When an amide is reduced with sodium/alcohol, it yields :

primary amine

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an alcohol

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secondary amine

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aldoxime

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Solution

The correct option is D primary amine
When an amide is reduced with sodium/alcohol, it yields primary amine.
$R C O N H_{2} N a / e t h a n o l - ------ \to R C H_{2} N H_{2} + H_{2} O$

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