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Reduction of nitrobenzene ( C₆H₅NO₂ ) with Sn/HCl forms anilinium chloride, which upon neutralization with a base yields aniline ( C₆H₅NH₂ ). This follows from NCERT principle where relation explains outcome clearly for students.
Ref: NCERT Chemistry Textbook for Class XI and XII, Chapter: Hydrocarbons, Topic: Aromatic hydrocarbons, benzene as example, Hückel rule and aromaticity.
Reduction of nitropropane (CH₃CH₂CH₂NO₂) with Sn/HCl yields propan-1-amine (CH₃CH₂CH₂NH₂), as per the PDF’s preparation methods.
Ref: NCERT Class 12 Chemistry > Chapter 13: Amines > Topic: Chemical Reactions of Amines - Mixed Concept Practice
Reduction of nitrobenzene (C₆H₅NO₂) with Sn/HCl yields aniline (C₆H₅NH₂), as per the PDF’s preparation methods.
Ref: NCERT Class 12 Chemistry > Chapter 13: Amines > Topic: Chemical Reactions of Amines - Mixed Concept Practice
Reduction of benzyl nitro compound (C₆H₅CH₂NO₂) with Sn/HCl yields benzylamine (C₆H₅CH₂NH₂), as per the PDF’s preparation methods.
Ref: NCERT Class 12 Chemistry > Chapter 13: Amines > Topic: Chemical Reactions of Amines - Mixed Concept Practice
Reduction of 1-nitropropane (CH₃CH₂CH₂NO₂) with Sn/HCl yields propan-1-amine (CH₃CH₂CH₂NH₂), as per the PDF’s preparation methods.
Ref: NCERT Class 12 Chemistry > Chapter 13: Amines > Topic: Chemical Reactions of Amines - Mixed Concept Practice
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