The reaction of acetone with bromine produces bromoacetone. — Physical Chemistry Chemistry Question
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
The chemical equation is:
CH3COCH3 + Br2 → CH3COCH2Br + H+ + Br- (or HBr)
[VISUAL] in solution shows the structural formulas of the reactants and products.
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]
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 kinetic curve is a straight line, therefore the process is zeroth order with respect to Br2.
[VISUAL] in solution shows the plot of [Br2] vs time, which is a straight line decreasing to zero.
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):
[VISUAL]
[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 a straight line [VISUAL]; 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 a straight line [VISUAL]; therefore the reaction is first-order with respect to H+.
Summary of orders: Br2: zeroth-order; Acetone: first-order; H+: first-order.
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
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:
[VISUAL]
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
This is not a straight line, so the process is not zeroth-order. Testing for first-order behavior by constructing a semilogarithmic graph [VISUAL] shows that the points fit to a reasonably straight line. Thus, the process is first-order with respect to Br2. (An alternative solution is estimating the half-life from various concentration pairs in the dataset, which yields a constant value.)
The half life of the limiting reagent was determined in a few experiments, and is independent of the concentration of the limiting reagent:
[VISUAL]
[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, the pseudo first-order rate constant (kobs) can be calculated as: kobs = ln2 / t½.
- Acetone concentration dependence at constant acidity (0.100 mol dm–3) [VISUAL]: A plot of kobs vs [C3H6O] is a straight line, meaning first-order with respect to acetone.
- Acidity dependence at constant acetone concentration (3.0 mmol dm–3) [VISUAL]: The pseudo first-order rate constant kobs is independent of acidity, indicating zeroth-order with respect to H+.
Summary of orders: Br2: first-order; Acetone: first-order; H+: zeroth-order.
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
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
The mechanism involves the following steps:
Step 1 (keto-enol tautomerism):
[VISUAL] showing: acetone + H+ ⇌ enol + H+ (forward rate constant k1, reverse rate constant k_-1)
Step 2 (bromination of enol):
[VISUAL] showing: enol + Br2 → bromoacetone + Br- + H+ (rate constant k2)
At high initial concentration of bromine, Step 1 is the rate-determining step, so: ka = k1.
At low initial bromine concentrations, Step 1 is a rapid pre-equilibrium, yielding kb = (k1 * k2) / k_-1.