Question

Treatment of 1-ethoxybutane with HI leads to both butyl iodide and ethyl iodide, along with the related alcohols. By contrast, when tert-butyl ethyl ether is treated the same way, only ethyl iodide and tert-butyl alcohol are formed. tert-Butyl iodide and ethyl alcohol are not produced. Explain. $$ \begin{gathered} \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2}-\mathrm{O}-\mathrm{CH}_{2} \mathrm{CH}_{3} \\ \text { | } \mathrm{HI} \\ \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{I}+\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{I} \\ \text { and } \\ \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OH}+\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH} \end{gathered} $$ $$ \left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}-\mathrm{O}-\mathrm{CH}_{2} \mathrm{CH}_{3} \stackrel{\mathrm{HI}}{\longrightarrow}\left(\mathrm{CH}_{3}\right)_{3} \mathrm{COH}+\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{I} $$

Short answer

1-Ethoxybutane forms butyl iodide and ethyl iodide due to possible cleavages. tert-Butyl ethyl ether only forms tert-butyl alcohol and ethyl iodide, favoring S_N1 mechanism with a stable carbocation intermediate.

Step-by-step solution

- Understand the Reactants

Identify the structures and names of the compounds involved. We are dealing with 1-ethoxybutane \( \text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{-O-}\text{CH}_2\text{CH}_3 \) and tert-butyl ethyl ether \( (\text{CH}_3)_3\text{C-O-}\text{CH}_2\text{CH}_3 \).

- Understand the Products

For 1-ethoxybutane with HI, we get butyl iodide and ethyl iodide \( \text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{I} \) and \( \text{CH}_3\text{CH}_2\text{I} \), along with butyl alcohol and ethyl alcohol \( \text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{OH} \) and \( \text{CH}_3\text{CH}_2\text{OH} \). For tert-butyl ethyl ether treated with HI, we get tert-butyl alcohol \( (\text{CH}_3)_3\text{COH} \) and ethyl iodide \( \text{CH}_3\text{CH}_2\text{I} \), with no tert-butyl iodide or ethyl alcohol.

- Analyze the Mechanism for 1-Ethoxybutane

This reaction proceeds through an S_N1 or S_N2 mechanism. Both butyl iodide and ethyl iodide form because the ethoxy group can cleave at either side, both forming stable cation/anion intermediates. This allows both side-products to form.

- Analyze the Mechanism for tert-Butyl Ethyl Ether

Tert-butyl group forms a stable carbocation when leaving, making S_N1 mechanism best. The formation of the tert-butyl carbocation \( (\text{CH}_3)_3\text{C}^+ \) is favored, leading to tert-butyl alcohol instead of tert-butyl iodide.

- Summarize Observations

In the case of tert-butyl ethyl ether, the reaction selectively favors the S_N1 mechanism due to the stability of the tert-butyl carbocation. Thus, no tert-butyl iodide or ethyl alcohol, only tert-butyl alcohol and ethyl iodide.

Key concepts

ether cleavage

Ether cleavage is the breaking of an ether molecule into two separate compounds when treated with hydroiodic acid (HI). In our exercise, we see two examples: 1-ethoxybutane and tert-butyl ethyl ether. When HI reacts with 1-ethoxybutane, the ether bond can break in two different places, leading to the formation of two sets of products: butyl iodide and ethyl iodide, as well as their corresponding alcohols. However, in tert-butyl ethyl ether, only ethyl iodide and tert-butyl alcohol are produced without the formation of tert-butyl iodide or ethyl alcohol. This difference is rooted in the mechanisms and stability of intermediates formed during the reaction.

S_N1 and S_N2 mechanisms

The terms S_N1 and S_N2 refer to different types of organic reaction mechanisms for nucleophilic substitution.

• S_N1 Mechanism: This is a two-step process where the leaving group departs first, forming a carbocation intermediate. It proceeds via a unimolecular rate-determining step.


• S_N2 Mechanism: This is a one-step process where the nucleophile attacks the carbon atom from the opposite side of the leaving group, leading to a simultaneous bond formation and bond breaking step. It proceeds via a bimolecular rate-determining step.

Understanding these mechanisms helps explain why different products form under seemingly similar conditions.

carbocation stability

Carbocation stability plays a crucial role in determining the pathway and products of ether cleavage. Carbocations are positively charged carbon atoms with three bonds and no lone pairs. Their stability depends on the number of alkyl groups attached:

• Primary (1°) Carbocation: Least stable with only one alkyl group attached.
• Secondary (2°) Carbocation: More stable with two alkyl groups attached.
• Tertiary (3°) Carbocation: Most stable with three alkyl groups attached.

In our case, the tert-butyl group forms a stable tertiary carbocation, favoring the S_N1 mechanism and thus leading to the production of tert-butyl alcohol.

product distribution

The products of ether cleavage depend on the stability of the intermediates formed during the reaction. For 1-ethoxybutane, both butyl iodide and ethyl iodide are produced because the cleavage can occur at either side of the ether bond, leading to both primary and secondary cation/anion intermediates. The stability difference is not significant, allowing both products to form in similar amounts.
However, for tert-butyl ethyl ether, only ethyl iodide and tert-butyl alcohol are produced. The stable tert-butyl carbocation prefers to form, preventing the formation of tert-butyl iodide.

organic reaction mechanisms

Understanding the organic reaction mechanisms helps in predicting and explaining the outcome of chemical reactions. In our cases:

• 1-Ethoxybutane: The reaction can proceed through either S_N1 or S_N2 mechanisms, forming a mix of products.
• Tert-butyl ethyl ether: The reaction predominantly follows the S_N1 pathway due to the formation of a stable tertiary carbocation, directing the product formation towards tert-butyl alcohol and ethyl iodide.

By analyzing the intermediate states, we can see why certain products form over others, illustrating the importance of mechanisms in organic chemistry.