Home
Class 12
CHEMISTRY
Hofmann mustard oil reaction....

Hofmann mustard oil reaction.

A

reaction of primary amine with `CHCl_(3)`

B

reaction of primary amine with `CHCl_(3)+KOH`

C

reaction of primary amine with `CS_(2)+HgCl_(2)`

D

reaction of aromatic amine with iodoform

Text Solution

Verified by Experts

The correct Answer is:
C
Doubtnut Promotions Banner Mobile Dark
|

Topper's Solved these Questions

  • AMINES

    VMC MODULES ENGLISH|Exercise IN CHAPTER EXERCISE -A|10 Videos
  • AMINES

    VMC MODULES ENGLISH|Exercise IN CHAPTER EXERCISE -B|10 Videos
  • AMINES

    VMC MODULES ENGLISH|Exercise SOLVED EXAMPLES|21 Videos
  • ALDEHYDES,KETONES & CARBOXYLIC ACIDS

    VMC MODULES ENGLISH|Exercise IN-CHAPTER EXERCISE-M|10 Videos
  • ATOMIC STRUCTURE

    VMC MODULES ENGLISH|Exercise JEE Advanced (Archive )|67 Videos

Similar Questions

Explore conceptually related problems

A compound (X) has the molecular formula C_3 H_7 NO . With Br_2 and KOH , (X) gives (Y) . (Y) responds to mustard oil reaction . (Y) upon treatment with HNO_2 evolves N_2 and gives an alcohol (Z) which gives iodoform test , (X) is likelty to be :

Which of the following undergoes mustard oil reaction ?

Knowledge Check

  • Hofmann broinamide reaction involves the intermediate formation of:

    A
    a species with electron deficient carbon
    B
    a species with electron deficient nitrogen
    C
    a species with negative charge
    D
    a free radical species
  • Similar Questions

    Explore conceptually related problems

    Hofmann elimination reaction.

    Hofmann elimination reaction.

    Hofmann elimination reaction.

    Hofmann's Bromamide reaction

    Hofmann's Bromamide reaction

    Hofmann's Bromamide reaction

    RCONH_2 is converted into RNH_2 by means of Hofmann bromamide degradation. In this RCONHBr is formed form which this reaction has derived its name. Electron-donating group at phenyl activities the reaction. Hofmann degradation reaction is an intramolecular reaction. Which is the rate-determining step in Hofmann bromamide degradation ?