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The reaction given below is known as
$C_{2} H_{5} ONa + {IC}_{2} H_{5} ⟶ C_{2} H_{5} {OC}_{2} H_{5} + NaI$

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Kolbe synthesis

Wurtz synthesis

Williamson synthesis

Grignard synthesis

answer is C.

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Detailed Solution

$C_{2} H_{5} ONa + {IC}_{2} H_{5} \to C_{2} H_{5} {OC}_{2} H_{5} + NaI$

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