Question 36 of 39intermediate💡 UnderstandLong Answer3 marks

Write the mechanism of the reaction of HI with methoxymethane.

Correct Answer

The reaction of methoxymethane with HI proceeds through the following mechanism:

Step 1: Protonation of ether The reaction starts with protonation of the ether molecule. Methoxymethane (CH₃-O-CH₃) reacts with H⁺ from HI to form a protonated oxonium ion. CH₃-O-CH₃ + H⁺ ⇌ CH₃-O⁺H-CH₃

Step 2: Nucleophilic attack (S_N2 mechanism) Iodide ion (I⁻), being a good nucleophile, attacks the methyl carbon of the oxonium ion and displaces methanol. Since both alkyl groups are methyl (primary), the reaction follows S_N2 mechanism. I⁻ + CH₃-O⁺H-CH₃ → CH₃I + CH₃OH

Step 3: Further reaction (with excess HI) When HI is in excess and the reaction is carried out at high temperature, methanol reacts with another molecule of HI to form methyl iodide. CH₃OH + HI → CH₃I + H₂O

Exercise: EXERCISES | Q: 7.30 | (Page Page 32)
For More Understanding

Explanation

The context describes the mechanism of ether cleavage by HI. For methoxymethane (dimethyl ether), both alkyl groups are identical methyl groups, so the product distribution is straightforward. The mechanism involves: (1) protonation of ether oxygen to form oxonium ion, (2) S_N2 attack by iodide on the methyl carbon to give CH₃I and CH₃OH, and (3) with excess HI, methanol further converts to methyl iodide. This follows the general pattern described in the context where primary alkyl groups undergo S_N2 cleavage.

Solution Steps

  1. Step 1: Protonation of methoxymethane by H⁺ from HI to form oxonium ion

  2. Step 2: Nucleophilic attack by I⁻ via S_N2 mechanism to form CH₃I and CH₃OH

  3. Step 3: With excess HI, methanol converts to CH₃I and H₂O

PreviousNext