Question

Predict the products likely to be formed on cleavage of the following ethers with hydroiodic acid: a. \(\mathrm{CH}_{2}=\mathrm{CH}-\mathrm{CH}_{2}-\mathrm{O}-\mathrm{CH}_{3}\) b. \(\mathrm{CH}_{3} \mathrm{CH}_{2}-\mathrm{O}-\mathrm{CH}=\mathrm{CH}_{2}\) c. \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CCH}_{2}-\mathrm{O}-\mathrm{CH}_{3}\) d. \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{COCH}_{3}\)

Short answer

(a) Allyl alcohol, methyl iodide. (b) Ethyl alcohol, vinyl iodide. (c) Isobutyl alcohol, methyl iodide. (d) Tert-butyl iodide, methanol.

Step-by-step solution

Understand Ether Cleavage with HI

When ethers react with hydroiodic acid, they undergo cleavage, forming alcohol and alkyl iodide. The reaction follows a nucleophilic substitution mechanism, typically involving the attack on the less hindered or more stable carbocation center.

Determine Products for Compound (a)

For compound (a), \(\mathrm{CH}_{2}=\mathrm{CH}-\mathrm{CH}_{2}-\mathrm{O}-\mathrm{CH}_{3}\), HI would preferentially cleave at the \(\mathrm{O}-\mathrm{CH}_{3}\) bond due to the formation of a primary alcohol and an alkyl iodide as major products. Thus, the products are \(\mathrm{CH}_{2}=\mathrm{CH}-\mathrm{CH}_{2}\text{-OH}\) (allyl alcohol) and \(\mathrm{CH}_{3}\mathrm{I}\) (methyl iodide).

Determine Products for Compound (b)

For compound (b), \(\mathrm{CH}_{3} \mathrm{CH}_{2}-\mathrm{O}-\mathrm{CH} =\mathrm{CH}_{2}\), HI would cleave at the \(\mathrm{CH}_{2}\)-\(\mathrm{O}\) bond, producing \(\mathrm{CH}_{3}\mathrm{CH}_{2}\text{-OH}\) (ethyl alcohol) and \(\mathrm{CH}_{2} = \mathrm{CH}\mathrm{I}\) (vinyl iodide).

Determine Products for Compound (c)

For compound (c), \(\left(\mathrm{CH}_{3}\right)_{3}\mathrm{CCH}_{2}-\mathrm{O}-\mathrm{CH}_{3}\), the cleavage occurs at the \(\mathrm{C}-\mathrm{O}-\mathrm{CH}_{3}\) bond. The more stable tertiary carbocation is not formed here directly; conversely, the \(\mathrm{CH}_{3}\mathrm{I}\) leaves, forming \(\left(\mathrm{CH}_{3}\right)_{3}\mathrm{CCH}_{2}\text{-OH}\) as the product, along with \(\mathrm{CH}_{3}\mathrm{I}\).

Determine Products for Compound (d)

For compound (d), \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{COCH}_{3}\), preferential cleavage occurs at the \(\mathrm{OCH}_{3}\) because the tert-butyl carbocation formed as an intermediate is very stable. Thus, the products are \(\left(\mathrm{CH}_{3}\right)_{3}\mathrm{C}\text{-I}\) (tert-butyl iodide) and \(\mathrm{CH}_{3}\mathrm{OH}\) (methanol).

Key concepts

Nucleophilic Substitution

Nucleophilic substitution is a fundamental reaction in organic chemistry where a nucleophile replaces a leaving group on a carbon atom. In the context of ether cleavage with hydroiodic acid (HI), this process typically follows one of two primary mechanisms:

• S_N1 mechanism involves a carbocation intermediate and is favored in tertiary systems where carbocation stability is high.
• S_N2 mechanism occurs when a nucleophile attacks the electrophilic carbon directly, leading to bond cleavage. This is more common in primary or less hindered systems.

When ethers are subjected to HI, the iodide ion (I^-) acts as the nucleophile, attacking the ether's carbon-oxygen bond. Simultaneously, the hydrogen ion (H⁺) protonates the oxygen, promoting the cleavage of the C-O bond. Whether the reaction proceeds via the S_N1 or S_N2 route depends on factors such as the nature of the carbon attached to the oxygen and steric hindrance.

Carbocation Stability

Understanding carbocation stability is crucial when predicting the outcome of electrophilic and nucleophilic reactions in organic chemistry, including ether cleavage. A carbocation is a positively charged carbon atom, and its stability greatly influences reaction pathways. The order of carbocation stability is typically:

• Tertiary (3°) carbocations are the most stable, due to hyperconjugation and the inductive effects of surrounding alkyl groups.
• Secondary (2°) carbocations are stable but less so than tertiary ones.
• Primary (1°) carbocations are the least stable and are rarely formed during reactions.

In the ether cleavage process, if a carbocation intermediate can form, the reaction is more likely to proceed via the S_N1 mechanism, particularly if a tertiary carbocation can be formed, as seen in compound (d). Here, the tert-butyl carbocation's stability facilitates the reaction.

Hydroiodic Acid Reaction

The reaction with hydroiodic acid (HI) is a classical method for the cleavage of ethers. HI is a strong acid that dissociates into H⁺ and I^- ions in solution.

• The hydrogen ion (H⁺) is responsible for protonating the ether oxygen, enhancing its leaving group's ability.
• The iodide ion (I^-) acts as the nucleophile, facilitating the cleavage of the C-O bond and forming an alkyl iodide.

Consider the example of compound (a). The HI attacks the less hindered methoxy group ( O- CH₃), resulting in the formation of allyl alcohol and methyl iodide. This cleavage mechanism may vary depending on the steric and electronic nature of the ethers involved.

Organic Chemistry Reactions

Organic chemistry involves numerous reactions, each with distinct mechanisms and pathways. Understanding ether cleavage in the context of organic chemistry reactions highlights essential principles like nucleophilic substitution and reaction conditions. These reactions are foundational in understanding more complex organic synthesis.

• Ether cleavage demonstrates the interplay of charge, reactivity, and reaction conditions in organic processes.
• Each reaction typically involves analyzing variables such as substituents, leaving groups, and the functional group environment to predict the products accurately.

In ether cleavage, we particularly explore how different structural elements and conditions, such as in the presence of hydroiodic acid, lead to varying products, emphasizing the broader applicability of organic reaction principles.