(2,4-DIBROMOANILINE) ---- most abundant product (4-BROMOANILINE) ---- second mos

Question

(2,4-DIBROMOANILINE) ---- most abundant product

(4-BROMOANILINE) ---- second most abundant product

(2-BROMOANILINE) --- least abundant product

Provide a theoretical reason for the relative abundance of the following product pairs.  

Why is it that the bromine at the ortho & para postion (2,4-DIBROMOANILINE) is more abundant and the bromo at the para position (4-BROMOANILINE) is the second most abundant?.

Why is it that the bromine at the ortho & para postion (2,4-DIBROMOANILINE) is more abundant and the bromine at the ortho position (2-BROMOANILINE) is the least abundant?

Explanation / Answer

Step I

Bromoaniline contains one NH2 group at the top which is a +R group. That means it provides a positive resonance effect since it contains one lone pair of electrons which can easily move into the phenyl ring and can provide stability.

Step II

Bromine shows -I effect that is negative inductive effect. When lone pair of electrons from amine group goes into the ring the -ve charge created on the carbon atom gets stabilised by the -ve inductive effect of bromine group. But the bromine group's inductive effect works only at ortho- and para- positions.

Step III

Now, the more the number of bromine groups on ortho and para positions, more the stability of negative charge in phenyl ring and hence more stable the product is. Therefore, 2,4-dibromoaniline will always be the more abundant product as compared to 2-bromoaniline and 4-bromoaniline.

Step IV

2-Bromoaniline is the least abundant since two big groups (-NH2 and -Br) are very close to each other they create steric hinderance and reduce the stability of product. Hence, 4-bromoaniline is more stable product than 2-bromoaniline.

Related Questions

Navigate

• Browse (All)
• Browse #
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.