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When anisole is heated with $HI$, the product is:
(a) Phenyl iodide and methyl iodide.
(b) Phenol and methanol.
(c) Phenyl iodide and methanol.
(d) Methyl iodide and phenol.

Last updated date: 21st Jul 2024
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Hint: We know that anisole, otherwise called as methoxybenzene, is a natural compound with the equation\[C{H_3}O{C_6}{H_5}\]. It is a dry fluid with a smell suggestive of anise seed, and truth be told large numbers of its subsidiaries are found in normal and counterfeit aromas. The compound is essentially made artificially and is a forerunner to other engineered compounds. It is ether. It tends to be set up by the Williamson ether amalgamation; sodium phenoxide is responded with a methyl halide to yield anisole.

Complete answer:
We need to remember that the anisole in response with Hydrogen iodide acquires hydrogen ions and structures methyl phenyl oxonium particles.
Due to reverberation, there's some incomplete twofold bond character among oxygen and carbon of benzene rings. Thus, it's a lot simpler for iodine-to assault methoxy bond than phenoxyl bond. Hence, anisole responds with Hydrogen iodide to give phenol and methyl iodide.

Hence option D is correct.

Now we can discuss about the reactivity of anisole is discussed below:
Anisole goes through electrophilic aromatic replacement response at a quicker speed than benzene, which thus responds more rapidly than nitrobenzene.
The methoxy group is an ortho/para coordinating gathering, which implies that electrophilic replacement specially happens at these three locales.
The upgraded nucleophilicity of anisole versus benzene mirrors the impact of the methoxy group, which delivers the ring more electron-rich.
The methoxy group emphatically influences the pi haze of the ring as a mesmeric electron giver, more so than as an inductive electron pulling out group regardless of the electronegativity of the oxygen.