$\begin{array}{c}{\text{CH}}_{\text{3}}\\ \text{|\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\\ {\text{H}}_{\text{3}}{\text{C-C-O-CH}}_{\text{3}}\\ \text{|\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\\ {\text{CH}}_{\text{3}}\end{array}+\text{HI}\stackrel{573K}{\to }$

Correct statements about the given reaction is/are

tert-butyl alcohol & methyl iodide as the products.

It follows ${\text{S}}_{{\text{N}}^{\text{1}}}$mechanism

It requires polar protic solvents

tert- butyl iodide and methyl alcohol as the products.