Acetyl bromide reacts with an excess of ${\text{CH}}_3\text{MgI}$ followed by treatment with a saturated solution of ${\text{NH}}_4\text{Cl}$ to give:

1. ${\text{CH}}_3\text{COBr}$ reacts with the first mole of ${\text{CH}}_3\text{MgI}$ to form acetone (${\text{CH}}_3{\text{COCH}}_3$) and $\text{MgBrI}$.
2. Since ${\text{CH}}_3\text{MgI}$ is in excess, a second mole reacts with the acetone to form a magnesium alkoxide.
3. Hydrolysis with ${\text{NH}}_4\text{Cl}$ converts the alkoxide to a tertiary alcohol: $2$ -methyl-$2$ -propanol (tert-butyl alcohol).