The product P in the reaction, 




Grignard Reaction with Nitriles: Ketone Formation
This transformation involves the nucleophilic addition of a Grignard reagent to a nitrile group, followed by acidic hydrolysis to yield a ketone.
R-C≡N + CH3MgBr → R-C(-N=MgBr)(CH3)
R-C(-N=MgBr)(CH3) + H3O+ → R-CO-CH3 + NH4+ + MgBrOH
Final Product: The reaction converts the nitrile (R-C≡N) to a methyl ketone (R-CO-CH3)
Therefore, the correct answer is D (a ketone with structure R-CO-CH3).
Think of the nitrile as a "carbon delivery system":
1. The Nitrile Group
The -C≡N group is linear and highly polarized, with partial positive charge on carbon. This makes it:
2. The Grignard Reagent
CH3MgBr provides:
Starting with 3-methoxybenzonitrile:
3-Methoxybenzonitrile
+ CH3MgBr →
3-Methoxyacetophenone
The methoxy group (-OCH3) remains untouched because:
Key Takeaway: Grignard + nitrile → ketone, every time.
Therefore, the correct product is D (3-methoxyacetophenone).


200 ml of an aqueous solution contains 3.6 g of Glucose and 1.2 g of Urea maintained at a temperature equal to 27$^{\circ}$C. What is the Osmotic pressure of the solution in atmosphere units?
Given Data R = 0.082 L atm K$^{-1}$ mol$^{-1}$
Molecular Formula: Glucose = C$_6$H$_{12}$O$_6$, Urea = NH$_2$CONH$_2$