Q.(i) Identify A and B in the following reaction: Anisole (a benzene ring bearing an –O–CH3 substituent) treated with HI → A + B

Anisole with HI cleaves at the O-CH3 bond (not the aryl-O bond) to give phenol and methyl iodide; hydroboration-oxidation adds B2H6 across an alkene anti-Markovnikov, then oxidises to the corresponding alcohol.

(i) Anisole + HI -> A + B:

Anisole (C6H5-O-CH3) is a mixed (aryl-alkyl) ether. When ethers are cleaved by HI/HBr, the halide ion (I-) attacks the carbon that is less hindered and where the C-O bond is weaker - normally the smaller alkyl group, via an SN2-like mechanism. In anisole, the aryl-O bond has partial double-bond character (due to resonance/conjugation of the ring with the oxygen lone pair) and is much stronger/harder to break than the O-CH3 bond, so cleavage occurs exclusively at the O-CH3 bond:

C6H5-O-CH3 + HI -> C6H5-OH (A, phenol) + CH3-I (B, methyl iodide)

So A = phenol and B = methyl iodide.

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