IIT-JEE 2010 · Paper 2 · Chemistry Q4

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ChemistrySingle correct+5 / -2Question 4 of 57

The compounds P, Q and S were separately subjected to nitration using $\ce{HNO3/H2SO4}$ mixture. The major product formed in each case respectively, is

P: 4-hydroxybenzoic acid; Q: 4-methylanisole (H₃C-C₆H₄-OCH₃, para); S: phenyl benzoate (C₆H₅-CO-O-C₆H₅).

P nitrated ortho to OH (4-hydroxy-3-nitrobenzoic acid); Q nitrated ortho to CH₃; S nitrated para on the benzoyl ring (phenyl 4-nitrobenzoate).

P nitrated ortho to COOH (4-hydroxy-2-nitrobenzoic acid); Q nitrated ortho to CH₃; S nitrated ortho on the benzoyl ring (phenyl 2-nitrobenzoate).

P nitrated ortho to OH (4-hydroxy-3-nitrobenzoic acid); Q nitrated ortho to OCH₃; S nitrated para on the phenoxy ring (4-nitrophenyl benzoate).

P nitrated ortho to OH; Q nitrated ortho to OCH₃; S nitrated meta on the phenoxy ring (3-nitrophenyl benzoate).

Answer and worked solution

Official answer: C

P, 4-hydroxybenzoic acid. OH (strongly activating, o/p) and COOH (deactivating, meta) both direct to the position ortho to OH: 4-hydroxy-3-nitrobenzoic acid.
Q, 4-methylanisole. Both groups activate, but $\ce{OCH3}$ (+M) is much stronger than $\ce{CH3}$ : nitration ortho to $\ce{OCH3}$ .
S, phenyl benzoate. The ring attached to the ester oxygen (phenoxy ring) is activated by the O lone pair; the benzoyl ring is deactivated by C=O. Nitration goes para on the phenoxy ring: 4-nitrophenyl benzoate.

(C)

Topics:Aromatic hydrocarbons and electrophilic substitution Electronic effects (inductive, resonance, hyperconjugation)

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IIT-JEE 2010 · Paper 2, Chemistry question 4