Edexcel A-Level Chemistry AS Paper 2, June 2018: Question 8

11 marks · Medium difficulty · Open Response

Properties, mass spectrometry, mean bond enthalpy calculations, incomplete combustion enthalpy changes, and oxidation reactions of 2-methylpropan-2-ol.

Practise this question

Question

An exam question about 2-methylpropan-2-ol consisting of multiple parts: (a) drawing the fully displayed formula; (b)(i) explaining the absence of a molecular ion peak in its mass spectrum; (b)(ii) giving the formula of the species for the peak at m/z = 59; (c)(i) calculating the enthalpy of combustion using mean bond enthalpies from a provided table; (c)(ii) explaining the effect of incomplete combustion (smoky flame) on experimentally determined enthalpy; (c)(iii) giving the reason for the difference between the calculated value and the Data Book value; and (d) a multiple-choice question on the observation when heated with acidified potassium dichromate(VI).
Question text

8 This question is about 2-methylpropan-2-ol.

(a) Draw the fully displayed formula of 2-methylpropan-2-ol.

(1)

(b) The mass spectrum of 2-methylpropan-2-ol is shown.

10 15 20 25 30 35 40 45 50 55 60 65 70 75

m / z

(i) The relative molecular mass of 2-methylpropan-2-ol is 74. Give a possible reason

why there is no molecular ion peak in the mass spectrum of 2-methylpropan-2-ol.

(1)

(ii) Write the formula for a species that could be responsible for the peak at m / z = 59.

20 (1)

*P51460RA02024*

(c) The equation for the complete combustion of 2-methylpropan-2-ol is

C4H10O(l) + 6O2(g) → 4CO2(g) + 5H2O(l)

(i) Using the bond enthalpies shown in the table, calculate a value for the

enthalpy change, in kJ mol−1, for the complete combustion of

2-methylpropan-2-ol.

(4)

Bond Mean bond enthalpy / kJ mol−1

C C 347

C H 413

C O 358

O H 464

O O 498

C O 805

(ii) 2-methylpropan-2-ol burns in air with a smoky flame. Explain how burning

with a smoky flame affects the value of the experimentally determined

enthalpy change of combustion.

(2)

(iii) A Data Book value for the enthalpy change of combustion of 2-methylpropan-2-ol

is −2643.8 kJ mol−1. Give the main reason for the difference between this value

and your answer to part 8(c)(i).

(1)

… *P51460RA02124*

(d) Which observation would be expected when 2-methylpropan-2-ol is heated with

potassium dichromate(VI) and dilute sulfuric acid?

(1)

A orange to green

B green to orange

C purple to colourless

D no change

(Total for Question 8 = 11 marks)

Mark scheme

Show the mark scheme The mark scheme provides the fully displayed structural formula for 2-methylpropan-2-ol in part (a), accepts answers referencing molecular ions being unstable/fragmenting in (b)(i), shows the fragment structure for m/z = 59 in (b)(ii), gives the full step-by-step calculation for enthalpy of combustion yielding -2512 kJ mol-1 in (c)(i), marks incomplete combustion and less exothermic enthalpy values in (c)(ii), references states of matter and conditions (298 K / gaseous states) in (c)(iii), and identifies option D as the correct answer for part (d) since tertiary alcohols are not oxidized.

Question Acceptable Answer Additional Guidance Mark

Number

8(a) display all three methyl groups

allow –OH

do not award C-H-O

(1)

Question Acceptable Answer Additional Guidance Mark

Number

8(b)(i) An answer that makes reference to one of the

following:

molecular ion/molecule fragments/is unstable

(1)

Question Acceptable Answer Additional Guidance Mark

Number

8(b)(ii)

allow + charge on any part of the ion/outside

33 the structure but + must be shown

allow displayed/structural/skeletal/ molecular

formulae or any combination of these.

(1)

Question Acceptable Answer Additional Guidance Mark

Number

8(c)(i) Example of calculation

calculation for bonds broken in the alcohol (*) (1) 3(C-C) + 9(C-H) + (C-O) + (O-H)

=(3x347) + (9x413) + 358 + 464 = (+)5580

(kJ mol-1)

calculation for bonds broken in oxygen

6(O=O) = (6 x 498) = (+)2988 (kJ mol-1)

and

total energy for bonds broken(**) (1)

total = + 5580 + 2988 = (+)8568 (kJ mol-1)

TE from ans * M1 + 2988

calculation for bonds made(***) (1)

= 8(C=O) + 10(O-H)

= (8x805) + (10x464) = -11080 (kJ mol-1)

calculation of cH (2-methylpropan-2-ol) with

sign (1) = +8568 – 11080 = -2512 (kJ mol-1)

allow TE for answer(**) + answer(***)

units not required but if given they must be

correct

correct final answer with no working scores 4

marks (4)

Question Acceptable Answer Additional Guidance Mark

Number

8(c)(ii) An explanation that makes reference to the following mark independently

points:

incomplete combustion (1)

cH (2-methylpropan-2-ol) will be less negative

/less exothermic than data book value (1) do not award just lower/smaller/decreases/

more positive

allow reduce the magnitude (of the value) (2)

Question Acceptable Answer Additional Guidance Mark

Number

8(c)(iii) An answer that makes reference to the following

points:

cH figures are at 298 K /data book bond energies

refer to gaseous state

and

water and/or 2-methylpropan-2-ol are/is (both) allow just liquid involved

liquid(s) (at 298 K)

do not award

data book bond energies are mean (values)/not

specific to 2-methylpropan-2-ol (1)

