The Unknown Alcohol's H NMR Spectra

Discussion & Error Analysis:
By examining the unknown alcohol’s H NMR spectra, it can be deduced that the alcohol is 1-(3-methoxyphenyl)ethanol. There are three peaks that have a shift of ~7.0-7.5, one with integration values of 1, 2, and 1.This shows that the alcohol is a meta-substituted benzene ring. The singlet peak with a chemical shift of~ 4.7 ppm with an integration of 1 is from the hydrogen on the OH group. The peak at ~ 1.7 ppm corresponds to the hydrogen of the carbon that is attached to the OH group. The peak with a chemical shift of 3.7 ppm with an integration of 3 comes from the hydrogens of the methoxy group. And finally, the peak with an integration of 3 with a shift of 1.4 ppm comes from the methyl group that neighbors the
In reaction B, the reaction involving the R-HBTM, the spot containing the ester product(top spot) was much more darker than the spot containing the remaining alcohol starting material(bottom spot). Because the created ester product is less polar than the starting material, it is reasonable that the the ester travelled further up the plate, leading to a higher Rf value than that of the alcohol(bottom spot). Comparing reaction B to reaction A, the one that used S-HBTM, the ester product spot was a lot less darker than that of reaction B’s ester product spot. Furthermore, for reaction A, the starting material(alcohol) spot was a lot darker than the product spot, which meant that relatively little starting material was reacted to make product. Using ImageJ, it is possible to quantify the integration of each spot. The qualitative data from observing the darkness of the spots was confirmed, that reaction B had a higher integration value for the product, whereas reaction A had a higher integration value for the starting material spot. Using this data, reaction B had a % conversion of 17.57% whereas reaction A had a conversion rate of 13.13%, suggesting that reaction B indeed was the faster reaction. Because reaction B formed more products than reaction A in the given amount of time, reaction B was the faster
This is because in lane B, the starting material spot was barely visible. This could have led to erroneous quantified data, because ImageJ integrates the spots based on how dark the spots are. Furthermore, the image that was used for ImageJ was blurry, which could also have affected ImageJ’s integration of the spots. In future experiments, one could run more TLC’s such that a decent TLC can be chosen out of multiple TLC’s, allowing for better quantification of data through ImageJ. Furthermore, a butanol can be used instead of an ethanol to test how a bulky alcohol will affect the reactivity of the reaction done in this