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An ether question usually comes down to one decision: when concentrated HI breaks the C–O bond, which carbon ends up holding the iodine? Get that wrong and both products are wrong. This post gives a three-question method that settles it, worked examples from methyl ethers to anisole, a mole calculation and the common slips.
Which exams test ether cleavage?
The JEE (Advanced) 2026 brochure lists, under ethers, "Preparation by Williamson's synthesis; C-O bond cleavage reactions." The JEE (Main) 2026 syllabus covers the general methods of preparation, properties and reactions of oxygen compounds, with ethers listed under "Structure", and the NEET (UG) 2026 syllabus lists "Ethers: Structure" in the same unit. The base text for all three is section 7.6.3 of the NCERT Class 12 chapter on alcohols, phenols and ethers, 2026-27 reprint. Check the documents for your own exam year.
For acidity and identification tests, see our overview of alcohols, phenols and ethers. This post goes deep on one reaction.
What happens, step by step, when HI meets an ether?
NCERT calls ethers "the least reactive of the functional groups". The C–O bond breaks only with concentrated HI or HBr at high temperature, and the order of reactivity is HI > HBr > HCl.
Protonation. The acid puts a proton on the ether oxygen, giving an oxonium ion. Now the leaving group is a neutral alcohol molecule, not an alkoxide.
The C–O bond breaks. Either iodide attacks a carbon from the back (SN2), or the alcohol leaves first to give a carbocation that iodide then captures (SN1).
Excess HI. If HI is in excess and the mixture is hot, the alcohol formed in step 2 reacts again and becomes a second alkyl iodide.
How do you decide which carbon gets the iodine?
Ask three questions, in this order.
Is one group aryl? If a benzene ring is on the oxygen, the O–aryl bond stays. NCERT explains that the ring carbon is sp² hybridised and the bond has partial double bond character. You get phenol and an alkyl iodide, and phenol does not go on to iodobenzene, because an sp² carbon cannot undergo the nucleophilic substitution needed.
Is one group tertiary? Then the reaction is SN1. The tertiary carbocation forms, so the tertiary group becomes the iodide and the other group becomes the alcohol.
Otherwise (methyl, primary or secondary groups), the reaction is SN2. In NCERT's words, "it is the lower alkyl group that forms alkyl iodide", because iodide attacks the less substituted carbon.
Worked examples: predicting the products
With one equivalent of HI, unless stated:
CH₃–O–CH₂CH₃ (methoxyethane): no aryl, no tertiary group, so SN2 at the methyl carbon: CH₃I + CH₃CH₂OH.
CH₃–O–CH(CH₃)₂ (2-methoxypropane): SN2 again, at the less hindered methyl: CH₃I + propan-2-ol.
(CH₃)₃C–O–CH₃ (2-methoxy-2-methylpropane): tertiary, so SN1: (CH₃)₃C–I + CH₃OH. This is NCERT's own example.
C₆H₅–O–CH₃ (anisole): aryl, so phenol + CH₃I, even with excess HI.
CH₃CH₂–O–CH₂CH₃ with excess HI and heat: first CH₃CH₂I + CH₃CH₂OH, then the ethanol becomes a second CH₃CH₂I. Overall, two ethyl iodides and water.
Notice the mirror image with Williamson's synthesis. NCERT notes that a tertiary halide with an alkoxide gives only an alkene, so to make a tert-butyl ether you pair a tert-butoxide with a primary halide. Breaking that same ether with HI puts the iodine on the tertiary carbon.
A worked mole calculation
Question. 8.8 g of 2-methoxy-2-methylpropane (C₅H₁₂O) is heated with excess concentrated HI. Find the theoretical mass of each iodide formed. Use C = 12, H = 1, O = 16, I = 127.
Molar mass of the ether: 5 × 12 + 12 × 1 + 16 = 88 g/mol, so 8.8/88 = 0.10 mol.
SN1 cleavage gives 0.10 mol (CH₃)₃C–I and 0.10 mol CH₃OH.
Excess HI turns the methanol into 0.10 mol CH₃I and water.
(CH₃)₃C–I, C₄H₉I: 48 + 9 + 127 = 184 g/mol, so 0.10 × 184 = 18.4 g.
CH₃I: 12 + 3 + 127 = 142 g/mol, so 0.10 × 142 = 14.2 g.
HI used: 0.20 mol in total.
A student who applies the SN2 rule here writes CH₃I and tert-butyl alcohol (74 g/mol), and gets "7.4 g of alcohol". The arithmetic is right; the chemistry is not.
Why draw the mechanism instead of memorising products?
Memorised products fail as soon as the ether changes. Drawing the oxonium ion and asking which bond breaks works for any ether. Many students skip that step: Grove, Cooper and Rush (2012) followed second-year organic chemistry students through a year and found "a large proportion of students who elect not to use the mechanistic convention in their work." The three questions above are a short way to keep the mechanism in view. Our post on SN1 and SN2 in haloalkanes covers the same two pathways from the other side.
Which mistakes cost the most marks?
Applying "the smaller group gets the iodine" to a tertiary ether.
Writing iodobenzene as a product from anisole.
Ignoring the word "excess": one equivalent stops at alcohol plus iodide.
Forgetting the conditions: concentrated HI or HBr, high temperature; HCl reacts least.
A practice routine
Answer the three questions first. For every ether, write "aryl? tertiary? else SN2" with your answers before drawing any product.
Draw one oxonium ion per problem, with the arrow for the bond that breaks.
Mix the cases. Build sets of eight ethers that mix methyl, secondary, tertiary and aryl groups, with and without excess HI. In a study of maths practice, Taylor and Rohrer (2010) found that mixing problem types lowered scores during practice but doubled scores on a test a day later; their error analysis suggested it helped students match each problem to its method.
If carbocation stability or the SN2 back-attack is still shaky, the Fermi Concept Library covers every concept in physics, chemistry, biology, mathematics and computing, so you can revisit the idea before the next set.
Today, take five ethers from the NCERT chapter and write the three answers for each before you draw a single product.
