11 public questions tagged with this topic.
Ethylamine (CH₃CH₂NH₂) forms ethyl isocyanide (CH₃CH₂NC) in the carbylamine reaction, as per the PDF’s chemical properties.
Ref: NCERT Class 12 Chemistry > Chapter 13: Amines > Topic: Nomenclature, Structure and Preparation - Ammonolysis and Gabriel Synthesis
Propan-1-amine (CH₃CH₂CH₂NH₂) forms propyl isocyanide (CH₃CH₂CH₂NC) in the carbylamine reaction, as per the PDF.
Ref: NCERT Class 12 Chemistry > Chapter 13: Amines > Topic: Nomenclature, Structure and Preparation - Ammonolysis and Gabriel Synthesis
Elimination (E₂) with alcoholic KOH favors tertiary halides due to stable alkene formation. (CH₃)3CBr reacts fastest.
Ref: NCERT Class 12 Chemistry > Chapter 6: Haloalkanes and Haloarenes > Topic: Polyhalogen Compounds - Chloroform Iodoform DDT Freons
Elimination (E₂) of CH₃CH₂CHBrCH₂CH₂CH₃ with alcoholic KOH forms CH₃CH₂CH=CHCH₂CH₃ (3-hexene) per Zaitsev's rule.
Ref: NCERT Class 12 Chemistry > Chapter 6: Haloalkanes and Haloarenes > Topic: Polyhalogen Compounds - Chloroform Iodoform DDT Freons
Elimination (E₂) of (CH₃)2CHCH₂CH₂Br with alcoholic KOH forms (CH₃)2C=CHCH₃ (2-methyl-2-butene) per Zaitsev's rule.
Ref: NCERT Class 12 Chemistry > Chapter 6: Haloalkanes and Haloarenes > Topic: Elimination Reactions and Reaction with Metals and Other Reactions
Elimination (E₂) of CH₃CH₂CH₂CH₂CH₂Br (1-bromopentane) with alcoholic KOH forms 1-pentene ( CH₃CH₂CH₂CH=CH₂ ).
Ref: NCERT Class 12 Chemistry > Chapter 6: Haloalkanes and Haloarenes > Topic: Elimination Reactions and Reaction with Metals and Other Reactions
Alcoholic KOH promotes elimination (E₂) in (CH₃)2CHCH₂Br , forming (CH₃)2C=CH₂ (2-methylpropene).
Ref: NCERT Class 12 Chemistry > Chapter 6: Haloalkanes and Haloarenes > Topic: Elimination Reactions and Reaction with Metals and Other Reactions
Primary halides undergo elimination (E₂) slowest due to less stable transition states. CH₃CH₂CH₂Cl reacts slowest among the options.
Ref: NCERT Class 12 Chemistry > Chapter 6: Haloalkanes and Haloarenes > Topic: Elimination Reactions and Reaction with Metals and Other Reactions
Alcoholic KOH promotes elimination (E₂) in CH₃CH₂CH₂CHBrCH₃ , forming CH₃CH₂CH=CHCH₃ (2-pentene) per Zaitsev's rule.
Ref: NCERT Class 12 Chemistry > Chapter 6: Haloalkanes and Haloarenes > Topic: Elimination Reactions and Reaction with Metals and Other Reactions
Alcoholic KOH promotes elimination (E₂) in (CH₃)2CHCl , forming CH₂=CHCH₃ (propene).
Ref: NCERT Class 12 Chemistry > Chapter 6: Haloalkanes and Haloarenes > Topic: Elimination Reactions and Reaction with Metals and Other Reactions
Alcoholic KOH promotes elimination (E₂) in CH₃CH₂CHClCH₃ , forming CH₃CH=CHCH₃ (2-butene) as the major product per Zaitsev's rule.
Ref: NCERT Class 12 Chemistry > Chapter 6: Haloalkanes and Haloarenes > Topic: Elimination Reactions and Reaction with Metals and Other Reactions
