Chemistry 9701/42 — October/November 2023

Cambridge A-Level · A Level Structured Questions · worked solutions for every part, with the mark scheme

questions

100

marks

120

minutes

Topics Equilibria · Transition Elements · Analytical Techniques · Reaction Kinetics · Electrochemistry · Chemical Energetics · +6 more

Q1Reaction KineticsEquilibriaFree sample

Propanone, $\text{CH}_3\text{COCH}_3$ , reacts with iodine, $\text{I}_2$ , in the presence of an acid catalyst.

\text{CH}_3\text{COCH}_3 + \text{I}_2 \rightarrow \text{CH}_3\text{COCH}_2\text{I} + \text{H}^+ + \text{I}^-

The rate equation for this reaction is shown.

\text{rate} = k[\text{CH}_3\text{COCH}_3][\text{H}^+]

(a)

Complete Table 1.1 to describe the order of the reaction.

Table 1.1

DifficultyEasy

Worked solution

Answer

	Value
Order with respect to $\text{CH}_3\text{COCH}_3$	1
Order with respect to $\text{I}_2$	0
Order with respect to $\text{H}^+$	1
Overall order	2

Final answer

1, 0, 1, 2

Detailed explanation

Background Concept

The rate equation expresses the relationship between the rate of a reaction and the concentrations of the reactants (and sometimes catalysts). The exponent of each concentration term gives the order of the reaction with respect to that species. The overall order is the sum of all individual orders. A species not appearing in the rate equation has an order of zero, meaning its concentration has no effect on the rate.

Understanding the Question

The rate equation is given as rate = k[CH₃COCH₃][H⁺]. We must read off the order with respect to each species and calculate the overall order. I₂ does not appear in the rate equation, so its order must be identified as zero.

Approach

Look at each species in the rate equation. The power to which its concentration is raised is the order with respect to that species. If a species does not appear, its order is zero. Sum all orders for the overall order.

Step-by-Step Reasoning

[CH₃COCH₃] appears to the first power → order = 1
[I₂] does not appear → order = 0
[H⁺] appears to the first power → order = 1
Overall order = 1 + 0 + 1 = 2

Key Takeaways

The order with respect to a species is the exponent of its concentration in the rate equation. Species absent from the rate equation have zero order. Overall order is the sum of individual orders.

Common Mistakes

Assuming I₂ must have a positive order because it is a reactant in the overall equation. The rate equation is determined experimentally, not from the stoichiometry.
Forgetting to include the zero order when calculating overall order (though it does not change the sum).

Things to Be Careful About

The order with respect to I₂ is zero, not absent — it must be stated as 0 in the table.
Overall order is the sum of all orders including zeros.

Techniques used

identify order from rate equationsum individual orders for overall order

(b)

An experiment is performed using a large excess of $\text{CH}_3\text{COCH}_3$ and a large excess of $\text{H}^+(\text{aq})$ . The initial concentration of $\text{I}_2$ is $1.00 \times 10^{-5}\text{ mol}\text{ dm}^{-3}$ . The initial rate of decrease in the $\text{I}_2$ concentration is $2.27 \times 10^{-7}\text{ mol}\text{ dm}^{-3}\text{ s}^{-1}$ .

(i)

Use the axes to draw a graph of $[\text{I}_2]$ against time for the first 10 seconds of the reaction.

Graph 1

DifficultyMedium-Easy

Worked solution

Working

With large excess of $\text{CH}_3\text{COCH}_3$ and $\text{H}^+$ , the rate is effectively constant (zero order in $\text{I}_2$ ).

Change in $[\text{I}_2]$ over 10 s:

\Delta[\text{I}_2] = 2.27 \times 10^{-7} \times 10 = 2.27 \times 10^{-6} \text{ mol dm}^{-3}

Final $[\text{I}_2]$ :

[\text{I}_2] = 1.00 \times 10^{-5} - 2.27 \times 10^{-6} = 7.73 \times 10^{-6} \text{ mol dm}^{-3}

Answer

A straight line from $1.00 \times 10^{-5}$ at $t = 0$ to $7.73 \times 10^{-6}$ (i.e. $0.773 \times 10^{-5}$ ) at $t = 10$ s.

