Objective: The aim of this experiment is to determine the absolute configuration of an unknown chiral secondary alcohol using the concurrent enantioselective conversion (CEC) method. This method uses both the R and S enantiomers of a chiral acyl transfer catalyst called homobenzotetramisole (HBTM), in separate parallel reactions, and thin layer chromatography to identify the stereochemistry of the secondary alcohol, which it is an R or S alcohol. enantiomer. Quantitative analysis was performed using a program called ImageJ after the appropriate photo was taken of the stained TLC plate. The molecular structure of the unknown alcohol was identified using 1H NMR spectroscopy by matching the hydrogens to the corresponding peak. Theory: The concurrent enantioselective conversion method uses each enantiomer of a kinetically resolving reagent, in this case R-HBTM and S-HBTM, in separate, parallel reactions, where the stereochemistry of the secondary alcohol is determined by the rate of the reactions. When using the CEC method, the enantiomer of the secondary alcohol will react with one enantiomer of the HBTM acyl transfer catalyst more quickly than with the other HBTM enantiomer. The mnemonic that identifies the absolute configuration of the secondary alcohol is the following: if the reaction is faster with S-HBTM, then the secondary alcohol has the R configuration. On the other hand, if the reaction is faster with R -HBTM, then the secondary alcohol has the S configuration. Thin layer chromatography will be used to find out which enantiomer of the HBTM reacts more quickly with the unknown secondary alcohol. The fast reaction corresponds to a higher Rf point (the ester) with a greater density and a slower reaction corresponds to a lower Rf point with a...... middle of paper ...... icted α-methyl-2-naphthalenemethanol. Probably the most obvious clue corresponding to this secondary alcohol was the seven hydrogens embedded in the aromatic region from 7.5 to 7.9 ppm. This compound was the only one to contain seven hydrogens belonging to naphthalene. The other two secondary alcohols, 3-methoxy-α-methylbenzyl alcohol and 4-bromo-α-methylbenzyl alcohol, have only four aromatic hydrogens. Conclusion: In conclusion, the unknown secondary alcohol α-methyl-2-naphthalenemethanol had the R-configuration because it reacted most quickly with S-HBTM and much slower with R-HBTM. TLC was a qualitative method and ImageJ served as a quantitative method to determine which reaction was the fastest esterification. Finally, 1H NMR helped identify the unknown from a finite list of possible alcohols by marking the hydrogens on the corresponding peaks..