Question Acceptable Answer Mark

Number

8(d) The only correct answer is D

A is not correct because tertiary alcohol is not oxidised

B is not correct because this is incorrect colour change for acidified dichromate

C is not correct because this is incorrect colour change for these reagents (1)

(Total for Question 8 = 11 marks)

How to answer it

Properties and Reactions of 2-methylpropan-2-ol

What this question tests

This comprehensive AS Level Chemistry question assesses core organic and physical chemistry concepts centered around 2-methylpropan-2-ol. Key skills tested include drawing fully displayed structural formulae, interpreting mass spectra (fragmentation and molecular ions), calculating enthalpy changes of combustion using mean bond enthalpies, understanding experimental limitations (incomplete combustion and state differences), and applying knowledge of alcohol oxidation states.

Question Part (a)

Fully Displayed Formula

✅ Correct Answer

Every single bond and atom must be explicitly shown. For 2-methylpropan-2-ol, the central carbon is bonded to three methyl groups (-CH₃) and one hydroxyl group (-OH).

    H
    |&
H—C—H
    |         H
H—C—C—O—H
    |         |
H—C—H  H
    |         
    H

❌ Common Errors

  • Condensing any of the methyl groups into -CH₃ instead of showing all individual C-H bonds.
  • Writing C-H-O connections instead of clearly bonding oxygen to carbon and hydrogen ( C-O-H ).
Maximum Marks: 1
Question Part (b)

Mass Spectrometry Analysis

(i) Absence of Molecular Ion Peak

✅ Correct Answer

The molecular ion ( M⁺ ) is too unstable and readily fragments immediately upon formation, meaning no ions reach the detector with the full relative molecular mass of 74.

💡 Key Knowledge

Tertiary alcohols undergo very rapid and extensive fragmentation in a mass spectrometer due to the high stability of the resulting tertiary carbocations formed after cleavage.

(ii) Species Responsible for Peak at m / z = 59

✅ Correct Answer

Loss of a methyl group ( CH₃ , mass 15) from the molecular ion (74 - 15 = 59):

[(CH₃)₂C(OH)]⁺ or [C₄H₉O]⁺ with a positive charge shown.

🧠 Exam Technique

Always remember to include the positive charge symbol ( + ) when writing formulas for species represented on a mass spectrum, as mass spectrometry detects ions only.

Maximum Marks: 1 + 1 = 2
Question Part (c)

Enthalpy of Combustion & Bond Energies

(i) Calculating Enthalpy of Combustion

📐 Step-by-Step Calculation

  1. Bonds broken in alcohol (C₄H₁₀O): 3(C—C) + 9(C—H) + 1(C—O) + 1(O—H)
    = (3 × 347) + (9 × 413) + 358 + 464 = +5580 kJ mol⁻¹
  2. Bonds broken in oxygen (6 O₂): 6(O=O) = 6 × 498 = +2988 kJ mol⁻¹
  3. Total energy input (bonds broken): 5580 + 2988 = +8568 kJ mol⁻¹
  4. Bonds made in products (4 CO₂ + 5 H₂O): Products contain 8(C=O) and 10(O—H)
    = (8 × 805) + (10 × 464) = 6440 + 4640 = -11080 kJ mol⁻¹
  5. Calculate ΔH: ΔH = Σ(Bonds broken) - Σ(Bonds made) = 8568 - 11080 = -2512 kJ mol⁻¹

❌ Common Calculation Traps

  • Forgetting to multiply individual bond values by the stoichiometric balancing coefficients from the combustion equation.
  • Omitting the negative sign on the final enthalpy change value (exothermic reactions require a minus sign).

🧠 Examiner Commentary

This is a standard 4-mark calculation. Even if you make an arithmetic error early on, follow-through marks (TE) are awarded if the structure of your calculation is correct.

(ii) Effect of a Smoky Flame

✅ Correct Answer

A smoky flame indicates incomplete combustion (forming carbon/soot instead of purely carbon dioxide). Less energy is released, meaning the experimentally determined enthalpy change will be less negative / less exothermic than the theoretical value.

(iii) Difference from Data Book Values

💡 Key Knowledge

Data book values refer to standard conditions (298 K) and standard states, whereas mean bond enthalpies are average values derived across a range of different gaseous molecules, not specifically tailored to liquid 2-methylpropan-2-ol and liquid water at 298 K.

Maximum Marks: 4 + 2 + 1 = 7
Question Part (d)

Oxidation of Alcohols

✅ Correct Answer: D (no change)

2-methylpropan-2-ol is a tertiary alcohol. Tertiary alcohols do not have a hydrogen atom attached to the carbon bearing the -OH group, meaning they are resistant to oxidation by acidified potassium dichromate(VI).

❌ Why other options are wrong

  • A (orange to green): This occurs with primary and secondary alcohols as chromium is reduced from Cr(VI) to Cr(III), but tertiary alcohols do not react.
  • B & C: These describe incorrect colour transitions not associated with this standard functional group test.
Maximum Marks: 1

Topics

Organic Chemistry · Physical Chemistry · Core Practicals · Topic 6: Organic Chemistry I · Topic 7: Modern Analytical Techniques I · Topic 8: Energetics I

Question and mark scheme from the Edexcel A-Level Chemistry examination, AS Paper 2, June 2018. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.