Question

In the presence of samarium iodide, the halogen atom of alkyl, alkenyl, and aryl halides is​

Answer 1

Answer:

In the presence of samarium iodide, the halogen atom of alkyl, alkenyl, and aryl halides is replaced by hydrogen. In tetrahydrofuran, the mechanism of alkyl halide reduction likely proceeds through radical intermediates. ... Protonation of this species then yields the reduced product.

Explanation:

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Answer 2

Explanation:

Reductions with samarium(II) iodide involve the conversion of various classes of organic compounds into reduced products through the action of samarium(II) iodide, a mild one-electron reducing agent.[1][2][3]

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