How are the following conversions carried out?
Methyl magnesium bromide → 2-Methylpropan-2-ol

The conversion you're referring to is a classic example of a Grignard reaction followed by a subsequent reaction to form an alcohol. Here's a step-by-step explanation of the process:

Formation of Methyl Magnesium Bromide (CH3MgBr):
Methyl magnesium bromide, often written as CH3MgBr, is prepared by reacting methyl bromide (CH3Br) with magnesium (Mg) metal in anhydrous ether or tetrahydrofuran (THF) solvent. The reaction typically occurs as follows:

CH3Br + Mg → CH3MgBr

This forms the Grignard reagent, which is an organomagnesium compound. It's important to conduct this reaction in anhydrous conditions because Grignard reagents react violently with water.

Reaction with Acetone (Propanone):
The next step involves reacting the methyl magnesium bromide with acetone (propanone). In this case, the acetone is 2-propanone, so the product will be a tertiary alcohol, specifically 2-methylpropan-2-ol. The reaction proceeds as follows:

CH3MgBr + (CH3)2CO → CH3(CH3)2C(OH) + MgBr(CH3)

The intermediate product is a magnesium alkoxide, which reacts with water present in the mixture to yield the alcohol:

CH3(CH3)2C(OH) + H2O → CH3(CH3)2CHOH + Mg(OH)Br

The magnesium byproduct usually remains in solution.

Isolation of the Alcohol:
After the reaction is complete, the mixture is typically quenched with a mild acid like dilute hydrochloric acid (HCl) or dilute sulfuric acid (H2SO4) to neutralize any remaining Grignard reagent and convert the magnesium alkoxide into the alcohol. The alcohol product can then be isolated through standard techniques like extraction or distillation.

So, by starting with methyl magnesium bromide and reacting it with acetone followed by hydrolysis, you would obtain 2-methylpropan-2-ol as the final product.