Question

\( (CH_3)_3C - OC_2H_5 \) on reaction with HI gives

• A. \( (CH_3)_3C - I \) and \( C_2H_5 - I \)
• B. \( (CH_3)_3C - OH \) and \( C_2H_5 - I \)
• C. \( (CH_3)_3C - I \) and \( C_2H_5 - OH \)
• D. \( (CH_3)_3C - OH \) and \( C_2H_5 - OH \)

Hint:

For primary/secondary alkyl ethers, the reaction follows \( S_N2 \) and iodide goes to the smaller alkyl group. For tertiary alkyl ethers, it follows \( S_N1 \) and iodide goes to the tertiary group.

Answer 1

The Correct Option is C

Step 1: Conceptual Understanding:
The reaction of ethers with hydrogen iodide (HI) is influenced by the nature of the alkyl groups attached to the oxygen atom in the ether. When one of the alkyl groups is tertiary (\( 3^\circ \)), the reaction follows an \( S_N1 \) mechanism. This is because a tertiary carbocation is more stable and can easily form during the reaction, facilitating the substitution process.
Step 2: Detailed Explanation:
In this case, the ether is tert-butyl ethyl ether. Upon reaction with HI, the ether undergoes protonation. The structure of the protonated molecule is \( (CH_3)_3C-O^+(H)-C_2H_5 \), where the oxygen atom is positively charged after accepting a proton from HI.
Due to the presence of the tert-butyl group, which is a highly stable \( 3^\circ \) carbocation, the C-O bond breaks via the \( S_N1 \) mechanism. The alkyl group \( (CH_3)_3C \) can stabilize the positive charge formed on the carbon atom, leading to the formation of the tert-butyl carbocation \( (CH_3)_3C^+ \). This carbocation is relatively stable due to inductive and hyperconjugation effects provided by the surrounding methyl groups.
After the carbocation formation, the iodide ion (\( I^- \)) attacks the stable carbocation, replacing the oxygen atom and forming tert-butyl iodide (\( (CH_3)_3C-I \)). In addition to this, ethanol (\( C_2H_5OH \)) is formed as the other product, coming from the ethyl group that was originally bonded to oxygen.
Step 3: Final Answer:
The products of the reaction are tert-butyl iodide (\( (CH_3)_3C-I \)) and ethanol (\( C_2H_5OH \)). These products result from the \( S_N1 \) mechanism where a stable tertiary carbocation is formed, leading to the substitution of the oxygen atom by an iodide ion.