A reaction of benzonitrile with one equivalent CH$$_3$$MgBr followed by hydrolysis produces a yellow liquid "P". The compound "P" will give positive _________ test.

We have benzonitrile whose molecular formula is $$\mathrm{C_6H_5CN}$$. When a nitrile reacts with one equivalent of a Grignard reagent ($$\mathrm{R-MgX}$$) in dry ether, the general rule is that the Grignard reagent adds to the carbon of the $$\mathrm{C\equiv N}$$ group and converts the nitrile into an imine magnesium halide. Later, on acidic hydrolysis, this imine is transformed into a ketone. Symbolically the rule can be stated as

$$\mathrm{R'-C\equiv N \;+\; R-MgX \;\longrightarrow\; R'-C(=N^-MgX)-R \xrightarrow[{\,\text{heat}\,}]{\text{H}_3\text{O}^+} R'-CO-R}$$

Now we apply this rule to the given reagents. In benzonitrile, $$\mathrm{R'} = \mathrm{C_6H_5}$$ (phenyl) and the Grignard reagent supplied is $$\mathrm{CH_3MgBr}$$ with $$\mathrm{R = CH_3}$$. Substituting these groups in the rule gives

$$\mathrm{C_6H_5-C\equiv N \;+\; CH_3MgBr \;\longrightarrow\; C_6H_5-C(=N^-MgBr)-CH_3 \xrightarrow[{\,\text{hydrolysis}\,}]{\text{H}_3\text{O}^+} C_6H_5-CO-CH_3}$$

Therefore the yellow liquid “P” obtained after hydrolysis is $$\mathrm{C_6H_5COCH_3}$$, whose common name is acetophenone and whose structural unit is

$$\mathrm{Ph-CO-CH_3}$$

We next decide which qualitative test is answered by this molecule. Acetophenone contains the $$\mathrm{-CO-CH_3}$$ functional group, i.e. the carbonyl carbon is directly attached to a methyl group. Organic chemistry tells us that

• The Iodoform test is positive for all compounds containing the $$\mathrm{-CO-CH_3}$$ (methyl ketone) group or $$\mathrm{CH_3-CH(OH)-}$$ group.
• Schiff’s, Ninhydrin’s and Tollen’s tests are meant for aldehydes or amino acids and are not given by simple methyl ketones.

Because acetophenone is a methyl ketone, it will liberate a yellow precipitate of $$\mathrm{CHI_3}$$ (iodoform) in the iodoform test, while it will fail the other three tests listed. Hence the liquid “P” gives a positive iodoform test.

Hence, the correct answer is Option A.