Experimental protocol is pictured above. \" .11 37% 1:23 PM LTE e handout and pr

Question

Experimental protocol is pictured above.

" .11 37% 1:23 PM LTE e handout and prepare a prelab in your noteboo EAS.pdf 2/3_ Nal/NaOCI HO HO OCH CH CH2OH/H2O OCH3 odination of vanillin Safety Sodium hypochlorite is household bleach; it may bleach and/or damage your clothing. Methanol and ethanol are flammable; avoid open flames. Experimental In a 100 mL round-bottomed flask containing a magnetic stir bar, dissolve 1.0 g vanillin in 20 mL of ethanol. To this solution, add 1.17 g of sodium iodide, then cool to 0 "C with an ice/water bath. Using a separatory funnel, add 11 mL of aqueous sodium hypochlorite solution (5.25% w/w) dropwise to the stirred reaction mixture over a period of 10 minutes. The color of the solution will turn from pale yellow to red-brown. Once the addition is complete, allow the mixture to warm to room temperature and continue to stir for 10 minutes Add 10 mL sodium thiosulfate solution (10% w/w), then acidify with hydrochloric acid (10% w/w). Use pH paper to monitor the acidity; generally, 6 mL or so of HCl solution are required. The aryl iodide should precipitate at this point Remove the ethanol from the suspension on a rotary evaporator. With gentle heating, this should require no more than 10 minutes. Note that water will still be present Cool the flask in an ice bath for 10 minutes, then collect the precipitate by vacuum filtration Wash well with ice-cold water, then with a small amount of cold ethanol. Continue to draw air through the crude product for several minutes to facilitate drying. Record the mass of the crude product. Recrystallize the crude product from ethyl acetate or aqueous 2-propanol (see instructions below). Collect the crystalline product by vacuum filtration. Weigh the product and determine its melting point (lit. 183-185 °C) Recrystallization from 2-propanol/H2O: Place the crude product in 100mL Erlenmeyer flask and with heating, add enough 2-propanol to dissolve it. While continuing to heat, gradually add hot water until the mixture becomes cloudy, then add enough 2-propanol to generate a clear (though colored) solution. Allow the hot solution to cool, then place the flask in an ice bath for a few minutes to ensure complete crystallization. imize File Preview

Explanation / Answer

The solvent (ethanol) was polar and therefore upon the addition of HCl a precipitate was formed due to the change in pH causing it to go from basic to acidic and therefore reducing solubility.

this reaction is probably run in a basic solution which allows for the relatively acidic -OH proton to be predominately in the ionized form -O. If your solvent is polar like water or ethanol then you should understand how a charge will dramatically increase solubility. However when you add acid (HCl) you remove the charge and the organic compound is no longer soluble

The chlorine attaches to the other ions, causing them to reform into salts and precipitate.

C8H8O3 + NaI/ NaOCl / H2O HCl -----> C8H7IO3 + Cl-

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