#amine alkylation

CHEMISTRY

What is the product when CH₃CH₂NHCH₂CH₃ reacts with excess CH₃I followed by NaOH ?

N-Ethylethanamine with excess CH₃I forms a quaternary salt ( CH₃CH₂N(CH₃)2CH₂CH₃+I- ), which with NaOH yields CH₃CH₂N(CH₃)CH₂CH₃ via elimination. This follows from NCERT principle where relation explains outcome clearly for students.

Ref: NCERT Chemistry Textbook for Class XI and XII, Chapter: Some Basic Concepts, Structure of Atom, Periodicity and relevant Chemistry topic, Topic: Mole concept and periodic trends.

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CHEMISTRY

What is the product when CH₃CH₂NH₂ reacts with excess CH₃CH₂I followed by AgOH ?

Ethanamine with excess CH₃CH₂I forms a quaternary salt ( (CH₃CH₂)4N+I- ), which with AgOH yields (CH₃CH₂)3N via elimination of CH₃CH₂I . This follows from NCERT principle where relation explains outcome clearly for students.

Ref: NCERT Chemistry Textbook for Class XI and XII, Chapter: Chemical Bonding and Molecular Structure, Topic: Covalent bonds in CH₃CH₂NH₂, counting bonds, Lewis structure.

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Nomenclature, Structure and Preparation - Ammonolysis and Gabriel Synthesis

Which amine produces a tertiary amine when treated with excess CH₃CH₂Cl?

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

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Nomenclature, Structure and Preparation - Ammonolysis and Gabriel Synthesis

The product of C₆H₅NH₂ reacting with CH₃CH₂Cl in the presence of a base is:

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

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Nomenclature, Structure and Preparation - Ammonolysis and Gabriel Synthesis

The product of CH₃CH₂CH₂NH₂ with CH₃CH₂Cl in the presence of a base is:

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

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Nomenclature, Structure and Preparation - Ammonolysis and Gabriel Synthesis

The product of CH₃CH₂NH₂ reacting with excess CH₃Cl is:

Ethylamine (CH₃CH₂NH₂) with excess CH₃Cl undergoes alkylation to form CH₃CH₂N(CH₃)₂ (N,N-dimethylethylamine), a tertiary amine, as per the PDF.

Ref: NCERT Class 12 Chemistry > Chapter 13: Amines > Topic: Nomenclature, Structure and Preparation - Ammonolysis and Gabriel Synthesis

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Nomenclature, Structure and Preparation - Ammonolysis and Gabriel Synthesis

The reaction of CH₃CH₂NH₂ with excess CH₃Cl yields:

Ethylamine (CH₃CH₂NH₂) with excess CH₃Cl forms N,N-dimethylethylamine (CH₃CH₂N(CH₃)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

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Nomenclature, Structure and Preparation - Ammonolysis and Gabriel Synthesis

An amine X reacts with CH₃CH₂Cl to form a tertiary amine. X could be:

A secondary amine like CH₃NHCH₃ reacts with CH₃CH₂Cl to add an ethyl group, forming a tertiary amine (CH₃CH₂N(CH₃)₂), as per the PDF’s alkylation.

Ref: NCERT Class 12 Chemistry > Chapter 13: Amines > Topic: Nomenclature, Structure and Preparation - Ammonolysis and Gabriel Synthesis

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