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In the following reaction,
Ethyl alcohol $+\mathrm{H}^{+} \quad \frac{413 \mathrm{~K}}{\longrightarrow}$
$\begin{array}{ll}\text { (in excess) } & \text { (from } \mathrm{H}_{2} \mathrm{SO}_{4} \text { ) }\end{array}$
product, the product is
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Ethyl alcohol $+\mathrm{H}^{+} \quad \frac{413 \mathrm{~K}}{\longrightarrow}$
$\begin{array}{ll}\text { (in excess) } & \text { (from } \mathrm{H}_{2} \mathrm{SO}_{4} \text { ) }\end{array}$
product, the product is
Solution:
2480 Upvotes
Verified Answer
The correct answer is:
diethyl ether
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