The reaction of 1-bromobutane with sodium azide followed by reaction with which reagent(s) affords n-butylamine?Select answer from the options belowFe, H3O+Fe, HClexcess NaOHexcess LiAlH4 followed by H2O
Question
The reaction of 1-bromobutane with sodium azide followed by reaction with which reagent(s) affords n-butylamine?Select answer from the options belowFe, H3O+Fe, HClexcess NaOHexcess LiAlH4 followed by H2O
Solution
The correct answer is Fe, H3O+.
Here's the step-by-step explanation:
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The reaction of 1-bromobutane with sodium azide (NaN3) will replace the bromine atom with an azide group (N3), resulting in butyl azide.
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The butyl azide is then reacted with iron (Fe) and hydronium (H3O+). This is a reduction reaction where the azide group (N3) is reduced to an amine group (NH2), resulting in n-butylamine.
So, the reagent that affords n-butylamine after the reaction of 1-bromobutane with sodium azide is Fe, H3O+.
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