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Date	May 2022	Marks available	2	Reference code	22M.2.hl.TZ1.5
Level	HL	Paper	2	Time zone	TZ1
Command term	Outline	Question number	5	Adapted from	N/A

Question

Organomagnesium compounds can react with carbonyl compounds. One overall equation is:

Compound B can also be prepared by reacting an alkene with water.

Iodomethane is used to prepare CH₃Mg $I$ . It can also be converted into methanol:

CH₃ $I$ + HO^– → CH₃OH + $I$ ^–

State the name of Compound B, applying International Union of Pure and Applied Chemistry (IUPAC) rules.

[1]

a(i).

Compound A and Compound B are both liquids at room temperature and pressure. Identify the strongest intermolecular force between molecules of Compound A.

[1]

a(ii).

State the number of $σ$ (sigma) and $π$ (pi) bonds in Compound A.

[1]

a(iii).

Deduce the hybridization of the central carbon atom in Compound A.

[1]

a(iv).

Identify the isomer of Compound B that exists as optical isomers (enantiomers).

[1]

a(v).

Draw the structural formula of the alkene required.

[1]

b(i).

Explain why the reaction produces more (CH₃)₃COH than (CH₃)₂CHCH₂OH.

[2]

b(ii).

Deduce the structural formula of the repeating unit of the polymer formed from this alkene.

[1]

b(iii).

Deduce what would be observed when Compound B is warmed with acidified aqueous potassium dichromate (VI).

[1]

Identify the type of reaction.

[1]

d(i).

Outline the requirements for a collision between reactants to yield products.

[2]

d(ii).

Explain the mechanism of the reaction using curly arrows to represent the movement of electron pairs.

[3]

d(iii).

The polarity of the carbon–halogen bond, C–X, facilitates attack by HO^–.

Outline, giving a reason, how the bond polarity changes going down group 17.

[1]

d(iv).

Markscheme

2-methylpropan-2-ol /2-methyl-2-propanol ✔

Accept methylpropan-2-ol/ methyl-2-propanol.

Do not accept 2-methylpropanol.

a(i).

dipole-dipole ✔

Do not accept van der Waals’ forces.

a(ii).

$σ$ : 9
AND
$π$ : 1 ✔

a(iii).

sp² ✔

a(iv).

butan-2-ol/CH₃CH(OH)C₂H₅ ✔

a(v).

b(i).

carbocation formed from (CH₃)₃COH is more stable / (CH₃)₃C⁺ is more stable than (CH₃)₂CHCH₂⁺ ✔

«because carbocation has» greater number of alkyl groups/lower charge on the atom/higher e^- density
OR
«greater number of alkyl groups» are more electron releasing
OR
«greater number of alkyl groups creates» greater inductive/+I effect ✔

Do not award any marks for simply quoting Markovnikov’s rule.

b(ii).

Do not penalize missing brackets or n.

Do not award mark if continuation bonds are not shown.

b(iii).

no change «in colour/appearance/solution» ✔

«nucleophilic» substitution
OR
SN2 ✔

Accept “hydrolysis”.

Accept SN1

d(i).

energy/E ≥ activation energy/E_a ✔

correct orientation «of reacting particles»
OR
correct geometry «of reacting particles» ✔

d(ii).

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

curly arrow showing I leaving ✔

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

Accept OH^- with or without the lone pair.

Do not allow curly arrows originating on H, rather than the -, in OH^-.

Accept curly arrows in the transition state.

Do not penalize if HO and I are not at 180°.

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

Award [2 max] if S_N1 mechanism shown.

d(iii).

decreases/less polar AND electronegativity «of the halogen» decreases ✔

Accept “decreases” AND a correct comparison of the electronegativity of two halogens.

Accept “decreases” AND “attraction for valence electrons decreases”.

d(iv).

Examiners report

Naming the organic compound using IUPAC rules was generally done well.

a(i).

[N/A]

a(ii).

Mediocre performance in stating the number of σ (sigma) and π (pi) bonds in propanone; the common answer was 3 σ and 1 π instead of 9 σ and 1 π, suggesting the three C-H σ bonds in each of the two methyl groups were ignored.

a(iii).

sp² hybridization of the central carbon atom in the ketone was very done well.

a(iv).

Mediocre performance; some identified 2-methylpropan-1-ol or -2-ol, instead butan-2-ol/CH₃CH(OH)C₂H₅ as the isomer that exists as an optical isomer.

a(v).

Good performance; some had a H and CH₃ group on each C atom across double bond instead of having two H atoms on one C and two CH₃ groups on the other.

b(i).

Poor performance, particularly in light of past feedback provided in similar questions since there was repeated reference simply to Markovnikov's rule, without any explanation.

b(ii).

Mediocre performance; deducing structural formula of repeating unit of the polymer was challenging in which continuation bonds were sometimes missing, or structure included a double bond or one of the CH₃ group was missing.

b(iii).

Mediocre performance; deducing whether the tertiary alcohol could be oxidized solicited mixed responses ranging from the correct one, namely no change (in colour, appearance or solution), to tertiary alcohol will be reduced, or oxidized, or colour will change will occur, and such.

Excellent performance on the type of reaction but with some incorrect answers such as alkane substitution, free radical substitution or electrophilic substitution.

d(i).

Good performance. For the requirements for a collision between reactants to yield products, some suggested necessary, sufficient or enough energy or even enough activation energy instead of energy/E ≥ activation energy/E_a.

d(ii).

Mechanism for SN2 not done well. Often the negative charge on OH was missing, the curly arrow was not going from lone pair/negative charge on O in -OH to C, or the curly arrow showing I leaving placed incorrectly and specially the negative charge was missing in the transition state. Formation of a carbocation intermediate indicating SN1 mechanism could score a maximum of 2 marks.

d(iii).

Good performance on how the polarity of C-X bond changes going down group 17.

d(iv).

Syllabus sections

Core » Topic 6: Chemical kinetics » 6.1 Collision theory and rates of reaction

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