Animated Solution for Chemistry - Organic Chemistry: Phenyl magnesium bromide reacts with methanol to give

The reaction between a Grignard reagent and a protic solvent is a fundamental concept in organic chemistry that often catches students off guard. Let's dive deep into the mechanics of this reaction and understand why it proceeds the way it does.

We are given two reactants: phenyl magnesium bromide ($\text{PhMgBr}$) and methanol (${\text{CH}}_3\text{OH}$).

Phenyl magnesium bromide is a classic Grignard reagent. The key to understanding its reactivity lies in the carbon-magnesium bond. Magnesium is highly electropositive compared to carbon. As a result, the carbon atom pulls the electron density of the bond towards itself, acquiring a significant partial negative charge (${\delta }^{-}$). This makes the phenyl group act as a powerful carbanion (${\text{Ph}}^{-}$). A carbanion is not only a strong nucleophile but also an exceptionally strong base.

On the other side, we have methanol. Methanol is a protic solvent, meaning it contains a hydrogen atom attached to a highly electronegative atom—in this case, oxygen. The oxygen atom pulls electron density away from the hydrogen, leaving the hydrogen with a partial positive charge (${\delta }^{+}$). This hydrogen is considered an acidic proton or an active hydrogen.

When these two molecules meet, a competition arises: will the Grignard reagent act as a nucleophile and attack the carbon atom of methanol, or will it act as a base and abstract the acidic proton?

The golden rule of organic chemistry kinetics provides the answer: Acid-base reactions are almost always faster than nucleophilic substitution or addition reactions.

Because the proton transfer involves the movement of a tiny, unhindered hydrogen nucleus, it happens almost instantaneously. The strong base (${\text{Ph}}^{-}$) will immediately abstract the acidic proton (${\text{H}}^{+}$) from methanol.

As the phenyl carbanion grabs the proton, it forms benzene (${\text{C}}_6{\text{H}}_6$). Simultaneously, the remaining methoxide ion (${\text{CH}}_3{\text{O}}^{-}$) coordinates with the magnesium cation (${\text{MgBr}}^{+}$) to form methoxy magnesium bromide (${\text{Mg(OCH}}_3\text{)Br}$).

Therefore, the reaction yields a mixture of benzene and methoxy magnesium bromide. This reaction is so reliable and quantitative that it is often used analytically to determine the number of active hydrogens in an unknown compound (Zerewitinoff determination). Always remember to check for acidic protons before predicting a nucleophilic attack with a Grignard reagent!