Answer :

When a nitrile is treated with 2 equivalents of a Grignard reagent, a ketone is typically formed as the final product.

Here's how the reaction proceeds step-by-step:

  1. Formation of the Imine Intermediate:

    • A nitrile, which has the general structure [tex]\text{RC} \equiv \text{N}[/tex], reacts with 1 equivalent of a Grignard reagent (let's represent it as [tex]\text{RMgX}[/tex] where [tex]\text{R}[/tex] is the alkyl group and [tex]\text{X}[/tex] is a halide) to form an imine ion through nucleophilic attack.
    • This step forms an intermediate called an iminium ion, [tex]\text{(R')(C=N)(MgX)}[/tex].

  2. Hydrolysis of the Imine Intermediate:

    • The iminium ion is then subjected to hydrolysis (typically with aqueous acid), which converts the imine into an amine group (this gives the intermediate ketone structure).

  3. Reaction With a Second Equivalent of Grignard Reagent:

    • In the next step, a second equivalent of the Grignard reagent is added to the intermediate.
    • This equivalent performs another nucleophilic attack, which ultimately leads to the formation of an alcohol.

  4. Hydrolysis to the Ketone:

    • Another step of hydrolysis returns the structure to a ketone.

This sequence of adding a Grignard reagent twice and then hydrolyzing results in a ketone.

This type of reaction is useful in organic synthesis for creating ketones, as it allows chemists to introduce complex alkyl groups.

Overall Reaction Example:

For a nitrile [tex]\text{R'-C} \equiv \text{N}[/tex], with 2 equivalents of a Grignard reagent [tex]\text{R-MgX}[/tex], under hydrolytic conditions, it will yield [tex]\text{R'COR}[/tex] (a ketone).

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Rewritten by : Barada