Question

Write the structure of (A), (B) and (C) in the following reaction: phenol (C₆H₅OH) with concentrated HNO₃ gives (C); phenol with dilute HNO₃ gives (A) + (B).

Hint:

OH is ortho/para directing: dilute HNO3 gives 2-nitrophenol (A) and 4-nitrophenol (B); concentrated HNO3 gives 2,4,6-trinitrophenol, picric acid (C).

Answer 1

Background: The \(-OH\) group of phenol is a strong activating, ortho/para directing group. So nitration puts \(-NO_2\) at the ortho and para positions.

Step 1: With dilute HNO₃ (mild conditions). Only one \(-NO_2\) group enters, going to the ortho and para positions. So a mixture of two products is formed:
(A) = ortho-nitrophenol = 2-nitrophenol (\(-OH\) at C1, \(-NO_2\) at C2).
(B) = para-nitrophenol = 4-nitrophenol (\(-OH\) at C1, \(-NO_2\) at C4).
\[ C_6H_5OH \xrightarrow{\text{dil. } HNO_3} o\text{-}O_2N\text{-}C_6H_4\text{-}OH + p\text{-}O_2N\text{-}C_6H_4\text{-}OH \]

Step 2: With concentrated HNO₃ (strong conditions). Three \(-NO_2\) groups enter, at positions 2, 4 and 6. So:
(C) = 2,4,6-trinitrophenol, commonly called picric acid.
\[ C_6H_5OH \xrightarrow{\text{conc. } HNO_3} 2,4,6\text{-}(O_2N)_3C_6H_2OH \]

Summary of structures:
(A) 2-nitrophenol: benzene ring with \(-OH\) on C1 and \(-NO_2\) on C2.
(B) 4-nitrophenol: benzene ring with \(-OH\) on C1 and \(-NO_2\) on C4.
(C) picric acid (2,4,6-trinitrophenol): benzene ring with \(-OH\) on C1 and \(-NO_2\) on C2, C4 and C6.
\[\boxed{A = 2\text{-nitrophenol},\ B = 4\text{-nitrophenol},\ C = 2,4,6\text{-trinitrophenol (picric acid)}}\]