step) OR Only C4H9Br is in rate equation / rate determining step (so OH is in a step other than rate determining step)<br />
Question Number 17 (c) Acceptable Answers First mark Choice of bromoalkane must be consistent with rate equation in (a)(iii). If [OH] is not in rate equation, secondary/tertiary bromoalkane. If [OH] is in rate equation, primary/secondary bromoalkane. (1) Second and third marks Either SN1 or SN2 mechanism can score 2 marks regardless of choice of bromoalkane. Mark Lone pairs not required Curly arrow from C-Br bond to Br (making Br) (1) Curly arrow from anywhere on OH / HO to C+ in correct intermediate (making alcohol) (1) OR Both curly arrows from OH and from C-Br bond to Br (may both be shown at start) (1) Transition state including minus charge (and product) (1) Do not penalise if C2H5 shown instead of C3H7. Bonds in transition state can be dotted. Do not penalise the missing H atoms in alkyl groups in mechanism.<br />
Question Number 17 (d) QWC Acceptable Answers Reject Mark (Primary and tertiary) carbocation intermediates have different stabilities (1) as (inductive effects of) alkyl groups stabilise tertiary carbocation (1) OR Steric hindrance differs for attack on primary and tertiary carbon (in the molecule) / less space available for attack by OH on tertiary carbon / more space for attack by OH on primary carbon (1) as bulky / three alkyl groups obstruct attack (1) Tertiary bromoalkanes react by SN1 without further explanation carbocation intermediates have different reactivity steric hindrance in carbocation<br />](https://learnah.org/wp-content/uploads/2018/08/June_2010_MS_-_Unit_4_Edexcel_Chemistry_A-level_17-0.jpg)
