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CH(3) - O - CH(2)CH(3) overset(HI//Delta...

`CH_(3) - O - CH_(2)CH_(3) overset(HI//Delta)underset("Excess")(to)( A) + (B)`
Product (A) and (B ) are

A

`CH_(3)OH + CH_(3) CH_(2)I`

B

`CH_(3)I +CH_(3) CH_(2) OH`

C

`CH_(3) I + CH_(3)CH_(2)I`

D

`CH_(3)OH +CH_(3)CH_(2) OH`

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem of what products (A) and (B) are formed when CH₃OCH₂CH₃ is treated with excess HI under heating, we can follow these steps: ### Step 1: Identify the Reactants The reactant is an ether, specifically ethyl methyl ether (CH₃OCH₂CH₃). ### Step 2: Understand the Reaction Conditions The ether is treated with excess hydrogen iodide (HI) and heated. HI is a strong acid that can cleave ethers. ### Step 3: Mechanism of Ether Cleavage When the ether is treated with HI, the oxygen atom in the ether can abstract a proton (H⁺) from HI, resulting in the formation of an oxonium ion (CH₃OH⁺CH₂CH₃). This step makes the ether more reactive. ### Step 4: Nucleophilic Attack The iodide ion (I⁻), which is generated from the dissociation of HI, can now act as a nucleophile. It can attack one of the carbon atoms adjacent to the oxygen atom in the oxonium ion. ### Step 5: Formation of Products 1. The iodide ion can attack the methyl group (CH₃) leading to the formation of methyl iodide (CH₃I) and ethanol (CH₃CH₂OH). 2. Alternatively, the iodide ion can attack the ethyl group (CH₂CH₃), leading to the formation of ethyl iodide (CH₃CH₂I) and methanol (CH₃OH). ### Step 6: Excess HI Effect Since HI is in excess, the alcohols formed (methanol and ethanol) can further react with HI: - Methanol (CH₃OH) can react with HI to form methyl iodide (CH₃I). - Ethanol (CH₃CH₂OH) can react with HI to form ethyl iodide (CH₃CH₂I). ### Step 7: Final Products The final products after all reactions are: - Product A: Methyl iodide (CH₃I) - Product B: Ethyl iodide (CH₃CH₂I) ### Conclusion Thus, the products (A) and (B) are: - A = CH₃I (Methyl iodide) - B = CH₃CH₂I (Ethyl iodide) ### Answer The correct answer is option C: CH₃I and CH₃CH₂I. ---
To solve the problem of what products (A) and (B) are formed when CH₃OCH₂CH₃ is treated with excess HI under heating, we can follow these steps: ### Step 1: Identify the Reactants The reactant is an ether, specifically ethyl methyl ether (CH₃OCH₂CH₃). ### Step 2: Understand the Reaction Conditions The ether is treated with excess hydrogen iodide (HI) and heated. HI is a strong acid that can cleave ethers. ...
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