Correct method to prepare $M e_{3} C - O - M e$ in good yield is/are:

M e_{3} C - C l + M e O N a ⟶

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M e_{2} C = C H_{2} (i) H g (O A C)_{2} + M e O H - ------------- \to (i i) N a B H_{4}

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M e_{3} C - O N a + M e C l ⟶

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M e_{2} C = C H_{2} H^{⨁} - --- \to M e O H

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Solution

The correct option is D $M e_{2} C = C H_{2} H^{⨁} - --- \to M e O H$
(a) $3^{o}$ halide with strong base favours elimination reaction over substitution.

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