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Chemistry 386J Nov 7th, 2015 Name:________________________________________ Signature:______________________________________
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Problem 1. Circle the molecule with the lowest pKa, and provide a brief (one word) explanation. (6 pts) a.
Carbon has more s character. b.
Induction or dipole c.
Poorer solvation or methyl destabilizes anion
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Problem 2. a. Give a rough estimation of the pKa for the following molecules. (3 pts)
b. The equilibrium constant of enolization for acetone is roughly Keq= 10-8 s-1 in water, estimate the pKa of the enol tautomer of acetone. (3 pts)
Enol pKa = __________12___________ c. Given your calculated pKa of the enol, is it more or less acidic than acetone? Explain why this makes sense by identifying the factors that make the enol more or less acidic. (2 pts) More, proton on oxygen that is adjacent to a sp2 carbon d. At pH 2, a particular acid exists in ratio of 1 : 105 (A- : HA), estimate the Ka of this acid. (1 pt) 107
10 16 20
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Problem 3. a. Please predict and explain (BRIEFLY) whether the pKa of acetic acid in isopropanol will increase or decrease compared to the pKa of acetic acid in aqueous media. (2 pts) Solvent change means poorer solvation of the acetate anion, which means the anion is not as well stabilized. Thus, not as acidic/ higher pKa. pKa= 11.3 (J. Biochem. Biophys. Methods 38, (1999) 123.) b. Is the circled proton acidity: More/ Less/ No Change, relative to an aqueous environment, please explain (BRIEFLY). (2 pts) The proton is in the vicinity of the cationic zinc, reducing the pKa to less than 5. You can find this example iIn one of the “Going Deeper Section” in Anslyn/Daugherty textbook.
(II)
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Problem 4. Below is a proposed route for the formation of a zinc centered dynamic assembly that you have seen on exam 1. Pyridine-2-carbaldehyde and dipicolylamine condense to form iminium 1. In the presence of Zinc(II) and water, the tris-pyridyl ligand coordinates to yield hemiaminal 2. With the introduction of an alcohol species, the alcohol gets incorporated into the assembly to form a hemiaminal ether 3.
To gain insight into the reaction mechanism, a substituent effect study was conducted with various para-substituted pyridine-2-carbaldehyde (substituent location shown above as X).
Below is a table showing the log(kx/kH) and para values for each substituent studied, where kx is the rate of formation for the hemiaminal ether assembly. Questions are on the next page.
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a. With the data provided, what charge build-up do you expect the rate determining step to have? (1 pt) Negative/ Positive/ No Charge b. Below is a proposed mechanism for the hemiaminal ether formation. Based on the observed data, circle the step(s) that could be the rate determining step. (2 pts)
c. Given your conclusion as to the rate determining step, would you expect + to give a worse or better correlation? If there are more than one step identified as potentially the rate determining step above, specify what you would expect for each case. (5 pts)
The correlation with + parameter would be better if the step of losing water is the true rds. Otherwise, if the rds is the step where the assembly loses zinc, we would expect a not too great
correlation with +.
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Problem 5. The hydrolysis of benzene arylsulfonates can go through either the stepwise (A) or concerted (B) mechanism (shown below).
The following Hammett plot was constructed by observing the rate of hydrolysis of benzene arylsulfonates with various substituted phenols.
ρ= 1.48
a. Which mechanism(s) above is(are) supported by this Hammett Plot? (1 pt) A B Neither Both More on next page!
slow
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The following Brønsted plot was constructed by observing the rate of base hydrolysis of benzene arylsulfonates with various substituted phenol leaving groups. Looking at the Brønsted plot, there are two βLG values that can be extracted, one for the high pKa phenols, and one for the low pKa phenols.
b. Rationalize the “break” in this Brønsted plot and interpret the βLG values. Also, identify the mechanism(s) from the previous page that are best supported by the different βLG values. (5 pts) The “break” in the plot above indicates that there is a change in mechanism at around pKa 8.5. The low βLG for the low pKa phenols means that there is a small effect of having a more stable leaving group (higher pKa phenol) in this mechanism. This description best fits the mechanism A shown above. On the other hand, the high βLG value for the high pKa phenols means the extent of leaving group departure is large at TS. Therefore, the second mechanism better fits this descriptor.
Going deeper… (don’t have to do anything here!) Notice the disagreement on reaction mechanism between the Brønsted and Hammett plots! Sometimes statistics play tricks on our results. In this case, with three more data points, the Brønsted plot is shown on the left!
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Problem 6. For the following two transformations, the carboxylic acid cyclizes to give you the cyclic anhydride product.
a. Would you expect the H‡ to be similar or different for the two reactions? Provide a brief explanation. (2 pts) Similar Different (circle one) Both reactions are forming two C-O bonds.
b. Would you expect the S‡ to be similar or different for the two reactions? Provide a brief explanation. (2pts) Similar Different (circle one) Reaction A is intramolecular while B is intermolecular.
c. What equation can we use to determine the H‡ and S‡ of the reaction? (1 pt) Eyring
d. Briefly describe how would you use this equation to determine the H‡ and S‡ of the reactions. Also, draw and label the axes of a plot you would generate with the data collected. (4 pts)
Measure the rate of reaction at various temperatures, then plot the temp-1 vs ln(k/t).
Temp-1
Ln(k/T)
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