CHEMISTRY 2020 Experiment 4: Kinetics of Nucleophilic Substitutions

CHEMISTRY 2020 Experiment 4: Kinetics of Nucleophilic Substitutions Chetna Raj June 14th, 2016 Chemistry 2020 Lab 3 Results 1. Determine the effect of varying [OH-] on the rate of the reaction. To do this, you should complete the following table: Table 1: Reaction Rates in Variations of [OH-] Experiment [tBuCl] (M) [OH-] (M) Reaction Time (s) Reaction Rate (M/s) Rate Constant k (s-1) 1 0.03 0.003 44 7.18 × 10-5 2.39 × 10-3 2 0.03 0.006 106 6.32 × 10-5 2.11 × 10-3 3 0.03 0.009 195 5.49 × 10-5 1.83 × 10-3 What is the dependence of [OH-] on the rate law? Since the rate constants remained relatively similar with varying [OH-] concentrations; in other words, it is independent of the rate law since it does not affect the rate to a great degree. In addition, because this is an SN1 mechanism, it takes a great deal of energy to ionize the tertiary halide into a carbocation, which is the slow step, hence the rate limiting step. The hydroxide ion is protic solvent and only helps to stabilize the carbocation. Calculations: [tBuCl] = C2 = C1V1 /V2 = (3 mL × 0.1 mol/L) ÷ 10 mL = 0.03 M [OH-] = 10% of 0.03M = 0.003 M k = 2.303 log10 [1 / (1-χt)] ÷ t = 2.303 log10 [1 / (1-0.1)] ÷ 44 = 2.39 × 10-3 s-1 rate = k[tBuCl] = (2.39 × 10-3)(0.03) = 7.18 × 10-5 Ms-1 2. Determine the effect of varying [tBuCl] on the reaction rate. Completing the following table will assist you. Table 2: Reaction Rates in Variations of [tBuCl] Experiment [tBuCl] (M) [OH-] (M) Reaction Time (s) Reaction Rate (M/s) Rate Constant k (s-1) 4 0.06 0.003 34 9.06 × 10-5 1.51 × 10-3 1 0.015 0.003 32 4.94 × 10-5 3.29 × 10-3 5 0.015 0.003 32 4.94 × 10-5 3.29 × 10-3 What is the dependence of [tBuCl] on the rate law? Since there is a large difference between the reaction rate of experiment 4 and experiment 1 and between their respective rate constants, it is evident that [tBuCl] affects the rate of the reaction. After performing calculations, tBuCl is determined to be first order in the rate law. Calculations: [tBuCl] = C2 = C1V1 /V2 = (3 mL × 0.2 mol/L) ÷ 10 mL = 0.06 M [OH-] = 0.3 mL × 0.1 M = 0.003 M k = 2.303 log10 [1 / (1-χt)] ÷ t = 2.303 log10 [1 / (1-0.05)] ÷ 34 = 1.51 × 10-3 s-1 rate = k[tBuCl] = (1.51 × 10-3)(0.06) = 9.06 × 10-5 Ms-1 Determining order of reaction: 9.06 × 10-5 = 1.51 × 10-3 [tBuCl] α 9.06 × 10-5 = 1.51 × 10-3 (0.06) α 4.94 × 10-5 = 3.29 × 10-3 [tBuCl] α 4.94 × 10-5 = 3.29 × 10-3 (0.015) α (4)α = 3.99595 α = ln (3.99595)/ln4 = 0.999 = 1 ∴ [tBuCl] is first order, therefore overall reaction is first order. 3. Based on the experimental rate law, what mechanism(s) do you think are happening in the solvolysis reaction of tBuCl? What product(s) do you predict for this reaction (draw the reaction mechanisms and product structures)? 1) Rate Determining Step: formation of Carbocation 2) Nucleophile Attacks Carbocation 3) Solvent gains a proton 4. Calculate the activation energy (Ea) for this reaction. To do this, you need to construct an Arrhenius plot. First, complete the following table. Table 3: Reaction Rates in Variations of Temperatures Experiment Temp (K) Reaction Time (s) Reaction Rate (M/s) Rate Constant k (s-1) 1 291.15 44 7.18 × 10-5 2.39 × 10-3 6, trial 1 280.15 333 9.49 × 10-6 3.16 × 10-4 6, trial 2 280.15 347 9.11 × 10-6 3.04 × 10-4 7, trial 1 302.15 29 1.09 × 10-4 3.63 × 10-3 7, trial 2 302.15 29 1.09 × 10-4 3.63 × 10-3 Arrehnius Plot: Hydrolysis of tBuCl ln k lnk Line of Best Fit 1/T (K) Figure 1: Arrhenius Plot of Solvolysis of tBuCl in Sodium Hydroxide Slope of Line of best fit = -0.7399 mol•K Y-intercept = -4.4667 Activation Energy (Ea) Calculation: Slope = m = -Ea/R Ea = m × R = -0.7399 mol•K × (8.314×10-3 kJ/mol•K) = 6.15 × 10-3 kJ Explain clearly the effect of temperature on the reaction rate, making reference to the