Kinetic Investigation of Unimolecular Solvolysis

Scott Humason

Chem 241: MWF: 8:30-9:50

Kinetic Investigation of Unimolecular Solvolysis

1.) How many millimoles of tert-butyl choloride and hydroxide ion are present in each flask prior to mixing?

Tert-butyl Chloride
3mL X .10 mol = .3mmol

Hydroxide Ion
.3ml X .10 mol = .03mml
+ 6.7 ml H20


2.) Does the acetone participate directly in the reaction? How about the water? Explain?

The water and acetone help stabilize the carbocation. The acetone also helps slow down the reaction by decreasing molar concentration, making it easier to measure the time.


3.) What is the relationship between rate constant for unimolecular reaction and percent reaction given by the expression?

Rate Constant:

[A]o [.1mol Tert-butyl Chloride]
-Ln -------- = [K(time)] Ln ------------------------------------ = K
[A]t [.09mol Hydroxide Ion]
----------------------------------------------
(time)

[.1mol Tert-butyl Chloride]o
1st reaction -Ln ------------------------------------ = [K(56)] K = -.002
[.09mol Hydroxide Ion]t

[.1mol Tert-butyl Chloride]o
2nd reaction -Ln ------------------------------------ = [K(78)] K = -.0013
[.09mol Hydroxide Ion]t

[.1mol Tert-butyl Chloride]o
3rd reaction -Ln ------------------------------------ = [K(82)] K = -.0012
[.09mol Hydroxide Ion]t


4.) What conclusion can you draw about effect of temp on Sn1 reaction rate constant? Do you think your results would be qualitatively true for other reactions?

I had every expectation that the colder the temp that the reaction would be slower and the warmer the temp the faster the reaction would take place. However, the reaction at room temp was the faster initially to get started but the duration of time from start to end was a little bit longer of a time span, with the warmer being faster between reactions.

Temp Graph:

[Fill in Data Here]

5.) Calculate rate constant for each temp in problem B

[.1mol Tert-butyl Chloride]o
@13 Celsius -Ln ------------------------------------ = [K(224)] K = 4.7 ^4
[.09mol Hydroxide Ion]t

[.1mol Tert-butyl Chloride]o
@11.5 Celsius -Ln ------------------------------------ = [K(278)] K = 3.78 ^4
[.09mol Hydroxide Ion]t

[.1mol Tert-butyl Chloride]o
@33 Celsius -Ln ------------------------------------ = [K(127)] K = 8.29 ^4
[.09mol Hydroxide Ion]t

[.1mol Tert-butyl Chloride]o
@34 Celsius -Ln ------------------------------------ = [K(103)] K = .001
[.09mol Hydroxide Ion]t


6.) Which is a more polar solvent, acetone or water?

Water, having 2 lone electron pairs, and no counter balancing atoms make for an overall electronegative molecule thus is more polar. The lone electron pairs also create a considerable electro negative force.


7.) Note that the rate determining step of this reaction involves the production of a positively charged particle. What effect would you expect a more polar medium to have on solvolysis rate? Explain. How does this expectation compare with results of the experiments?

The positively charged particle is the cation intermediate that is created when the Chlorine is removed. To remove the Cl-, a suitable leaving environment is needed for both the anion of the leaving group and the carbocation. A more polar medium would slow down the leaving group and ultimately have no effect on the carbocation. Since the first step, the leaving group, is the slow step, a more polar medium will slow down the reaction rate considerably.


8.) Summarize the leaving group and alkyl group influence rate constant for unimolecular solvolysis.

Chlorine is not the best leaving group, second to Fluorine in its affinity to be in a bond (due to short bond length). Being a tertiary reaction makes it difficult/impossible to use a nucleophile to push off the leaving group, so a more polar positive and/or hindered base will make for a better leaving environment and the stronger the nucleophile the quicker the cation will be attacked by a nucleophile.