Q. Product on monobromination of this compound is

A) [image:10_image54.emf]
B) [image:10_image55.emf]
C) [image:10_image56.emf]
D) [image:10_image57.emf]
Explanation

(b)
[image:10_image120.emf]
It is electrophilic substitution, so electrophile must be attacked on o/p-position due to higher electron density on this position. In this ring, the attached –NH- group will have high electron density due to resonance and $ortho$ position is blocked, so electrophile is attached on $para$ position.

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.