Edexcel International A Level (IAL) Chemistry (YCH11) - Unit 4 - 11.6 Iodination of propanone-Study Notes - New Syllabus

Edexcel International A Level (IAL) Chemistry (YCH11) -Unit 4 – 11.6 Iodination of propanone- Study Notes- New syllabus

Edexcel International A Level (IAL) Chemistry (YCH11) -Unit 4 – 11.6 Iodination of propanone- Study Notes -International A Level (IAL) Chemistry (YCH11) – per latest Syllabus.

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Edexcel International A Level (IAL) Chemistry (YCH11) -Concise Summary Notes- All Topics

11.6  Iodination of Propanone

(i) Determining Orders in Acid-Catalysed Iodination of Propanone

The rate of the reaction between propanone and iodine in acidic conditions is investigated to determine the order with respect to propanone, iodine, and hydrogen ions.

Reaction

\( \mathrm{CH_3COCH_3 + I_2 \rightarrow CH_3COCH_2I + HI} \)

(acid acts as a catalyst)

Method

  • Use iodine solution (brown colour)
  • Measure decrease in colour over time
  • Use colorimetry or timing disappearance of colour
  • Carry out multiple experiments
  • Change one concentration at a time:
    • \( \mathrm{[CH_3COCH_3]} \)
    • \( \mathrm{[I_2]} \)
    • \( \mathrm{[H^+]} \)
  • Keep all other variables constant

Obtaining Rate Data

  • Measure initial rate using change in iodine concentration
  • Rate ∝ decrease in iodine concentration
  • Use color intensity or time for colour to fade

Deducing Orders

  • Propanone:
    • Increasing \( \mathrm{[CH_3COCH_3]} \) increases rate
    • First order with respect to propanone
  • Iodine:
    • Changing \( \mathrm{[I_2]} \) has no effect on rate
    • Zero order with respect to iodine
  • Hydrogen ions:
    • Increasing \( \mathrm{[H^+]} \) increases rate
    • First order with respect to \( \mathrm{H^+} \)

Overall Rate Equation

\( \mathrm{rate = k[CH_3COCH_3][H^+]} \)

(zero order in \( \mathrm{I_2} \))

Key Insight

  • Rate-determining step does not involve iodine
  • Iodine reacts after slow step
  • Therefore, iodine concentration does not affect rate

Example 1

In an experiment, doubling the concentration of propanone doubles the rate, while doubling iodine concentration has no effect. Deduce the order with respect to each reactant.

▶️ Answer/Explanation

Propanone: rate doubles → first order

Iodine: no change → zero order

Reaction depends only on propanone (and acid)

Example 2

The rate equation is found to be independent of iodine concentration. Explain this in terms of the reaction mechanism.

▶️ Answer/Explanation

Iodine is not involved in the rate-determining step

Slow step involves propanone and \( \mathrm{H^+} \)

Iodine reacts in a fast step after this

Therefore, changing iodine concentration does not affect rate

(ii) Predicting Species in the Rate-Determining Step

The rate equation obtained from experimental data can be used to identify which species are involved in the rate-determining (slow) step of a reaction mechanism.

Key Principle

  • The rate-determining step controls the overall rate
  • Only species involved in this step appear in the rate equation
  • Species not in rate equation are not involved in the slow step

Using Experimental Data

  • Determine rate equation from experiments
  • Identify reactants affecting rate
  • Match these species to rate-determining step

Application to Iodination of Propanone

Experimental data shows:

\( \mathrm{rate = k[CH_3COCH_3][H^+]} \)

  • Propanone and \( \mathrm{H^+} \) affect rate → in slow step
  • Iodine does not affect rate → not in slow step

Key Insight

  • Slow step involves formation of intermediate
  • Fast step involves iodine reacting with intermediate
  • Therefore, iodine is excluded from rate equation

Example 1

The rate equation for a reaction is:

\( \mathrm{rate = k[A][B]} \)

Reactant C is present but does not appear in the rate equation. Predict which species are involved in the rate-determining step.

▶️ Answer/Explanation

A and B are involved in the rate-determining step

C does not affect rate → not in slow step

C may react in a later fast step

Example 2

In the iodination of propanone, experimental data shows that changing iodine concentration does not affect the rate. Explain what this indicates about the mechanism.

▶️ Answer/Explanation

Iodine is not involved in the rate-determining step

Slow step involves propanone and \( \mathrm{H^+} \)

Iodine reacts in a fast step after intermediate is formed

Therefore, iodine concentration does not affect rate

(iii) Deducing a Possible Mechanism

A reaction mechanism can be proposed using experimental rate data by identifying the steps involved and ensuring consistency with the rate equation and rate-determining step.

Key Principles

  • Mechanism consists of a sequence of elementary steps
  • One step is the rate-determining (slow) step
  • Rate equation must match the slow step
  • Intermediates are formed and then consumed

Method to Deduce Mechanism

  • Use experimental rate equation
  • Identify species in rate-determining step
  • Propose slow step using these species
  • Add fast steps to complete overall reaction
  • Ensure mechanism sums to overall equation

Application: Iodination of Propanone

Experimental rate equation:

\( \mathrm{rate = k[CH_3COCH_3][H^+]} \)

Proposed Mechanism

  • Step 1 (slow):

\( \mathrm{CH_3COCH_3 + H^+ \rightarrow intermediate} \)

  • Formation of enol intermediate
  • Step 2 (fast):

\( \mathrm{intermediate + I_2 \rightarrow CH_3COCH_2I + HI} \)

Explanation

  • Slow step involves propanone and \( \mathrm{H^+} \) → matches rate equation
  • Iodine appears only in fast step → zero order
  • Intermediate is formed and then consumed

Example 1

A reaction has rate equation:

\( \mathrm{rate = k[A][B]} \)

Propose a two-step mechanism consistent with this data.

▶️ Answer/Explanation

Step 1 (slow): \( \mathrm{A + B \rightarrow X} \)

Step 2 (fast): \( \mathrm{X \rightarrow products} \)

Slow step contains A and B → matches rate equation

X is an intermediate

Example 2

The rate equation for a reaction is:

\( \mathrm{rate = k[A]} \)

Reactant B is present but does not affect rate. Suggest a mechanism and explain the role of B.

▶️ Answer/Explanation

Step 1 (slow): \( \mathrm{A \rightarrow X} \)

Step 2 (fast): \( \mathrm{X + B \rightarrow products} \)

B does not appear in rate equation → not in slow step

B reacts in fast step

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