An organic compound ‘A’ (C₉H₁₀O) when treated with conc. HI undergoes cleavage to yield compounds ‘B’ and ‘C’. ‘B’ gives yellow precipitate with AgNO₃ where as ‘C’ tautomerizes to ‘D’. ‘D’ gives positive iodoform test. ‘A’ could be

Solution

Chemistry ma’am · 4 steps

Work backwards: iodoform-positive D is acetaldehyde, so C is vinyl alcohol and B is benzyl iodide.

1. Work back from D. A positive iodoform test needs a CH₃CO⁻ group. The simplest compound formed by tautomerism is acetaldehyde, CH₃CHO, from the enol vinyl alcohol CH₂=CH-OH. So the ether must carry an O-CH=CH₂ group.
2. Which bond breaks. The vinyl C–O bond has partial double-bond character, so HI does not break it. HI breaks the other O–C bond, and that carbon ends up as the iodide B.
3. Identify B. B must give a yellow precipitate of AgI, so its C–I bond must ionise easily. Benzyl iodide does this because the ring stabilises the benzyl carbocation. An aryl group cannot work: aryl ethers give a phenol, and a phenol gives no precipitate.

   Ether	Products with HI	Yellow ppt?	Iodoform from tautomer?
   (a) p-tolyl vinyl ether	p-cresol and vinyl alcohol	No (phenol)	Yes
   (b) allyl phenyl ether	phenol and allyl iodide	Yes	No (phenol does not tautomerise)
   (c) benzyl vinyl ether	benzyl iodide and vinyl alcohol	Yes	Yes, CH₃CHO
   (d) phenyl prop-1-enyl ether	phenol and CH₃CH=CHOH	No (phenol)	No, gives CH₃CH₂CHO

4. So. A is benzyl vinyl ether, C₆H₅CH₂-O-CH=CH₂ (C₉H₁₀O), which is option (c). It gives benzyl iodide (B) and vinyl alcohol (C), and C tautomerises to acetaldehyde (D).

The idea to keep. In HI cleavage, vinyl and aryl C–O bonds stay intact. The halide forms on the alkyl or benzyl side, and a vinyl side leaves as an enol that tautomerises to a carbonyl compound.

Answer from the NTA final answer key. Solution by PrepWiser.