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Ammonolysis of CH₃CH₂Cl with excess NH₃ favors the formation of the primary amine CH₃CH₂NH₂ (ethylamine), as per the PDF’s preparation methods.
Ref: NCERT Class 12 Chemistry > Chapter 13: Amines > Topic: Reaction with Nitrous Acid and Diazonium Salts - Part 2
A primary amine like CH₃NH₂ can be dialkylated with excess CH₃CH₂Cl to form CH₃N(CH₂CH₃)₂, a tertiary amine, as per the PDF’s alkylation.
Ref: NCERT Class 12 Chemistry > Chapter 13: Amines > Topic: Nomenclature, Structure and Preparation - Ammonolysis and Gabriel Synthesis
Aniline (C₆H₅NH₂) can be alkylated with CH₃CH₂Cl and a base to form N-ethylaniline (C₆H₅NHCH₂CH₃), as per the PDF’s alkylation.
Ref: NCERT Class 12 Chemistry > Chapter 13: Amines > Topic: Nomenclature, Structure and Preparation - Ammonolysis and Gabriel Synthesis
Propan-1-amine (CH₃CH₂CH₂NH₂) reacts with CH₃CH₂Cl and a base to form N-ethylpropan-1-amine (CH₃CH₂CH₂NHCH₂CH₃), as per the PDF’s alkylation.
Ref: NCERT Class 12 Chemistry > Chapter 13: Amines > Topic: Nomenclature, Structure and Preparation - Ammonolysis and Gabriel Synthesis
This is a Wurtz reaction variant. CH₃CH₂Cl and CH₃Cl couple with Na to form CH₃CH₂CH₃ (propane) as one product.
Ref: NCERT Class 12 Chemistry > Chapter 6: Haloalkanes and Haloarenes > Topic: Elimination Reactions and Reaction with Metals and Other Reactions
This is a Wurtz reaction, where two alkyl halides couple to form an alkane. CH₃CH₂Cl forms CH₃CH₂CH₂CH₃ (butane).
Ref: NCERT Class 12 Chemistry > Chapter 6: Haloalkanes and Haloarenes > Topic: Classification and Nomenclature and Isomerism of Haloalkanes
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