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Date	November 2019	Marks available	1	Reference code	19N.2.hl.TZ0.3
Level	HL	Paper	2	Time zone	TZ0
Command term	State	Question number	3	Adapted from	N/A

Question

Propene is an important starting material for many products. The following shows some compounds which can be made from propene, C₃H₆.

Propene (C₃H₆) → C₃H₇Cl → C₃H₈O → C₃H₆O

Consider the conversion of propene to C₃H₇Cl.

An experiment was carried out to determine the order of reaction between one of the isomers of C₃H₇Cl and aqueous sodium hydroxide. The following results were obtained.

State the type of reaction.

[1]

a(i).

State the IUPAC name of the major product.

[1]

a(ii).

Outline why it is the major product.

[1]

a(iii).

Write an equation for the reaction of the major product with aqueous sodium hydroxide to produce a C₃H₈O compound, showing structural formulas.

[1]

a(iv).

Determine the rate expression from the results, explaining your method.

[3]

b(i).

Deduce the type of mechanism for the reaction of this isomer of C₃H₇Cl with aqueous sodium hydroxide.

[1]

b(ii).

Sketch the mechanism using curly arrows to represent the movement of electrons.

[4]

b(iii).

Write an equation for the complete combustion of the compound C₃H₈O formed in (a)(iv).

[1]

c(i).

Determine the enthalpy of combustion of this compound, in kJ mol⁻¹, using data from section 11 of the data booklet.

[3]

c(ii).

State the reagents for the conversion of the compound C₃H₈O formed in (a)(iv) into C₃H₆O.

[1]

d(i).

Explain why the compound C₃H₈O, produced in (a)(iv), has a higher boiling point than compound C₃H₆O, produced in d(i).

[2]

d(ii).

Explain why the ¹H NMR spectrum of C₃H₆O, produced in (d)(i), shows only one signal.

[1]

d(iii).

Propene is often polymerized. Draw a section of the resulting polymer, showing two repeating units.

[1]

Markscheme

«electrophilic» addition ✔

NOTE: Do not accept “nucleophilic addition” or “free radical addition”.
Do not accept “halogenation”.

a(i).

2-chloropropane ✔

a(ii).

secondary carbocation/carbonium «ion» is more stable
OR
carbocation/carbonium «ion» stabilized by two/more alkyl groups ✔

a(iii).

CH₃CHClCH₃ (l) + OH⁻ (aq) → CH₃CH(OH)CH₃ (aq) + Cl⁻ (aq)
OR
CH₃CHClCH₃ (l) + NaOH (aq) → CH₃CH(OH)CH₃ (aq) + NaCl (aq) ✔

a(iv).

Rate = k [C₃H₇Cl] [OH⁻] ✔

«[OH⁻] held constant and» [C₃H₇Cl] triples AND rate triples «so first order wrt C₃H₇Cl» ✔

[C₃H₇Cl] doubles AND [OH⁻] doubles AND rate quadruples «so first order wrt OH⁻» ✔

b(i).

SN2 ✔

NOTE: Accept ‘bimolecular nucleophilic substitution.’

b(ii).

curly arrow going from lone pair on O/negative charge on OH^– to C ✔

curly arrow showing C–Cl bond breaking ✔

representation of transition state showing negative charge, square brackets and partial bonds ✔

formation of CH₃CH(OH)CH₃ AND Cl^– ✔

NOTE: Do not allow arrow originating on H in OH^–.

Allow curly arrow going from bond between C and Cl to Cl in either reactant or transition state.

Do not award M3 if OH–C bond is represented.

Accept formation of NaCl instead of Cl^–.

b(iii).

2C₃H₈O (l) + 9O₂ (g) → 6CO₂ (g) + 8H₂O (g)
OR
C₃H₈O (l) + 4.5O₂ (g) → 3CO₂ (g) + 4H₂O (g) ✔

c(i).

bonds broken:
7(C–H) + C–O + O–H + 2(C–C) + 4.5(O=O)
OR
7(414 «kJ mol⁻¹») + 358 «kJ mol⁻¹» + 463 «kJ mol⁻¹» + 2(346 «kJ mol⁻¹») + 4.5(498 «kJ mol⁻¹») / 6652 «kJ» ✔

bonds formed:
6(C=O) + 8(O–H)
OR
6(804 «kJ mol⁻¹») + 8(463 «kJ mol⁻¹») / 8528 «kJ» ✔

«ΔH = bonds broken − bonds formed = 6652 – 8528 =» −1876 «kJ mol⁻¹» ✔

NOTE: Award [3] for correct final answer.

c(ii).

K₂Cr₂O₇/Cr₂O₇^2–/«potassium» dichromate «(VI)» AND acidified/H⁺
OR
«acidified potassium» manganate(VII) / «H⁺ and» KMnO₄ / «H⁺ and» MnO₄– ✔