Final answer

Straight line from 1.00 × 10⁻⁵ to 7.73 × 10⁻⁶ mol dm³ over 10 s

Detailed explanation

Background Concept

When a reactant is present in large excess, its concentration remains effectively constant throughout the reaction. This means the rate does not depend on that reactant's concentration, and the reaction appears zero-order with respect to it. A zero-order reaction has a constant rate, so the concentration of the limiting reactant decreases linearly with time.

Understanding the Question

CH₃COCH₃ and H⁺ are in large excess, so their concentrations are essentially constant. The rate equation rate = k[CH₃COCH₃][H⁺] therefore becomes rate = k' (a constant). Since the rate of decrease of [I₂] equals this constant rate, [I₂] falls linearly. We need to plot [I₂] against time for 10 seconds.

Approach

Recognise that with excess CH₃COCH₃ and H⁺, the rate is constant → zero order in I₂ → linear decrease.
Calculate the total decrease in [I₂] over 10 s using the given initial rate.
Subtract from initial [I₂] to find the final value.
Draw a straight line between the two points.

Step-by-Step Reasoning

Initial [I₂] = 1.00 × 10⁻⁵ mol dm³
Rate of decrease = 2.27 × 10⁻⁷ mol dm³ s⁻¹ (constant because rate depends only on [CH₃COCH₃] and [H⁺], both in large excess)
Decrease over 10 s = 2.27 × 10⁻⁷ × 10 = 2.27 × 10⁻⁶ mol dm⁻³
Final [I₂] = 1.00 × 10⁻⁵ − 2.27 × 10⁻⁶ = 7.73 × 10⁻⁶ mol dm⁻³ = 0.773 × 10⁻⁵ mol dm⁻³
The graph is a straight line (not curved) because the rate is constant.

Key Takeaways

Large excess of other reactants → pseudo-zero-order behaviour with respect to the monitored species.
Zero order → constant rate → linear concentration-time graph.
The gradient of a concentration-time graph gives the rate (negative for reactants).

Common Mistakes

Drawing a curve (exponential decay) instead of a straight line. The reaction is zero-order in I₂, not first-order.
Calculating the final concentration incorrectly (e.g. forgetting to convert 2.27 × 10⁻⁶ to the same power as 10⁻⁵).
Starting the line at 0 instead of at 1.00 × 10⁻⁵.

Things to Be Careful About

The line must be straight (constant gradient), not curved.
The endpoint must be at 0.773 × 10⁻⁵ (or equivalently 7.73 × 10⁻⁶), which is between 0.70 × 10⁻⁵ and 0.80 × 10⁻⁵ on the given axis.

Techniques used

recognise pseudo-zero-order conditionscalculate concentration change from constant rateplot a linear concentration-time graph

(ii)

State whether it is possible to calculate the numerical value of the rate constant, $k$ , for this reaction from your graph. Explain your answer.

DifficultyMedium-Easy

Worked solution

Answer

No, it is not possible to calculate $k$ because the concentrations of $\text{CH}_3\text{COCH}_3$ and $\text{H}^+$ (which appear in the rate equation) are not known.

Final answer

No; the concentrations of CH₃COCH and H⁺ in the rate equation are not known.

Detailed explanation

Background Concept

The rate constant k can be calculated from rate = k[A]ᵐ[B]ⁿ only if the rate and all concentrations appearing in the rate equation are known. The rate can be obtained from the gradient of a concentration-time graph, but if any concentration term in the rate law is unknown, k cannot be isolated.

Understanding the Question

We are told CH₃COCH₃ and H⁺ are in 'large excess' but their actual concentrations are not given. The rate equation depends on these concentrations. Can we still find k?

