preview

Preparation of P-Nitroaniline

Better Essays

Introduction An electrophile is a reagent attracted to electrons and accepts an electron pair in order to bond to a nucleophile. Electrophiles will attack benzene and result in hydrogen substitution. However, this is not thermodynamically favoured because a sp3 hybridized carbon is generated, which disrupts the cyclic conjugation. In order to regenerate the aromatic ring, a proton is lost at the sp3 hybridized carbon. Thus, p-Nitroaniline can be prepared by means of electrophilic aromatic substitution. To begin, nitric acid needs to be activated as it has little electrophilic power. Thus, concentrated sulfuric acid is added to protonate the nitric acid. Dehydration produces the nitronium ion, which is a strong electrophile and has most …show more content…

The solution became cloudy with solid particles forming in the centre of the flask.
• After the final vacuum suction, the end product appeared as tiny fibers shaped as needles that were shiny light yellow brown.

Table 1: Observed Results for p-Nitroaniline end product
Weight of watch glass 28.21g
Weight of watch glass + products 29.95g
Theoretical Yield 1.839g
Actual Yield 1.74g
% Yield Actual yield x 100% = 1.74 g x 100% = 94.62%
Theoretical yield 1.839g
Appearance of final product • Shiny, light-brown/dark yellow color
• Precipitate took the form of tiny fibers that look liked needles
Experimentation Boiling point 146 ºC - 150 ºC

Calculations
The balanced equation:

PhNHCOCH3 + HNO3 + H2SO4  p-Nitroaniline

Table 2: Physical Properties of Compounds Used in Experiment PhNHCOCH3 HNO3 H2SO4 p-Nitroaniline

Mol Wt (g/mol) 135.16 --- --- 138.12
Concentration --- 16M 18M ---
Amount Used 1.8g 1.6ml 7.0ml ---
Moles 0.01332 mol 0.0256 mol 0.126 mol 0.01332 mol Limiting reagent In excess In excess Theoretical yield: 1.839g

Calculations for the moles of each reagent used:
PhNHCOCH3 (acetanilide): 1.8 g x 1 mol = 0.01332 mol 135.16 g

HNO3 (nitric acid): 16.0 mol x 0.0016 L = 0.0256 mol L

H2SO4 (sulphuric acid): 18.0 mol x 0.007 L = 0.126 mol L

Theoretical yield of p-Nitroaniline:

Get Access