Discussion
A mono-substituted aromatic compound P decolourizes bromine water, gives a purple red colour with aqueous FeCl3 and is also soluble in aqueous NaOH. Compound P undergoes acetylation with reagent Q in the presence of a base to give R as the major product. Reagent Q displays two characteristic bands at 1830 cm−1 and 1750 cm−1 in its IR spectrum. P, Q and R are
- A.
- B.
- C.
- D.
Answer: C
The compounds are phenol (P), acetic anhydride (Q) and phenyl acetate (R), which is option (C).
A mono-substituted aromatic compound that decolourises bromine water, gives a purple-red colour with aqueous FeCl3 and dissolves in aqueous NaOH is a phenol. Aniline does not give the last two. The reagent with two C=O stretching bands at about 1830 and 1750 cm−1 is an acid anhydride, whose symmetric and asymmetric C=O stretches give a pair of bands. An acid chloride shows a single band near 1800 cm−1. Acetylation of the phenolic OH with acetic anhydride and a base gives the ester, phenyl acetate.
Options (A) and (B) use acetyl chloride, which does not fit the two-band IR spectrum, and (B) starts with aniline. Option (D) has the right reagents but a ring C-acylated product, which is the Fries product and needs a Lewis acid, not a base.
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