University:
Augusta University
Course:
CHEM 1441 | General Chemistry
Academic year:

2022
Views:

41

Pages:

3
Author:

sommatosmby

Directive Effects in the Bromination of Vanillin Purpose: The purpose of this experiment is to understand and perform an electrophilic bromination reaction of vanillin and to figure out the directive effects by deducing the identity of the major product with its melting point. Not Formed Formed Not Formed Results: We used 1.52 g of vanillin, which is equal to 0.01 mol of vanillin. There will be a 1:1 molar reaction between vanillin reactant and 5-bromovanillin product. Thus, the theoretical product should be 0.01 mol of 5-bromovanillin, or 2.31 g. 1. Crude Yield = 3.31 g / 2.31 g = 143% Crude Yield We used 0.76g of the crude product for recrystallization and obtained 0.60 g of recrystallized product. This is a 78.9% recrystallization. Recrystallized Yield = 143 % Crude Yield * 78.9 % Recrystallization = 113 % Recrystallized Yield overall w/ a 78.9% recrystallization from crude product. 2. Our melting point of our recrystallized product was 165.3-166.9 ℃. This indicates our bromination product was 5-bromovanillin by the following table: Bromination Product Melting Point 2-Bromovanillin 155℃ 5-Bromovanillin 166℃ 6-Bromovanillin 178℃ Discussion: 1. Based on our melting point, the alcohol directed ortho due to being the strongest electron-donating group and the aldehyde (slightly) directed meta due to being an electron withdrawing group. The resultant product was 5-bromovanillin. This matches our prediction due to our knowledge of resonance and electrophilic aromatic substitution. The other two products did not form because -OH is the stronger electron donating group relative to the -OMe. For example, 2-bromovanillin is only directed by the -OMe ortho direction and 6-bromovanillin is only directed by the -OMe para direction. The 5-bromovanillin is favored by both the alcohol and the aldehyde. 2. Given Bromine’s electrophilicity and the fact that there are two electron donating group substituents on vanillin, the reactivity of vanillin is appreciably high enough relative to benzene that no Lewis acid is needed. 4. 2-Bromovanillin will have two doublets and two singlets as opposed to 4 singlets for 5- and 6-bromovanillin because its aromatic hydrogens are adjacent to one another. As for distinguishing 5-bromovanillin from 6-bromovanillin, 5-bromovanillin’s hydrogen adjacent to bromine will be downfield relative to 6-bromovanillin’s hydrogen adjacent to bromine due to the aldehyde electron-withdrawing effect de-shielding it. Pre-Laboratory Assignment: 1. In 2-bromovanillin, the bromination is directed by the -OMe electron-donating group to the ortho position. In 5-bromovanillin, the bromination is directed to the meta position by the electron-withdrawing aldehyde group and to the ortho position by the electron-donating alcohol group. In 6-bromovanillin, the bromination is directed to the para position by the electron-donating -OMe group. 2. We may predict that 5-bromovanillin will be formed because it has both the EWG aldehyde group and the EDG alcohol group directing the bromination to the 5-position. 3. 8HBr + 2KBrO3 + 2CH3CO2H --> 5 Br2 + 2CH3CO2K + 6H2O