Chemistry · Alcohols, Phenols And Ethers · NEET
It gives PHENOL + CH3I. Anisole is C6H5-O-CH3 (methyl phenyl ether). The bond that breaks is between the oxygen and the methyl carbon (the alkyl side). Iodide (I-) attacks the CH3 carbon and forms CH3I. The oxygen stays on the benzene ring, so you get phenol (C6H5-OH). You do NOT get iodobenzene, because the aryl C-O bond does not break. This exact question came in NEET 2020 and 2017.
In phenol and anisole, the oxygen lone pair goes into the benzene ring by resonance. This gives the aryl C-O bond a partial double-bond character. A double bond is stronger and shorter than a normal single bond, so it is hard to break. The iodide ion cannot do an SN2 attack on the aromatic carbon either, because the ring is flat and blocks the back-side attack. So iodide always attacks the ALKYL carbon instead.
For a normal dialkyl ether, iodide attacks the SMALLER (less bulky, less substituted) alkyl group by SN2. Example: for CH3-O-C2H5, you get CH3I + C2H5OH, because iodide prefers the less hindered methyl carbon. BUT if one group is a tertiary (3 degree) group like tert-butyl, the rule flips: the tert-butyl side leaves as a stable 3 degree carbocation (SN1) and picks up iodide to give tert-butyl iodide. So bulky tertiary groups follow SN1, small groups follow SN2.
Step 1: The ether oxygen is protonated by HI to form an oxonium ion (positive oxygen). This makes the leaving group better. Step 2: Iodide (a good nucleophile) attacks the less substituted carbon by SN2 and pushes out an alcohol molecule. This gives one alkyl iodide + one alcohol. Order of reactivity of acids: HI > HBr > HCl (HI works best). The reaction needs concentrated HI at high temperature.
Yes. With limited HI you get one alkyl iodide + one alcohol. But with EXCESS HI, the alcohol formed also reacts further with HI and turns into a second alkyl iodide. Example: propyl tert-butyl ether + excess HI gives 1-iodopropane + tert-butyl iodide (both iodides). Exception: if the alcohol is PHENOL (from an aryl ether), phenol does NOT react further with HI, so phenol stays as phenol. That is why anisole always gives phenol, never iodobenzene, even with excess HI.
C6H5-CH2-O-C6H5 gives benzyl iodide (C6H5CH2I) + phenol (C6H5OH). The molecule has one alkyl-type carbon (the benzylic CH2) and one aryl carbon (the ring). Iodide attacks the benzylic carbon (alkyl side), so benzyl iodide forms. The aryl C-O bond stays, so the ring side becomes phenol. You do NOT get iodobenzene. This was asked in NEET 2023.
Anisole on cleavage with HI gives
The heating of phenyl-methyl ether (anisole) with HI produces
Consider the following reaction and identify the products A and B: C6H5-CH2-O-C6H5 (benzyl phenyl ether) + HI, heat, gives A + B
Try the real previous-year questions from this chapter — each with the answer and a full solution.
Phenol. Anisole (C6H5-O-CH3) + HI gives phenol (C6H5-OH) + methyl iodide (CH3I). The aryl C-O bond does not break, so iodobenzene never forms.
The bond on the alkyl (less bulky) side breaks. For an alkyl-aryl ether like anisole, only the alkyl C-O bond breaks; the aryl C-O bond stays because it has partial double-bond character.
Usually SN2 - iodide attacks the smaller, less substituted carbon. But when one group is tertiary (like tert-butyl), that side follows SN1 through a stable 3 degree carbocation.
HI is best. The order is HI greater than HBr greater than HCl. The reaction needs concentrated HI (or HBr) at high temperature.
Yes, if it is an alcohol. With excess HI the alcohol reacts further to give a second alkyl iodide. But phenol does not react further, so aryl ethers always stop at phenol.
