Preparation/Synthesis of 1,1'-Bi-2-naphthol

                                                    Synthesis of 1,1'-Bi-2-naphthol

Apparatus; Beakers,Round Bottom Flask,Laboratory Stand

Chemicals;2-Nephthol(C₁₀H₇OH),Ferric Chloride(FeCl₃)

Chemical Equation;

2 C₁₀H₇OH  + 2FeCl₃(aq)    ----------->   C₂₀H₁₄O₂   + 2FeCl₂ + 2HCl

Procedure

Dissolve 3.6g 2-Nephthol in 150 mL water and heat on water heater

Dissolve 7g FeCl₃ in 40 mL water

Add FeCl₃ Solution in hot 2-Nephthol solution

Oily drops of 2-Nephthol disappear and flaks of  1,1'-Bi-2-naphthol appeared

Heat further for 10 min

Filter hot suspension and wash with hot water

Dry the product and weigh

Theoratical Yeild

Mass of  2-Nephthol = 3.6g

Molar Mass of 2-Nephthol = 144.17gmol^-1

Moles of 2-Nephthol = 3.6g/144.17gmol^-1 = 0.02497 mol

Mass of FeCl₃ = 7g

Molar Mass of FeCl₃ = 270.3 gmol^-1

Moles of of FeCl₃ = 0.025

2 mol of 2-Nephthol produce = 1 mol of  1,1'-Bi-2-naphthol

1 mol of 2-Nephthol produce = ½ mol of  1,1'-Bi-2-naphthol

0.0249 mol of 2-Nephthol produce = ½ ×  0.0249 mol of  1,1'-Bi-2-naphthol

0.0249 mol of 2-Nephthol produce = 0.01249

2 mol of FeCl₃ produce = 1 mol of  1,1'-Bi-2-naphthol

1 mol of FeCl₃ produce = ½ mol of  1,1'-Bi-2-naphthol

0.025 mol of FeCl₃ produce = ½ × 0.025 mol of  1,1'-Bi-2-naphthol

                                             = 0.0125 mol of  1,1'-Bi-2-naphthol

2-Nephthol is a limiting reagent

Mass of 1,1'-Bi-2-naphthol = Moles × Molar Mass

                                            = 0.01249 mol × 286.32 gmol^-1

                                            = 3.58 g

Practical Yeild = 3.88 g

Disscusion

Experimently Obtained yeild is nearly greater than theoratical yeild .this happen due to the presence FeCl2 as biproduct.

The actual colour of crystalls of 1,1'-Bi-2-naphthol is white but we obtain greenishblack product.

This indicate presence of FeCl₂ with greenishblack colour in our product.