Question

Regarding Grignard synthesis, let's say you start with 1.25 g of 4-bromo-N,N-dimethylaniline, 13 mL anhydrous THF, 0.20g magnesium turnings, and the product is p-dimethylamino phenyl magnesium bromide. 1 or 2 tiny crystals of iodine also is also added. The product can be completely destroy by water.

1/ Calculate mass and volume of water needed to destroy p-dimethylamino phenyl magnesium bromide. And what will be the product of this hydrolysis?

2/ What is the role of iodine in this reaction?

Appreciate all helps with specific answers/explanation. Thanks

Answer 1

Grignard reagent is a strong nucleophile and strong base. In the presence of water (acidic protons), grignard reagent will break its orgnaometallic bond and the carbon atom with negative charge wll extract proton and become a hydrocarbon.

A small amount of water will destroys the grignard reagent.

For every 1 mol p-dimethylamino phenyl magnesium bromide we need one mole water to react.

Moles of p-dimethylamino phenyl magnesium bromide produced from 4-bromo-N,N-dimethylaniline =

=6.25*10^-3 mol p-dimethylamino phenyl magnesium bromide

Moles of  p-dimethylamino phenyl magnesium bromide produced from Mg turnings =

=8.23*10^-3 mol p-dimethylamino phenyl magnesium bromide

So here 4-bromo-N,N-dimethylaniline is the limiting reagent

Mass of water required to destroy p-dimethylamino phenyl magnesium bromide

Mass of H₂O required = 0.1125 g H₂O

Density of water = 1 g/mL

Volume of water required = 0.1125 g * 1 mL/g = 0.1125 mL

2)

Iodine has two functions

i) It acts as an indicator, it colors the solution if Mg is unreacted with phenyl bromide.

It gets decolorize when Mg reacts with phenyl bromide

ii) I₂ acts as a activator. It chemically cleans the surface of Mg and activates the Mg.