Intermediate II Energy Intermediate I Scheme 6 Reaction Energy Diagram ring can

Question

Intermediate II Energy Intermediate I Scheme 6 Reaction Energy Diagram ring can be even more complicated. is the among the Predicting the substitution when there is more than substituent that controls the regioselectivity. It is poSS The reaction today will convert potassium iodide more complicated. Just remember the strongest electron-donating group cause very low yields of several different products. potassium iodide to the activated electrophile substituen ty. It is possible that high competition ts present will using sodium hypochlorite. This type of activation elim using sodium hypochlorite. This type of activation et has relied on. iron that the traditional el lelectrophilic aromatic substitution has relied on. This has owiathn your prelab pregparation ry to ye of activation eliminates the need for aluminum and significant financial and environmental benefits. In your prelab scovery a inancial and eerophilic aromati son eliminates the need for aluminum and d what is the iscovery a mechanism for the oxidation of iodide to the electrophile an identity of the electrophile. has many advantages over the traditional electrophilic aromatic substitution. This reaction is efficient and selec monoiodinated products than traditional reactions. The reaction also uses environmentally benign reagents (sodium h (aqueous alcohol instea is efficient and selective aqus alcohol instead of halogenated solvents). This reaction will convert vanillin to the monoiodinated product (Scheme 7) ally benign reagents (sodium hypochlorite versus nitric acid) and solvents CoG KNaOCI CH3OH/H-o Scheme the reaction to be done in lab today Procedure In a 100 mL round bottom flask containing a magnetic stir bar, dissolve vanillin (1.00 g, 6.57 mmol) in ethanol (20 mL). To this solution add potassium iodide (1.17 g 7.05 mmol) and cool to 0 °C. Using an addition funnel add sodium hypochlorite (11.0 mL of a freshly opened commercial Ultra Chlorox bleach solution) dropwise over a 10 minute period. Upon completion of the addition let the reaction stir for an additional 10 minutes at room temperature.

Explanation / Answer

the electrophile is I-Cl,.

KI becomes K+ and I-, and NaOCl becomes Na+ and OCl-, which quickly becomes HOCl. I- attacks HOCl and the result is I-Cl and OH-

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