Melissa Ramessar 212762654
Organic Chem 2020 Lab 4

1.)
Variation of [OH]

Experiment [tBuCl] (M) [OH]
(M) Reaction time (s) Rate Constant k (s-1) Reaction rate (M/s)
1 0.03 0.003 27 3.903×10-3 1.171×10-4
2 0.03 0.006 91 2.453×10-3 7.358×10-5
3 0.03 0.009 135 2.643×10-3 7.928×10-5

According to the experiments, the rate law does not depend on [OH]. When increasing the [OH] in experiment 2, the rate of reaction was lowered tremendously; however that could be due to other reason such as errors rather than the rate law depending on [OH]. These results are predicted for a reaction being a zero order reaction. Using the equation below, it is confirmed that the reactions are zero order.
(experiment1/experiment2)=(k1[tBuCl]^A [OH]^B)/(k2[tBuCl]^A [OH]^B )
((1.171×10^-4)/(7.358×10^-5))=(〖3.903×10^-3 [0.03]〗^A [0.03]^B)/(2.453×10^-3 [0.03]^A [0.06]^B ) 1.591 = 1.591 (0.5) B …show more content…
The heat allows for the molecules to proceed with the reaction faster than the usually do in the transition state.

5.)
According to the table 1, when the different solvent compositions changed, so did the reaction. If there was a lower acetone concentration and more water, the reaction was faster. The fact that was is known as a protic solvent and acetone is aprotic is shown in the 90:10, being the highest and 70:30 being the lowest which affected the reaction mechanism.
Since this reaction is a SN1 mechanism, the carbocation prefer protic solvents as a result of the hydrogen bond with the chloride, which helps it to leave and get rid of the negative charge. This helps to stabilize the carbocation in the transition state. It can be concluded that as the amount of water increases in acetone mixture, the reaction rate will also increase.