Question

\(Z\) is an aromatic compound with substituents \(P\) and \(Q\). What are \(P\) and \(Q\)?

• A. \(-OH,\ -SO_3H\)
• B. \(-CHO,\ -NO_2\)
• C. \(-SO_3H,\ -NO_2\)
• D. \(-COOH,\ -NO_2\)

Hint:

Alkyl benzenes on oxidation with \(KMnO_4\) give benzoic acid. The \(-COOH\) group is deactivating and meta-directing, so nitration gives meta-nitrobenzoic acid.

Answer 1

The Correct Option is D

Step 1: Aromatization of alkane.
The given alkane undergoes aromatization in the presence of \[ Cr_2O_3,\ 773K,\ 10-20\ atm \] This gives an alkyl benzene compound.
Here, the product \(X\) is toluene: \[ C_6H_5CH_3 \]

Step 2: Oxidation of toluene.
Toluene is oxidized using \[ KMnO_4,\ OH^- \] followed by \[ H_3O^+ \] The methyl group of toluene is oxidized to carboxylic acid group: \[ C_6H_5CH_3 \rightarrow C_6H_5COOH \] Thus, product \(Y\) is benzoic acid.
So, one substituent is \[ P=-COOH \]

Step 3: Nitration of benzoic acid.
Benzoic acid undergoes nitration with \[ Conc.\ HNO_3+H_2SO_4 \] The \(-COOH\) group is a meta-directing group.
Therefore, nitration gives mainly meta-nitrobenzoic acid.
So, the second substituent is \[ Q=-NO_2 \]

Step 4: Final conclusion.
Hence, the substituents \(P\) and \(Q\) are \[ \boxed{-COOH,\ -NO_2} \] Therefore, the correct option is \[ \boxed{(4)} \]