Aromaticity Chemistry Of Aromatic Compounds

Step 1: Ph–Cl + Mg in dry ether → Ph–MgCl (Grignard reagent)
Step 2: Ph–MgCl + CO2} → Ph–COOMgCl → H3}O\textsuperscript{+} → Benzoic acid (intermediate X)
Step 3: Benzoic acid + LiAlH4} → Benzyl alcohol (reduction of –COOH to –CH2}OH)

This reaction is a Sandmeyer reaction followed by deuteration. Here's the step-by-step: 1. Aniline is converted into a diazonium salt using at 273 K. 2. CuCl is used to replace diazonium with Cl (if present). 3. In this reaction, D₂O is used, meaning diazonium is decomposed and D (deuterium) replaces the diazonium group. 4. The D atom substitutes at the same position as the amine was, and the methyl remains in the para position.

Step (i): Friedel–Crafts acylation — CH CO group attaches at para position.
Step (ii): Oxidation with KMnO — CH CO group and any methyl groups convert to COOH.
Step (iii): Acidification gives the final dicarboxylic acid — terephthalic acid.

- Statement I: Chlorobenzene undergoes nitration to form 1-chloro-4-nitrobenzene as the major product due to the electron-withdrawing effect of the chlorine atom, which directs the nitro group to the para position. Therefore, this statement is correct.
- Statement II: Chlorobenzene undergoes nitration more slowly than benzene because the chlorine atom, being electron-withdrawing, deactivates the ring and makes it less reactive toward electrophilic substitution. Hence, this statement is also correct. Thus, the correct answer is option (1) Statements I, II are correct.

The given transformation involves the oxidation of an alkyl side chain (–CH(CH ) ) on a benzene ring to a carboxylic acid group (–COOH). This type of oxidation is best carried out using strong oxidizing agents like potassium permanganate (KMnO ) under basic conditions followed by acidification. KMnO oxidizes all benzylic alkyl groups (regardless of their length or branching) to benzoic acid, as long as there is at least one benzylic hydrogen.