Approach

Identify what information the graph provides (the rate) and what information the rate equation requires (concentrations of CH₃COCH₃ and H⁺). Check whether all required values are known.

Step-by-Step Reasoning

From the graph, we can determine the rate (gradient = 2.27 × 10⁻⁷ mol dm⁻³ s⁻¹).
The rate equation is rate = k[CH₃COCH₃][H⁺].
To find k, we need: k = rate / ([CH₃COCH₃][H⁺]).
The actual concentrations of CH₃COCH₃ and H⁺ are not given (only that they are in 'large excess').
Therefore k cannot be calculated.

Key Takeaways

'Large excess' tells us the concentration is approximately constant, but does not give its numerical value.
To calculate k, every quantity in the rate equation must have a known numerical value.

Common Mistakes

Saying 'yes' because the rate can be read from the graph, forgetting that the concentrations in the rate law are still needed.
Saying 'no' without explaining why — the mark requires both the statement and the reason.

Things to Be Careful About

The answer must include both parts: (1) it is not possible, and (2) the reason (concentrations in the rate law are unknown).

Techniques used

identify which concentrations are needed to calculate krecognise unknown concentrations prevent calculation

(c)

The experiment is repeated at a different temperature. The initial concentrations of $\text{H}^+$ ions, $\text{I}_2$ and $\text{CH}_3\text{COCH}_3$ are all $0.200\text{ mol}\text{ dm}^{-3}$ .

The value of $k$ at this temperature is $2.31 \times 10^{-5}\text{ mol}^{-1}\text{ dm}^3\text{ s}^{-1}$ .

Calculate the initial rate of this reaction.

\text{rate} = \text{..................................... mol}\text{ dm}^{-3}\text{ s}^{-1}

DifficultyEasy

Worked solution

Working

\text{rate} = k[\text{CH}_3\text{COCH}_3][\text{H}^+]

\text{rate} = (2.31 \times 10^{-5}) \times (0.200) \times (0.200)

\text{rate} = 9.24 \times 10^{-7} \text{ mol dm}^{-3} \text{ s}^{-1}

Answer

$9.24 \times 10^{-7}$ mol dm $^{-3}$ s $^{-1}$

Final answer

9.24 × 10⁻⁷ mol dm⁻³ s⁻¹

Detailed explanation

Background Concept

The rate equation relates the rate of a reaction to the concentrations of the species involved, raised to powers equal to their respective orders. Given the rate constant and all concentrations, the rate can be calculated by direct substitution.

Understanding the Question

All values needed are given: k = 2.31 × 10⁻⁵ mol¹ dm³ s⁻¹, [CH₃COCH] = 0.200 mol dm⁻³, [H⁺] = 0.200 mol dm⁻³. The rate equation is rate = k[CH₃COCH₃][H⁺]. Simply substitute and multiply.

Approach

Substitute the three known values into the rate equation and compute the product.

Step-by-Step Reasoning

rate = k × [CH₃COCH] × [H⁺]
rate = (2.31 × 10⁻⁵) × (0.200) × (0.200)
0.200 × 0.200 = 0.0400
2.31 × 10⁻⁵ × 0.0400 = 2.31 × 4.00 × 10⁻⁷ = 9.24 × 10⁻⁷ mol dm⁻³ s⁻¹

Key Takeaways

When all concentrations and k are known, the rate is a straightforward multiplication.
Check units: mol⁻¹ dm³ s⁻¹ × mol dm⁻³ × mol dm⁻³ = mol dm⁻³ s⁻¹ ✓

Common Mistakes

Arithmetic errors with the powers of ten.
Including [I₂] in the calculation (it does not appear in the rate equation).

Things to Be Careful About

The answer should be given to 3 significant figures to match the precision of the data.

Techniques used

substitute values into rate equationmultiply concentrations and rate constant

(d)

The experiment is repeated using an excess of $\text{H}^+(\text{aq})$ . The new rate equation is shown.

