DP Chemistry (last assessment 2024)

Question 20N.2.hl.TZ0.2

Date	November 2020	Marks available	[Maximum mark: 14]	Reference code	20N.2.hl.TZ0.2
Level	hl	Paper	2	Time zone	TZ0
Command term	Calculate, Deduce, Draw, Justify, Predict, State, Suggest	Question number	2	Adapted from	N/A

[Maximum mark: 14]

20N.2.hl.TZ0.2

Compound A is in equilibrium with compound B.

(a)

Predict the electron domain and molecular geometries around the oxygen atom of molecule A using VSEPR

[2]

Markscheme

Electron domain geometry: tetrahedral ✔

Molecular geometry: bent/V-shaped ✔

Examiners report

The majority of students got at least one of electron domain geometry or molecular geometry correct.

(b)

State the type of hybridization shown by the central carbon atom in molecule B.

[1]

Markscheme

${sp}^{2}$ ✔

Examiners report

The vast majority of students could identify the hybridization around a central carbon atom.

(c)

State the number of sigma ( $σ$ ) and pi ( $π$ ) bonds around the central carbon atom in molecule B.

[1]

Markscheme

$σ$ -bonds: $3$
AND
$π$ -bonds: $1$ ✔

Examiners report

The vast majority of students could identify BOTH sigma and pi bonds in a molecule.

(d)

The IR spectrum of one of the compounds is shown:

Deduce, giving a reason, the compound producing this spectrum.

[1]

Markscheme

B AND $C = O$ absorption/ $1750 « {cm}^{- 1} »$
OR
B AND absence of $O - H / 3200 - 3600 « {cm}^{- 1} absorption »$ ✔

Accept any value between $1700 - 1750 c m^{- 1}$ .

Examiners report

Many candidates identified B having C = O and a peak at 1750.

(e)

Compound A and B are isomers. Draw two other structural isomers with the formula $C_{3} H_{6} O$ .

[2]

Markscheme

Accept any two $C_{3} H_{6} O$ isomers except for propanone and propen-2-ol:

✔✔

Penalize missing hydrogens in displayed structural formulas once only.

Examiners report

A surprising number of candidates drew propanone here as an option, either failing to read the question or perhaps finding the structural formulae provided difficult to understand.

(f(i))

The equilibrium constant, $K_{c}$ , for the conversion of A to B is $1.0 \times 10^{8}$ in water at $298 K$ .

Deduce, giving a reason, which compound, A or B, is present in greater concentration when equilibrium is reached.

[1]

Markscheme

$B$ AND $K_{c}$ is greater than $1$ /large ✔

Examiners report

Most candidates identified B, the product, as being in greater concentration at equilibrium however some lost the mark because they did not include a reason.

(f(ii))

Calculate the standard Gibbs free energy change, $∆ G^{⦵}$ , in $kJ {mol}^{- 1}$ , for the reaction (A to B) at $298 K$ . Use sections 1 and 2 of the data booklet.

[1]

Markscheme

$« Δ G^{Θ} = - R T \ln K = 0.00831 kJ {mol}^{- 1} K^{- 1} (298 K) (\ln 1.0 \times 10^{8}) = »$
$- 46 « kJ {mol}^{- 1} »$ ✔

Examiners report

Most candidates could apply the formula for Gibbs free energy change, ΔG^Θ, correctly however some did not get the units correct.

(g(i))

Propanone can be synthesized in two steps from propene. Suggest the synthetic route including all the necessary reactants and steps.

[3]

Markscheme

$H_{2} O$ /water «and $H^{+}$ » ✔

${CH}_{3} CH (OH) {CH}_{3}$ /propan-2-ol ✔

$K_{2} {Cr}_{2} O_{7}$ /«potassium» dichromate(VI) AND $H^{+}$
OR
${KMnO}_{4}$ /«acidified potassium» manganate(VII) ✔

Accept $H_{3} O^{+}$ .

Examiners report

The mean mark was ⅔ for the required synthetic route. Some candidates failed to identify water as a reagent in the hydration reaction, or note that dichromate ion oxidation requires acidic conditions. This was also the question with most No Response.

(g(ii))

Propanone can be synthesized in two steps from propene.

Suggest why propanal is a minor product obtained from the synthetic route in (g)(i).

[2]

Markscheme

primary carbocation «intermediate forms»
OR
minor product «of the water addition would be» propan-1-ol
OR
anti-Markovnikov addition of water ✔

primary alcohol/propan-1-ol oxidizes to an aldehyde/propanal ✔

Examiners report

This question regarding the formation of a minor product was not well answered. Many candidates struggled to explain the formation of propan-1-ol and to then oxidize it to propanal.