Electrophilic+Aromatic+Iodination

Merissa Honey & Heather Browning Electrophilic Aromatic Iodination



I think this model required you to turn it by hand. ;) Funny!


 * __ Introduction: __**

The purpose of this laboratory experiment is for an aromatic compound to undergo an electrophilic substitution reaction. To carry this out, our method combines sodium iodide and common bleach as the oxidizing agent in aqueous alcohol as the solvent. These reations occur via a two-step-addition-elimination mechanism in which the **π** electrons of the aromatic ring attack the electrophilic agent, forming a cationic intermediate to form the substituted product. In the synthesis of chemical compounds, nucleophilic substitution and elimination reactions are heavily used. However, when one wishes to add an electrophile to an aromatic molecule, the reaction that is needed involves electrophilic substitution. This reaction involves a two step mechanism of both addition and elimination which results in the substitution of a hydrogen for the electrophile (__Green Organic Chemistry: Strategies, Tools, and Laboratory Experiments__, Doxsee and Hutchison Pages 182). The initial addition of the electrophile occurs through breaking a pi bond in the aromatic ring and creating a carbocation. This is followed by the elimination of the hydrogen (or other leaving group) which then restores the pi bond and the aromatic ring. This type of reaction can be carried out with a halogen as the leaving group, which can be an important step in the synthesis of larger molecules.

This lab is the iodination of the aromatic compound, Vanillin. Iodine is unreactive with many aromatic systems so a mixture of oxidizing agents will be used for iodination. The iodination reaction combines a salt, sodium iodide, with NaOCl as the oxidizing agent in aqueous ethanol. This combination allows for a efficient, selective yeild of monoiodinated product and "enviromentally benign solvent.


 * __ Procedure: __**

1. In a 100 mL round bottom flask dissolve ≈ 1.0 grams of 4-hydroxyacetonphenone (Vanillin) in 20 mL of ethanol. 2. To this solution add 1.17 grams equivalents of sodium iodide and cool in ice bath to 0 degrees C. 3. Use a separatory funnel to slowly (drop wise) add 11 mL of sodium hypochlorite solution (NaOCl) (5.25% w/w) to the stirred reaction mixture with in the flask over 10 minutes. (Slow addition of the bleach will result in higher yield and easier purification.) The color should turn from pale yellow to red-brown. 4. Stir for 10 minutes and allow to come to room temperature. 5. Add 10 mL of sodium thiosulfate and then acidify with hydrochloric acid( use pH paper, generally 6 mL of HCL are required). The aryl iodide should precipitate at this point. 6. Attach the flask to a rotary evaporator to remove ethanol from the suspension. (The reduced pressure, decreased boiling point, heat and spinning motion help to separate out the solvent. This process should be approximately 10 minutes. Note that water will be present. 7. Cool the flask in an ice bath for 10 minutes then collect precipitate by vacuum filtration.  8. Recrystalize product by placing it in an Erlenmeyer flask and with heating add enough 2-propanol to dissolve it. Then vacuum filtrate.  9. Weigh your product and determine a melting point. The melting point given in the literature is 155-156 °C


 * // Citation: //** Gilbertson, Robert, Kathryn Parent, Lallie McKenzie, and Jum Hutchinson. "Electrophilic Aromatic Iodination of --hydroxyacetonphenone." //Green Chem// . N.p., 2002. Web. 14 Feb 2011. .

Appearance of reactant: Vanillin was small and clear similar to sugar crystals
 * __ Data: __**

Theoretical Yield: C8H8O3 = 120.142g (vanillin) C8H7O3I = 278.0353 g/mol 1.005g vanillin x __1mol vanillin__ x __1mol 5-iodovanillin__ x _278.0353 g_ = 2.3257g = 2.326g 5-iodovanillin ...............................120.142g.........1 mol vanillin .........1 mol 5-iodovanillin

Percent Yield: __2.073g actual product__ x 100% = **89.31%** 2.321g theoretical product

Actual Yield of Product: **2.073 g**

Melting Point: Sweating-Melted: 173.7-174.5° C Literature: 183-185° C

Purified Product Appearance: Shiny light yellow, metallic film tint, creamy colored

When bleach was a dramatic color change was noted. Reactants changed to dark rust color. Rate of addition of NaOCl = 22 drops per minute.
 * **Vanillin** || **Sodiumiodide** || **Dried mass of product prior to crystallization** || **Crystal Mass of Final Product (4-hydroxy-3-iodoacetophenone)** ||
 * 1.005 grams || 1.206 grams || 2.596 grams || 2.073 grams ||




 * [[image:Mechanism_of_Reaction.jpg width="942" height="605" align="center"]] ||

__**Conclusion:**__

Iondination of the aromatic benzyne ring of Vanillin was accomplished by oxidation of Iodine with sodium bicarbonate. sodium hypochlorite. With green chemistry in mind for the halogenation of aromatic molecules, iodine is the choice halogen especially when compared to bromine, which is both volatile and dangerous to health, and chlorine, which is also damaging to respiratory health. Although iodine is “easy to work with,” it does require an oxidizing agent as a catalyst for the reaction (__Green Organic Chemistry: Strategies, Tools, and Laboratory Experiments__, Doxsee and Hutchison Pages 183). In this lab, potassium iodide ws combined with common, household bleach (NaOCl; the oxidizing agent) to react with vanillin (4-hydroxy-3-methoxybenzaldehyde), an aromatic compound, and create 4-hydroxy-3-iodoacetophenone, This is not the name of your product...? an essentially odorless compound.

A point of error in lab may have been that while adjusting the separatory funnel in the NaOCl addition the rate of addition was sped up and ≈ 2 mL of NaOCl was added faster than 22 drops per minute. This did not appear to effect the product but is noted here.Other sources of error include that some product was lost on the filter paper of the vacuum filtration apparatus.

The recovered product 4-hydroxy-3-iodoacetophenone was tiny crystalline structures that were metallic yellow. The amount of product obtained via the reaction was 2.073g wich resulted with an 89.31% recovery. The crude product was a yellow, whitish powder with the consistency of flour. After the addition of 2-propanol to purify the product one of the instances of possible error occurred. what was it? Despite the recovery percentage, the melting point of the product was 173.5-174.5 degree C which is lower than that of the literature value provided by the instructor. This range indicates that the product was not completely purified. One degree C is a very narrow range, and indicates good purity. Marking the sample as extremely pure if not effectively recovered.

Scores: format (2/2) style (1.5/2) data (3/3) quality of result (1/1) quality of reported data (1/1) conclusion (1/2) error (0.5/1) Post-lab Q (0/2) Where is the post-lab Q? for a total of 10/14.