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Essay Preparation of 4-Bromoaniline

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Preparation of 4-bromoaniline
Introduction
Aromatic compounds tend to undergo electrophilic aromatic substitutions rather than addition reactions. Substitution of a new group for a hydrogen atom takes place via a resonance-stabilized carbocation. As the benzene ring is quite electron-rich, it almost always behaves as a nucleophile in a reaction which means the substitution on benzene occurs by the addition of an electrophile. Substituted benzenes tend to react at predictable positions. Alkyl groups and other electron-donating substituents enhance substitution and direct it toward the ortho and para positions. Electron-withdrawing substituents slow the substitution and direct it toward the meta positions.
Aromatic compounds also undergo …show more content…

Aniline 93.1 10 0.107 Yes
Ethanoic anhydride 102.1 12 0.118 No
Glacial ethanoic acid 60.1 25 0.416 No

No. moles produced by aniline = 0.107 moles
Theoretical mass of N-phenylethanamide = n*MR = 0.107 × 135 = 14.44g
Percentage yield = actual yield (g)/theoretical yield (g) × 100 = (11.9/14.44) × 100 = 82%
For preparation of N-(4-bromophenyl) ethanamide:
Product RMM Volume used/ g Volume used/ mL Moles used Limiting reagent?
N-phenylethanamide 135.2 5 0.0370 No
Bromine 159.8 2.1 0.0131 Yes

No. moles produced by bromine = 0.0131 moles
Theoretical mass of N-(4-bromophenyl) ethanamide = n*MR = 0.0131 × 159.8 = 2.10 g
Percentage yield = actual yield (g)/theoretical yield (g) × 100 = (5.00/2.10) × 100 = 238%
For preparation of 4-bromoaniline:
Product RMM Volume used/ g Volume used/ mL Moles used Limiting reagent?
N-(4-bromophenyl) ethanamide 214.1 5 0.0234 Yes
Hydrochloric acid 36 50 5.00 No

No. moles produced by N-(4-bromophenyl) ethanamide = 0.0234 moles
Theoretical mass of 4-bromoaniline= n*MR = 0.0234 × 214.1 = 5.01 g
Percentage yield = actual yield (g)/theoretical yield (g) × 100 = (1.60/5.01) × 100 = 32%

Discussion
The preparation of N-phenylethanamide from aniline was the first step of the experiment.

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