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Question 19N.2.hl.TZ0.3

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Date November 2019 Marks available [Maximum mark: 21] Reference code 19N.2.hl.TZ0.3
Level hl Paper 2 Time zone TZ0
Command term Apply, Deduce, Determine, Draw, Explain, Formulate, Outline, Show, Sketch, State, Write Question number 3 Adapted from N/A
3.
[Maximum mark: 21]
19N.2.hl.TZ0.3

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.

(a(i))

State the type of reaction.

[1]

Markscheme

«electrophilic» addition ✔

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

(a(ii))

State the IUPAC name of the major product.

[1]

Markscheme

2-chloropropane ✔

(a(iii))

Outline why it is the major product.

[1]

Markscheme

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

(a(iv))

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

[1]

Markscheme

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

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.

(b(i))

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

[3]

Markscheme

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(ii))

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

[1]

Markscheme

SN2 ✔

NOTE: Accept ‘bimolecular nucleophilic substitution.’

(b(iii))

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

[4]

Markscheme

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–.

(c(i))

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

[1]

Markscheme

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

(c(ii))

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

[3]

Markscheme

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.

(d(i))

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

[1]

Markscheme

K2Cr2O7/Cr2O72–/«potassium» dichromate «(VI)» AND acidified/H+
OR
«acidified potassium» manganate(VII) / «H+ and» KMnO4 / «H+ and» MnO4– ✔

 

NOTE: Accept “H2SO4” or “H3PO4” for “H+”.
Do not accept HCl.
Accept “permanganate” for “manganate(VII)”.

(d(ii))

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

[2]

Markscheme

C3H8O/propan-2-ol: hydrogen-bonding AND C3H6O/propanone: no hydrogen bonding/«only» dipole–dipole/dispersion forces ✔

hydrogen bonding stronger «than dipole–dipole» ✔

(d(iii))

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

[1]

Markscheme

only one hydrogen environment
OR
methyl groups symmetrical «around carbonyl group» ✔

NOTE: Accept “all hydrogens belong to methyl groups «which are in identical positions»”.

(e)

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

[1]

Markscheme

✔

 

NOTE: Continuation bonds must be shown.

Methyl groups may be drawn on opposite sides of the chain or head to tail.

Ignore square brackets and “n”.

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