activation energy value you obtained. Note: the control referred in the text below is the reaction taking place at room temperature, with the same reagent concentrations. As the temperature increases, the reaction goes to completion at a faster rate than when the temperature is decreased, which may explain the relatively low activation energy. This was evident from the data collected; when the temperature was brought down to 7˚C, the indicator took much longer to change color (333 seconds) relative to the control. When the temperature was raised to 29˚C, the indicator took much less time to change color relative to the control (29 seconds). Evidently, the activation energy of the reaction being relatively lower reflects on the faster reaction rate; since high temperatures already excite electrons such that they are more delocalized, it also becomes easier to form the carbocation intermediate. Hence, lower energy is required than in cooler temperatures. Temperature is a significant factor in the solvolysis reaction in that a greater temperature excites the electrons more and it becomes easier to reach the activation energy required for the reaction to go to completion. 5. Comment on how the reaction rate varied when you adjusted the solvent composition, making reference to the mechanism(s) you proposed in question 3. In this experiment, the hydroxide ion was used as the protic solvent. This is an effective solvent because it forms a complex with the carbocation created in the first step of the reaction mechanism, stabilizing it. The carbocation is highly unstable, and would just combine back with the chlorine halide it lost to stabilize again rather than be used in the second step of the mechanism where the nucleophile attacks. This is was prevented by the hydroxide ions from sodium hydroxide because the positively charged carbocation easily formed a complex with the negatively charged hydroxide ion, stabilizing the carbocation so the reaction can go to completion. 6. Comment on how the reaction rate varied when a different substrate (i PrCl) was used. Justify your answer by making reference to the mechanism(s) you proposed in question 3. iPrCl is less bulky relative to tBuCl because it has two methyl groups and a hydrogen about the central carbon rather than three. However, the carbocation of iPrCl is also less substituted than the one formed from tBuCl for this reason of having less alkyl groups. When performing solvolysis with iPrCl, the indicator never changed color, hence, no reaction was observed. Therefore, the rate determining step that forms the carbocation by releasing a halide ion (in this case chlorine) does not occur because the carbocation formed would be secondary (2˚) and unstable due to being less substituted. It would either never form or combine back with the halide before the nucleophile (water) gets a chance to attack. 7. In this experiment you used relatively low concentrations of OH- . If [OH-] was increased, only one product should be observed upon reaction with tBuCl. What mechanism do you think would occur? Draw the structure of the predicted product based on the mechanism you proposed. Hydroxide ions create a basic environment that neutralizes the acid product (HCl). However, it is also a strong base in a solution containing a strong substrate (due to the tertiary carbon atom). Hence, rather than having an SN1 reaction, excess hydroxide ions may trigger an E2 elimination reaction, where the strong base (hydroxide ion) removes a proton from the alkyl halide to form water and an alkene. Since this process is bimolecular, the rate of this reaction depends on the concentration of both the substrate and the base (hydroxide ions). Figure 2: Structure of products 1,1-dimethylalkene and water from E2 elimination reaction in excess hydroxide ions.