Question

t-butyl methyl ether on reaction with anhydrous $HI$ gives an alcohol (A) and halide (B). Which of the following statement is correct?

• A. Alcohol (A) performs Lucas test immediately at room temperature
• B. Halide (B) undergoes $S_{N}1$ reaction most readily
• C. Alcohol (A) on heating with copper gives formaldehyde
• D. Halide (B) on reaction with $KCN$ undergoes $S_{N}2$ reaction

Answer 1

Answer: (B), (C)

The correct options are
B Alcohol (A) on heating with copper gives formaldehyde
C Halide (B) undergoes $S_{N}1$ reaction most readily

Since the tertiary carbocation formed which is more stable so reaction will follow $S{N}^1$ mechanism.

So the tertiary butyl iodide and methnol will formed as product.

Methanol will oxidised to give formaldehyde.

Options B and C are correct.