How many of the following ethers cannot be prepared by Williamson’s synthesis?
$C H_{3} O C H_{2} C H_{3}, C_{6} H_{5} O C H_{3}, {(C_{6} H_{5})}_{2} O, {(C H_{3})}_{3} C O C H_{3,} C_{6} H_{5} C H_{2} O C H_{3}, {(C_{2} H_{5})}_{2} O, C_{6} H_{5} C H_{2} O C_{6} H_{5}, (C H_{3}) C H_{2} O C H_{2} C H_{3}, {(C H_{3})}_{3} C - O - C {(C H_{3})}_{3}$

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answer is 2.

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Detailed Solution

Williamson’s synthesis requires that the alkyl halide should be $1^{\circ}$ and alkoxide may be $1^{\circ}, 2^{\circ}$ or $3^{\circ} .$ Thus, the two ethers which cannot be prepared by Williamson’s synthesis are: ${(C_{6} H_{5})}_{2} O, {(C H_{3})}_{3} C - O - C {(C H_{3})}_{3}$