Preview

Rxn of Iodoethane with Sodium Saccharin

Powerful Essays
Open Document
Open Document
954 Words
Grammar
Grammar
Plagiarism
Plagiarism
Writing
Writing
Score
Score
Rxn of Iodoethane with Sodium Saccharin
Title: Reaction of Iodoethane with Sodium Saccharin- Ambient Nucleophile
Dates Performed: February 21 + 28, 2013
Date Submitted: March 14, 2013

Abstract:

The product ratio of N-ethylsaccharin to O-ethylsaccharin that occurred due to alkylation with iodoethane at 80 oC was determined to be 81.5% to 18.5%, respectively, based on an analysis of the 1H NMR spectrum that was collected. The melting point range of 87.8-94.7 oC also indicated that the mixture was largely composed of N-ethylsaccharin. The more prevalent product structure is:

C2H5I
C2H5I

And/Or
And/Or

N-ethylsaccharin product O-ethylsaccharin product

Experimental Procedure:

Sodium saccharin (0.446 g) was dissolved in N-N-dimethylformamide (DMF) at 80oC. Iodoethane (0.16 mL) was added and heated at constant temperature (80oC) for ten minutes. The product was separated by adding water to the room temperature solution, further cooled in ice and the solid collected and dried via vacuum filtration. The following week, the dried product (0.442g) was analyzed by melting point (87.8-94.7oC), and a 1H NMR spectrum.

Data and Calculations:

Table 1: Reactant and Product Masses and Properties | Mol mass (g/mol) | Bp (oC) | Mp (oC) | Density (g/mL) | Amount to use | Sodium saccharin | 223 | | | | 0.002 mol, 0.446 g | Iodoethane | 156.0 | 72 | | 1.95 | 0.002 mol, 0.78 g | N-N-Dimethylforammide (DMF) | 73.1 | 153 | | 0.945 | 1.0 mL | N-ethylsaccharin | 211.2 | | 95 | | | O-ethylsaccharin | 211.2 | | 211 | | |

Table 2: 1H NMR Ha | 3.8 ppm, quartet | Hb | 4.7 ppm, quartet |

Theretical Yield=0.002 mol sodium saccharin1 mol product1 mol sodium saccharin=0.002 mol x 223g1 mol=0.446 g

% yield= actual masstheoretical mass=0.442g0.446g=0.9910x 100%=99.1% yield

Vial with product: 7.619 g
Empty vial -7.177 g 0.442 g product yielded

NMR Data:

You May Also Find These Documents Helpful

  • Powerful Essays

    Ap Biology Unit 1 Summary

    • 2660 Words
    • 11 Pages

    * Monosaccharides can be distinguished by the carbonyl group they possess- aldehyde or ketone- and the number of atoms in their carbon backbone…

    • 2660 Words
    • 11 Pages
    Powerful Essays
  • Satisfactory Essays

    How many C atoms in 60.5 g of sucrose? C12H22O11 molar mass = 342.34 g/mol…

    • 753 Words
    • 3 Pages
    Satisfactory Essays
  • Powerful Essays

    CHEM 120L Exam Notes

    • 3033 Words
    • 13 Pages

    Know how to calculate percent yield (i.e. CuSO4 was obtained from the reaction of CuO with sulfuric acid. If 2.5 g of CuSO4 was obtained from 5.0 g of CuO, what is the percent yield?)…

    • 3033 Words
    • 13 Pages
    Powerful Essays
  • Better Essays

    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 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 form from a 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…

    • 1080 Words
    • 5 Pages
    Better Essays
  • Powerful Essays

    During this experiment, we used recrystallization methods in order to help in the purification of the unknown solid, as well as drying and vacuum filtration. We then ground the unknown and combined it with different chemicals (such as acetanilide or phenacetin) and used the melting point ranges to determine the identity. We used the Mel-Temp method in order to measure the melting points. (It is important to remember that if a chemical is mixed with a differing chemical, the melting point is decreased drastically, but if it is mixed with a component very similar in chemical make up, the melting point will not differ much from the expected.)…

    • 1399 Words
    • 6 Pages
    Powerful Essays
  • Powerful Essays

    Chem Ia Lead

    • 990 Words
    • 4 Pages

    The theoretical yield stated above (4.60 g) was calculated by finding the appropriate amount of moles of the substance in a balanced equation (1 mole) and using stoichiometry to find out the number of moles that are in the actual equation with the quantities used. This value came out as 4.60 g and is used for the theoretical yield.…

