The reaction between 1-chloro-1-phenylethane and hydroxide ions to produce 1-phenylethanol is: C₆H₅CH(Cl)CH₃ + OH⁻ → C₆H₅CH(OH)CH₃ + Cl⁻ 1-chloro-1-phenylethane 1-phenylethanol The rate of this reaction can be studied by measuring the amount of hydroxide ions that remain in solution at a given time. The reaction can effectively be stopped if the solution is diluted with an ice-cold solvent.
- (a)Describe a suitable method for studying the rate of this reaction at a temperature of 40°C, given the following. • a solution of 0.10 mol dm⁻³ 1-chloro-1-phenylethane, labelled A • a solution of 0.10 mol dm⁻³ sodium hydroxide, labelled B • 0.10 mol dm⁻³ HCl • volumetric glassware • ice-cold solvent • stopclock • access to standard laboratory equipment and chemicals[4]
- (b)(i)The rate of this reaction was measured at different initial concentrations of the two reagents. The table shows the results obtained. ``` experiment [C₆H₅CH(Cl)CH₃] /mol dm⁻³ [OH⁻] /mol dm⁻³ relative rate 1 0.05 0.10 0.5 2 0.10 0.20 1.0 3 0.15 0.10 1.5 4 0.20 0.15 to be calculated ``` Deduce the order of reaction with respect to each of [C₆H₅CH(Cl)CH₃] and [OH⁻]. Explain your reasoning. order with respect to [C₆H₅CH(Cl)CH₃] ______________ order with respect to [OH⁻] ______________[2]
- (b)(ii)Write the rate equation for this reaction, stating the units of the rate constant, k. rate = ______________ mol dm⁻³ s⁻¹ units of k = ______________[1]
- (b)(iii)Calculate the relative rate for experiment 4. relative rate for experiment 4 = ______________[1]
- (c)(i)Use your answers in (b)(i) to help you to draw the mechanism for the reaction of 1-chloro-1-phenylethane with hydroxide ions, including the following. • all relevant lone pairs and dipoles • curly arrows to show the movement of electron pairs • the structures of any transition state or intermediate[3]
- (c)(ii)This reaction was carried out using a single optical isomer of 1-chloro-1-phenylethane. Use your mechanism in (i) to predict whether the product will be a single optical isomer or a mixture of two optical isomers. Explain your answer.[1]
- (d)The proton NMR spectrum of a sample of 1-phenylethanol shows four peaks: a multiplet for the C₆H₅ protons and three other peaks as shown in the table. When the sample is shaken with D₂O and the proton NMR spectrum recorded, fewer peaks are seen. Complete the table for the proton NMR spectrum of 1-phenylethanol, C₆H₅CH(OH)CH₃. Use of the Data Booklet might be helpful. [Figure X.X][4]
This question uses a diagram from the original paper, which isn't shown here. Open the full paper from the past papers archive to see it.
Mark scheme
Marking points from the official mark scheme (May/June 2017). The mark or mark type is shown in brackets.
- [4] Any of the three methods possible. Any 4 of the 5 points for each method available for maximum 4 marks.
- [1] Ensure both solutions (A and B) at 40 °C before mixing
- [1] mix known volumes of A and B and start the clock
- [1] at known time take out a sample / X and add it to ice-cold solvent
- [1] titrate against HCl
- [1] repeat at time at known time intervals
- [1] from 1 and 3: when [RCl] is trebled, so is rate, so order w.r.t. [RCl] = 1
- [1] from 1 and 2: when both concentrations are doubled, rate doubles so [OH¯] has no effect on rate, so order w.r.t.[OH] = 0
- [1] rate = k[RC1] AND units: sec¯¹ 1/s
- [1] relative rate = 2.0
- [1] C-Cl dipole and first curly arrow
- [1] intermediate cation
- [1] OH with lone pair and curly arrow
- [1] If SN1: racemate / mixture of / two optical isomers will be formed, because: the intermediate is planar / has a plane of symmetry / OH¯ can approach from top or bottom or from any direction
- [1] If SN2: one optical isomer because attack always from fixed direction / or from same side / the "configuration” always inverts / there is an asymmetric transition state
- [1] the three groups are in their correct places wrt the δ values
- [1] no. of H atoms for each peak agrees with group column
- [1] splitting patterns doublet, singlet and quartet are assigned to correct groups
- [1] peak identified as OH disappears with D₂O, no other peak disappears
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