$ROH \to_{low temp}^{Conc. H_{2} {SO}_{4}} R - O - R$
The ether that can be obtained in best yield is

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$C_{2} H_{5} - O - C_{2} H_{5}$

${({CH}_{3})}_{3} C - O - C {({CH}_{3})}_{3}$

${CH}_{3} - O - C_{2} H_{5}$

${({CH}_{3})}_{2} CH - O - CH {({CH}_{3})}_{2}$

answer is D.

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Detailed Solution

The mechanism is $S_{N} 2$ Symmetrical ethers containing 10 alkyl groups can be prepared in good yields. Mixed ethers are not obtained in good yields.

3⁰ alcohols & 2⁰ alcohols give elimination products.

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