Step 1: Understanding the reduction of ketoxime.
Ketoxime (R\(_2\)C=NOH) can be reduced by sodium in ethanol to form a primary amine (R\(_2\)NH). Sodium reduces the oxime group (-NOH) to an amine group (-NH\(_2\)).
Step 2: Analyzing the options. (A) 1\(^\circ\) amine: Correct — The reduction of ketoxime with sodium in ethanol forms a primary amine. (B) 2\(^\circ\) amine: This is incorrect. A secondary amine is not formed in this reaction. (C) 1\(^\circ\) and 2\(^\circ\) amine: This is incorrect. Only a primary amine is formed. (D) 3\(^\circ\) amine: This is incorrect. A tertiary amine is not formed in this reaction.
Step 3: Conclusion.
The correct product is a 1\(^\circ\) amine, so the answer is (A) 1\(^\circ\) amine.