The reaction of acetone with bromine produces bromoacetone. — Physical Chemistry Chemistry Question
Theoretical Problem 27: Reaction of acetone with bromine
The reaction of acetone with bromine produces bromoacetone.
Give the chemical equation of the reaction assuming that acetone is in excess.
Model Answer
[VISUAL]
Reaction Equation:
CH3COCH3 + Br2 → CH3COCH2Br + H+ + Br-
In a mechanistic study, the reaction was followed in several kinetic experiments at 25 °C in aqueous solution by measuring the concentration of Br2 using a spectrophotometric method. The following kinetic curve was recorded when the initial concentrations were [Br2]0 = 0.520 mmol dm–3, [C3H6O]0 = 0.300 mol dm–3, and [HClO4]0 = 0.050 mol dm–3.
[VISUAL]
Table 1:
t (min) | 0 | 2 | 4 | 6 | 8 | 10 | 12 | 14
[Br2] (µmol dm–3) | 520 | 471 | 415 | 377 | 322 | 269 | 223 | 173
t (min) | 16 | 18 | 20 | 22 | 24 | 26 | 28 | 30
[Br2] (µmol dm–3) | 124 | 69 | 20 | 0 | 0 | 0 | 0 | 0
Which is the limiting reagent in this experiment?
Model Answer
Br2 is the limiting reagent.
What is the order of reaction with respect to the limiting reagent?
Model Answer
The plot of the kinetic curve:
[VISUAL]\nThe kinetic curve is a straight line, therefore the process is zeroth order with respect to Br2.
The time where the characteristic break point occurs on the kinetic curve is called the reaction time and was determined in aqueous solution at 25 °C. The following table gives the reaction time in several different experiments (′ denotes minutes, ″ denotes seconds):
Table 2:
[Br2]0 (mmol dm-3) | [C3H6O]0 (mmol dm-3) | [HClO4]0 (mmol dm-3) | reaction time
0.151 | 300 | 50 | 5′ 56″
0.138 | 300 | 100 | 2′ 44″
0.395 | 300 | 100 | 7′ 32″
0.520 | 100 | 100 | 30′ 37″
0.520 | 200 | 100 | 15′ 13″
0.520 | 500 | 100 | 6′ 09″
0.520 | 300 | 200 | 4′ 55″
0.520 | 300 | 400 | 2′ 28″
Determine the orders of reaction with respect to all three components.
Model Answer
As the process is of the zeroth-order with respect to Br2, and all the other reagents are in large excess, the rate is constant in each experiment. It can be simply calculated as:
v = [Br2]0 / tbreak
where tbreak is the reaction time. The dependence of the rate on the reagent concentrations can be studied directly using this formula.
Plotting the rate as a function of acetone concentration at constant acidity (0.100 mol dm–3) gives:
[VISUAL]
This is a straight line; therefore the reaction is first-order with respect to acetone.
Plotting the rate as a function of acid concentration at constant acetone concentration (0.300 mol dm–3) gives:
[VISUAL]
This is a straight line, therefore the reaction is first-order with respect to H+.
What is the rate equation of the process?
Model Answer
v = ka [C3H6O] [H+]
What is the value and unit of the rate constant?
Model Answer
The rate constant can be determined by dividing the rates calculated in each experiment with both the acetone and acid concentrations. From the average of the 8 measurements shown in the table:
ka = 2.86×10−5 dm3 mol–1 s–1 (second-order rate constant with second order unit).
A different, electrochemical method allowed detection of much smaller concentrations of Br2. A kinetic curve, the initial concentrations for which were [Br2]0 = 1.80 µmol dm–3, [C3H6O]0 = 1.30 mmol dm–3, and [HClO4]0 = 0.100 mol dm–3, is given in the following table:
Table 3:
t (s) | 0 | 10 | 20 | 30 | 40 | 50 | 60 | 70
[Br2] (µmol dm–3) | 1.80 | 1.57 | 1.39 | 1.27 | 1.06 | 0.97 | 0.82 | 0.73
t (s) | 80 | 90 | 100 | 110 | 120 | 130 | 140 | 150
[Br2] (µmol dm–3) | 0.66 | 0.58 | 0.49 | 0.45 | 0.39 | 0.34 | 0.30 | 0.26
Which is the limiting reagent in this experiment?
Model Answer
Br2 is the limiting reagent.
What is the order of reaction with respect to the limiting reagent?
Model Answer
The plot of the kinetic curve:
[VISUAL]
This is not a straight line; the process is not zeroth-order. Testing for first-order behavior is possible by constructing a semilogarithmic graph. This plot looks like:
[VISUAL]
The points fit to a reasonably straight line. Thus, the process is first-order with respect to Br2.
An alternative solution: estimating the half life from various concentration pairs in the dataset gives a constant value. Therefore the decay of Br2 follows first-order kinetics.
The half life of the limiting reagent was determined in a few experiments, and is independent of the concentration of the limiting reagent:
Table 4:
[Br2]0 (µmol dm-3) | [C3H6O]0 (mmol dm-3) | [HClO4]0 (mol dm-3) | t½ (s)
1.20 | 3.0 | 0.100 | 24
1.50 | 3.0 | 0.100 | 23
1.50 | 1.0 | 0.100 | 71
1.50 | 0.4 | 0.100 | 177
1.50 | 3.0 | 0.030 | 23
1.50 | 3.0 | 0.400 | 24
Determine the order of reaction with respect to all three components.
Model Answer
The process is first-order with respect to the limiting reagent Br2. From the half-lives of the first order curves, a pseudo first-order rate constant (kobs) can be calculated as follows:
kobs = ln2 / t½
The dependence of kobs on the concentrations of reagents in large excess reveals the kinetic orders with respect to the remaining two reagents.
The acetone concentration dependence at constant acidity (0.100 mol/dm3):
[VISUAL]
This is a straight line; therefore the reaction is first-order with respect to acetone.
For the [H+] dependence at constant acetone concentration (3.0 mmol/dm3) we get:
[VISUAL]
The pseudo first-order rate constant kobs is practically independent of acidity, therefore the reaction is zeroth-order with respect to H+.
What is the rate equation of the process?
Model Answer
v = kb[C3H6O][Br2]
What is the value and unit of the rate constant?
Model Answer
The rate constant can be determined by dividing the pseudo first-order rate constants calculated in each experiment by the acetone concentration. From the average of the 6 measurements shown in the table:
kb = 9.82 dm3 mol−1 s−1 (second-order rate constant with second order unit)
Suggest a detailed mechanism to interpret the experimental findings.
Model Answer
Mechanism scheme:
[VISUAL]
At high initial concentration of bromine, Step 1 is rate determining, therefore:
ka = k1
At low initial bromine concentrations, Step 1 is a rapid pre-equilibrium, therefore:
kb = k1 * k2 / k_-1