Animated Solution for Chemistry - Organic Chemistry: To synthesis 1.0 mole of 2-methylpropan-2-ol from ethylethanoate ...... equivalents of ${\text{CH}}_3\text{MgBr}$ reagent will be required. (Integer value)

Imagine you are tasked with synthesizing a specific molecule: 2-methylpropan-2-ol, which is a classic tertiary alcohol. Your starting material is ethyl ethanoate, a common ester. The tool at your disposal is the highly versatile Grignard reagent, specifically methyl magnesium bromide (${\text{CH}}_3\text{MgBr}$).

The core question we need to answer is: exactly how many equivalents of this Grignard reagent are required to drive this transformation to completion? To answer this, we must dive into the step-by-step mechanism of how Grignard reagents interact with esters.

Ester to Ketone

Recall the standard behavior of esters when they encounter Grignard reagents. The Grignard reagent acts as a powerful nucleophile because the carbon-magnesium bond is highly polarized, giving the methyl group a strong partial negative charge.

When we introduce the first equivalent of ${\text{CH}}_3\text{MgBr}$, the nucleophilic methyl group attacks the electrophilic carbonyl carbon of ethyl ethanoate. This pushes the pi electrons up onto the oxygen. However, because we have a good leaving group (the ethoxy group, $-{\text{OC}}_2{\text{H}}_5$), the carbonyl double bond reforms, and the leaving group is expelled.

The result of this first step is an intermediate product: Acetone (${\text{CH}}_3{\text{COCH}}_3$). So, we have successfully used one equivalent of the Grignard reagent.

Ketone to Alcohol

But the reaction does not stop there! Ketones are actually more reactive towards nucleophilic attack than esters. As soon as acetone is formed in the reaction mixture, it becomes a prime target for any remaining Grignard reagent.

The second equivalent of ${\text{CH}}_3\text{MgBr}$ will immediately attack the carbonyl carbon of the newly formed acetone. This time, there is no leaving group to expel. The pi electrons move up to the oxygen, forming a stable alkoxide intermediate.

Following a final workup step involving hydrolysis with water (${\text{H}}_2\text{O}$), the alkoxide is protonated, and we finally obtain our target molecule: 2-methylpropan-2-ol.

Let's count them up.

1. One equivalent was used to convert the initial ester into a ketone. 2. A second equivalent was used to convert that intermediate ketone into our final tertiary alcohol.

Therefore, exactly 2 equivalents of the Grignard reagent are required. It is a beautifully straightforward mechanism, but it highlights a crucial rule in organic synthesis: when reacting esters with Grignard reagents, you will almost always form a tertiary alcohol (unless you start with a formate ester, which yields a secondary alcohol), and it will always consume two equivalents of the reagent.