Animated Solution for Chemistry - Organic Chemistry: Acetyl bromide reacts with excess of ${\text{CH}}_3\text{MgI}$ followed by treatment with a saturated solution of ${\text{NH}}_4\text{Cl}$ gives

When dealing with Grignard reagents, one of the most critical details to watch out for is the stoichiometry. The phrase "in excess" completely changes the trajectory of the reaction. Let's break down exactly what happens when acetyl bromide meets an excess of methyl magnesium iodide.

Nucleophilic Acyl Substitution

We start with acetyl bromide, an acid halide. Acid halides are highly reactive electrophiles because the halogen atom (bromine, in this case) is an excellent leaving group. When the Grignard reagent (${\text{CH}}_3\text{MgI}$) is introduced, it acts as a powerful source of the methyl carbanion (${\text{CH}}_3^{-}$).

This nucleophilic methyl group attacks the electrophilic carbonyl carbon of acetyl bromide. The intermediate formed quickly collapses, kicking out the bromide ion. This process is known as nucleophilic acyl substitution, and it results in the formation of a ketone:

The product formed here is acetone.

Nucleophilic Addition

If we had only used one equivalent of the Grignard reagent, the reaction might have stopped at acetone. However, the problem explicitly states that the Grignard reagent is in excess.

Ketones are highly susceptible to nucleophilic attack. Therefore, a second molecule of ${\text{CH}}_3\text{MgI}$ will immediately attack the newly formed acetone. This time, there is no leaving group. The methyl carbanion attacks the carbonyl carbon, and the pi electrons are pushed entirely onto the oxygen atom, forming a stable tertiary alkoxide intermediate:

Mild Hydrolysis

To isolate the final alcohol, we need to protonate the alkoxide oxygen. This is done using a saturated solution of ammonium chloride (${\text{NH}}_4\text{Cl}$).

You might wonder, why not use a strong acid like $\text{HCl}$? Tertiary alcohols are notoriously prone to dehydration in the presence of strong acids, which would lead to the formation of an alkene. Ammonium chloride provides a perfectly mild acidic environment to safely protonate the alkoxide without risking dehydration.

The final product is a tertiary alcohol with four carbon atoms: 2-methyl-2-propanol (also known as tert-butyl alcohol).