Essay about Reaction Iodoethane with Saccharin, an Ambident Nucleophile

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Discussion: In this experiment, we alkylate sodium saccharin to N-ethylsaccharin with iodoethane in an aprotic solvent N,N dimethylformamide. Nucleophiles in this experiment will react better in an aprotic solvent. Aprotic solvents have dipoles due to its polar bonds but they do not have H atoms that can be donated into a H-bond. The anions which are the O- and N- of sodium saccharin are not solvated therefore are “naked” and the reaction is not inhibited and preceded in an accelerated rate. The reaction was an SN2 reaction. Since the Oxygen and Nitrogen are more electronegative than the carbon on which they’re attached electrons are pulled towards O- and N- attracting the ethane from Iodoethane. Iodine being more electronegative …show more content…
N-ethylsaccharin has a melting point of 95°C and O-ethylsaccharin has a melting point of 211°C. If our product is a mixture, there should be a wide melting point range and partial melting would be observed close to 95°C and some solids would still be present until everything melts when the temperature is close to 211°C. The average melting point observed was 79.5°C- 83.5°C which is close to the expected melting point 95°C of N-ethylsaccharin.
This indicates that the product formed was N-ethylsaccharin. But since there are about 10+ degree difference from 95°C, it indicates melting point depression. Possibly because unreacted material or other impurities were mixed with the final product. For product confirmation, an HNMR spectrum was handed by instructor. The signal given off at (= 1.3 ppm) indicates that there are Sp3 C- H’s and signal given off at (=7.7ppm) is an indication that there is a benzene ring. The signal in the HNMR spectrum for the methylene protons of an OCH2CH3 group appears farther downfield at (=4.5ppm) because oxygen atom has a stronger deshielding effect on Hydrogen than what a nitrogen atom will do. The signal at (= 3.8ppm) corresponds to NCH2CH3 group. Since the methylene protons have three adjacent hydrogens around, both signals would have 4 splittings or a quartet. Percentages of of N-ethylsaccharin and O-ethylsaccharin in the final product can be determined by