The problem asks for the major product of the nitration of three different substituted benzene compounds: P (anisole), Q (acetanilide), and S (tert-butylbenzene) using a nitrating mixture ($HNO_3 / H_2SO_4$). This reaction is an electrophilic aromatic substitution (EAS), where the electrophile is the nitronium ion ($NO_2^+$). The position of substitution on the benzene ring is determined by the electronic and steric effects of the substituent already present.

Step 1: Analyze the nitration of compound P (Anisole)

1. Identify the substituent: Compound P is anisole, which has a methoxy group ($-OCH_3$) attached to the benzene ring.
2. Determine its directing effect: The $-OCH_3$ group is an activating group. The oxygen atom donates electron density to the ring via resonance (+R effect), which is stronger than its electron-withdrawing inductive effect (-I effect). Activating groups are ortho, para-directing because they increase the electron density at the ortho and para positions, making them more reactive towards electrophiles.
3. Predict the major product: Both ortho and para products are formed. However, the para position is sterically less hindered than the two ortho positions. Therefore, the para-isomer (4-nitroanisole) is the major product.

$\text{Anisole (P)} \xrightarrow{\mathrm{HNO}_3 / \mathrm{H}_2\mathrm{SO}_4} \text{p-nitroanisole (Major Product)} + \text{o-nitroanisole (Minor Product)}$

Step 2: Analyze the nitration of compound Q (Acetanilide)

1. Identify the substituent: Compound Q is acetanilide, with an acetamido group ($-NHCOCH_3$) on the ring.
2. Determine its directing effect: The nitrogen atom's lone pair can be delocalized into the benzene ring (+R effect), making the group activating and ortho, para-directing. Although the activating effect is moderate because the lone pair is also in resonance with the adjacent carbonyl group, it still directs incoming electrophiles to the ortho and para positions.
3. Predict the major product: The $-NHCOCH_3$ group is quite bulky. This steric hindrance significantly disfavors electrophilic attack at the nearby ortho positions. Consequently, the electrophile $NO_2^+$ preferentially attacks the sterically accessible para position. The major product is p-nitroacetanilide (4-nitroacetanilide).

$\text{Acetanilide (Q)} \xrightarrow{\mathrm{HNO}_3 / \mathrm{H}_2\mathrm{SO}_4} \text{p-nitroacetanilide (Major Product)} + \text{o-nitroacetanilide (Minor Product)}$

Step 3: Analyze the nitration of compound S (tert-Butylbenzene)

1. Identify the substituent: Compound S is tert-butylbenzene, with a tert-butyl group ($-C(CH_3)_3$).
2. Determine its directing effect: Alkyl groups, like the tert-butyl group, are weakly activating due to inductive effects (+I) and hyperconjugation. They are ortho, para-directing.
3. Predict the major product: The tert-butyl group is extremely large and bulky. This causes severe steric hindrance at the two adjacent ortho positions, making it very difficult for the incoming $NO_2^+$ group to attack there. Therefore, substitution occurs almost exclusively at the sterically unhindered para position. The major product is p-nitro-tert-butylbenzene (4-nitro-tert-butylbenzene).

$\text{tert-Butylbenzene (S)} \xrightarrow{\mathrm{HNO}_3 / \mathrm{H}_2\mathrm{SO}_4} \text{p-nitro-tert-butylbenzene (Major Product)}$

Conclusion

• Nitration of P (Anisole) gives p-nitroanisole as the major product.
• Nitration of Q (Acetanilide) gives p-nitroacetanilide as the major product.
• Nitration of S (tert-Butylbenzene) gives p-nitro-tert-butylbenzene as the major product.

Comparing these results with the given options, Option C correctly shows these three para-substituted compounds as the respective major products.