PREPARATORY PROBLEM 34 (PRACTICAL) The nucleophilic substitution reaction between a tertiary amine a — Organic Chemistry Chemistry Question

PREPARATORY PROBLEM 34 (PRACTICAL)

The nucleophilic substitution reaction between a tertiary amine and an alkyl halide is known as the Menshutkin reaction. This experiment investigates the rate law for the reaction between the amine known as DABCO (1,4-diazabicyclo[2.2.2]octane) and benzyl bromide:

[VISUAL]

It is possible for the second nitrogen in the DABCO molecule to react with a second benzyl bromide. However, in this experiment the DABCO will always be in excess so further reaction is unlikely. The reaction could proceed by either the SN₁ or the SN₂ mechanism. In this experiment, you will confirm that the order with respect to benzyl bromide is 1 and determine the order with respect to DABCO. This should enable you to distinguish between the two possible mechanisms.

As the reaction proceeds neutral species, DABCO and benzyl bromide, are replaced by charged species, the quaternary ammonium ion and Br–. Therefore the electrical conductivity of the reaction mixture increases as the reaction proceeds and so the progress of the reaction can be followed by measuring the electrical conductivity as a function of time.

Benzyl bromide is a lacrymator. This experiment should be performed in a fume cupboard.

The Method in Principle
The rate law for the reaction can be written as:
-d[RBr]/dt = d[Br–]/dt = k [RBr] [DABCO]^α [1]
where we have assumed that the order with respect to the benzyl bromide, RBr, is 1 and the order with respect to DABCO is α.

In the experiment, the concentration of DABCO is in excess and so does not change significantly during the course of the reaction. The term [DABCO]^α * k on the right-hand side of Eqn. [1] is thus effectively a constant and so the rate law can be written:
-d[RBr]/dt = kapp [RBr] where kapp = k [DABCO]^α [2]
kapp is the apparent first order rate constant under these conditions; it is not really a rate "constant", as it depends on the concentration of DABCO.

To find the order with respect to DABCO we measure kapp for reaction mixtures with different excess concentrations of DABCO. From Eqn. [2], and taking logs, we find:
ln kapp = ln k + α ln [DABCO] [3]
So a plot of ln kapp against ln [DABCO] should give a straight line of slope α.

The constant kapp may be found by measuring the conductance at time t, G(t), and at time infinity, G∞. It is shown that a graph of ln[G∞ – G(t)] against t should be a straight line with slope -kapp.

In practice it is rather inconvenient to measure conductance at time infinity but this can be avoided by analysing the data using the Guggenheim method. In this method each reading of the conductance at time t, is paired up with another at time t + ∆, G(t⁺∆), where ∆ is a fixed time interval that needs to be at least a half-life. A plot of ln[G(t⁺∆)–G(t)] against time should be a straight line of slope – kapp. For example, suppose we take measurements at fixed regular intervals, say each 30 s and choose an appropriate value of ∆, say 3 minutes (180 s). The plot made is of the points {x,y} = {0, ln[G(180)^- G (0)]}, {30, ln[G(210)^- G(30)]}, {60, ln[G(240)^- G(60)]}, ...

The Apparatus
Cheap conductivity meters are commercially available, for example the Primo₅ conductivity stick meter from Hanna instruments works well with this practical. These simply dip into the solution and the conductance of the solution can be read off the digital display. (www.hannainst.co.uk/product/PRIMO^5-Conductivity-stick-meter/PRIMO5/)

Chemica ls
• DABCO (1,4-diazabicyclo[2.2.2]octane), three solutions in ethanol with a concentration of 0.15, 0.20 and 0.25 mol dm–3, respectively
• benzyl bromide, solution in ethanol, c = 0.6 mol dm–3

Procedure
You are provided with the following solutions, all in ethanol: 0.15, 0.20 and 0.25 mol dm–3 DABCO, and approx. 0.6 mol dm–3 benzyl bromide (this must be freshly made up). You should measure kapp for each of these solutions by measuring the conductance as a function of time and then analysing the data using the Guggenheim method. From the three values of kapp, the order with respect to DABCO can be found by plotting ln kapp against ln [DABCO], as shown by Eqn. [3].
Ideally we ought to keep the reagents and the reaction mixture in a thermostat. However, as the heat evolved is rather small, the temperature will remain sufficiently constant for our purposes.

