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Question #44394

Difficulty: 8/10 | 🟢 Free

The Question

Consider the following reaction sequences starting from phenol:

Reaction I: Phenol →125∘C, 5 atm1. NaOH,CO2→2. H3O+P\xrightarrow[125^\circ\text{C, 5 atm}]{\text{1. } NaOH, CO_2} \xrightarrow{\text{2. } H_3O^+} \mathbf{P} (Major Product)

Reaction II: Phenol →150∘C, 5 atm1. KOH,CO2→2. H3O+Q\xrightarrow[150^\circ\text{C, 5 atm}]{\text{1. } KOH, CO_2} \xrightarrow{\text{2. } H_3O^+} \mathbf{Q} (Major Product)

Let cc and dd be the position numbers of the carbon atoms attached to the −COOH-COOH group in the aromatic rings of P\mathbf{P} and Q\mathbf{Q} respectively, strictly assigning the carbon attached to the −OH-OH group as position 1.

When P\mathbf{P} and Q\mathbf{Q} are treated separately with excess aqueous bromine (Br2/H2OBr_2/H_2O), they undergo electrophilic aromatic substitution to form products X\mathbf{X} and Y\mathbf{Y} respectively. Let aa and bb be the total number of bromine atoms present in one molecule of X\mathbf{X} and Y\mathbf{Y} respectively.

Calculate the value of the expression: (c×d)+(a×b)(c \times d) + (a \times b).

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 Na+Na^+ is required for ortho selectivity, whereas the larger K+K^+ 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 −COOH-COOH group on a highly activated ring acts as a leaving group (CO2CO_2), leading to decarboxylative polybromination rather than simple substitution around the existing groups.

Quick Facts About This Question

Question Information

SubjectChemistry
ChapterAlcohols, Phenols, and Ethers - Class 12
ConceptKolbe-Schmitt Reaction Mechanism
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree