In the cleavage mechanism of the unsymmetrical ether methoxyethane () using exactly one equivalent of cold , the iodide nucleophile selectively attacks the less sterically hindered carbon atom. Exactl

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?

Prashant Kotian · Accepted Answer

Correct answer: 1. The oxonium intermediate is . The bulky nucleophile attacks the least hindered alkyl group via . The methyl group () is less hindered than the ethyl group (). Therefore, the attack occurs at the methyl carbon, cleaving the bond to form methyl iodide () and ethanol. Methyl iodide contains exactly 1 carbon atom.

💡 Solution & Explanation

Practice 22,000+ questions like this