Acidic Cleavage Of Which Ether Will Yield The Following Products

Acidic Cleavage of Ethers: Which Ether Produces the Specified Products?

When an ether is treated with a strong acid such as hydroiodic acid (HI), hydrobromic acid (HBr), or hydrochloric acid (HCl), the C–O–C bond undergoes heterolytic cleavage. The reaction proceeds through a carbocation intermediate, and the final products depend on the substituents attached to the oxygen atom. Understanding which ether yields a particular pair of products requires knowledge of carbocation stability, steric factors, and the nucleophilicity of the halide ion.

This article walks you through the mechanistic basis of acidic ether cleavage, shows how to predict the outcome, and provides a step‑by‑step strategy for identifying the ether that generates a given set of products.

1. Fundamentals of Acid‑Catalyzed Ether Cleavage

1.1. General Reaction

The generic equation for the acid‑induced cleavage of an ether is:

R–O–R'  +  HX   →   R–X   +   R'–OH   (or   R'–X   +   R–OH)

• R and R' are alkyl or aryl groups attached to the oxygen.
• HX is a strong acid (HI > HBr > HCl).
• The halide (X⁻) acts as the nucleophile that attacks the more substituted carbon, while the –OH group remains attached to the less substituted carbon.

1.2. Why the Reaction Occurs

1. Protonation of the ether oxygen makes the C–O bond more electrophilic.
2. Departure of water generates a carbocation at the carbon that can best stabilize the positive charge (tertiary > secondary > primary).
3. Nucleophilic attack by the halide ion on the carbocation yields the corresponding alkyl halide.
4. The other fragment retains the –OH group, producing an alcohol.

Italic terms such as carbocation and protonation are highlighted to aid comprehension.

2. Predicting the Major Products

2.1. Carbocation Stability Dictates the Site of Cleavage

• Tertiary carbocation → most favored → yields the corresponding tertiary alkyl halide.
• Secondary carbocation → intermediate favorability → secondary alkyl halide.
• Primary carbocation → least favored; often the reaction proceeds via an SN2 pathway instead, giving a primary halide and an alcohol.

2.2. Steric Effects

Even when a secondary carbocation is possible, a less hindered primary carbon may be attacked if the secondary carbon is heavily substituted with bulky groups.

2.3. Influence of the Halide

• Iodide (I⁻) is the best nucleophile among the three halides, making HI the most efficient reagent for ether cleavage.
• Bromide (Br⁻) works similarly but is slightly less reactive.
• Chloride (Cl⁻) is a poorer nucleophile; HCl often leads to substitution only with highly activated substrates.

3. Step‑by‑Step Strategy to Identify the Starting Ether

When presented with a set of products, follow these steps to pinpoint the ether that would generate them under acidic conditions.

1. Write the products (e.g., tert‑butyl bromide and ethanol).
2. Determine the carbon skeleton of each product.
3. Re‑assemble the original ether by joining the carbon fragments through an oxygen atom.
4. Check for possible rearrangements (e.g., hydride or alkyl shifts) that could alter the carbocation intermediate.
5. Validate with steric and electronic considerations to ensure the predicted cleavage aligns with the most stable carbocation pathway.

4. Worked Example

4.1. Given Products

• 2‑bromo‑2‑methylpropane (tert‑butyl bromide)
• Methanol

4.2. Reconstructing the Ether

1. The alkyl halide product originates from a tertiary carbocation.
2. The alcohol product originates from the less substituted carbon that retained the –OH group.
3. Therefore, the original ether must have been tert‑butyl methyl ether (CH₃–O–C(CH₃)₃).

4.3. Mechanism 1. Protonation of the ether oxygen.

2. Cleavage of the C–O bond at the tert‑butyl side, generating a tert‑butyl carbocation.
3. Nucleophilic attack of I⁻ on the carbocation → tert‑butyl iodide (or bromide if HBr is used).
4. The remaining fragment, methanol, retains the –OH group.

Thus, tert‑butyl methyl ether is the ether that yields the specified products.

5. Multiple‑Choice Scenarios Often exam questions present four candidate ethers and ask which one produces a given pair of products. Use the following checklist:

• Does the alkyl halide correspond to a tertiary/secondary/primary carbon?
• Is the alcohol derived from a primary carbon?
• Are there any competing rearrangements that would alter the carbon skeleton?

Sample Question

Which of the following ethers, when treated with HI, gives ethyl bromide and methanol?

A. Methyl ethyl ether (CH₃–O