Animated Solution for Chemistry - Organic Chemistry: A reaction of benzonitrile with one equivalent ${\text{CH}}_3\text{MgBr}$ followed by hydrolysis produces a yellow liquid P. The compound P will give positive

Imagine you are a molecular architect, and your task is to build a specific structure from a simple blueprint. Our starting material is benzonitrile, a benzene ring adorned with a cyanide group. This cyanide group is like a locked door, waiting for the right key to open it. Enter the Grignard reagent, methyl magnesium bromide.

Let's decode this reaction step by step. We are starting with benzonitrile, and we are treating it with one equivalent of methyl magnesium bromide, which is our Grignard reagent. The nucleophilic methyl group from the Grignard reagent attacks the electrophilic carbon of the cyanide group.

This breaks one of the pi bonds, forming an intermediate imine salt. The reaction can be represented as:

Next, we perform acid hydrolysis. The imine salt is highly reactive with water. The carbon-nitrogen double bond is completely cleaved and replaced by a carbon-oxygen double bond.

And there we have it! Our yellow liquid product P is acetophenone. Notice the structure carefully. It has a methyl group directly attached to the carbonyl carbon. It is a methyl ketone.

Now, the question asks which test this compound will respond to. Since acetophenone is a methyl ketone, it contains that specific acetyl group required to give a positive iodoform test.

When we treat acetophenone with iodine and sodium hydroxide, the methyl group gets fully halogenated and then cleaved off, forming a yellow precipitate of iodoform, along with sodium benzoate.

This yellow precipitate confirms the presence of the methyl ketone group. Therefore, compound P will give a positive iodoform test. The correct option is A.