Buy custom Reaction Iodoethane with Saccharin, An Ambident Nucleophile

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Topic: Science
Number of pages / Number of words: 4 / 908
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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 their polar bonds but they do not have H atoms that can be donated into an 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 at an accelerated rate. The reaction was an SN2 reaction.

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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 breaks off from ethane and joins the Na+. Since the Oxygen of sodium saccharin is more electronegative than the nitrogen, therefore, this gives oxygen a higher partial negative charge, therefore, an attack on Oxygen will give a product that is formed faster; this can be called a kinetic product. The transition state energy is lower than a product formation by thermodynamic control. At thermal equilibrium at 80°C, a more stable product is formed from higher transitional state energy.

The nucleophiles in the molecule sodium saccharin are O- and N- and the major product formed depending on which oh the nucleophile was attacked most in the reaction. Nucleophilic attack by nitrogen will yield N-ethylsaccharin and nucleophilic attach by oxygen will yield O-ethylsaccharin. “N-saccharin is more stable than O- ethylsaccharin because the Ethyl group is attached to the Nitrogen giving the same spacial configuration for the five-membered ring (which is flat or planar).” (Richard y.a.). The carbonyl carbon is sp2 and flat. This has little ring strain and is stable. The first bond between carbon and oxygen in a carbonyl group is created by overlapping an sp2 hybrid orbital from carbon with an sp2 hybrid orbital from oxygen (sigma bond). The second bond between carbon and oxygen is created by overlapping a p orbital from carbon with a p orbital from oxygen (pi bond). The two remaining sp2 hybrid orbitals on oxygen are used to hold oxygen's lone pairs(bruice). O-ethylsaccharin is then less stable because the Ethyl group is attached to the Oxygen that used to be a carbonyl group, giving the Carbon a sp3 configuration (joined to two other carbons, the Oxygen with the Ethyl group and a Hydrogen).

This puts a strain on the ring, and therefore is less stable.”(Richard y.a.). Upon mixing the reactants, sodium saccharin slightly dissolved producing a clear colorless liquid. When placed at an 80°C hot bath, the solution completely dissolved and turned yellowish-green. Iodoethane is a clear and colorless liquid, but when exposed to light and air the Iodide dissociates from ethane and gives off a yellow color as a sign of decomposition. The solution was covered to prevent this from happening. But, as iodide dissociates from CH3CH2 that then gave off its yellowish color which shows SN2 reaction taking place. SN2 reaction happened fast The limiting Reagent is Iodoethane, as the alkylating agent; it was not used in excess and dictated how far the reaction went. The Na+ binds with I-(noted disappearance of the yellow color, as I- binds with Na+) then the ethane group bonds with either the Oxygen of saccharin or the Nitrogen of saccharin. The final product after vacuum filtration had some unreacted material, indicated by some yellow-green solid formation.

The actual product was homogenous white powder. Therefore, one cannot tell if two products were formed by visual inspection only. In order to identify whether the product was majorly N-ethylsaccharin or O- ethylsaccharin, a melting point was done to a sample. 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 is 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, and HNMR spectrum was handed by the instructor. The signal is given off at (?= 1.3 ppm) indicates that there are Sp3 C- H’s and the 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 N-ethylsaccharin and O-ethylsaccharin in the final product can be determined by integration. But since integration was not shown, approximation by the ratio of peaks can tell the ratio between N-ethylsaccharin and O-ethylsaccharin in the product. The signal at (?=3.8 ppm) which corresponds to ?? NCH2CH3 group is three times the length of the signal given off by ?? OCH2CH3 group. This indicates that there are about 75% N-ethylsaccharin and 25% O-ethylsaccharin in the final product according to the HNMR spectrum.

General points of the essay

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