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SolutionsOrganometallic Reagents

1. Prepare the reagents

  1. Ethylmagnesium bromide from bromoethane.
  2. Butyllithium from 1-bromobutane.
  1. CH3CH2Br + Mg (dry ether) → CH3CH2MgBr
  2. CH3CH2CH2CH2Br + 2 Li (ether or hexane) → CH3CH2CH2CH2Li + LiBr

2. A drop of water

What happens if ethylmagnesium bromide comes into contact with water? Why must one work under anhydrous conditions?

It is destroyed: CH3CH2MgBr + H2O → CH3CH3 + Mg(OH)Br. The carbon of the organometallic is an extremely strong base (the pKa of ethane is above 50) and violently removes the proton from water.

3. Synthesis of alcohols

Which alcohol is obtained by treating CH3MgBr with each compound and then with H3O+?

  1. Formaldehyde
  2. Acetaldehyde
  3. Acetone
  1. Ethanol, CH3CH2OH (primary).
  2. 2-Propanol, (CH3)2CHOH (secondary).
  3. 2-Methyl-2-propanol, (CH3)3COH (tertiary).

4. An impossible reagent

Why can a Grignard reagent not be prepared from 2-bromoethanol, HOCH2CH2Br?

Because the OH group is acidic: as soon as the organomagnesium compound formed, its carbon would remove the OH proton (of the same or another molecule) and the reagent would be destroyed.

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