    • 990 Words
    • 4 Pages
    Powerful Essays
  • Satisfactory Essays

    Diels Alder Lab Results

    • 250 Words
    • 1 Page

    ,2.beta. ,5. alpha.)‐ Retention Structural Time Formula (Mins) 9.312 C15H24 9.312 C11H14O CAS Number 87‐44‐5 129‐130 N/A 32763‐64‐7 251.775 N/A 0.901 1.178 7.407 7.407 C9H16 C9H16O 4551‐51‐3 160.999 N/A 900193‐87‐ 209.252 N/A 0.883 1.041 5.666 C12H23O2Cl 6974‐05‐6 0.972 5.666 C16H25O2F7 900215‐97‐ 296.979 N/A CAS Number Boiling Point Melting Point 274.482 35‐36 Boiling Point Density (g/mL) 1.141 Retention Structural Time Formula (mins) 3.870 C10H16 Melting Density Point (g/mL) 3760‐14‐3 74 N/A 0.867 3.870 5.506 C10H16 C8H16 3760‐14‐3 692‐96‐6 176‐177…

    • 250 Words
    • 1 Page
    Satisfactory Essays
  • Satisfactory Essays

    Borohydride Reduction

    • 528 Words
    • 3 Pages

    Introduction: The purpose of this lab was to reduce vanillin to vanillyl alcohol. This lab report recaps the procedure and results of the lab. The chemical process studied in this lab was reduction, the process of reducing the number of bonds to oxygen and increasing the number of bonds to hydrogen. Other chemical processes included in the lab were recrystallization, melting point, and extraction.…

    • 528 Words
    • 3 Pages
    Satisfactory Essays
  • Satisfactory Essays

    diels alder reaction

    • 279 Words
    • 2 Pages

    The experimental product yield is 0.28g (some solid left on the filter paper and some for the melting point measurement.)…

    • 279 Words
    • 2 Pages
    Satisfactory Essays
  • Satisfactory Essays

    4. The reaction of the unknown with bromine gave a negative result meaning the solution did not turn clear, but rather, maintained an orangish brown color. This is interpreted to mean that bromine in not adding to either side of an alkene bond, so there is no C=C bond in the unknown compound.…

    • 482 Words
    • 2 Pages
    Satisfactory Essays
  • Satisfactory Essays

    4. Calculate the number of grams of CaCO3 that are expected to be produced. This is your theoretical yield.…

    • 505 Words
    • 3 Pages
    Satisfactory Essays
  • Good Essays

    Purpose: The purpose of this lab is to become familiar with the separations of mixtures and solids.…

    • 718 Words
    • 3 Pages
    Good Essays
  • Satisfactory Essays

    4.08 Percent Yield

    • 231 Words
    • 2 Pages

    The product’s mass would be greater making the calculations adjust. If my actual yield is…

    • 231 Words
    • 2 Pages
    Satisfactory Essays
  • Better Essays

    Nmr of Red Bull Post Lab

    • 2244 Words
    • 9 Pages

    Viewed as one of the most significant tools in analytical chemistry, the Nuclear Magnetic Resonance (NMR) is a spectroscopy tool often used by chemists to determine the molecular profile – quality and quantity of the major components - of any given organic substance. Operating on basis of spectroscopy (study of the interaction between matter and radiated energy - Harris), the NMR can be either a one (1H NMR) or two (1H/13C NMR) dimensional spectral analysis that can be used identify and quantify the major components of a given substance by examining spectral patterns that derive from excitation of matter with radiated energy in the form of radio waves. In this experiment we will examine the spectral patterns derived from NMR scans of Red Bull and sugar-free Red Bull energy drinks. By examining the spectral splitting patterns of the major substance components as derived from experimental NMR scans against predicted spectrums, we will be able to identify both the quality and quantity of many of the major components of Red Bull and sugar-free Red Bull. As will be seen, some of the components of the two energy drinks will be identified with minimal difficulty. The experimental concentrations of DMSO, Caffeine and Sodium Citrate were 0.0014 M, 0.0021 M and 0.0017M respectfully for Red Bull and 0.0014 M, 0.0018 M and 0.0019 M, respectfully for Sugar Free Red Bull. The concentrations of the other major components of Red Bull and Sugar Free Red Bull were not obtained because analysis of those compounds required a higher degree of expertise than can be offered for this lab and will be identified by process of comparison, elimination and two dimensional spectral analysis where both protons and carbon 13 are examined to identify and quantify a compound that does not show up on a one dimensional NMR spectrum. This experiment will allow students to gain a better understanding of a Nuclear Magnetic Resonance,…

    • 2244 Words
    • 9 Pages
    Better Essays
  • Good Essays

    The product was placed in a beaker, and enough aqueous methylated spirit (IMS) was added whilst the mixture was heated to dissolve the product. The mixture was then left to cool and was placed in ice to aid crystallization. The product was vacuum filtrated after a few minutes of standing in ice and the crystals were collected and dried in a vacuum oven at 40 degrees celsius. The mass of the crystals was calculated and the melting point was determined.…

    • 652 Words
    • 3 Pages
    Good Essays