Edexcel A-Level Chemistry AS Paper 2, June 2016: Question 5
14 marks · Medium difficulty · Open Response
Describe and calculate aspects of the preparation of 2-chloro-2-methylpropane from 2-methylpropan-2-ol, including separation, purification, distillation, percentage yield, and mechanism.
Practise this questionQuestion
Question text
5 The following procedure may be used to prepare 2-chloro-2-methylpropane.
Step 1 Place 15cm3 of 2-methylpropan-2-ol in a separating funnel and slowly add
30cm3 of concentrated hydrochloric acid (an excess), while swirling the
funnel.
Step 2 When all the hydrochloric acid has been added, leave the mixture to stand
for 20 minutes, shaking it gently at intervals.
Step 3 Once the organic and aqueous layers have completely separated, discard the
aqueous layer.
Step 4 Add saturated sodium hydrogencarbonate solution, a little at a time, to the
organic layer. After each addition, invert the separating funnel and open the tap.
Step 5 Discard the aqueous layer.
Step 6 Transfer the organic layer to a small flask, add a solid drying agent and swirl
the flask.
Step 7 Decant the liquid into a clean flask and distil it to collect pure
2-chloro-2-methylpropane.
Some data on the organic reactant and product are given in the table.
Data 2-methylpropan-2-ol 2-chloro-2-methylpropane
molar mass / g mol–1 74.0 92.5
boiling temperature / oC 82 51
density / g cm–3 0.79 0.84
(a) Draw a diagram of a separating funnel, labelling the aqueous layer and the layer
of 2-chloro-2-methylpropane that would be observed at the end of Step 2.
(2)
(b) Give the reason why sodium hydrogencarbonate solution is added to the organic
layer in Step 4 and why it is important to open the tap after adding this solution.
(2)
(c) Which one of these anhydrous compounds may be used as a drying agent in
Step 6?
(1)
A sodium chloride *P49838A01428*
B sodium hydroxide
C sodium nitrate
D sodium sulfate
(d) A student set up this apparatus for distillation in Step 7 as shown.
water
in
water
anti-bumping out
granules
electric
heater
(i) Describe three ways in which this apparatus must be modified for safe and
efficient use. Assume the apparatus is suitably clamped.
(3)
… 15
… *P49838A01528*
(ii) Give a suitable temperature range over which to collect the final product
during the distillation.
(1)
(e) In the preparation, 15cm3 of 2-methylpropan-2-ol produced
6.9 cm3 of 2-chloro-2-methylpropane.
The equation for the reaction is
(CH3)3COH + HCl o (CH3)3CCl + H2O
Calculate the percentage yield of 2-chloro-2-methylpropane, using data from the
16 table.
*P49838A01628*
Data 2-methylpropan-2-ol 2-chloro-2-methylpropane
molar mass / g mol–1 74.0 92.5
boiling temperature / oC 82 51
density / g cm−3 0.79 0.84
(3)
(f ) The mechanism for the reaction is in three stages.
CH3 CH3
Stage 1
H3C C CH3 H3C C CH3
O: H+ O+
H H
H
CH3 CH3
Stage 2
H C C CH H C C+ CH3 + H O
33 3 2
O+
H H
CH3 CH3
Stage 3
H C C+ CH H3C C CH3
:Cl– Cl
Add curly arrows to the reactants in Stages 2 and 3 to complete the mechanism.
(2)
(Total for Question 5 = 14 marks)
Mark scheme
Show the mark scheme
How to answer it
Preparation of 2-chloro-2-methylpropane
What this question tests
This question assesses core organic synthesis and practical chemistry skills, including separation techniques, purification methods (neutralisation and drying), distillation apparatus assembly, stoichiometry and percentage yield calculations, and curly arrow reaction mechanisms (SN1 pathway).
Separating Funnel Layers
✅ Correct Answer
A diagram of a separating funnel showing two distinct layers with correct labels:
- Top layer labelled: 2-chloro-2-methylpropane (organic layer)
- Bottom layer labelled: aqueous layer (or hydrochloric acid)
💡 Key Knowledge
Layers separate based on density. Using the data table: 2-chloro-2-methylpropane has a density of 0.84 g cm⁻³ , whereas the aqueous hydrochloric acid solution has a higher density ( ~1.0 g cm⁻³ ). Therefore, the organic product floats on top.
