In the cleavage mechanism of the unsymmetrical ether methoxyethane () using exactly one equivalent o — Alcohols Phenols and Ethers Chemistry Question
Question
In the $S_N2$ cleavage mechanism of the unsymmetrical ether methoxyethane ($CH_3-O-CH_2CH_3$) using exactly one equivalent of cold $HI$, the iodide nucleophile selectively attacks the less sterically hindered carbon atom. Exactly how many carbon atoms are present in the resulting alkyl halide product molecule?
Answer: 1
💡 Solution & Explanation
The oxonium intermediate is $CH_3-O^+(H)-CH_2CH_3$. The bulky $I^-$ nucleophile attacks the least hindered alkyl group via $S_N2$. The methyl group ($CH_3$) is less hindered than the ethyl group ($CH_2CH_3$). Therefore, the attack occurs at the methyl carbon, cleaving the bond to form methyl iodide ($CH_3I$) and ethanol. Methyl iodide contains exactly 1 carbon atom.
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