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Question #44394
The Question
Consider the following reaction sequences starting from phenol:
Reaction I: Phenol (Major Product)
Reaction II: Phenol (Major Product)
Let and be the position numbers of the carbon atoms attached to the group in the aromatic rings of and respectively, strictly assigning the carbon attached to the group as position 1.
When and are treated separately with excess aqueous bromine (), they undergo electrophilic aromatic substitution to form products and respectively. Let and be the total number of bromine atoms present in one molecule of and respectively.
Calculate the value of the expression: .
Options
- A.
12
- B.
13
- C.
17
- D.
25
Check Your Answer
Correct Answer
Step-by-Step Solution
Explanation
How the Examiner Thinks
Exception_Exposure
Learn to identify edge cases and boundary exceptions that JEE examiners hide in math problems.
Concepts Behind This Question
Kolbe-Schmitt Reaction Mechanism|Basicity-Solvation Competition|Aniline Resonance Pitfalls
Why This Question Is Tricky
Common Mistakes
The question tests two major exceptions/hidden variables in electrophilic aromatic substitution.
First, it challenges the heuristic that 'larger cations = more reactive nucleophiles'.
In the Kolbe-Schmitt reaction, the tight chelation of is required for ortho selectivity, whereas the larger fails to chelate, shifting the regioselectivity to the thermodynamically favored para product.
Second, it tests the ipso-substitution trap during bromination in aqueous media, where students frequently forget that the group on a highly activated ring acts as a leaving group (), leading to decarboxylative polybromination rather than simple substitution around the existing groups.
Quick Facts About This Question
Question Information
| Subject | Chemistry |
| Chapter | Alcohols, Phenols, and Ethers - Class 12 |
| Concept | Kolbe-Schmitt Reaction Mechanism |
| Difficulty | 8 |
| Cognitive Load | 7 |
| Target Solve Time | 160 sec |
| Access | Free |