\text{rate} = k_1[\text{CH}_3\text{COCH}_3]

(i)

The value of $k_1$ is $1.1 \times 10^{-3}\text{ s}^{-1}$ . Calculate the value of the half-life, $t_{\frac{1}{2}}$ .

t_{\frac{1}{2}} = \text{...................................................... s}

DifficultyMedium-Easy

Worked solution

Working

The rate equation rate = $k_1$ [CH₃COCH] is first order in CH₃COCH₃.

t_{\frac{1}{2}} = \frac{\ln 2}{k_1} = \frac{0.693}{1.1 \times 10^{-3}}

t_{\frac{1}{2}} = 630 \text{ s}

Answer

$t_{\frac{1}{2}} = 630$ s

Final answer

630 s

Detailed explanation

Background Concept

For a first-order reaction, the half-life is constant and independent of the initial concentration. The relationship between half-life and the rate constant is t₁/₂ = ln2 / k. This is a defining feature of first-order kinetics.

Understanding the Question

The new rate equation is rate = k₁[CH₃COCH₃], which is first order in CH₃COCH₃ (overall order = 1). Given k₁ = 1.1 × 10⁻³ s¹, calculate the half-life.

Approach

Identify the order (first order from the single concentration term to the power 1), then apply t₁/₂ = ln2/k₁.

Step-by-Step Reasoning

rate = k₁[CH₃COCH₃] → first order in CH₃COCH₃
For first order: t₁/₂ = ln2 / k₁
ln2 = 0.693
t₁/₂ = 0.693 / (1.1 × 10⁻³) = 630 s

Key Takeaways

First-order reactions have a constant half-life independent of concentration.
The units of k₁ (s⁻¹) confirm first-order kinetics (rate has units mol dm⁻³ s¹, concentration has units mol dm⁻³, so k must have units s⁻¹ for first order).

Common Mistakes

Using the zero-order half-life formula t₁/₂ = [A]₀/(2k) instead.
Forgetting to convert ln2 ≈ 0.693 correctly.
Arithmetic error: 0.693/1.1 = 0.63, then × 10³ = 630 (not 63).

Things to Be Careful About

The answer should be given to 2 significant figures (matching k₁ = 1.1 × 10⁻³), giving 630 s.

Techniques used

apply the first-order half-life formulasubstitute rate constant into t₁/₂ = ln2/k

(ii)

Use your answer to (i) to draw a graph of $[\text{CH}_3\text{COCH}_3]$ against time for this reaction. The initial value of $[\text{CH}_3\text{COCH}_3]$ on your graph should be $0.200\text{ mol}\text{ dm}^{-3}$ . The final value of $[\text{CH}_3\text{COCH}_3]$ on your graph should be $0.0250\text{ mol}\text{ dm}^{-3}$ .

Graph 2

DifficultyMedium-Easy

Worked solution

Answer

A smooth exponential decay curve starting at 0.200 mol dm⁻³, halving three times at equal intervals of 630 s:

$t = 0$ : $[\text{CH}_3\text{COCH}_3] = 0.200$
$t = 630$ s: $[\text{CH}_3\text{COCH}_3] = 0.100$
$t = 1260$ s: $[\text{CH}_3\text{COCH}_3] = 0.050$
$t = 1890$ s: $[\text{CH}_3\text{COCH}_3] = 0.025$

The curve is smooth and concave up, with each halving taking the same time (630 s).

Final answer

Smooth exponential decay curve with three halvings at 630 s intervals from 0.200 to 0.025 mol dm⁻³

Detailed explanation

Background Concept

A first-order reaction shows exponential decay of reactant concentration with time. The defining feature is that the half-life is constant: the time for the concentration to halve is the same regardless of the starting concentration. On a concentration-time graph, this produces a smooth curve that gets progressively shallower, with equal horizontal distances between successive halvings.

