The reaction proceeds via an \( S_N1 \) mechanism where the chlorine group leaves, forming a carbocation.
After the formation of the carbocation, a 1,2-hydride shift occurs to stabilize the carbocation, resulting in a more stable tertiary carbocation.
Subsequent attack by hydroxide ion (\( \text{OH}^- \)) leads to the formation of the major product:
The major product is a tertiary alcohol.
20 mL of sodium iodide solution gave 4.74 g silver iodide when treated with excess of silver nitrate solution. The molarity of the sodium iodide solution is _____ M. (Nearest Integer value) (Given : Na = 23, I = 127, Ag = 108, N = 14, O = 16 g mol$^{-1}$)