Purification & Pressure Release
✅ Correct Answer
- Step 4 reason: To react with and neutralise any unreacted/excess hydrochloric acid.
- Tap opening reason: To release the carbon dioxide gas produced and relieve the build-up of pressure.
❌ Common Errors
Students often lose marks by stating the tap is opened "because pressure builds up" without explaining why (due to carbon dioxide gas generation from the reaction between sodium hydrogencarbonate and excess acid). Never state that hydrogen gas is released.
Selecting a Drying Agent
✅ Correct Answer
D (sodium sulfate)
🧠 Exam Technique
An effective drying agent must be an anhydrous inorganic salt that chemically removes traces of water from the organic layer without reacting with the organic product itself. Sodium hydroxide is a strong base and would react; sodium chloride is not a drying agent; sodium nitrate is incorrect.
Distillation Apparatus Modifications
✅ Correct Answer
Three essential modifications for safe/efficient distillation:
- The bulb of the thermometer must be positioned directly opposite the side-arm opening to the condenser.
- The water flow in the condenser must be reversed (water in at the bottom, out at the top) to ensure the jacket fills completely.
- The apparatus must be left open (put a vent after the condenser / leave a gap before the receiver / use an open conical flask) to prevent a closed system and an explosion hazard.
❌ Common Errors
Sealing the collection vessel entirely is a major safety hazard. Students also frequently reverse the water jacket flow direction incorrectly or place the thermometer bulb too high or too low in the still head.
Collection Temperature Range
✅ Correct Answer
50 - 52°C (Accept any range between 49°C and 53°C provided it includes the boiling temperature of 51°C).
❌ Common Errors
Do not give a single fixed temperature value like 51°C ; distillation is carried out over a narrow temperature window to collect the pure fraction effectively.
Percentage Yield Calculation
📐 Step-by-Step Calculation
- Calculate mass and moles of alcohol used:
Mass = Volume × Density = 15.0 cm³ × 0.79 g cm⁻³ = 11.85 g
Moles of 2-methylpropan-2-ol = 11.85 g / 74.0 g mol⁻¹ = 0.16014 mol - Determine theoretical yield:
Since the stoichiometric ratio is 1:1, theoretical moles of 2-chloro-2-methylpropane = 0.16014 mol
Theoretical mass = 0.16014 mol × 92.5 g mol⁻¹ = 14.8125 g
*(Alternatively, theoretical volume = 14.8125 g / 0.84 g cm⁻³ = 17.634 cm³ )* - Calculate percentage yield:
Actual volume produced = 6.9 cm³
Percentage Yield = (6.9 / 17.634) × 100 = 39.1%
*(Using moles/mass yields the same result: 39.1% )*
❌ Calculation Traps
- Forgetting to convert volume to mass using density ( mass = volume × density ) before finding moles.
- Mixing up molar masses for reactant and product.
Mechanism Curly Arrows
✅ Correct Answer
- Stage 2: Draw a curly arrow starting from the C–O single bond and pointing towards the oxygen atom (breaking the C–O bond to form a carbocation and water).
- Stage 3: Draw a curly arrow starting from the lone pair of electrons on the chloride ion ( :Cl⁻ ) and pointing to the positively charged carbon atom ( C⁺ ).
❌ Common Errors
- Drawing curly arrows starting in the middle of space instead of explicitly originating from a bond or a lone pair of electrons.
- Using incorrect single-headed arrows instead of double-headed curly arrows representing movement of electron pairs.
Topics
Organic Chemistry · Core Practicals · Physical Chemistry · Core Practical 6: Investigating chlorination of 2-methylpropan-2-ol · Core Practical 7: Identify unknown organic liquids and inorganic solids · Topic 5: Formulae, Equations and Amounts of Substance · Topic 6: Organic Chemistry I
Question and mark scheme from the Edexcel A-Level Chemistry examination, AS Paper 2, June 2016. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.