The major products $A$ and $B$ formed in the following reaction sequence are :

  • A
    $A$ = Acetanilide,$B$ = m-bromoacetanilide
  • B
    $A$ = Acetanilide,$B$ = p-bromoacetanilide
  • C
    $A$ = p-aminoacetophenone,$B$ = $3-$bromo$-4-$aminoacetophenone
  • D
    $A$ = p-aminoacetophenone,$B$ = $3,5-$dibromo$-4-$aminoacetophenone

Explore More

Similar Questions

In the following reaction sequence,$C_2H_5NH_2$ $\xrightarrow{HNO_2} A$ $\xrightarrow{PCl_5} B$ $\xrightarrow{NH_3} C$,the compound '$C$' is

State True or False for the following statements:
$(i)$ Aniline becomes polar in its resonance structure.
$(ii)$ The separation of charge occurs in the resonance structure of aniline.
$(iii)$ The $NH_2$ group of aniline is electron donating.
$(iv)$ The $NH_2$ group of aniline is electron attracting in resonance $(-R)$.

Arrange the following compounds according to increasing order of boiling points: $n-C_4H_9OH$ $(A)$, $n-C_4H_9NH_2$ $(B)$, $n-C_4H_{10}$ $(C)$, and $C_2H_5NHC_2H_5$ $(D)$.

Given below are two statements: one is labelled as Assertion $(A)$ and the other is labelled as Reason $(R)$.
Assertion $(A)$: Experimental reaction of $CH_{3}Cl$ with aniline and anhydrous $AlCl_{3}$ does not give $o-$ and $p-$methylaniline.
Reason $(R)$: The $-NH_{2}$ group of aniline becomes deactivating because of salt formation with anhydrous $AlCl_{3}$ and hence yields $m-$methylaniline as the product.
In the light of the above statements,choose the most appropriate answer from the options given below.

Among the following,which one does not act as an intermediate in the Hofmann rearrangement?

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo