Product (A) and (B) in above reaction is :

O^{-} - O | | S - O - H, O^{-} - O | | S - O - C H_{3}

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O^{-} - O | | S - O - H, O^{-} - O_{⨁} | O^{-} - O - C H_{3}

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O^{-} - O | | S - O - C H_{3}, O^{⊖} - O_{⨁} | C H_{3} - H

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O^{-} - O_{⨁} | H - O, O^{-} - O | | S | | C H_{3} - O^{-}

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Solution

The correct option is A $O^{-} - O | | S - O - H, O^{-} - O | | S - O - C H_{3}$
$O^{-}$ act as a strong nucleophilic. Upon reaction with methyl iodide,it gives substitution product.And it get proportional in acidic condition.
A is the correct answer.

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Similar questions

(x)

C H_{3} - O | | S | | - O O - H

(y)

C H_{3} - O | | C - O - H

(z)

C H_{3} - O H

p K_{a}

C H_{3} - O | | C - O - C H_{2} - C H_{3} + H - O \to (O = O_{18})

C H_{3} - O | | C - C H_{3} (A), C H_{3} - O | | C - H (B), C H_{3} - O | | C - C H_{2} - C l (C), H - O | | C - H (D), C l - C H_{2} - O | | C - H (E)

^{3}

C F_{3} - O | | S | | O - O^{⊖}

C_{6} H_{5} - O | | S | | O - O^{⊖}

C_{6} H_{5} - O^{⊖}

C H_{3} - O | | C - O^{⊖}

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