Kinetic Runs
1. Rinse the conductivity dipping electrode with ethanol from a wash bottle, catching the waste in a beaker. Allow the excess ethanol to drain off and gently dry the electrode with tissue.
2. Transfer 10 cm₃ of the DABCO solution to a clean dry boiling tube.
3. Add 100 µl of the benzyl bromide solution.
4. Insert and withdraw the dipping electrode of the conductance meter a few times in order to mix the solution and then, with the electrode in place, start the stop-watch.
5. Record the conductance at 30 second intervals (it is essential to make the measurements at regular intervals), starting with the first reading at 30 seconds and continuing until there is no further significant change in the conductance, or for 10 minutes, whichever is the shorter time.
6. From time to time, gently lift the electrode in and out so as to stir the solution.
7. Once the measurements have been made, remove the electrode, discard the solution and clean the electrode as in step 1.
8. Make the measurements for the 0.15 mol dm–3 solution of DABCO, and then for the 0.20 and 0.25 mol dm–3 solutions.

Data Analysis
For each run determine kapp using the Guggenheim method – three minutes is about right for the fixed interval ∆. Then plot ln kapp against ln [DABCO] and hence determine the order with respect to DABCO.

Supplementary Information
The key to this experiment is how to use the measured conductance of the reaction mixture to determine the first order rate constant, kapp. The first stage is simply to integrate the rate law; to do this we note that for each benzyl bromide molecule that reacts one bromide ion is generated so that at any time [Br–] = [RBr]init – [RBr], where [RBr]init is the initial concentration of benzyl bromide. Thus the rate equation can be written in terms of [Br–] by putting [RBr] = [RBr]init – [Br–]; integration is then straightforward:
d[Br–]/dt = kapp ([RBr]init – [Br–])
∫ d[Br–]/([RBr]init – [Br–]) = ∫ kapp dt
i.e. -ln([RBr]init – [Br–]) = kapp * t + const
The constant can be found by saying that at time zero, [Br–] = 0, hence:
-ln([RBr]init) = const
hence -ln([RBr]init – [Br–]) = kapp * t – ln([RBr]init)
which can be written:
[Br–] = [RBr]init (1 – exp[– kapp * t]) [4]

When the reaction has gone to completion, at time infinity, the concentration of bromide is equal to the initial concentration of RBr so Eqn. [4] can be written:
[Br–] = [Br–]∞ (1 – exp[– kapp * t]) [5]
where [Br–]∞ is the concentration of Br– at time infinity. Equation [5] says that the concentration of Br– approaches a limiting value of [Br–]∞ with an exponential law. A similar relationship can be written for the other product, the quaternary ammonium ion, whose concentration will be written [R₄N⁺]:
[R₄N⁺] = [R₄N⁺]∞ (1 – exp[– kapp * t]) [6]

We will assume that the conductance of the reaction mixture, G, is proportional to the concentration of the charged species present:
G = λ_Br– [Br–] + λ_R₄N⁺ [R₄N⁺]
where λ are simply the constants of proportionality.
Using Eqns. [5] and [6] to substitute for the concentration of Br– and R₄N⁺ we find:
G = λ_Br– [Br–]∞ (1 – exp[– kapp * t]) + λ_R₄N⁺ [R₄N⁺]∞ (1 – exp[– kapp * t])
= (λ_Br– [Br–]∞ + λ_R₄N⁺ [R₄N⁺]∞) (1 – exp[– kapp * t])
= G∞ (1 – exp[– kapp * t]) [7]
where we have recognised that λ_Br– [Br–]∞ + λ_R₄N⁺ [R₄N⁺]∞ is the conductance at time infinity, G∞.

Equation [7] can be rearranged to give a straight line plot:
G/G∞ = 1 - exp[– kapp * t]
ln(1 - G/G∞) = – kapp * t
or ln(G∞ – G) / G∞ = – kapp * t
or ln(G∞ – G) = – kapp * t + ln G∞
Hence a plot of ln(G∞ – G) against t should be a straight line with slope -kapp.

The Guggenheim Method
From the conductance relationship, the conductance at time t, G(t), can be written:
G(t) = G∞ (1 – exp[– kapp * t])
At some time (t + ∆) later the conductance is G(t + ∆):
G(t + ∆) = G∞ (1 – exp[– kapp * (t + ∆)])
The difference G(t + ∆) – G(t) is:
G(t + ∆) – G(t) = G∞ (exp[– kapp * t] – exp[– kapp * (t + ∆)])
= G∞ exp[– kapp * t] (1 – exp[– kapp * ∆])
Taking logarithms of both sides gives:
ln[G(t + ∆) – G(t)] = ln[G∞ (1 – exp[– kapp * ∆])] – kapp * t
This implies that a plot of ln[G(t⁺∆) – G(t)] against time should be a straight line of slope – kapp; to make this plot there is no need to know the value of the conductance at infinite time, G∞, and this is the main advantage of the Guggenheim method.