Chemistry 238 - lab report # 5 PDF

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Ethan Fronefield Chem 238 – 513 September 23, 2020 Iodination of Salicylamide Results: Table 1: Product weight and yield Section

Salicylamide (g)

Assignment 513-107 1.097 Table 2: Benzene substitution IR

Sodium Iodide

Product melting

Product Yield

(g) 1.403

point (oC) 245-255

(g) 1.244

Substitution Pattern 1,2 - Disubstituted 1,2,3 - Trisubstituted 1,4 - Disubstituted 1,2,4 - Trisubstituted Table 3: Functional group IR

IR value (cm-1) 748.08 - Lab Manual Range (770-730) 778.71 - Lab Manual Range (790-750) 815.14 - Lab Manual Range (870-800) 846.63 - Lab Manual Range (850-800)

Functional Group Aromatic ring N-H bend C=O stretch O-H stretch N-H stretch

IR value (cm-1) 1572.58 1623.10 1675.66 3213.60 3458.09

In this experiment 1.097 g of salicylamide and 1.403 g of sodium iodide were used to produce 1.244 g of the resulting product. IR spectroscopy was used on the product and the peaks from the report are listed in the table above.

Discussion:

Scheme 1: Mechanism for an electrophilic aromatic substitution reaction In this mechanism, an electron pair from the aromatic ring reacts with the electrophile producing a carbocation which is then resonance stabilized by delocalization of the charge surrounding the ring. The carbocation then loses a proton, regenerating the aromatic ring and yielding the substituited aromatic compound. When predicting I+ should add using directing groups, I know I have an OH and an amide functional group. OH is an ortho and para diecting group and will make 2 products and amide is a meta directing gorup but will not form any product due to the size of iodine, para substitution izs favored due to less steric hinderance. I predict I+ will add to the para position next to the hydroxyl group because it has the highest electron density, and this will form 5iodosalicylamide.

Based on my IR data, my prediction was correct because the IR data shows a trisubstituted product with the 1,2,4 trisubstituted peak having the highest intensity. It also shows a disubstituted product with 1,2 disubstituted having the highest intensity. Both of these show that I+ added to the para position. Theoretical yield = ((1.097 g)/(137.138 g/mol)) = 0.008 – limiting reagent ((1.403 g)/(149.89 g/mol)) = 0.0094 0.008 mol * 263.03 g/mol = 2.104 Percent yield = 1.244/ 2.104 = 68.44 I could not find the melting point for my product but based on the theoretical yield the product formed was not pure because the percent yield was 68.44%. This means there is either extra starting material in the product or other products formed in the reaction are in the product.

Scheme 2: Resonance structures showing amide as an electron withdrawing, meta- directing group. In this scheme amide functions as a deactivating and meta- directing group, it reduces electron density at the ortho- and para- positions, shown by the positive charges at both orthoand 1 para- position.

Scheme 3: Resonance structures showing hydroxyl as an activating group and directs to the ortho- and para- positions. In this scheme hydroxyl functions as an activating ortho- and para- directing group, it increases the electron density at the ortho- and para- positions, shown by the negative charges at both ortho- and 1 para- position.

Possible Errors: Possible errors in this experiment would be adding too little or not enough HCL when acidifying the solution. Too much HCL would make the solution too acidic and not enough would not make it acidic enough. If the solution was too much of either the amount of product formed would be altered and inaccurate. Conclusion: In this experiment the iodination of salicylamide resulted in the 1,2,4 trisubstituted product, this product is 5 – iodosalicylamide, this is shown in the IR data from 700-900 cm-1. The I+ added to this position because of the para directing of the hydroxyl group. This position is favored due to the higher electron density and there is no steric hinderance....