Understanding the Question

We need to draw [CH₃COCH₃] against time showing decay from 0.200 to 0.025 mol dm⁻³. Since 0.200 → 0.100 → 0.050 → 0.025 is three halvings, and each takes 630 s, the total time is 1890 s. The x-axis has no numbered values, so we must label the time values ourselves.

Approach

Calculate the concentration after each half-life: 0.200, 0.100, 0.050, 0.025.
Calculate the times: 0, 630, 1260, 1890 s.
Draw a smooth exponential curve through these points, ensuring the time intervals between halvings are equal.

Step-by-Step Reasoning

Start: [CH₃COCH₃] = 0.200 at t = 0
After 1st half-life (630 s): [CH₃COCH₃] = 0.100
After 2nd half-life (1260 s): [CH₃COCH₃] = 0.050
After 3rd half-life (1890 s): [CH₃COCH] = 0.025
The curve must be smooth (not straight lines between points), concave upward, showing the characteristic exponential shape.
Equal time intervals (630 s) between each halving confirm first-order behaviour.

Key Takeaways

First-order decay: constant half-life, exponential curve.
Three halvings from 0.200 to 0.025 (0.200/2³ = 0.025).
The graph must show equal time intervals between successive halvings.

Common Mistakes

Drawing straight-line segments instead of a smooth curve.
Making the time intervals unequal (which would suggest non-first-order behaviour).
Not labelling the x-axis with time values (the axis has no pre-printed numbers).
Drawing the curve going to zero rather than to 0.025.

Things to Be Careful About

The times marked must agree with the answer to (d)(i) (630 s per half-life).
The curve should be smooth and show the characteristic exponential shape.
Label the time axis clearly with 0, 630, 1260, 1890 s.

Techniques used

plot exponential decay with constant half-lifemark successive halvings at equal time intervals

(e)

A four-step mechanism is suggested for the overall reaction.

\text{CH}_3\text{COCH}_3 + \text{I}_2 \rightarrow \text{CH}_3\text{COCH}_2\text{I} + \text{H}^+ + \text{I}^- \quad \text{rate} = k[\text{CH}_3\text{COCH}_3][\text{H}^+]

Part of this mechanism is shown.

\begin{aligned} \text{step 1:}& \quad \text{CH}_3\text{COCH}_3 + \text{H}^+ \rightarrow \text{CH}_3\text{C}^+(\text{OH})\text{CH}_3 \\ \text{step 2:}& \quad \text{CH}_3\text{C}^+(\text{OH})\text{CH}_3 \rightarrow \text{CH}_3\text{C}(\text{OH})=\text{CH}_2 + \text{H}^+ \\ \text{step 3:}& \quad \rightarrow \\ \text{step 4:}& \quad \text{CH}_3\text{C}^+(\text{OH})\text{CH}_2\text{I} \rightarrow \text{CH}_3\text{COCH}_2\text{I} + \text{H}^+ \end{aligned}

(i)

Write an equation for step 3.

DifficultyMedium

Worked solution

Answer

\text{CH}_3\text{C(OH)}=\text{CH}_2 + \text{I}_2 \rightarrow \text{CH}_3\text{C}^+(\text{OH})\text{CH}_2\text{I} + \text{I}^-

Final answer

CH₃C(OH)=CH₂ + I₂ → CH₃C⁺(OH)CH₂I + I⁻

Detailed explanation

Background Concept

In a multi-step mechanism, the sum of all elementary steps must give the overall reaction. Each step must be balanced in both atoms and charge. If a step is missing, it can be deduced by identifying what the previous step produces and what the next step consumes, then using the overall equation to determine any other reactants or products needed.

Understanding the Question

Step 2 produces CH₃C(OH)=CH₂ (the enol form of propanone). Step 4 begins with CH₃C⁺(OH)CH₂I. The overall reaction consumes I₂. So step 3 must convert CH₃C(OH)=CH₂ into CH₃C⁺(OH)CH₂I using I₂.

