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Write the mechanism of the reaction of HI with methoxy benzene.

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Hint: The organic compound anisole, also known as methoxy benzene, has the formula. It's a colourless liquid with an anise seed aroma, and many of its derivatives can be used in natural and synthetic fragrances. The compound is mostly synthesised and serves as a precursor to other synthetic compounds.

Complete answer:
Methoxy benzene forms phenol and iodomethane as it reacts with hydroiodic acid HI.
We can represent this reaction as,
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We know that methoxy benzene is alkyl aryl ether. So, methoxy benzene has two bonds that are joined by an oxygen atom. The oxygen-methyl group (O-alkyl bond) is one, and the oxygen-aryl group is the other (O-Aryl bond). Because of the partial double bond character of the Oxygen-Aryl bond, it is the most stable of the three.
That is, oxygen-methyl bond is cleaved in this reaction, resulting in methyl iodide and phenol.

The reaction between HI and methoxy benzene occurs in two stages and they are;

Stage 1: By trapping $H^+$ from HI, the oxygen in ether is protonated (added a proton), hence forming a protonated ether molecule known as oxonium ion.

Stage 2: Since Iodide ion (I-) is a strong nucleophile, the \[S{N_2}\] reaction occurs in this step. To form methyl iodide and phenol, the I- ion preferentially targets the protonated ether molecule's least substituted carbon (in this case, the methyl group).

Note:
We should also remember that phenol is an aromatic organic compound with the molecular formula. The chemical compound with the formula \[C{H_3}I\] is iodomethane, also known as methyl iodide and commonly abbreviated "MeI."