KETO-ENOL TAUTOMERISM

INTRODUCTION

Tautomerism refers to an equilibrium between two different structure of the same compound. Usually the tautomers differ in the point of attachment of a hydrogen atom. One of the most common examples of a tautomeric system is the equilibrium between a ketone and its enol form,



where Ke is the equilibrium constant. For ordinary ketones Ke is usually very small, but in -dicarbonyls the possibility of intramolecular H-bonding can make enol formation much more favorable.



The value of Ke depends greatly on the identity of the two groups R and R'. In this experiment two different -dicarbonyls, a -diketone and a -ketoester, will be studied. In acetylacetone, a -diketone whose systematic name is 2,4-pentanedione, R and R' are both methyl groups. The other compound, ethylbenzoylacetate, is a -ketoester, in which R is a benzene ring and R' is the OCH2CH3 group. Because of the spacing provided by the oxygen atom in the OCH2CH3 group, steric hindrance between the benzene ring and CH2CH3 is not appreciable. However, this spacing is not present in acetylacetone, and in the keto form interactions between the methyl groups is pronounced. In general for -diketones the enol form separates the R and R' groups and is favored, but the effect is much less pronounced for -ketoesters.

Solvent polarity is also very important in determining the magnitude of Ke. The enol form is less polar than the keto tautomer because of the intramolecular hydrogen bonding. Thus, an increase in solvent polarity favors the more polar keto form.

The value of Ke can be determined quite easily from 1H NMR measurements. The rate of interconversion is slow enough that separate signals can be observed for the protons in the two different forms. The ratio of their concentrations can be evaluated from peak integration data and pertinent stoichiometic relationships. Each compound will be studied in two solvents: C6D6 (nonpolar) and CDC13 (somewhat polar).

PROCEDURE (NOTE: The sample may already have been prepared by the instructor. If so, skip steps 1-3.)

1. Weight 4 clean #507 NMR tubes with caps.

2. To two of the tubes use a Pasteur pipet to add one drop if acetylacetone. Repeat with ethylbenzoylacetate for the other two tubes. Reweigh the tubes to make sure the samples are in the 10-20 mg range. The exact weight is not critical.

3. For each compound fill one tube to a height of 60 mm with CDC13. Likewise, for each compound fill the other tube to 60 mm with d6-benzene. (Deuterated solvents are EXPENSIVE. Please do not waste them. Chloroform and benzene are also toxic, and inhalation of vapors should be avoided.) If the solvent does not contain TMS, use a syringe to add about 2 mL TMS vapor to the respective tubes. Cap the tubes and invert several times to mix the contents.

4. With the aid of the instructor, obtain 1H NMR spectra and peak integrations for all four samples. If possible, obtain spectra at several temperatures.

5. Clean and return all tubes to the instructor.

DATA ANALYSIS

1. Using both chemical shift and integration data assign all peaks in the four spectra.

2. Use the peak integration data and the stoichiometric relationship of the number of protons of each type to calculate Ke. For each sample use several different combinations of integrals. Calculate the average Ke and 95% confidence limits for each sample. If directed by the instructor, also calculate the 95% confidence limits propagated from the uncertainties in the integration values.

3. If data were obtained at several temperatures, prepare a van't Hoff plot, and calculate Ho and So for the equilibrium.

4. In the report discuss the peak assignments in terms of the observed chemical shifts. Discuss the relative value of Ke for the two compounds in terms of their molecular structures. Also discuss the effects of solvent polarity.

REFERENCES

1. Drexler, E.J.; Field, K.W. J. Chem. Educ. 53, 593 (1976).

2. Burdett, J.L.; Rogers, M.T. J. Amer. Chem. Soc. 86, 2105 (1964).

3. Rogers, M.T.; Burdett, J.L. Can. J. Chem. 43, 1516 (1965).