Approach

Identify the product of step 2: CH₃C(OH)=CH₂
Identify the reactant of step 4: CH₃C⁺(OH)CH₂I
Note that I₂ is consumed in the overall reaction but does not appear in steps 1, 2, or 4.
Therefore I₂ must be a reactant in step 3.
The difference between CH₃C(OH)=CH₂ and CH₃C⁺(OH)CH₂I is the addition of I to the terminal carbon and loss of the C=C double bond, with a positive charge on the carbon bearing OH.
I₂ provides one I atom (as I⁺ equivalent) and the other becomes I⁻.

Step-by-Step Reasoning

CH₃C(OH)=CH₂ has the formula C₃H₆O (as the enol)
CH₃C⁺(OH)CH₂I has the formula C₃H₆IO⁺
Difference: +I and charge goes from 0 to +1
I₂ → I (added) + I⁻ (released)
So: CH₃C(OH)=CH₂ + I₂ → CH₃C⁺(OH)CH₂I + I⁻
Check atom balance: C₃H₆O + I₂ → C₃H₆IO + I ✓
Check charge balance: 0 + 0 → +1 + (−1) = 0 ✓

Key Takeaways

Missing mechanism steps can be deduced by comparing adjacent steps and the overall equation.
Always check both atom and charge balance.
I₂ acts as an electrophilic iodine source (I⁺) in this step, generating I⁻ as the leaving species.

Common Mistakes

Writing I⁺ as a product instead of I⁻.
Forgetting the positive charge on the carbon in CH₃C⁺(OH)CH₂I.
Not including I₂ as a reactant (it must appear somewhere in the mechanism since it's in the overall equation but not in steps 1, 2, or 4).

Things to Be Careful About

The charge on CH₃C⁺(OH)CH₂I must be shown as + on the carbon (or the whole species as a cation).
The equation must be balanced in both atoms and charge.

Techniques used

deduce missing step from surrounding stepsbalance atoms and charge across mechanism steps

(ii)

Suggest the slowest step of the mechanism. Explain your answer.

DifficultyMedium-Easy

Worked solution

Answer

Step 1 is the slowest step (rate-determining step) because it is the only step that involves both $\text{CH}_3\text{COCH}_3$ and $\text{H}^+$ , which are the two species appearing in the rate equation.

Final answer

Step 1; it involves both CH₃COCH₃ and H⁺ which appear in the rate equation.

Detailed explanation

Background Concept

The rate-determining step (RDS) is the slowest step in a reaction mechanism. The species that appear in the rate equation are those involved in or before the RDS. If a species appears in the rate equation, it must be a reactant in the RDS or in a fast pre-equilibrium step before the RDS. The simplest case is when the RDS directly involves the species in the rate law.

Understanding the Question

The rate equation is rate = k[CH₃COCH₃][H⁺]. We need to identify which of the four steps is the RDS. The RDS must involve both CH₃COCH₃ and H⁺ as reactants (since both appear in the rate law).

Approach

Examine each step to see which one has both CH₃COCH₃ and H⁺ as reactants.

Step-by-Step Reasoning

Step 1: CH₃COCH₃ + H⁺ → CH₃C⁺(OH)CH₃ — involves both species in the rate equation ✓
Step 2: CH₃C⁺(OH)CH₃ → CH₃C(OH)=CH₂ + H⁺ — does not involve CH₃COCH
Step 3: CH₃C(OH)=CH₂ + I₂ → ... — involves neither CH₃COCH₃ nor H⁺
Step 4: CH₃C⁺(OH)CH₂I → ... — involves neither

Only step 1 involves both species that appear in the rate equation, so step 1 must be the RDS.

Key Takeaways

The rate equation reflects the composition of the transition state of the RDS.
Species in the rate equation must be present in or before the RDS.
The RDS is the step whose reactants match the rate equation.

Common Mistakes

Choosing step 3 because it involves I₂ (which is in the overall equation but NOT in the rate equation — it has zero order).
Saying 'step 1 because it's first' without linking to the rate equation.

