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Ethyl bromide is industrially prepared from

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Ethyl alcohol $+ HBr$

Ethanol+ ${Br}_{2}$

Alcohol + HBr

None of these

answer is A.

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Detailed Solution

Ethyl bromide prepared from ethyl alcohol by refluxing alcohol through constant boiling with HBr (48%) in presence of little concentrated sulphuric acid.
$C_{2} H_{5} OH + HBr \frac{H_{2} {SO}_{4} (conc.)}{reflux} C_{2} H_{5} Br + H_{2} O$

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