Consider the three aromatic molecules (P, Q and R) whose structures have been given below :

The correct order regarding the reactivity of these compounds with 

under optimum but slightly acidic medium is :

The reaction described is an azo-coupling reaction (electrophilic aromatic substitution) with a benzenediazonium ion. For this reaction to occur efficiently, the benzene ring must be highly activated with high electron density, primarily provided by the strong mesomeric ($$+M$$) donating effect of the nitrogen's lone pair.

• Compound P (N,N-dimethylaniline): There is no steric hindrance around the $$-\text{NMe}_2$$ group. The lone pair on the nitrogen atom is fully coplanar with the benzene ring, allowing for maximum resonance ($$+M$$ effect). This strongly activates the ring, making it the most reactive.
  
• Compound Q (2-methyl-N,N-dimethylaniline): The single bulky methyl group at the ortho-position introduces steric crowding, pushing the $$-\text{NMe}_2$$ group slightly out of the benzene ring's plane. This decreases the overlap of the nitrogen lone pair with the $$\pi$$-system (partial SIR effect), lowering its reactivity compared to P.
  
• Compound R (2,6-dimethyl-N,N-dimethylaniline): With two bulky methyl groups flanking the $$-\text{NMe}_2$$ group on both sides, the steric crowding is extreme. To minimize repulsion, the $$-\text{NMe}_2$$ group rotates completely out of the plane of the benzene ring. Because it is perpendicular, resonance is entirely inhibited (Steric Inhibition of Resonance / SIR effect). The lone pair cannot activate the ring at all, making it the least reactive.

As the number of ortho-methyl groups increases, resonance decreases, dropping the electron density on the ring and reducing its reactivity toward electrophiles:

$$\text{P} > \text{Q} > \text{R}$$

Therefore, the correct choice is [A] P > Q > R.