Things to Be Careful About

The explanation must explicitly state that step 1 involves both species in the rate equation. Simply naming the step is not enough.

Techniques used

link rate equation to rate-determining stepidentify species in RDS from rate law

(iii)

Identify one conjugate acid-conjugate base pair in the mechanism.

\text{conjugate acid .......................................... conjugate base ..........................................}

DifficultyMedium-Easy

Worked solution

Answer

Conjugate acid: $\text{CH}_3\text{C}^+(\text{OH})\text{CH}_3$ Conjugate base: $\text{CH}_3\text{COCH}_3$

(These differ by one $\text{H}^+$ , as seen in step 1.)

Final answer

Conjugate acid: CH₃C⁺(OH)CH₃; Conjugate base: CH₃COCH

Detailed explanation

Background Concept

A conjugate acid-base pair consists of two species that differ by exactly one proton (H⁺). The acid has the extra proton; the base is what remains after the acid donates H⁺. In the Brønsted-Lowry theory, every acid-base reaction involves two such pairs. In a mechanism, proton transfer steps create conjugate pairs between the species before and after the transfer.

Understanding the Question

We need to find any pair of species in the mechanism that differ by one H⁺. Looking at the steps:

Step 1: CH₃COCH₃ gains H⁺ to become CH₃C⁺(OH)CH₃ → these are a conjugate pair
Step 2: CH₃C⁺(OH)CH₃ loses H⁺ to become CH₃C(OH)=CH₂ → these are a conjugate pair
Step 4: CH₃C⁺(OH)CH₂I loses H⁺ to become CH₃COCH₂I → these are a conjugate pair

Any one of these is acceptable.

Approach

Scan the mechanism for pairs of species that differ by exactly one H⁺ (i.e., one proton transfer between them).

Step-by-Step Reasoning

In step 1: CH₃COCH₃ + H⁺ → CH₃C⁺(OH)CH₃. The product has one more H⁺ than the reactant. So CH₃C⁺(OH)CH₃ is the conjugate acid of CH₃COCH₃.
In step 2: CH₃C⁺(OH)CH₃ → CH₃C(OH)=CH₂ + H⁺. The reactant has one more H⁺ than the organic product. So CH₃C⁺(OH)CH₃ is the conjugate acid of CH₃C(OH)=CH₂.
In step 4: CH₃C⁺(OH)CH₂I → CH₃COCH₂I + H⁺. So CH₃C⁺(OH)CH₂I is the conjugate acid of CH₃COCH₂I.

Key Takeaways

Conjugate acid-base pairs differ by exactly one H⁺.
In a mechanism, any proton transfer step creates a conjugate pair between the species on either side.
The acid is the species with the extra proton; the base is the deprotonated form.

Common Mistakes

Pairing species that differ by more than one H⁺ or by something other than a proton (e.g., pairing CH₃COCH₃ with CH₃C(OH)=CH₂ — these differ by H⁺ but are not directly related in a single proton transfer in this mechanism... actually they could be argued, but the mark scheme specifically lists the three valid pairs).
Confusing conjugate pairs with the same species appearing in different steps.

Things to Be Careful About

The pair must differ by exactly one H⁺ (one proton), not one hydrogen atom.
Only the three pairs listed in the mark scheme are accepted.

Techniques used

identify conjugate acid-base pairs differing by one protonlocate proton transfer steps in mechanism

The rest of this paper

8 more questions

Q2Equilibria11M
Q3Electrochemistry · Chemical Energetics18M
Q4Transition Elements8M
Q5Group 2 · Equilibria6M
Q6Transition Elements10M
Q7Hydrocarbons · Analytical Techniques · Carboxylic Acids and Derivatives11M
Q8Polymerisation · Analytical Techniques12M
Q9Nitrogen Compounds · Introduction to A Level Organic Chemistry14M

Loading the full paper…