Question #2fdbe

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Question #2fdbe

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Answer 1 · 2017-09-22T00:11:10

Add $N H_{3}$ $NH_"3"$ under alcoholic conditions.

Explanation:

The molecular formula for Ethyl Chloride is $C H_{3} C H_{2} C l$ $CH_"3"CH_"2"Cl$

To convert a haloalkane to a amine you add $N H_{3}$ $NH_"3"$ under alcoholic conditions.

So in this reaction $N H_{2}$ $NH_"2"$ is substituted for the halogen ( $C l$ $Cl$ ), as the Chlorine is removed it reacts with the Hydrogen that was lost from the $N H_{3}$ $NH_"3"$ , this results in a $H C l$ $HCl$ being produced.

The resulting reaction is:

$C H_{3} C H_{2} C l + N H_{3} \to C H_{3} C H_{2} N H_{2} + H C l$ $CH_"3"CH_"2"Cl + NH_"3" -> CH_"3"CH_"2"NH_"2" + HCl$

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Question #2fdbe

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