Aim: Preparation of tribromobenzene (symmetrical) from aniline

Requirements:

  • Aniline
  • Glacial acetic acid
  • Bromine + Glacial acetic acid
  • Alcohol
  • Sulphuric acid
  • Benzene
  • Sodium nitrite

Procedure:

Step 1: Preparation of 2,4,6 tribromoaniline from aniline

Dissolve 5.0 mL distil aniline in 19.4 mL gla. Acetic acid shaking well then take 8.6 ml bromine solution in 20.4 mL gla. Acetic acid in small dropping funnel then add bromine solution in aniline solution dropwise this reaction exothermic reaction cool the flask and the final product should be color yellow by adding little more bromine in cold water and get a product recrystallize in methanol.

Step 2: Preparation of tribromobenzene (symmetrical) from 2,4,6 tribromoaniline

Dissolve 5.0 g of 2,4,6 tribromoaniline in 15 mL rectified sprite and 3.7 mL of benzene by heating on a water bath and add 1.5 mL sulphuric acid to the heat solution attach reflux condenser to the flask and heat on a water bath detach the condenser, remove the flask from water bath add 1.0 g sodium nitrite shake the flask vigorously allow the solution to cool for 10 minutes and then immerse the flask in an ice bath a mixture crystalized out filter with suction on Buchner funnel wash with a small quantity of alcohol recrystallized the crude from methylated spirit the yield of symmetrical tri bromo benzene m. p. of 122 °C.

Results:

Step 1.

  • Theoretical Yield: ______________g.
  • Practical Yield: ________________g.
  • % Yield of the product: _____________%.
  • M.P of 2,4,6 tribromoaniline: ___________°C

Results:

Step 2.

  • Theoretical Yield: ______________g.
  • Practical Yield: ________________g.
  • % Yield of the product: _____________%.
  • M.P of symmetrical tribromobenzene: ___________°C

Chemical Reaction:

Mechanism:

Calculation:

Step 1

  • Theoretical Yield

Aniline: C8H9NO (93 g.) Density = 1.02 g/mL 2,4,6 tribromoaniline: C6H4Br3N (330.0 g.)

From Chemical Reaction

93.0 g. Aniline                                   =                      330.0 g. 2,4,6 tribromoaniline

∴ 5.1 g. Aniline                                 =                      (?) g 2,4,6 tribromoaniline

                        = \frac{5.1 \times 330.0}{93.0}

                        = 18.09 g.

  • Practical Yield:

Actual weight of the dried product = ________________g.

  • % of Yield =\frac{\text { Practical Yield } \times 100}{\text { Theoretical Yield }}

                             = _______________%

Step 2

  • Theoretical Yield

2,4,6 tribromoaniline: C6H4Br3N (330.0 g.) symmetrical tribromobenzene: C6H3Br3 (315 g.)

From Chemical Reaction

330.0 g. 2,4,6 tribromoaniline          =                      315 g. symmetrical tribromobenzene

∴ 5.0 g. 2,4,6 tribromoaniline           =                      (?) g symmetrical tribromobenzene

                        = \frac{5.0 \times 315.0}{330.0}

                        = 4.87 g.

  • Practical Yield:

Actual weight of the dried product = ________________g.

  • % of Yield =\frac{\text { Practical Yield } \times 100}{\text { Theoretical Yield }}

                             = _______________%