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Question:

(CH3)3COCH3 and CH3OC2H5 are treated with hydroiodic acid. The fragments obtained after reactions are respectively:

(CH3)3CI+CH3OH;CH3OH+C2H5I

(CH3)3COH+CH3I;CH3OH+C2H5I

(CH3)3CI+CH3OH;CH3I+C2H5OH

CH3I+(CH3)3COH;CH3I+C2H5OH

Solution:

When mixed ethers are used, the formation of alkyl iodide depends on the nature of alkyl groups. Methyl iodide is formed when one group is methyl and the other a primary or secondary alkyl group. Here reaction follows SN2 mechanism and because of the steric effect of the larger group, I⁻ attacks the smaller (methyl) group.
CH3OC2H5 + HI → CH3I + C2H5OH