Cleavage of Ethers NEET 2023: How Is the HI Reaction Solved?
The cleavage of ethers NEET 2023 question should be approached by marking the carbon directly bonded to oxygen on each side. Benzyl methyl ether is heated with HI:
The single-correct question asks you to identify products A and B. It is a hard-tagged NEET 2023 question with an expected solving time of 60 seconds. Choose only after checking protonation, bond cleavage and carbocation stability.
How Do You Solve the Cleavage of Ethers NEET 2023 Question?
The official SN1 reasoning requires cleavage of the benzylic carbon–oxygen bond. Protonation makes oxygen a good leaving group. Heterolytic cleavage then forms a resonance-stabilised benzyl carbocation and methanol. Iodide captures the carbocation to complete the reaction.
What Is the Ether Given in the Reaction?
Step 1: Identify the substrate.
The substrate is unsymmetrical benzyl methyl ether because the two groups attached to oxygen are different. One side is benzyl and the other is methyl.
The carbon attached to oxygen on the left is benzylic. Oxygen is not bonded directly to the benzene ring.
Why Is Oxygen Protonated First?
Step 2: Protonate oxygen using HI.
The ether oxygen accepts a proton from HI through one of its lone pairs. This forms an oxonium ion and an iodide ion.
Protonation converts the neutral ether oxygen into a good leaving group, allowing carbon–oxygen bond cleavage.
Which Carbon–Oxygen Bond Breaks?
Step 3: Cleave the benzylic carbon–oxygen bond heterolytically.
The bonding electron pair remains with oxygen. Methanol leaves, and a benzyl carbocation forms.
This pathway is favourable because the benzyl carbocation is resonance-stabilised. A methyl carbocation is not viable.
How Is Benzyl Iodide Formed?
Step 4: Allow iodide to trap the carbocation.
Iodide acts as a nucleophile and attacks the positively charged benzylic carbon.
The overall transformation is:
Therefore, A is benzyl iodide and B is methanol. This corresponds to option 3, labelled C in the supplied option list.
How Can You Solve This Ether Cleavage in 60 Seconds?
For the cleavage of ethers NEET 2023 problem, use an atom-first check instead of judging which group looks larger. Mark the two atoms directly attached to oxygen, compare the possible intermediates and preserve both carbon fragments in the products.
- Circle the atoms directly attached to oxygen. They are the benzylic methylene carbon and the methyl carbon.
- Classify the left side correctly. The phenyl group is separated from oxygen by a methylene group. It is benzylic, not an aryl–oxygen bond.
- Compare possible carbocations. The benzyl carbocation is resonance-stabilised. A methyl carbocation is not viable.
- Preserve the carbon skeleton. The original phenyl–methylene unit and methyl unit must remain in the products.
- Apply the final product check. Iodine must be on the benzylic carbon, while the oxygen-containing fragment must be methanol.
What Carbon Count Confirms the Products?
The reactant contains eight carbon atoms: six in the phenyl ring, one benzylic carbon and one methyl carbon.
Benzyl iodide contains seven carbons, while methanol contains one.
The carbon count is conserved. Any proposed answer that removes the benzylic methylene carbon fails this check.
Why Is Treating a Benzyl Ether as an Aryl Ether Wrong?
The error is reading benzyl methyl ether as though oxygen were bonded directly to the benzene ring. The substrate contains a benzylic carbon–oxygen bond at an sp3 carbon, not an aryl carbon–oxygen bond at an sp2 carbon.
Compare the two structures:
- Benzyl methyl ether:
- Anisole, an aryl methyl ether:
Predictions involving phenol, iodobenzene or loss of the benzylic methylene group fail the carbon-skeleton check.
- Phenol would place oxygen directly on the ring, although the original oxygen was separated from the ring by a methylene group.
- Iodobenzene would place iodine directly on the ring and remove the benzylic carbon position.
- Ring-substituted phenol or aryl iodide products result from misreading a benzyl ether as an aryl ether.
The position of one methylene group changes the reaction class. Check the atom directly bonded to oxygen before applying a cleavage rule.
How Do You Solve Three Related Ether-Cleavage Questions?
The same bond-selection test handles aryl, benzylic, tertiary and ordinary primary ethers. First classify the atoms attached directly to oxygen. Then decide whether a stable carbocation can form or iodide must attack an unhindered carbon through SN2.
Question 1: What Are the Products When Anisole Reacts With Hot HI?
Anisole gives phenol and methyl iodide.
The aryl carbon–oxygen bond at the sp2 carbon is not cleaved by ordinary SN1 or SN2 attack. Iodide attacks the methyl carbon, so the aryl fragment becomes phenol.
Question 2: What Are the Products When Tert-Butyl Methyl Ether Reacts With HI?
The products are tert-butyl iodide and methanol.
Cleavage forms a favourable tertiary carbocation. Iodide then captures that carbocation.
Question 3: What Are the Final Products When Diethyl Ether Is Heated With Excess HI?
The final products are two molecules of ethyl iodide and water.
Ethanol forms after the first cleavage, but excess HI converts it into ethyl iodide. Ethanol is only an intermediate.
Use Alcohols Phenols Ethers Practice Questions NEET: 15 MCQs to apply this classification to more structures.
What Quick Doubts Should You Clear About Ether Cleavage?
Benzyl and aryl compounds differ in the atom directly bonded to oxygen. Benzyl iodide retains a methylene group between the ring and iodine. Iodobenzene does not. The same direct-atom check determines when phenol can form and whether SN1 or SN2 reasoning applies.
What Is the Difference Between Benzyl Iodide and Iodobenzene?
Benzyl iodide has iodine bonded to the benzylic methylene carbon.
Iodobenzene has iodine bonded directly to an aryl carbon.
They are different compounds and cannot be used interchangeably.
When Does Phenol Form During Ether Cleavage?
Phenol forms from aryl–alkyl ethers such as anisole, where oxygen is bonded directly to an aryl carbon.
It does not form from benzyl methyl ether under the stated SN1 mechanism because oxygen is attached to benzylic methylene, not directly to the ring.
When Should You Use SN1 Instead of SN2 Reasoning?
SN1 is favoured when cleavage can produce a resonance-stabilised benzylic carbocation or another stable carbocation, such as a tertiary carbocation. SN2 commonly occurs at an unhindered methyl or primary carbon when such a stable carbocation is unavailable.
In the exam, mark the two atoms directly bonded to oxygen first. That step identifies the bond as benzylic, aryl, methyl, primary or tertiary.
Frequently asked questions
What are the products of benzyl methyl ether with HI?
Benzyl methyl ether gives benzyl iodide and methanol on heating with HI. Protonation is followed by benzylic C–O bond cleavage and capture of the resonance-stabilised benzyl carbocation by iodide.
Why does phenol not form in the NEET 2023 ether question?
Phenol does not form because oxygen is bonded to a benzylic methylene carbon, not directly to the benzene ring. Phenol forms during cleavage of aryl–alkyl ethers such as anisole.
Is cleavage of benzyl methyl ether with HI SN1 or SN2?
The stated mechanism follows an SN1 pathway because cleavage produces a resonance-stabilised benzyl carbocation. Iodide then attacks the benzylic carbocation to form benzyl iodide.
What does diethyl ether give with excess HI?
Diethyl ether gives two molecules of ethyl iodide and one molecule of water with excess HI. Ethanol forms after the first cleavage but reacts further with HI, so it is only an intermediate.