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Step-by-Step Solution
Step 1: Reacting Benzonitrile with CH₃MgBr
Benzonitrile ($\text{C}_6\text{H}_5\text{CN}$) reacts with one equivalent of methylmagnesium bromide (CH₃MgBr). In this step, the nucleophile (CH₃⁻ from the Grignard reagent) attacks the electrophilic carbon in the nitrile group, forming an intermediate magnesium salt of an imine.
Step 2: Hydrolysis of Intermediate
Upon acid hydrolysis, the intermediate imine salt is converted into a ketone. Specifically, the product formed is acetophenone ($\text{C}_6\text{H}_5\text{COCH}_3$), which possesses a methyl group adjacent to the carbonyl carbon.
Step 3: Recognizing the Methyl Ketone Functional Group
Acetophenone contains the structure $-\text{COCH}_3$, classifying it as a methyl ketone. Methyl ketones are known to give a positive iodoform test (also known as the haloform reaction).
Step 4: Reason for the Positive Iodoform Test
The iodoform test involves the halogenation of the methyl group adjacent to a carbonyl, followed by base-induced cleavage, producing iodoform (CHI₃), a yellow precipitate. Acetophenone, having the $-\text{COCH}_3$ group, readily undergoes this reaction and gives a yellow precipitate of iodoform.
